VYSOKÁ ŠKOLA BÁŇSKÁ TECHNICKÁ UNIVERZITA OSTRAVA


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1 VYSOKÁ ŠKOLA BÁŇSKÁ TECHNICKÁ UNIVERZITA OSTRAVA FAKULTA METALURGIE A MATERIÁLOVÉHO INŽENÝRSTVÍ APPLIED MECHANICS Study Support Leo Václavek Ostrava 2015
2 Title:Applied Mechanics Code: Author: doc. Ing. Leo Václavek, CSc. Edition: first, 2015 Number of pages: 56 Academic materials for the Advanced Engineering Materials study programme at the Faculty of Metallurgy and Materials Engineering. Proofreading has not been performed. Execution: VŠB  Technical University of Ostrava ii
3 CONTENTS 1 INTRODUCTION 1 2 STRESS VECTOR 2 3 UNIAXIAL STRESS STATE AXIALLY LOADED BAR STRESS STATE ON AN INCLINED PLANE IN AN AXIALLY LOADED BAR 7 4 PLANE STRESS STATE THIN PLATE LOADED IN THE MIDPLANE STRESS TRANSFORMATION FOR PLANE STRESS STATE MOHR`S CIRCLE FOR PLANE STRESS STATE Principal Stresses Normal Stresses on Orthogonal Faces HOOKE`S LAW FOR PLANE STRESS STATE; THE RELATIONSHIP INVOLVING E, ν,g Shear Strain, Hooke`s Law for Shear Strain Hooke`s Law for Plane Stress Pure Shear, Relationship Involving E, ν, and G 22 iii
4 5 TRIAXIAL STATE OF STRESS STATE OF STRESS AT A POINT PRINCIPAL STRESSES AND PRINCIPAL DIRECTIONS THE METHOD OF MOHR`S CIRCLES OF STRESS GENERALIZED HOOKE`S LAW FOR ISOTROPIC MATERIALS Volumetric Strain and Bulk Modulus OCTAHEDRAL STRESS, SPHERICAL AND DEVIATOR STRESS TENSOR DIFFERENTIAL EQUATIONS OF EQUILIBRIUM 43 6 STRAIN ENERGY 47 7 DEFORMATION STRAIN DISPLACEMENT RELATIONS STRAIN COMPATIBILITY RELATIONS 54 REFERENCES 56 iv
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6 1 INTRODUCTION Applied mechanics has two major branches, that deal with the behavior of solids and with the behavior of fluids. This textbook is an introduction to the mechanics of solid deformable bodies. We consider bodies at rest, that is whose acceleration and velocity are both zero. The external loads acting on a deformable body are surface forces and body forces. Surface forces are concentrated loads (they have units of force, e.g. newtons (N)), line loads that have units of force per unit length (e.g. N/m) and surface loads (have units force per unit area, e.g. N/m 2 ). Body forces are proportional to the body`s mass. Examples of these include gravitationalweight forces and magnetic forces. Body force per unit volume is body force density (has units e.g. N/m 3 ). When a body is exposed to external loadings, e.g. as in Fig. 1, internal forces are induced inside the body. Study of the distribution of these internal forces inside the solid leads to the definition and use of the stress vector and stress tensor. FIGURE 1 External loads acting on a deformable body. 1
7 2 STRESS VECTOR Time needed to study approximately 30 minutes. Aim of this chapter is to introduce the concepts of internal force, stress vectors and stress components. A general body subjected to external loadings is shown in Fig. 2(a). To investigate the internal forces, a section is made through the body as shown in Fig. 2(b). The resultant internal force F is acting on small area A with unit normal vector n, as illustrated in Fig. 2(c). FIGURE 2 Sectioned solid under external loading. M(x M,, y M, z M ) is an inner point of the body in plane ρ. The force F depends on the location of point M, on the area A and the orientation of the plane ρ. The magnitude of the average force per unit area is F A. This ratio is called average stress acting on an area A. The limiting ratio F A as A 2
8 goes to zero, with M remaining an interior point of A, defines stress vector f n (also called the traction vector) f nx F f n = lim A 0 A, f n = { f ny }. (2.1) f nz The stress vector f n with components f nx, f ny, and f nz, in the x, y and z directions has the unit (N/m 2 ). It lies along the limiting direction of the force vector F and, in general, is neither perpendicular nor tangent to plane ρ with unit normal n, as illustrated in Fig. 2(d). Review Stress vector f n is defined as the limiting ratio of internal force F acting on an area A as A goes to zero. Direction of the stress vector, in general, is neither perpendicular nor tangent to A. Questions 1. What do average stress and the stress vector depend on? 2. What do the components of the stress vector depend on? 3
9 3 UNIAXIAL STRESS STATE Time needed to study approximately 5 hours. Aim of this chapter is to introduce the concepts of stress and strain in the simple case of a uniaxial stress state. A straight bar undergoes axial loading within the limits of linearly elastic behavior of material. 3.1 AXIALLY LOADED BAR We begin with the very simple case of the body shown in Fig. 2. Let`s have a straight homogenous bar of a constant cross section A, undergoing axial loading, as shown in Fig. 3(a). FIGURE 3 A straight bar undergoing axial loading. The bar, sectioned into two parts by a fictitious cut perpendicular to the longitudinal axis of the bar, is shown in Fig. 3(b). The resultant magnitude of normal internal force F is F = P, or F = P. When the section A = bh (see Fig. 3(a)) is far from the point of application of force P (ends of bar), we can suppose that the internal force F is uniformly distributed over the cross section. Then, instead of eqn (2.1) we can simply introduce normal stress vector σ, as internal force intensity, that is force (perpendicular to area) per unit area σ = F A. (3.1) 4
10 Magnitude of σ is normal stress σ, σ = F A = F A = P A. (3.2) The units of stress are units of force divided by units of area. In the SI system of units stress is specified using the basic units of force (newton) and length (meter) as newton per meter squared (N/m 2 ). This unit is called pascal (1Pa = 1N/m 2 ). In engineering work stress is normally expressed in megapascals (1MPa = 10 6 N/m 2 ), or gigapascals (1GPa = 10 9 N/m 2 ). The sign convention for normal stress is as follows: A positive value for σ indicates tensile stress, a force F pulls on the area on which it acts. A negative value for σ indicates compressive stress. Fig. 4 illustrates an original the undeformed shape and final deformed shape of the bar, subjected to axial load P. The original crosssectional area of the bar is A = bh and the initial length FIGURE 4 The deformation of a homogenous isotropic bar under axial loading. of the unloaded bar is l. When force P is not too heavy, the relationship between elongation l of the bar and the magnitude of axial force P = P is linear, and is called Hooke`s law l = Pl EA. (3.3) The constant E is the modulus of elasticity, also called Young`s modulus. This constant depends on material and temperature. Typical units of E are MPa or GPa. When a deformable body is subjected to an axial tensile force, not only does it elongate, but it also contracts laterally. Likewise, a compressive force acting on a body causes it to contract in the direction of the force and yet its sides expand laterally. The elongation or contraction of a line segment per unit of the length is referred to as normal strain. For the bar shown in Fig. 4, the strain in longitudinal direction is ε l = l l (3.4) and strains in the lateral direction are 5
11 ε b = b b, ε h = h h. (3.5) Notice that strains are dimensionless quantities. If the bar stretches (i.e. l > 0), the strain ε l is positive and is called tensile strain. A shortening of the bar (i.e. l < 0) results in a negative value for ε l and is referred to as compressive strain. Within the elastic range the ratios of lateral strains and longitudinal strains are constants. For homogenous and isotropic material transverse contraction is related to the longitudinal elongation by or for the orientation of axes in Fig. 4 The constant ν, ε b = ε h = νε l, (3.6) ε y = ε z = νε x. (3.7) ν = ε lateral ε longitudinal (3.8) is referred to as Poisson`s ratio. The negative sign is used here since longitudinal elongation (positive strain) causes lateral contraction (negative strain) and vice versa. Notice that this lateral strain is the same in all lateral directions. Using eqns (3.2) and (3.3) we obtain l = σ l, (3.9) E and eqns (3.4) and (3.9) yield the linear relationship between stress and strain σ = Eε l, σ x = Eε x, (3.10) when uniaxial stress is applied to the homogenous (the same properties throughout), isotropic (the same properties in every direction) member oriented along the x axis, see Fig. 5. FIGURE 5 The deformation (much exaggerated) of a specimen under uniaxial stress. The equation (3.10) is called Hooke`s law. The subscripts on σ and ε identify the axis of a particular stress and strain. 6
12 3.2 STRESS ON AN INCLINED PLANE IN AN AXIALLY LOADED BAR Let us consider, how the stress on an oblique plane is related to axial stress σ x. As illustrated in Fig. 6(a), unit vector n is perpendicular to an oblique plane and the angle between normal n and the direct axis x of the bar is φ. Unit vector t is perpendicular to n and n x is the unit vector along the axis x. FIGURE 6 The force resultants and stresses on an oblique section through an axial tension bar. Stress vector σ x and normal stress σ x on the cross section in Fig. 6(b) is, see Fig. 3 and eqns (3.1), (3.2), To satisfy the equilibrium of the free body in Fig. 6(c) we get σ x = F A = σ xn x, (3.11) σ x = F A = F A = P A. (3.12) P + F = 0, F = P, (3.13) where F is the resultant internal force in an oblique section. Area A n of the oblique section is related to the crosssectional area A of the bar (see Fig. 6(a)) by the following relation A n = A/ cos φ. (3.14) 7
13 Thus, the stress vector f n (also called the traction vector), see Fig. 6(d), is given by using eqns (3.14), (3.11) f n = F A n = F A cos φ = σ x cos φ. (3.15) The force F, whose magnitude is F, may be resolved into components N = F cos φn and T = F sin φt along normal n and tangent t to the oblique section, respectively, as shown in Fig. 6(e). Similarly eqn (3.11) the ratios N/A n and T/A n define normal stress vector σ n and the shear stress vector τ n, given by equations σ n = N A n = N A cos φ = F A cos φ2 n = σ x cos φ 2 n, (3.16) τ n = T A n = T A cos φ = F A sin φ cos φ t = σ x sin φ cos φt. (3.17) Vectors σ n and τ n are perpendicular and tangent, respectively, see Fig. 6(f), to the inclined plane, σ n = σ n n, τ n = τ n t. (3.18), (3.19) These vectors are components of stress vector f n, such that f n = σ n + τ n. (3.20) The magnitude of the stress vector σ n is σ n = σ n and is called normal stress, the analogously magnitude of stress vector τ n is τ n = τ n and is called shear stress. From eqns (3.16) (3.19) we get for normal stress σ n and shear stress τ n σ n = σ x cos φ 2 (3.21) τ n = σ x sin φ cos φ. (3.22) Using trigonometric identities, we can write σ n and τ n as functions of the doubleangle 2φ. The resulting equations σ n = σ x (1 + cos 2φ), (3.23) 2 τ n = σ x sin 2φ. (3.24) 2 When φ = 0, see Fig. 6(f), we get σ n = σ x = P A, τ n = 0. If angle φ varies, normal stress σ n decreases and shear stress τ n arises on an oblique section. The maximum shear stress of magnitude σ x 2 occurs on the planes oriented at 45 to the axis x of a bar undergoing axial deformation with axial stress σ x. When angle φ = 90, we get σ n = 0, τ n = 0, i.e. on sections parallel to the axis of the bar is zero stress, which is typical for the uniaxial stress state. Notice that longitudinal fibers are not mutually affected by stress and deform separately. The angle φ can be eliminated from eqns (3.23), (3.24). The result is the equation of a circle in the σ n,τ n plane (σ n 0.5σ x ) 2 + τ n 2 = 0.25σ x 2. (3.25) This circle, shown in Fig. 7, is a particular case of a more general Mohr`s mapping of a stress state. 8
14 FIGURE 7 Mohr`s circle for a uniaxial stress state. Review Section 3.1 defines normal stress σ in a section perpendicular to the longitudinal axis of the axially loaded bar. A positive value of σ indicates tensile stress, a negative value indicates compressive stress. Hooke`s law and Poisson`s ratio are discussed in the case of the linearly elastic behavior of homogenous, isotropic materials. Section 3.2 introduces you to stress distribution on an inclined sectioningplane. Normal and shear stress on an oblique plane depend on the orientation of the plane on which the stress acts. Questions 1. How is normal strain that goes with normal stress defined? 2. Express mathematically the relationship between normal stress and the strain in a longitudinal and lateral direction. What is the constant of proportionality? 3. On what plane does maximum shear stress in the bar undergoing axial loading occur? 4. How are longitudinal fibers in an axially loaded bar mutually affected? 9
15 4 PLANE STRESS STATE Time needed to study approximately 12 hours. Aim of this chapter is to introduce the stressstrain analysis under the conditions of the plane stress state. You will be able to use stresstransformation equations and their graphical representation. You will know how to find principal stresses and their directions. You will be capable of using Hooke`s law for the plane stress state. 4.1 THIN PLATE LOADED IN THE MIDPLANE The two dimensional state of stress, the plane state of stress, is the case which is of great practical significance. Its state of stress occurs for instance in a thin plate subjected to forces acting in the midplane of the plate, as shown in Fig. 8(a), (b). FIGURE 8 Thin plate loaded in the midplane. The plate has a constant thickness. 10
16 In Fig. 8(c), (d) there are shown the freebody diagrams of small elements. This diagram is a sketch of the outlined shape of the body which is represented by being isolated or free from its surroundings. In this sketch there are shown all the forces, in this case stress vectors that the surroundings exert on the freed body element. As illustrated in Fig. 8(a), there is no load on the surface of the plate, the normal of which is in the zdirection. Then the normal and shear stress components will be zero on the face of an element that lies on the surface. Consequently, the corresponding stress components on the opposite face will also be zero, and so the material at arbitrary point M will be subjected to the plane stress state. The nonzero normal and shear stress vector components σ x, σ y, τ xy, τ yx on the faces of the freed element are oriented along the x,y axes, as shown in Fig. 8(c). The subscript notation x, y in normal stress vectors σ x, σ y is used to reference the direction of the outward normal line of the area. Two subscripts are used for the shearstress vector components τ xy, τ yx. The first indices specify the orientation of the area, and second indices refer to the direction lines for the shear stress vectors. The normal stress vector components σ x, σ y act perpendicular to the area, and the shear stress vector components τ xy, τ yx act within the plane of the area, as shown in Fig. 8(c), (d). The components τ xy, τ yx (the magnitudes of τ xy, τ yx ) can be related by satisfying moment equilibrium for the element. To show the relationship between these components, we will consider a freebody diagram of the element, Fig. 9. FIGURE 9 Free body diagram of the element. For simplicity, there are not labeled the dashed forces acting on the hidden sides of the element in Fig. 9(a). The force components on the hidden sides of the element are designated with stresses having primes in Fig. 9(b). Let`s assume, that stresses are continuous functions of coordinates x, y, z.,, Then letting side x 0, we get σ x = σ x, τ xy = τ xy. In a similar manner, if instead y 0, we get,, σ y = σ y, τ yx = τ yx. 11
17 Let`s consider summing the moments about centerline c through centroid C, parallel to the z axis, see Fig. 9(a),(b). Due to the nonuniformity of the distribution of normal it is obvious that forces vanish in the limit. Therefore we have M c = 0 ; τ xy y z x τ yx x z y = 0. If we divide through by x y z, then it is necessary that τ xy = τ yx. (4.1) Hence for two perpendicular sides of a cubic element the magnitudes of shearing stress vectors perpendicular to the line of the intersection of these sides are equal. This couple of shearing stresses is called complementary shearing stresses. The shear stress vectors direct to the line of intersection, or away from this line, see Figs. 8(c), (d) and Fig. 9 (b), points A, B respectively. The three quantities, two normal stress components σ x, σ y and one shear stress component τ xy, which act on the four faces of the element as shown in Figs. 8, 9, are therefore sufficient to describe the plane stress state at any point M of a plate. Note stress area force arm moment In case of a plane stress state there exists one system of parallel planes on which stress is equal to zero. (E.g. planes parallel with x,y coordinate axes in Fig. 8.) In the case of a uniaxial stress state there exists an infinite number of such systems of parallel planes with zero stress. The lines of intersection of such nonparallel plane systems define the stress state axis. (e.g. the axis parallel with coordinate axis x in Figs. 3, 4, 5, 6.) 4.2 STRESS TRANSFORMATION FOR PLANE STRESS STATE Let`s assume that the stresses σ x,σ y and τ xy are known, see Figs. 8, 10(a) and the stresses σ n, τ n, shown in Fig. 10(b), (c) are to be determined. The free body diagram of the triangular element in Fig. 10(c) is shown in Fig. 11(b), the areas of the various faces are given in Fig. 11(a). By writing the equilibrium equations for the free body in Fig. 11(b) and substituting the areas from Fig. 11(a) F n = 0 : σ n A σ x ( A cos φ) cos φ σ y ( A sin φ) sin φ + τ xy ( A cos φ) sin φ + τ yx ( A sin φ) cos φ F t = 0 : = 0. τ n A σ x ( A cos φ) sin φ + σ y ( A sin φ) cos φ τ xy ( A cos φ) cos φ + τ yx ( A sin φ) sin φ = 0. 12
18 FIGURE 10 The relationship of an arbitrarily oriented n face to the reference (x,y) faces. FIGURE 11 The freebody diagram based on a triangular plane stress element. Dividing these equations by A and using the fact that τ xy = τ yx (eqn (4.1)), we get σ n = σ x cos φ 2 + σ y sin φ 2 2τ xy sin φ cos φ, (4.2) τ n = (σ x σ y ) sin φ cos φ + τ xy (cos φ 2 sin φ 2 ). (4.3) Equations (4.2), (4.3) can be expressed in more convenient form by using trigonometric identities 13
19 2 sin φ cos φ = sin 2φ, sin φ 2 = (1 cos 2φ) 2, cos φ 2 = (1 + cos 2φ) 2. Then, from eqns (4.2), (4.3) the stresstransformation equations for plane stress state become σ n = σ x+ σ y 2 + σ x σ y 2 cos 2φ τ xy sin 2φ, (4.4) τ n = σ x σ y MOHR`S CIRCLE FOR PLANE STRESS STATE sin 2φ + τ xy cos 2φ. (4.5) Mohr`s circle is an ingenious graphical representation of the plane stress transformation equations, eqns (4.4), (4.5). Let`s rewrite eqns (4.4), (4.5) in the form σ n σ avg = σ x σ y 2 cos 2φ τ xy sin 2φ, τ n = σ x σ y 2 sin 2φ + τ xy cos 2φ. Squaring both sides of each of these equations and adding the resulting squares, we get or where (σ x σ avg ) 2 + τ 2 n = ( σ x σ y ) τ 2 xy, (4.6) (σ x σ avg ) 2 + τ n 2 = R 2, (4.6) σ avg = σ x+ σ y 2, (4.7) R = ( σ x σ y ) 2 + τ2 2 xy. (4.8) Apparently, stresstransformation equations (4.4), (4.5) are just the parametric equations of the circle with the parameter being φ and with the coordinates of point N on the circle, representing the normal stress σ n and shear stress τ n. Let`s suppose that stresses σ x, σ y and τ xy are given, see Figs. 8 and 10. When angle φ shown in Fig. 10(b), (c) is φ = 0 and φ = 90, we get from eqns (4.4), (4.5) the coordinates σ n, τ n of the image points X(σ x, τ xy ) and Y(σ y, τ xy ) respectively in stressplane (σ n, τ n ), as shown in Fig. 12(a). The connecting line of these image points cut off the center C(σ avg, 0) of the Mohr`s circle on the σ n axis. Now we can apply the compass in point C and draw the circle through image points X, Y. As illustrated in Fig. 12(a), σ avg is in accordance with eqn (4.7) and radius R of the circle, the hypotenuse of the shaded triangle, is in agreement with eqn (4.8). Now we can determine normal stress σ n and shear stress τ n on the n face, as shown in Fig. 10(b), (c). To get along Mohr`s circle from image point X(σ x, τ xy ) (x face in Fig. 10(b), (c) is represented by X point on the circle) to image point N(σ n, τ n ) (n face is represented by N point), the same orientation must be given to the central angle 2φ in the stress plane in Fig. 12(b), as angle φ admits, if in the freebody diagram as shown in Fig. 10(b), (c) we 14
20 FIGURE 12 Mohr`s circle for plane stress state. change from direction x (normal of x face, reference surface of σ x ) to the direction of n (the normal of n face, the direction of reference surface of σ n ). To proceed with Mohr`s method, a special plus or minus convention must be observed. It is: Positive values σ n in the stress plane yield tensile stress component vectors in the freebody diagram; positive values τ n in the stress plane lead to shearing stress component vectors, for which the interior of the body which is affected by τ n, in the field of vision of the shearingstress lies on the right hand side Principal Stresses From eqns (4.4), (4.5) there can be seen that σ n and τ n depend on the angle of inclination φ of the planes on which these stresses act. It is often important to determine the orientation of the planes that causes the normal stress to be a maximum and a minimum. The maximum and minimum normal stresses at a point are called the principal stresses. The planes on which the principal stress act are called the principal planes and the mutually perpendicular axes that are normal to the principal planes (hence, that coincide with the principal stress directions) are called principal axes. Let`s employ a graphical representation of the plane stress transformation equations, eqns (4.4), (4.5) by Mohr`s circle. As illustrated in Fig. 13(a), the image points P 1 (σ 1,0) and P 2 (σ 2,0), where the circle intersects the σ n axis, two values of the principal stresses σ 1 and σ 2 are given. Note that shear stress is zero at these points. From the construction in Fig. 13(a) σ 1 = σ avg + R, σ 2 = σ avg R, (4.9) or using eqns (4.7), (4.8) 15
21 σ 1 = σ x+σ y 2 + ( σ x σ y ) 2 + τ2 2 xy, σ 2 = σ x+σ y 2 ( σ x σ y ) 2 + τ2 2 xy. (4.10) FIGURE 13 Determination of the principal stresses and corresponding principal planes and principal axes In our case, in the stress plane (σ n, τ n ) we arrive at the image point P 1 (σ 1,0) from image point X(σ x, τ xy ) on Mohr`s circle about angle 2φ P1 in a clockwise direction, as shown in Fig. 13(a). Therefore, we must mark off in the freebody diagram angle φ P1 with the same orientation of the x direction in order to obtain principal axis 1 (direction 1), as illustrated in Fig.13(c). Principal axis 2 (direction 2) is perpendicular to it. The inclination of the principal plane (or principal axis 1) and angle φ P1 can be found setting τ n = 0 in eqn (4.5) or using the shaded triangle in Fig. 13(a): tan 2φ P1 = Normal stresses on orthogonal faces 2τ xy (σ x σ y ). (4.11) If the stress state is defined by the components σ x, σ y and τ xy oriented along x, y axes, Fig. 14(a), we can obtain the components σ, x, σ, y, and τ xy oriented along x,, y, axes, see Fig 14(b), so that they represent the same state of stress at the point. The angle φ, Fig 14(b), is measured from the positive x to positive x, axis in a counterclockwise direction. Using eqns (4.4), (4.5) we obtain, 16
22 FIGURE 14 The stresses on orthogonal faces. σ x, = σ x+ σ y 2 + σ x σ y 2 cos 2φ τ xy sin 2φ, (4.12) τ x, y, = σ x σ y 2 sin 2φ + τ xy cos 2φ. (4.13) If normal stress acting in the y, direction is needed, it can be obtained by simply substituting (φ = φ + 90 ) for φ into eqn (4.12), see Fig. 14(b). The result is σ y, = σ x+ σ y 2 σ x σ y 2 cos 2φ + τ xy sin 2φ. (4.14) By adding eqns (4.12) and (4.14), and eventually eqn (4.10), we get it independently of angle φ, σ x + σ y = σ x, + σ y, = σ 1 + σ 2. (4.15) Thus, the sum of the normal stresses on orthogonal faces is a constant, which is called a stress invariant. Note It should be emphasized, that the state of stress at the point of the body is the physical quantity, which is independent of the chosen system of coordinates. The plane stress state at a point is characterized by a combination of two normal stress components σ x, σ y and one shear stress component τ xy, which act on the four faces of an element of material located at the point, as shown in Fig. 15(a). The three stress components σ, x, σ,, y, and τ xy representing the same state of stress at a point on an element oriented in x,, y, directions will be different, Fig. 15(b). We have shown how to transform the stress components from one orientation of an element to an element having a different orientation by using eqns (4.12), (4.13), (4.14). 17
23 FIGURE 15 The plane stress state. The maximum and minimum inplane normal stresses σ 1 and σ 2, i.e. the principal stresses, given by eqn (4.10), acting at a point on the element are shown in Fig. 15(c). The element in this position is not subjected to shear stress. The angle φ P1 is given by eqn (4.11). 4.4 HOOKE`S LAW FOR PLANE STRESS STATE; THE RELATIONSHIP INVOLVING E, ν, G Shear strain, Hooke`s law for shear strain We have shown in Sect. 4.1 that pairs of shear stresses on adjacent faces of the rightangled element must have an equal magnitude and be directed either toward or away from the corners of the element, as in the manner shown in Fig. 9. Let`s consider the shear strain that is associated with shear stress under the conditions of the pure shear stress state, as illustrated in Fig. 16(a), (b). If the material is homogenous and isotropic, then the shear stress will distort the element uniformly as shown in Fig. 16(c). The angle γ xy is called shear strain. It measures the angular distortion of the element relative to the sides originally along the x and y axes. Since shear strains are usually very small in magnitude, we can use the smallangle approximation tan γ γ = (BB 1 ) (AB), where BB 1 and AB are shown in Fig. 16(d). 18
24 FIGURE 16 Illustrations for a definition of shear strain, an element in a pure shear stress state. For most engineering materials the elastic behavior is linear. Then the displacement BB 1, Fig. 16(d), of the upper face of the element with respect to the bottom face AC is proportional to shear stress τ. The shear stress for the element in Fig. 16 is related to shear strain by Hooke`s law for shear τ xy = Gγ xy, or τ = Gγ. (4.16) Here the constant of proportionality, G, is called the shear modulus of elasticity (simply shear modulus) or the modulus of rigidity. Notice, that the eqn (4.16) is similar to eqn (3.10), Hooke`s law, which is formulated for a uniaxial stress state. The units of G are the same as those for E (MPa, GPa), since γ is measured in radians, a dimensionless quantity. Remark a twisted shaft, a practical case of pure shear. What happens when a circular bar, either solid or hollow, is subjected to the equal opposite torques of magnitude T that twist one end relative to the other can be illustrated in Fig. 17. FIGURE 17 The deformation and the stress state of a circular cylinder subjected to twisting about its axis. 19
25 When the torque T is applied, the circles and longitudinal grid lines originally marked on the shaft, Fig. 17(a), tend to distort as shown in Fig. 17(b). Twisting causes the circles to remain circles and longitudinal grid lines deforms into a helix (with elevation γ), that intersect the circles at equal angles. As a result, the square marked in Fig. 17(a) changes into a rhomb, marked in Fig. 17(b). The element shown in Fig. 17(c) is subjected to shear strain γ. Normal stresses are equal to zero, i.e. σ x = σ y = 0. Each element of the twisted shaft is under the conditions of the pure shear stress state Hooke`s law for plane stress If the material at a point is subjected to a uniaxial stress state σ x (recall sect. 3.1), Fig 18(a), or to a plane stress state σ x, σ y, Fig 18(b), associated normal strains ε x, ε y, ε z are developed in the material. For a uniaxial stress state, σ x, Fig 18(a), the linear relationship between stress and strain is given in Sect. 3 by Hooke`s law, eqn (3.10): ε x = σ x E. (3.10) repeated Poisson`s ratio, eqns (3.7), (3.8) and Hooke`s law, eqn (3.10), relates the transverse strain ε y, ε z, to ε x and to normal stress σ x ε y = ε z = νε x = ν σ x E. (3.7) repeated If the material at a point is subjected to plane stress state, Fig. 18(b), the stresses σ x, σ y can be related to the strains ε x, ε y, ε z by using the principal of superposition. When σ x is applied, the element elongates in the x direction and strain ε x (σ x ) in this direction is in accordance with eqn (3.10) FIGURE 18 Illustration of deformation (normal strains ε x, ε y, ε z ) of an element which is subjected to normal stress. ε x (σ x ) = σ x E. 20
26 Application of σ y causes the element to contract with strain ε x (σ y ) in the x direction according to eqn (3.7) ε x (σ y ) = ν σ y E. When these two normal strains are superimposed, the normal strain ε x, Fig. 18(b) is determined ε x (σ x, σ y ) = ε x (σ x ) + ε x (σ y ) = σ x E ν σ y E. Similar equations can be developed for the normal strains in the y and z directions. The final result can be written as ε x = 1 E (σ x νσ y ), ε y = 1 E (σ y νσ x ), (4.17) ε z = ν E (σ x + σ y ). These three equations express Hooke`s law for normal strains in the case of a plane stress state. They are valid only if the principle superposition applies. It requires the linearelastic behavior of the material and small deformations. Since the material is isotropic, the element in Fig. 18(b) will remain rectangular when subjected to normal stresses. If we now apply shear stress τ xy to the element, Fig. 16, the material will deform only due to a shear strain γ xy, τ xy and will not cause other strains in the material. Recall that Hooke`s law for shear stress and shear strain can be written as γ xy = τ xy G. (4.16) repeated The stresses σ x, σ y can be determined from the first two eqns (4.17) when normal strains ε x, ε y are given σ x = E 1 ν 2 (ε x + νε y ), σ y = E 1 ν 2 (ε y + νε x ). (4.18) These equations are often used for the determination of stresses when the normal strains on the free surface of the body are determined by measurement. Remark The edges of the element shown in Fig. 19 don`t change directions owing to the action of normal stress. FIGURE 19 Stress on an element in plane stress state. 21
27 The element remains rectangular when shearing stresses are zero. But the planes when shear stresses are zero are principal planes with principal stresses. So it is obvious, that the principal axes of stress coincide with the principal axes of strain if a material is isotropic Pure shear, relationship involving E, ν, and G In Sect we specified that for most engineering materials Hooke`s law for shear can be written as τ = Gγ. (4.16) repeated Here G is the shear modulus of elasticity, γ is shear strain and τ is shear stress. Now we will show how there are related the three material constants, E, ν (see Sect. 3.1) and G. Fig. 20(a). Let`s consider an element of the material subjected to pure shear (σ x = σ y = 0), as shown in FIGURE 20 Element of the material subjected to pure shear. Mohr`s circle for a pure shear stress state is shown in Fig. 20(b). The center C of the circle is at σ avg = (σ x + σ y ) 2 = 0, recall eqn (4.7). The image points A and B have coordinates A(0, τ xy = τ), B(0, τ xy = +τ), in accordance with the plus and minus convention established in Sect Hence, the radius of this circle is τ. The image points P 1 (σ 1, 0), P 2 (σ 2, 0) give the two values of principal stresses σ 1 = +τ, σ 2 = τ. Since 2φ P1 = 90 counterclockwise, Fig. 20(b), the element must be oriented φ P1 = 45 counterclockwise from the x axis in order to define the direction of the plane on which principal stress σ 1 acts, as shown in Fig. 20(c). Now the principal strains ε 1, ε 2 in the direction of principal stresses σ 1, σ 2 respectively can be related to the shear stress τ if the first two eqns (4.17) of the Hooke`s law are used: ε 1 = 1 (σ E 1 νσ 2 ) = 1 1+ν (τ ν( τ)) = τ, E E 22
28 Note ε 2 = 1 E (σ 2 νσ 1 ) = 1 E 1+ν (( τ) ντ) = τ. (4.19) E From the first two eqns (4.17) it`s obvious, since σ x = σ y = 0, then ε x = ε y = 0, Fig. 20(a). Deformation of the element in pure shear is illustrated in Fig. 21. FIGURE 21 Illustration of deformation on the element in a pure shear stress state. The undeformed shape of the square element is shown in Fig. 21(a). The length of the segment A 1 B 1 of the deformed element, Fig. 21(b), is equal to the length of the segment AB of the undeformed element, Fig. 21(a), since ε x = ε y = 0, as shown therein before. The angle BAC before deformation is π 4, and angle B 1 A 1 C 1 after deformation is (π 4 γ 2). The original right angle at A is decreased by γ and angle at point B is increased by γ. Small angle γ is called shear strain. The shape of the rhomb A 1 B 1 D 1 E 1, thus also the change of the originally right angle at point A and B, shown in Fig. 21(b) is fully given by its diagonals. The change in lengths of diagonals can be calculated using eqns (4.19), as is obvious from Figs. 21(b), (c). So we can make the result that shear modulus G, eqn (4.16), can be expressed using the modulus of elasticity E and Poisson`s ratio ν, eqns (4.19). The procedure may be as follows. For the triangle A 1 B 1 C 1 in Fig. 21(b) we can write tan ( π γ ) = (d d) (d+ d) 2 = 1 d d = 1 ε 1, 1+ d d 1+ε 1 where the last term is apparent from the sketch in Fig. 21(c). Since shearing strain γ is a very small quantity, it holds tan γ 2 γ 2, thus we have (trigonometric identity is used) tan ( π γ ) = tan π 4 tanγ tan π = 1 tan γ 2 4 tanγ 2 1+tan γ = 1 γ γ 2 (a). (b) 23
29 Equating the right hand sides of eqns (a), (b) we obtain ε 1 = γ 2. (4.20) Combining eqn (4.160, the first of eqns (4.19) and eqn (4.20), we get the desired result G = E 2(1+ν). (4.21) Review Section 4.1 illustrates a two dimensional state of stress, which is a case of great practical significance. In Section 4.2 normal stress σ n and shear stress τ n on the oblique section, whose normal n is counterclockwise at angle φ from the xaxis, are given by derived stress transformation equations for plane stress. Mohr`s graphical representation of plane stress transformation equations, it`s geometry and properties, is described in Section 4.3. There is shown how to calculate principal stresses, and the directions of principal axes. In Section 4.4 Hooke`s law for plane stress and isotropic linearly elastic material is given. The relationship involving Young`s modulus, Poisson`s ratio and the shear modulus of elasticity is derived. Questions 1. How are the shear stress vectors and the components of shear stress on two perpendicular faces of a cubic element related? 2. Knowledge in which quantities is sufficient to describe the plane stress state? 3. What is the geometrical representation of the stress transformation equations for a plane stress state? 4. Which image points on a Mohr`s circle for a plane stress state corresponds to stress components which act on the mutually perpendicular planes at a point in a body? 5. How can we determine the principal stresses and orientation of the principal axes at a point of a body for the conditions of a plane stress state? 6. How are the normal stresses on a pair of arbitrarily oriented orthogonal faces at a point in a body related? 7. How is pure shear stress state characterized? 8. How is shear strain and shear stress related? What is the constant of proportionality? 9. How are stresses and strains related if a material at a point is subjected to the plane stress state? 10. How are Young`s modulus E (modulus of elasticity) and the shear modulus of elasticity G (modulus of rigidity) related? 24
30 5 TRIAXIAL STATE OF STRESS Time needed to study approximately 20 hours. Aim of this chapter is to generalize developments gained in the foregoing Chapter 4 (the plane stress state) to the general case of stress distribution in three dimensions. Since failure moduli are often related to maximum normal stresses or to maximum shear stresses, the means of determining these extreme stresses are defined in this chapter. Octahedral stresses and the decomposition of the stress tensor into spherical and deviatoric stress tensors are defined, as they are directly related to the distortional strain energy (defined in Chapter 6), which is often used in failure theories for ductile materials. 25
31 The generalization of explanation concerned with plane stress state in Chapter 4 will bring us to the concept of triaxial stress state. Although stress state can be conveniently expressed and manipulated as a tensor (see Chandrasekharaiah 1994, Saad 2009, Ottosen 2005, Okrouhlík 1995), the more conventional strength of material approach will be followed here. 5.1 STATE OF STRESS AT A POINT Stress is a term used to define the intensity and direction of the internal forces acting at a given point on a particular plane. A complete description of the magnitudes and directions of stresses on all possible planes through a point constitute the state of stress at the point. The stress vector f n defined in Chapter 2 by eqn (2.1) is related to a surface with the outer unit normal vector n, as shown in Fig. 2. It is obvious that the stress vector will, in general, be different when other sections through the same point M in Fig. 2 are considered. Let`s consider some special stress vectors, obtained when sections perpendicular to the coordinate axes through the inner point M of the loaded body are taken, as illustrated in Fig. 22. FIGURE 22 Illustration of stress vectors and stress components. Assume that the outer vector n (see Fig. 2) is taken in the direction of the x axis. The corresponding stress vector in Fig. 22(a) is denoted by f x and we can resolve this vector into components along the coordinate axes (cf. eqn (2.1)), i.e. σ x f x = σ x e 1 + τ xy e 2 + τ xz e 3 = { τ xy } = {f x }, (5.1) τ xz where e 1, e 2 and e 3 are unit vectors along each coordinate direction x, y and z, or f x = [σ x τ xy τ xz] T = {f x }, (5.2) where σ x, τ xy and τ xz denote the components of f x in the x, y and z direction respectively. These components are illustrated in Fig. 22(a). Likewise the corresponding stress vectors f y, f z f y = [τ yx σ y τ yz] T = {f y }, (5.3) f z = [τ zx τ zy σ z] T = {f z }, (5.4) 26
32 if the outer normal unit vectors are taken in the direction of the y and z axis, as shown in Figs. 22(b), (c). The components σ x, σ y, σ z given by eqns (5.2)  (5.4) are called normal stresses, whereas τ xy, τ xz, τ yx, τ yz, τ zx, τ zy are referred to as shear stresses. For example τ zy is the ycomponent of the stress vector for a surface with an outer unit vector in the z direction. matrix format Using the special stress vectors considered above, we can define the quantity σ ij written in σ x τ xy τ xz T σ = [σ ij ] = [ τ yx τ zx σ y τ zy τ yz ]. σ z (5.5) The matrix (5.5) is symmetric, τ xy = τ yx, τ xz = τ zx, τ yz = τ zy. It can be shown in the same manner as in case of the plane stress state, see Fig. 9, the following text and resulting eqn (4.1). Consequently, to completely define the most general state of stress at a point requires the specification of six components of stress, three normal stresses σ x, σ y and σ z and three shearing stresses τ xy, τ xz and τ yz. The positive directions of each stress component are illustrated in Fig. 23. The infinitesimal cube of dimensions dx, dy and dz of the material surrounding inner point M (see Figs. 22, 23) is freed from the loaded body (cf. Fig. 8). FIGURE 23 A three dimensional state of stress referred to rectangular Cartesian axes at an inner point M of the externally loaded body. Positive stresses are shown in Fig. 23(b) only on the positive faces; oppositely directed positive stresses act on the negative (hidden) faces of the cube. Regardless of the coordinate system, positive normal stress always acts in tension out of the face. The sign of the shear stress depends on coordinate system orientation. For example, on a plane with a normal in the positive y direction, positive τ yx acts in the positive x direction and positive τ yz acts in the positive z direction. The quantity T σ defined by eqn (5.5) is called the stress tensor. (Symmetric secondorder tensor.) The stress tensor T σ contains all the information necessary to determine the stress vector f for arbitrary oblique sections through the point in question, as demonstrated in the following. Consider the 27
33 small tetrahedron shown in Fig. 24. Arbitrary plane ABC is passed through the infinitesimal cube shown in Fig. 23(b). Oblique plane ABC will in the limit pass through the origin O (Fig. 24). FIGURE 24 Unit vector n and stress vector f n on an oblique plane ABC. Referring to Fig. 24(a) we find the direction cosines for plane ABC defined by its normal unit vector n We can write cos α = n x, cos β = n y, cos γ = n z. (5.6) n = n x e 1 + n y e 2 + n z e 3, (5.7) where e 1, e 2 and e 3 are unit vectors along coordinate axes x, y and z (as shown in Fig. 22). If matrix notation is used, then n x cos α {n} = [n x n y n z] T = { n y } = { cos β}. (5.8) n z cos γ If the area of triangle ABC is A, then from purely geometrical considerations, the areas of triangles BOC (the face of tetrahedron perpendicular to the x), COA and AOB are, respectively A x = A cos α, A y = A cos β, A z = A cos γ. (5.9) The components f nx, f ny and f nz of the stress vector f n which acts on an oblique section ABC are shown in Fig. 24(b). The positive stress components on the three hidden faces of the tetrahedron (a cut corner of a stress cube in Fig. 23(a)) are shown in Fig. 25(a). The stress components f nx, f ny and f nz of the stress vector f n which act on an oblique plane ABC, see Fig. 24(b), can be computed using the following three equations of force equilibrium in the three coordinate directions 28
34 Af nx A x σ x A y τ yx A z τ zx = 0, Af ny A x τ xy A y σ y A z τ zy = 0, (5.10) Af nz A x τ xz A y τ yz A z σ z = 0. FIGURE 25 Stress components on the faces of the volume element cut by arbitrary plane ABC. For example the stress components in x direction, shown in Fig. 25(b), multiplied by an appropriate area enter into the equilibrium equation of forces acting in an x direction, i.e. first eqn (5.10). Analogously the second and third eqns (5.10). Substituting from eqn (5.9) into eqn (5.10) and dividing out the common term A, we can write the expression for the three components of stress vector f n acting on inclined plane ABC as or in matrix notation or simply f nx = σ x cos α + τ yx cos β + τ zx cos γ, f ny = τ xy cos α + σ y cos β + τ zy cos γ, (5.11) f nz = τ xz cos α + τ yz cos β + σ z cos γ, f nx σ x τ yx τ zx cos α {f n } = { f ny } = [ τ xy σ y τ zy ] { cos β}, (5.12) f τ nz xz τ yz σ z cos γ {f n } = [σ ij ]{n}, (5.13) where eqns (5.5) and (5.8) are used. The magnitude f n of the stress vector f n is f n = f n = f 2 nx + f 2 ny + f2 nz. (5.14) 29
35 Stress vector f n can be divided into two components, σ n and τ n, parallel and perpendicular to n, as shown in Fig. 26(a). As f n = f nx + f ny + f nz, the normal stress vector σ n is the sum of projections f nx, f ny and f nz on the normal to the plane ABC, see Fig. 26(b). FIGURE 26 Normal stress vector σ n and shear stress vector τ n on an inclined plane ABC. σ n = σ n n = ( f nx cos α + f ny cos β + f nz cos γ)n. (5.15) Magnitude (with plus or minus sign) of the stress vector σ n is termed normal stress σ n, or in matrix notation Substituting eqn (5.13) into (5.17) yield σ n = f nx cos α + f ny cos β + f nz cos γ, (5.16) cos α σ n = [f nx f ny f nz ] { cos β} = {f n } T {n}. (5.17) cos γ σ n = {n} T [σ ij ] T {n} = {n} T [σ ij ]{n}. (5.18) Recall: σ ij is symmetric, σ ij = σ ji, so [σ ij ] T = [σ ij ]. Substituting expressions (5.5) and (5.8) into (5.18) yield σ n = σ x cos α 2 + σ y cos β 2 + σ z cos γ 2 + 2(τ xy cos α cos β + τ xz cos α cos γ + τ yz cos β cos γ). (5.19) The magnitude of stress vector τ n, see Fig. 26(a), is termed shear stress τ n (also called tangential stress τ n, since it is the magnitude of the stress vector component which is tangential to plane ABC). We read from Fig. 26(a) that is valid. τ n = f n 2 σ n 2 (5.20) 30
36 Note The direction of stress vector τ n can be found from the condition that line of the shearing stress action is the line of intersection of the plane ABC and the plane containing stress vectors f n and σ n, see Fig. 26(a). 5.2 PRINCIPAL STRESSES AND PRINCIPAL DIRECTIONS Let`s have a look at Fig. 26 once more and ask the following question: Is it possible to find such a plane (or more planes) through the inner point of the body, when the stress vector f n is collinear with the unit vector n and identical with normal stress vector σ n? In this situation shear stress vector τ n = 0, as shown in Fig. 27. FIGURE 27 Normal stress vector σ n is collinear with stress vector f n. As in this case σ n and f n have the same direction cosines cos α, cos β and cos γ, then f nx = f n cos α = σ cos α, f ny = f n cos β = σ cos β, (5.21) f nz = f n cos γ = σ cos γ. The notation σ for the magnitude of normal stress vector σ n is in eqns (5.21) used. Substituting (5.21) into (5.11) yields σ cos α = σ x cos α + τ yx cos β + τ zx cos γ, σ cos β = τ xy cos α + σ y cos β + τ zy cos γ, (5.22) σ cos γ = τ xz cos α + τ yz cos β + σ z cos γ, and after rearranging we have (σ x σ) cos α + τ yx cos β + τ zx cos γ = 0, τ xy cos α + (σ y σ) cos β + τ zy cos γ = 0, (5.23) τ xz cos α + τ yz cos β + (σ z σ) cos γ = 0. 31
37 The three direction cosines cos α, cos β and cos γ are geometrically related by cos α 2 + cos β 2 + cos γ 2 = 1. (5.24) The eqn (5.24) simply states that the three direction cosines can`t all be zero simultaneously, and the definition of any two uniquely determines the third. The set of three homogenous linear equations (5.23) in unknowns cos α, cos β and cos γ gives the real values for the direction cosines when the determinant of their coefficients is equal zero. Hence (σ x σ) τ yx τ zx τ xy (σ y σ) τ zy = 0. (5.25) τ xz τ yz (σ z σ) Finally, expanding the determinant and using the symmetry of the matrix (5.5), we obtain the cubic equation σ 3 σ 2 (σ x + σ y + σ z ) + σ(σ x σ y + σ y σ z + σ z σ x τ 2 xy τ 2 xz τ 2 yz ) (σ x σ y σ z + 2τ xy τ xz τ yz σ x τ 2 yz σ y τ 2 xz σ z τ 2 xy ) = 0. (5.26) The equation (5.26) is a socalled characteristic equation of the stress state. We can write it in the form where σ 3 I σ σ 2 + II σ σ III σ = 0, (5.27) I σ = σ x + σ y + σ z, (5.28) II σ = σ x σ y + σ y σ z + σ z σ x τ 2 xy τ 2 xz τ 2 yz = σ x τ xy τ yx σ y + σ y τ yz τ zy σ z + σ z τ zx τ xz σ x, (5.29) III σ = σ x σ y σ z + 2τ xy τ xz τ yz σ x τ 2 yz σ y τ 2 xz σ z τ 2 xy = σ x τ yx τ zx τ xy τ xz σ y τ yz τ zy. σ z (5.30) Due to the symmetry of the matrix (5.5), the characteristic equations (5.26), (5.27) have three real roots (eigenvalues of the matrix (5.5)). These three roots, σ 1, σ 2 and σ 3, the values of σ in eqns (5.26), (5.27), are the magnitudes of principal stress vectors in the given point of the body and are called principal stresses. Note Two principal stresses are equal to zero in the case of a uniaxial stress state, see Chapter 3. This implies II σ = III σ = 0 in eqn (5.27). In the case of a plane stress state, see Chapter 4, III σ = 0 and one of the roots of the characteristic equation (5.27) is equal to zero, thus one of the principal stresses is equal to zero. Principal stresses act on principal planes. Zero shear stress is, by definition, on the principal plane, as shown in Fig
38 Having determined the three principal stress values σ 1, σ 2 and σ 3, see Fig. 28, for a given triaxial stress state, we can determine the directions in which these stresses act as given by their respective direction cosines (or given eigenvectors of matrix (5.5)). If one of the known principal stress values, say σ 1, is substituted in two of the three eqns (5.23) together with Cartesian stress components (σ x, σ y, τ yz ) and when we append eqn (5.24), the system of three equations is obtained, from which results the unknown direction cosines cos α 1, cos β 1 and cos γ 1 for the principal stress σ 1. Angles α 1, β 1 and γ 1 are outlined in Fig. 28(b). This procedure can be repeated, substituting the other principal stress values σ 2 and σ 3, in turn, to produce the direction cosines (eigenvectors) of these stresses, principal directions. A system of coordinate axes chosen along the principal directions of the stress is referred to as the principal axes of stress. Principal axes of stress are depicted as p 1, p 2 and p 3 in Fig. 28(b). It can be deduced that the three principal planes may always be taken to be mutually perpendicular. FIGURE 28 A three dimensional state of stress at a point in a loaded body. Stresses are shown only on the visible positive faces of rectangular element. It is of great importance that the largest of the three principal stresses is the maximum normal stress that can occur on any plane passing through the point. It is also important to recognize that the minimum principal stress is the minimum normal stress that can occur on any plane passing through the point. It should be recognized that for any given equilibrium system of forces applied to a body, the principal stresses and principal planes at a point in body are uniquely determined and must be independent of the orientation of the Cartesian coordinate axes x, y and z, see Fig. 28(a). For this reason the coefficients I σ, II σ and III σ in characteristic equation (5.27) are nonvarying quantities, or invariants. That means that I σ, II σ and III σ given by eqns (5.28), (5.29) and (5.30) are invariants of stress and must have the same magnitudes for all choices of coordinate axes x, y and z. Relative to principal axes p 1, p 2 and p 3, see Fig. 28(b), the stress invariants may be written in terms of the principal stresses as 33
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