STRESS TENSOR 3.1. STRESSES

Transcription

1 3 STRESS TENSOR The definitions of stress vector and stress components will be given and the equations of equilibrium will be derived. We shall then show how the stress components change when the frames of reference are changed from one rectangular Cartesian frame of reference to another, and in this way we will prove from physical standpoint that the stress components transform according to the tensor transformation rules. The symmetry of the stress tensor will then be discussed, and the consequences of the symmetry property will be derived. The principal stresses, the stress deviations, the octahedral stresses, and finally, the stress tensor in general curvilinear coordinates, form the material for the remainder of the chapter. Except for Sec et seq., we shall use only rectangular Cartesian frames of reference, whose coordinate axes will be denoted by x 1, x, x 3 and are rectilinear and orthogonal to each other. We shall use subscripts for all components unless stated otherwise STRESSES Consider a configuration occupied by a body B at some time (Fig. 3.1:1). Imagine a closed surface S within B. We would like to know the interaction Fig. 3.1:1. Stress principle. between the material exterior to this surface and that in the interior. In this consideration, there arises the basic defining concept of continuum mechanics the stress principle of Euler and Cauchy. Consider a small surface element of area S on our imagined surface S. Let us draw a unit vector normal to S, with its direction outward from the interior of S. Then we can distinguish the two sides of S according to the direction of. Consider the part of material lying on the positive side of the normal. This part exerts a force F on the other part, which is situated on the negative side of the normal. The force F is a function of 66

2 Sec. 3.1 STRESSES 67 the area and the orientation of the surface. We introduce the assumption that as S tends to zero, the ratio F/ S tends to a definite limit df/ds and that the moment of the forces acting on the surface S about any point within the area vanishes in the limit. The limiting vector will be written as (1) T = df ds, where an overhead is introduced to denote the direction of the normal of the surface S. The limiting vector T is called the stress vector, or traction, and represents the force per unit area acting on the surface. The assertion that there is defined upon any imagined closed surface S in the interior of a continuum a stress vector field whose action on the material occupying the space interior to S is equipollent to the action of the exterior material upon it, is the stress principle of Euler and Cauchy. Consider now a special case in which the surface S k is parallel to one of the coordinate planes. Let the normal of S k be in the positive direction of the x k -axis. Let the stress vector acting on S k be denoted by T,withthreecom- k ponents T k 1, T k, T k 3 along the direction of the coordinate axes x 1, x, x 3,respectively. The index i of T k i denotes Fig. 3.1:. Notations of stress components. the components of the force, and the symbol k indicates the surface on which the force acts. In this special case, we introduce a new set of symbols for the stress components, () k T 1 = τ k1, k T = τ k, k T 3 = τ k3. If we arrange the components of the tractions acting on the surfaces k =1, k =,k = 3 in a square matrix, we obtain Components of Stresses 1 3 Surface normal to x 1 τ 11 τ 1 τ 13 Surface normal to x τ 1 τ τ 3 Surface normal to x 3 τ 31 τ 3 τ 33

3 68 STRESS TENSOR Chap. 3 This is illustrated in Fig. 3.1:. The components τ 11, τ, τ 33 are called normal stresses, and the remaining components τ 1, τ 13, etc., are called shearing stresses. Each of these components has the dimension of force per unit area, or that of [Mass] [Length] 1 [Time]. A great diversity in notations for stress components exists in the literature. The most widely used notations in American literature are, in reference to a system of rectangular Cartesian coordinates x, y, z. (3) σ x τ xy τ xz, τ yx σ y τ yz, τ zx τ zy σ z. Love writes X x, Y x for σ x and τ xy, and Todhunter and Pearson use xx, xy. In this book we shall use both τ ij and σ ij. We use τ ij to denote stress tensors in general, and we use σ ij to denote the physical components of stress tensors in curvilinear coordinate (see p. 86). In rectangular Cartesian coordinates, the tensor components and the physical components coincide. Hence, we use σ ij for Cartesian stress tensors. Although the lack of uniformity may seem awkward, little confusion will arise, and in many instances different notations actually result in clarity. It is important to emphasize again that a stress will always be understood to be the force (per unit area) which the part lying on the positive side of a surface element (the side on the positive side of the outer normal) exerts on the part lying on the negative side. Thus, if the outer normal of a surface element points in the positive direction of the x 1 -axis and τ 11 is positive, the vector representing the component of normal Fig. 3.1:3. Senses of positive stresses. stress acting on the surface element will point in the positive x 1 -direction. But if τ 11 is positive while the outer normal points in the negative x 1 -axis direction, then the stress vector acting on the element also points to the negative x 1 -axis direction (see Fig. 3.1:3). Similarly, positive values of τ 1, τ 13 will imply shearing stress vectors pointing to positive x, x 3 -axes if the outer normal agrees in sense with

4 Sec. 3. LAWS OF MOTION 69 x 1 -axis, whereas they point to the negative x, x 3 -direction if the outer normal disagrees in sense with the x 1 -axis, as illustrated in Fig. 3.1:3. A careful study of the figure is essential. Naturally, these rules agree with the usual notions of tension, compression, and shear. 3.. LAWS OF MOTION The fundamental laws of mechanics for bodies of all kinds are Euler s equations, which extend Newton s laws of motion for particles. Let the coordinate system x 1, x, x 3 be an inertial frame of reference. Let the space occupied by a material body at any time t be denoted by B(t). Let r be the position vector of a particle with respect to the origin of the coordinate system. Let V be the velocity vector of a particle at the point x 1, x, x 3.Then (1) P = Vρdv, B(t) is called the linear momentum of the body in the configuration B(t), and let () H = r Vρdv, B(t) is called the moment of momentum. In these formulas ρ is the density of the material and the integration is over the volume B(t). Newton s laws, as stated by Euler for a continuum, assert that the rate of change of linear momentum is equal to the total applied force F acting on the body, (3) Ṗ = F, and that the rate of change of moment of momentum is equal to the total applied torque L, (4) Ḣ = L. The torque L is taken with respect to the same point as the origin of the position vector r. It is easy to verify that if (3) holds, then if (4) holds for one choice of origin, it holds for all choices of origin. It is assumed that force and torque are quantities about which we have a priori information in certain frames of reference. On material bodies considered in mechanics of continuous media, two types of external forces act: The derivatives Ṗ and Ḣ are material derivatives; i.e., the time rate of change of Pand H of a fixed set of particles (cf. Secs. 5. and 5.3).

5 70 STRESS TENSOR Chap Body forces, acting on elements of volume of the body.. Surface forces or stresses, acting on surface elements. Examples of body forces are gravitational forces and electromagnetic forces. Examples of surface forces are aerodynamic pressure acting on a body and pressure due to mechanical contact of two bodies. To specify a body force, we consider a volume bounded by an arbitrary surface S (Fig. 3.:1). The resultant force vector contributed by the body force is assumed to be representable in the form of a volume integral Xdv. Fig. 3.:1. Body forces. B The three components of X, namely, X 1, X, X 3, all of dimensions force per unit volume M(LT ), are called the body force per unit volume. For example, in a gravitational field, X i = ρg i, where g i are components of a gravitational acceleration field and ρ is the density (mass per unit volume) at a given point of the body. The surface force acting on an imagined surface in the interior of a body is the stress vector conceived in Euler and Cauchy s stress principle. According to this concept, the total force acting upon the material occupying the region B interior to a closed surface S is (5) F = S TdS + B Xdv, where T is the stress vector acting on ds whose outer normal vector is. Similarly, the torque about the origin is given by the expression (6) L = r TdS + r Xdv. S In the following section we shall make some elementary applications of these equations to obtain the fundamental properties of the stress tensor. B

6 Sec. 3.3 CAUCHY S FORMULA CAUCHY S FORMULA With the equations of motion, we shall first derive a simple result which states that the stress vector T (+) representing the action of material exterior to a surface element on the interior is equal in magnitude and opposite in direction to the stress vector T ( ) which represents the action of the interior material on the exterior across the same surface element: (1) T ( ) = T (+). To prove this, we consider a small pill box with two parallel surfaces of area S and thickness δ, as shown in Fig. 3.3:1. When δ shrinks to zero, while S remains small but finite, the volume forces and the linear momentum and its rate of change with time vanish, as well as the contribution of surface forces on the sides of the pill box. The equation of motion (3.:3) implies, therefore, for small S, T (+) S + T ( ) S =0. Equation (1) then follows. Another way of stating this result is that the stress vector is a function of Fig. 3.3:1. Equilibrium of a pill box across a surface S. the normal vector to a surface. When the sense of direction of the normal vector reverses, the stress vector reverses also. Nowweshallshowthatknowing the components τ ij, we can write down at once the stress vector acting on any surface with unit outer normal vector whose components are 1,, 3. This stress vector is denoted by T,with components T 1, T, T 3 given by Cauchy s formula () T i = j τ ji, which can be derived in several ways. We shall give first an elementary derivation. Let us consider an infinitesimal tetrahedron formed by three surfaces parallel to the coordinate planes and one normal to the unit vector (see Fig. 3.3:). Let the area of the surface normal to be ds. Then the area

7 7 STRESS TENSOR Chap. 3 Fig. 3.3:. Surface tractions on a tetrahedron. of the other three surfaces are ds 1 = ds cos(, x 1 ) = 1 ds = area of surface to the x x 3 -plane, ds = ds = area of surface to the x 3 x 1 -plane, ds 3 = 3 ds = area of surface to the x 1 x -plane, and the volume of the tetrahedron is dv = 1 3 hds, where h is the height of the vertex P from the base ds. The forces in the positive direction of x 1 acting on the three coordinate surfaces can be written as ( τ 11 + ε 1 )ds 1, ( τ 1 + ε )ds, ( τ 31 + ε 3 )ds 3, where τ 11, τ 1, τ 31 are the stresses at the point P. The negative sign is obtained because the outer normals to the three surfaces are opposite in sense with respect to the coordinate axes, and the ε s are inserted because the tractions act at points slightly different from P. If we assume that the stress field is continuous, then ε 1, ε, ε 3 are infinitesimal quantities. On the other hand, the force acting on the triangle normal to has a

8 Sec. 3.4 EQUATIONS OF EQUILIBRIUM 73 component ( T 1 + ε)ds in the x 1 -axis direction, the body force has an x 1 - component equal to (X 1 +ε )dv, and the rate of change of linear momentum has a component ρ V L dv. Here T 1 and X 1 refer to the point P and ε, ε are again infinitesimal. The first equation of motion is thus ( τ 11 + ε 1 ) 1 ds +( τ 1 + ε ) ds +( τ 31 + ε 3 ) 3 ds +( T 1 + ε)ds +(X 1 + ε ) 1 3 hds = ρ V hds. Dividing through by ds and taking the limit h 0, one obtains (3) T 1 = τ τ 1 + τ 31 3, which is the first component of Eq. (). Other components follow similarly. Cauchy s formula assures us that the nine components of stresses τ ij are necessary and sufficient to define the traction across any surface element in a body. Hence the stress state in a body is characterized completely by the set of quantities τ ij. Since T i is a vector and Eq. () is valid for an arbitrary vector j, it follows from the quotient rule (Sec..10) that τ ij is atensor. Henceforth τ ij will be called a stress tensor. We note again that in the theoretical development up to this point we have assumed, first, that stress can be defined everywhere in a body, and, second, that the stress field is continuous. The same assumption will be made later with respect to strain. These are characteristic assumptions of continuum mechanics. Without these assumptions we can do very little indeed. However, in the further development of the theory, certain mathematical discontinuities will be permitted often they are very useful tools but one should always view these discontinuities with great care against the general basic assumptions of continuity of the stress and strain fields EQUATIONS OF EQUILIBRIUM We shall now transform the equations of motion (3.:3), (3.:4) into differential equations. This can be done elegantly by means of Gauss theorem and Cauchy s formula. But we shall pursue here an elementary course to assure physical clarity. Consider the static equilibrium state of an infinitesimal parallelepiped with surfaces parallel to the coordinate planes. The stresses acting on the various surfaces are shown in Fig. 3.4:1. The force τ 11 dx dx 3 acts on the left-hand side, the force (τ 11 + τ11 x 1 dx 1 )dx dx 3 acts on the right-hand side, etc. These expressions are based on the assumption of continuity of the stresses. The body force is X i dx 1 dx dx 3.

9 74 STRESS TENSOR Chap. 3 σ 3 σ 31 σ Fig. 3.4:1. Equilibrating stress components on an infinitesimal parallelepiped. Now, the equilibrium of the body demands that the resultant forces vanish. Consider the forces in the x 1 -direction. As shown in Fig. 3.4:, we have six components of surface forces and one component of body force. The sum is ( τ 11 + τ ) 11 dx 1 dx dx 3 τ 11 dx dx 3 x ( τ 1 + τ 1 dx x ( τ 31 + τ 31 dx 3 x 3 ) dx 3 dx 1 τ 1 dx 3 dx 1 ) dx 1 dx τ 31 dx 1 dx + X 1 dx 1 dx dx 3 =0. Dividing by dx 1 dx dx 3,weobtain τ 11 + τ 1 + τ 31 (1) + X 1 =0. x 1 x x 3 A cyclic permulation of subscripts 1,, 3 leads to similar equations of equilibrium of forces in x, x 3 -directions. The whole set, written concisely, is () This is an important result. Sec τ ji + X i =0. x j A shorter derivation will be given later in

10 Sec. 3.4 EQUATIONS OF EQUILIBRIUM 75 Fig. 3.4:. Components of tractions in x 1 direction. The equilibrium of an element requires also that the resultant moment vanishes. If there do not exist external moments proportional to a volume, the consideration of moments will lead to the important conclusion that the stress tensor is symmetric, (3) τ ij = τ ji. This is demonstrated as follows. Referring to Fig. 3.4:3 and considering the moment of all the forces about the axis Ox 3, we see that those components Fig. 3.4:3. Components of tractions that contribute moment about Ox 3 -axis.

11 76 STRESS TENSOR Chap. 3 of forces parallel to Ox 3 or lying in planes containing Ox 3 do not contribute any moment. The components that do contribute a moment about the Ox 3 - axis are shown in Fig. 3.4:3. Therefore, properly taking care of the moment arm, we have ( τ 11 + τ ) 11 dx 1 dx dx 3 dx x 1 + τ 11dx dx 3 dx ( τ 1 + τ ) ( 1 dx 1 dx dx 3 dx 1 τ 1 + τ ) 1 dx dx 1 dx 3 dx x 1 x ( τ + τ ) dx dx 1 dx 3 dx 1 x τ dx 1 dx 3 dx 1 ( τ 3 + τ ) 3 dx 3 dx 1 dx dx 1 x 3 τ 3dx 1 dx dx 1 ( τ 31 + τ ) 31 dx 3 dx 1 dx dx x 3 + τ 31dx 1 dx dx X 1 dx 1 dx dx 3 dx + X dx 1 dx dx 3 dx 1 =0. On dividing through by dx 1 dx dx 3 and passing to the limit dx 1 0, dx 0, dx 3 0, we obtain (4) τ 1 = τ 1. Similar considerations of resultant moments about Ox 1, Ox lead to the general result given by Eq. (3). Again a shorter derivation will be given later in Sec It should be noted that if an external moment proportional to the volume does exist, then the symmetry condition does not hold. For example, if there is a moment c 3 dx 1 dx dx 3 about the axis Ox 3, then we obtain in place of Eq. (4) the result (5) τ 1 τ 1 + c 3 =0. Maxwell pointed out that nonvanishing body moments exist in a magnet in a magnetic field and in a dielectric material in an electric field with different planes of polarization. If the electromagnetic field is so intense and the stress level is so low that τ 1 and c 3 are of the same order of magnitude, then, according to (5), τ 1 cannot be equated to τ 1. In this

12 Sec. 3.4 EQUATIONS OF EQUILIBRIUM 77 case, we have to admit the stress tensor τ ij as asymmetric. If c 3 is very much smaller in comparison with τ 1, then we can omit c 3 in (5) and consider (4) as valid approximately. In developing a physical theory, particularly for the purpose of engineering, one of the most important objectives is to obtain the simplest formulation consistent with the desired degree of accuracy. A decision of whether or not we shall treat the stress tensor as symmetric must be based on the purpose of the theory. Since electromagnetic fields pervade the universe, the stress tensor is in general unsymmetric. But, if the theory is formulated for the purpose of a structural or mechanical engineer who studies the stress distribution in a structure or a machine with a view towards assessing its strength, stability, or rigidity, then a stress is important when it is of the order of the yielding stress of the material. Even for a structure which is designed primarily on the basis of stability, such as a column, an arch, or a thin-walled shell, a good design should produce a critical stress of the order of the yielding stress under the critical conditions, for otherwise the material is not economically used. When τ ij is comparable to the yielding stress in magnitude (of order 10,000 to 100,000 lb/sq in. or 70 to 700 mpa for a steel, or 50 to 5,000 lb/sq in. or 0.35 to 35 mpa for a concrete), there are few circumstances in which the assumption of symmetry in stress tensor should cause concern. However, if one wants to study the influence of a strong electromagnetic field on the propagation of elastic waves, or such influence on some highfrequency phenomenon in the material, then the stress level may be very low and the body moment may be significant. In such problems the stress tensor may not be assumed symmetric. In the rest of this book the stress tensor will be assumed to be symmetric unless stated otherwise. Notes on Couple-stresses If, following Voigt, we assume that across any infinitesimal surface element in a solid the action of the exterior material upon the interior is equipollent to a force and a couple (in contrast to the assumption made in Sec. 3.1) then in addition to the traction T that acts on the surface we must have also a couple-stress vector M. These two vectors T and M, together, are now equipollent to the action of the exterior upon the interior. Similarly, one might have body couples as pointed out by Maxwell, i.e., couple per unit mass, c, with components c i,(i =1,, 3). If we accept these possibilities, then we must define a couple-stress tensor, M ij, in addition to the stress tensor τ ij. The tensor M ij is related to the couple-stress vector

13 78 STRESS TENSOR Chap. 3 by a linear transformation like Eq. (3.3:): M i = M ji j. An analysis of the angular momentum then leads to the equation M ji + ρc i = e ijk τ jk, x j i.e., M xx + M yx + M zx + ρc x = τ yz τ zy, etc. x y z Thus, the antisymmetric part of the stress tensor is determined by the body couples and the divergence of the couple-stress tensor. When couples of both kinds are absent, the stress tensor must be symmetric. Couple-stresses and body couples are useful concepts in dealing with submicron structures and molecular mechanics of materials, and in the dislocation theory of metals TRANSFORMATION OF COORDINATES In the previous section, the components of stress τ ij are defined with respect to a rectangular Cartesian system x 1, x, x 3. Let us now take a second set of rectangular Cartesian coordinates x 1, x, x 3 with the same origin but oriented differently, and consider the stress components in the new reference system (Fig. 3.5:1). Let these coordinates be connected by the linear relations x k = β ki x i = x k (1) x i, k =1,, 3, x i where β ki =cos(x k, x i) are the direction cosines of the x k-axis with respect to the x i -axis. Since τ ij is a tensor (Sec. 3.3) we can write down the transformation law at once. However, in order to emphasize the importance Fig. 3.5:1. Transformation of stress components under rotation of coordinates system.

14 Sec. 3.6 PLANE STATE OF STRESS 79 of the result we shall insert an elementary derivation based on Cauchy s formula derived in Sec. 3.3, which states that if ds is a surface element whose unit outer normal vector has components i, then the force per unit area acting on ds is a vector T with components T i = τ ji j. If the normal is chosen to be parallel to the axis x k,sothat 1 = β k1, = β k, 3 = β k3, then denoting T i as T k i,wehave T i = T k i = τ ji β kj. Since the component of the vector k T(= T) in the direction x m is τ km, i.e., k T = τ km g m, then k T = τ km g m = T k i g i = τ ji β kj g i, where g m and g i are the unit base vectors of the two coordinate systems. Using the relation g i = β mi g m [Eq. (.14:16)] and Eq. (1) for rectangular Cartesian coordinates, we obtain i.e., k T = τ km g m = τ jiβ kj β mi g m, (3) τ km = τ x k x m jiβ kj β mi = τ ji. x j x i If we compare Eq. (3) and Eq. (.5:) we see that the stress components transform like a Cartesian tensor of rank two. Thus, the physical concept of stress which is described by τ ij agrees with the mathematical definition of a tensor of rank two in a Euclidean space PLANE STATE OF STRESS A state of stress in which (1) τ 33 = τ 31 = τ 3 =0, Fig. 3.6:1. Change of coordinates in plane state of stress. is called a plane state of stress in the x 1 x -plane. In this case, the direction

15 80 STRESS TENSOR Chap. 3 cosines between two systems of rectangular Cartesian coordinates can be expressed in terms of a single angle θ, as shown in Fig. 3.6:1. We have, cos θ sin θ 0 (β ij )= sin θ cos θ Writing x, y and x, y in place of x 1, x and x 1, x ; σ x for τ 11 ; τ xy for τ 1, etc., we have Since we may also write σ x = σ x cos θ + σ y sin θ +τ xy sin θ cos θ, σ y = σ x sin θ + σ y cos θ τ xy sin θ cos θ, τ xy =( σ x + σ y )sinθ cos θ + τ xy (cos θ sin θ). sin θ = 1 (1 cos θ), cos θ = 1 (1 + cos θ), σ x = σ x + σ y + σ x σ y cos θ + τ xy sin θ, (3) σ y = σ x + σ y σ x σ y cos θ τ xy sin θ, Note that τ xy = σ x σ y sin θ + τ xy cos θ. (4) (5) σ x + σ y = σ x + σ y, σ x θ =τ xy, τ xy =0 when tanθ = τ xy (6). σ x σ y The directions given by the particular values of θ given by (6) are called the principal directions; the corresponding normal stresses are called the principal stresses (see Sec. 3.7). Following (5) and (6), the principal stresses are extreme values of the normal stresses, (7) } σ max = σ x + σ y ± σ min (σx σ y ) + τ xy.

16 Sec. 3.6 PLANE STATE OF STRESS 81 Fig. 3.6:. Mohr s circle in plane state of stress. Differentiating τ xy with respect to θ and setting the derivative to zero, we can find the angle θ at which τ xy attends its extreme value. This angle is easily seen to be ±45 from the principal directions given by (6), and the maximum value of τ xy is (8) τ max = σ max σ min = (σx σ y ) + τ xy. Figure 3.6: is a geometric representation of the relations above. It is the well-known Mohr s circle. Problem 3.1. Find the transformation law for the moments of inertia and products of inertia of an area about a set of rectangular Cartesian coordinates in aplane, I xx = y da, I xy = xyda, I yy = x da, with respect to rotation of the coordinate axes about the origin. Mohr s circle was invented in 1887 for the transformation of the inertia tensor. Problem 3.. Let v i, i =1,, 3, be the velocity vector field of a continuum, and let v iv j be the average value of the product v iv j over a period of time. Show that the correlation function v iv j,withcomponentsu, uv, uw, v, etc., in unabridged notations, is a symmetric tensor of the second order. Problem 3.3. Show that the mass moment of inertia of a set of particles, I ij = e ipqe jkq x px k dm, i, j =1,, 3, is a tensor, where dm is an element of mass and the integration is extended over the entire set of particles. Write out the matrix of the inertia tensor I ij. Show

17 8 STRESS TENSOR Chap. 3 that I ii (i not summed) is the moment of inertia about the x i axis, whereas I ij (i j) is equal to the negative of the product of inertia about the axes x i and x j. Show that for a rigid body rotating at an angular velocity ω j, the angular momentum vector of the body is I ijω j PRINCIPAL STRESSES The results of the previous section are restricted to the plane state of stress. Let us now generalize them to general three dimensional problems. In a general state of stress, the stress vector acting on a surface with outer normal depends on the direction of. Let us ask in what direction the stress vector becomes normal to the surface, on which the shearing stress vanishes. Such a surface is called a principal plane, its normal a principal axis, and the value of the normal stress acting on the principal plane is called a principal stress. Let define a principal axis and let σ be the corresponding principal stress. Then the stress vector acting on the surface normal to has components σ i. On the other hand, this same vector is given by the expression τ ji j. Hence, writing i = δ ji j, we have, on equating these two expressions and transposing them to the same side, (1) (τ ji σδ ji ) j =0. The three equations, i =1,, 3, are to be solved for 1,, 3. Since is a unit vector, we must find a set of nontrivial solutions for which = 1. Thus, Eq. (1) poses an eigenvalue problem. Since τ ij as a matrix is real and symmetric, we need only to recall a result in the theory of matrices to assert that there exist three real-valued principal stresses and a set of orthonormal principal axes. Because of the importance of these results, we shall give the details of the reasoning below. Equation (1) has a set of nonvanishing solutions 1,, 3 if and only if the determinant of the coefficients vanishes, i.e., () τ ij σδ ij =0. Equation () is a cubic equation in σ; its roots are the principal stresses. For each value of the principal stress, a unit normal vector can be determined. On expanding Eq. (), we have τ 11 σ τ 1 τ 13 (3) τ ij σδ ij = τ 1 τ σ τ 3 τ 31 τ 3 τ 33 σ = σ 3 + I 1 σ I σ + I 3 =0,

18 Sec. 3.7 PRINCIPAL STRESSES 83 where (4) I 1 = τ 11 + τ + τ 33, τ I = τ 3 τ 3 τ 33 + τ 11 τ 13 τ 31 τ 33 + τ 11 τ 1 τ 1 τ, τ 11 τ 1 τ 13 I 3 = τ 1 τ τ 3. τ 31 τ 3 τ 33 On the other hand, if σ 1, σ, σ 3 are the roots of Eq. (3), which can be written as (σ σ 1 )(σ σ )(σ σ 3 )=0, it can be seen that the following relations between the roots and the coefficients must hold: (5) I 1 = σ 1 + σ + σ 3, I = σ 1 σ + σ σ 3 + σ 3 σ 1, I 3 = σ 1 σ σ 3. Since the principal stresses characterize the physical state of stress at a point, they are independent of any coordinates of reference. Hence, Eq. (3) and the coefficients I 1, I, I 3 are invariant with respect to the coordinate transformation; I 1, I, I 3 are the invariants of the stress tensor. The importance of invariants will become evident when physical laws are formulated (see, for example, Chapter 6). We shall show now that for a symmetric stress tensor the three principal stresses are all real and that the three principal planes are mutually orthogonal. These important properties can be established when the stress tensor is symmetric, (6) τ ij = τ ji. The proof is as follows. Let 1,, 3, be unit vectors in the direction of the principal axes, with components 1 i, i, 3 i (i =1,, 3) which are the solutions of Eq. (1) corresponding to the roots σ 1, σ, σ 3, respectively; (τ ij σ 1 δ ij ) 1 j =0, (7) (τ ij σ δ ij ) j =0, (τ ij σ 3 δ ij ) 3 j =0.

19 84 STRESS TENSOR Chap. 3 Multiplying the first equation by i, the second by 1 i, summing over i and subtracting the resulting equations, we obtain (8) (σ σ 1 ) 1 i i =0, on account of the symmetry condition (6), which implies that (9) τ ij 1 j i = τ ji 1 j i = τ ij j 1 i. The last equality is obtained by interchanging the dummy indices i and j. Now, if we assume tentatively that Eq. (3) has a complex root, then, since the coefficients in Eq. (3) are real-valued, a complex conjugate root must also exist and the set of roots may be written as σ 1 = α + iβ, σ = α iβ, σ 3, where α, β, σ 3 are real numbers and i stands for the imaginary number 1. Inthiscase,Eqs.(7)showthat 1 j and j are complex conjugate to each other and can be written as 1 j a j + ib j, j a j ib j, in which a j and b j are real numbers and at least one of them is not zero. Therefore, 1 j j =(a j + ib j )(a j ib j ) = a 1 + a + a 3 + b 1 + b + b 3 0. It follows from (8) that σ 1 σ =iβ =0orβ = 0. This contradicts the original assumption that the roots are complex. Thus, the assumption of the existence of complex roots is untenable, and the roots σ 1, σ, σ 3 are all real. When σ 1 σ σ 3, Eq. (8) implies (10) 1 i i =0, i 3 i =0, 3 i 1 i =0; i.e., the principal vectors are mutually orthogonal to each other. If σ 1 = σ σ 3, we can determine an infinite number of pairs of orthogonal vectors 1 i and i and define 3 i as a vector orthogonal to 1 i and i.ifσ 1 = σ = σ 3, then any set of orthogonal axes may be taken as the principal axes.

20 Sec. 3.8 SHEARING STRESSES 85 If the reference axes x 1, x, x 3 are chosen to coincide with the principal axes, then the matrix of stress components becomes σ (11) (τ ij )= 0 σ σ SHEARING STRESSES We have seen that on an element of surface with a unit outer normal, ( i ), there acts a traction T,( T i = τ ji j ). The component of T in the direction of is the normal stress acting on the surface element. Let this normal stress be denoted by σ (n). Since the component of a vector in the direction of another vector is given by the scalar product of the two vectors, we obtain (1) σ (n) = τ ij i j. The magnitude of the resultant shearing stress on a surface element having the normal i is given by the equation () τ = T i σ (n), (see Fig. 3.8:1). Let the principal axes be chosen as the coordinate axes, Fig. 3.8:1. Notations. and let σ 1, σ, σ 3 be the principal stresses. Then T i =(σ 1 1 ) +(σ ) +(σ 3 3 ), and, from Eq. (1) σ(n) =[σ 1( 1 ) + σ ( ) + σ 3 ( 3 ) ].

21 86 STRESS TENSOR Chap. 3 On substituting into Eq. () and noting that ( 1 ) ( 1 ) 4 =( 1 ) [1 ( 1 ) ]=( 1 ) [( ) +( 3 ) ], we see that (3) τ =( 1 ) ( ) (σ 1 σ ) +( ) ( 3 ) (σ σ 3 ) +( 3 ) ( 1 ) (σ 3 σ 1 ). If 1 = =1/ and 3 =0,thenτ = ± 1 (σ 1 σ )andσ n = 1 (σ 1 + σ ). Problem 3.4. Show that τ max = 1 (σmax σmin) and that the plane on which τ max acts makes an angle of 45 with the directions of the largest and the smallest principal stresses MOHR S CIRCLES Let σ 1, σ, σ 3 be the principal stresses at a point. The stress components acting on any other surface elements can be obtained by the tensor transformation laws, Eq. (3.5:3). Otto Mohr, in papers published in 188 and 1900, has shown the interesting result that if the normal stress σ (n) and the shearing stress τ acting on any surface element be plotted on a plane, Fig. 3.9:1. Mohr s circles. with σ and τ as coordinates as shown in Fig. 3.9:1, the locus necessarily falls in a closed domain represented by the shaded area bounded by the three semicircles with centers on the σ-axis. A detailed proof can be found in Westergaard, 1. Elasticity and Plasticity, pp ; or Sokolnikoff, 1. Elasticity, p. 5. The practical problem of graphical construction of Mohr s circle from strain-gage data is discussed in Biezeno and Grammel, 1. Engineering Dynamics, Vol. 1, p. 45; Pearson, 1.4 Theoretical Elasticity, p. 64. See Bibliography on pp

22 Sec STRESS DEVIATIONS STRESS DEVIATIONS The tensor (1) τ ij = τ ij σ 0 δ ij, is called the stress deviation tensor, whereδ ij is the Kronecker delta and σ 0 is the mean stress () σ 0 = 1 3 (σ 1 + σ + σ 3 )= 1 3 (τ 11 + τ + τ 33 )= 1 3 I 1, where I 1 is the first invariant of Sec. 3.7 and τ ij specifies the deviation of the state of stress from the means stress. The first invariant of the stress deviation tensor always vanishes: (3) I 1 = τ 11 + τ + τ 33 =0. To determine the principal stress deviations, the procedure of Sec. 3.7 may be followed. The determinental equation (4) τ ij σ δ ij =0, may be expanded in the form (5) σ 3 J σ J 3 =0. It is easy to verify the following equations relating J, J 3 to the invariants I, I 3 defined in Sec. 3.7, (6) (7) J =3σ 0 I, J 3 = I 3 I σ 0 +σ 3 0 = I 3 + J σ 0 σ 3 0, and the alternative expressions below on account of Eq. (3), (8) J = τ 11 τ τ τ 33 τ 33 τ 11 +(τ 1) +(τ 3 ) +(τ 31 ) = 1 [(τ 11 ) +(τ ) +(τ 33 ) ]+(τ 1 ) +(τ 3 ) +(τ 31 ) = 1 6 [(τ 11 τ ) +(τ τ 33 ) +(τ 33 τ 11 ) ] +(τ 1 ) +(τ 3 ) +(τ 31 ) = 3 τ 0.

23 88 STRESS TENSOR Chap. 3 The τ 0 in the last equation is the octahedral stress, which will be defined in the next section. We note also the simple expressions (9) J = 1 τ ij τ ij, (10) J 3 = 1 3 τ ij τ jk τ ki. It can be easily shown that the principal stress deviations are (11) σ i = σ i σ 0. Problem 3.5. Show that the principal stresses as given by the three roots of Eq. (5) can be written as ( σ 1 = τ 0 cosα, σ = τ 0 cos α + π ) (, σ 3 = τ 0 cos α π ), 3 3 where cos 3α = J 3 /τ 3 0,andJ =3τ 0 / OCTAHEDRAL SHEARING STRESS The octahedral shearing stress τ 0 is the resultant shearing stress on a plane that makes the same angle with the three principal directions. Such a plane is called an octahedral plane; eight such planes can form an octahedron. See Fig. 3.11:1. The direction cosines i of a normal to the octahedral Fig. 3.11:1. Octahedral planes.

24 Sec OCTAHEDRAL SHEARING STRESS 89 plane relative to the principal axes are such that Hence, Eq. (3.8:3) gives ( 1 ) =( ) =( 3 ) = τ 0 =(σ 1 σ ) +(σ σ 3 ) +(σ 3 σ 1 ), which is proportional to the sum of the areas of Mohr s three semicircles. From Eqs. (3.10:6) and (3.10:8), it can be easily verified that the octahedral stress can be expressed in terms of the two invariants I 1 and I of Sec. 3.7, 9τ 0 =I 1 6I. The square of the octahedral stress happens to be proportional to the second invariant J of the stress deviation, Eqs. (3.10:6) and (3.10:8). In 1913, Richard von Mises proposed the hypothesis that yielding of some of the most important materials occurs at a constant value of the quantity J. Nadai then introduced the interpretation of J as proportional to the octahedral shearing stress. In this way, J or τ 0 enters into the basic equations of plasticity. Note that the normal stress on the octohedral plane equals the mean stress. Problem 3.6. If σ 1 >σ >σ 3 and σ 1, σ 3 are given, at what values of σ does τ 0 attain its extreme values? P3.8 Problem 3.7. Let σ x = 5c, σ y = c, σ z = c, τ xy = c, τ yz = τ zx =0, where c = 1, 000 lb/sq in. Determine the principal stresses, the principal stress deviations, the direction cosines of the principal directions, the greatest shearing stress, and the octahedral stress. Problem 3.8. Consider a horizontal beam as shown in Fig. P3.8. According to the usual elementary theory of bending, the fiber stress is σ xx = 1My/bh 3,

25 90 STRESS TENSOR Chap. 3 where M is the bending moment which is a function of x. Assume this value of σ xx, and assume further that σ zz = σ zx = σ zy = 0, that the body force is absent, that σ xy = 0 at the top and bottom of the beam (y = ±h/), and that σ yy =0 at the bottom. Derive σ xy and σ yy from the equations of equilibrium. Compare the results with those derived in elementary mechanics of materials STRESS TENSOR IN GENERAL COORDINATES So far we have discussed the stress tensor in rectangular Cartesian coordinates, in which there is no necessity to distinguish the contravariant and covariant transformations. The necessary distinction arises in curvilinear coordinates. Just as a vector in three-dimensional Euclidean space may assume either a contravariant or a covariant form, a stress tensor can be either contravariant, τ ij,ormixed,τj i, or covariant, τ ij. The tensors τ ij, τj i,andτ ij are related to each other by raising or lowering of the indices by forming inner products with the metric tensors g ij and g ij : (1) τj i = g αj τ iα = g αj τ αi, τ ij = g iα τj α, τ ij = g iα τα j. The correctness of these tensor relations are again seen by specializing them into rectangular Cartesian coordinates. We have seen that τ ij is symmetric in rectangular Cartesian coordinates. It is easy to see from the tensor transformation law that the symmetry property remains invariant in coordinate transformation. Hence, τ ij = τ ji in all admissible coordinates, and we are allowed to write the mixed tensor as τ i j and not τ i j or τ j i. But what is the physical meaning of each of these components? To clarify the meaning of the stress components in an arbitrary curvilinear coordinates system, let us first consider a geometric relationship. Let us form an infinitesimal tetrahedron whose edges are formed by the coordinate curves PP 1, PP, PP 3 and the curves P 1 P, P P 3, P 3 P 1,asshown in Fig. 3.1:1. Let us write, vectorially, PP 1 = r 1, PP = r, PP 3 = r 3. Then P 1 P = r r 1, P 1 P 3 = r 3 r 1, P P 3 = r 3 r, Similarly, the contravariant stress tensor τ ij is symmetric, τ ij = τ ji.itdoesnotmake sense, however, to say that the mixed tensor τj i is symmetric, since an equation like τj i = τ j i, with the indices switching roles on the two sides of the equation, is not a tensor equation.

26 Sec. 3.1 STRESS TENSOR IN GENERAL COORDINATES 91 Fig. 3.1:1. Geometric relationship. and we have () P 1 P P 1 P 3 =(r r 1 ) (r 3 r 1 ) = r 1 r 3 r r 1 + r r 3 = r r 3 + r 3 r 1 + r 1 r. Now the vector product A B of any two vectors A and B is a vector perpendicular to A and B, whose positive sense is determined by the righthand screw rule from A to B, and whose length is equal to the area of a parallelogram formed by A, B as two sides. Hence, if we denote by, 1,, 3 the unit vectors normal to the surfaces P 1 P P 3, PP P 3, PP 3 P 1, PP 1 P, respectively, and by ds, ds 1, ds, ds 3 their respective areas, Eq. () may be written as (3) ds = 1 ds 1 + ds + 3 ds 3. Now let us recall that in Sec..14, we defined the reciprocal base vectors g 1, g, g 3 which are perpendicular to the coordinate planes and are of length g 11, g, g 33, respectively. We see that the unit vectors 1,, 3 are exactly g 1 / g 11, g / g, g 3 / g 33, respectively. Hence, (4) ds = 3 i=1 ds i g ii gi.

27 9 STRESS TENSOR Chap. 3 If the unit normal vector is resolved into its covariant components with respect to the reciprocal base vectors, then (5) = i g i. We see from the last two equations that (6) i g ii ds = ds i, i not summed. This is the desired result. Let us now consider the forces acting on the external surface of the infinitesimal tetrahedron. In Sec. 3.1, the inside and outside of a volume are distinguished by drawing an outward pointing normal vector. The stress vector T is defined as the limit of force acting on the outside surface with a normal divided by the area of the surface. On the other side, the stress vector is T, see Sec. 3.3, Fig. 3.3:1. Now, for the tetrahedron shown in Fig. 3.1:1, the normal vector of the triangle P 1 P P 3 is outward, the stress vector is T, the area is ds, the force is TdS. On the triangle PP P 3, the normal vector i points inward, the stress vector is T, i the area is ds i, the force is, therefore, TdS i i as shown in Fig. 3.1:1. The forces on the other surfaces are determined similarly. The equation of motion of this infinitesimal tetrahedron is, in the limit, (7) TdS = i TdS i. Volume forces and inertia (mass acceleration) forces acting on the tetrahedron do not enter into this equation, because they are of higher order of smallness than the surface forces. On substituting (6) into (7) and canceling the nonvanishing factor ds, we have (8) T = 3 i T i g ii. i=1 If the coordinates x i are changed to a new set x i while the surface P 1 P P 3 and the unit outer normal remain unchanged, the stress vector T is invariant, but the vectors T i will change because they will be associated with the new coordinate surfaces. (See Fig. 3.1:, in which only one pair of vectors T, T are shown.) The covariant components i will change also. Inasmuch as T is invariant and i is a covariant tensor, Eq. (8) shows that T i g ii transforms according to a contravariant type of transformation.

28 Sec. 3.1 STRESS TENSOR IN GENERAL COORDINATES 93 Fig. 3.1:. Tractions referred to two different elementary tetrahedron with a common surface. Resolving the vectors i T g ii into their components with respect to the base vectors g i and the reciprocal base vectors g i,wehave (9) g ii i T = τ ij g j = τ i j gj, i not summed. On the other hand, the components of T may be written as (10) T = T j g j = T j g j. Substitution of (9) and (10) into (8) shows that (11) T j = τ ij i, T j = τ i j i. The tensorial character of τ ij and τj i are demonstrated both by (9) and by (11) according to the quotient rule. Following the tensor transformation rules for τ ij, τj i and i, we can rewrite Eq. (11) in the form (1) T j = τ j i i, T j = τ ij i where i are the contravariant components of. Let us recapitulate the important points. With the scaling factor g ii, the vectors i T g ii (i =1,, 3, not summed) transform according to the

29 94 STRESS TENSOR Chap. 3 contravariant rule. When the vectors T i g ii are resolved into components along the base vectors g j, the tensor components τ ij are obtained. When the resolution is made with respect to the reciprocal base vectors g j,then the mixed tensor τj i is obtained. On the other hand, if the stress vector T (acting on the triangle P 1 P P 3 with outer normal ) isresolvedinto the contravariant components T j along the base vectors g j and the covariant components T j along the reciprocal base vectors g j, then we obtain Eq. (11). The contravariant stress tensor τ ij and the mixed stress tensor τj i are related to the stress vectors T i by the Eq. (9). The covariant stress tensor τ ij defined in Eq. (1) cannot be so simply related to the vectors T i and are therefore of less importance PHYSICAL COMPONENTS OF A STRESS TENSOR IN GENERAL COORDINATES If we write Eq. (3.1:9) as (1) i T = 3 j=1 gjj j ij g τ gii = gjj 3 j=1 σ ij g j gjj, i not summed. Then, since g j / g jj (j not summed) are unit vectors along the coordinate curves, the components σ ij are uniform in physical dimensions and represent the physical components of the stress vector T i in the direction of the unit vectors g j / g jj, () σ ij = gjj g ii τ ij, i,j not summed. But σ ij is not a tensor. On the other hand, if we use the mixed tensor τ i j (3) i T = 3 j=1 τ i j g j g ii = 3 j=1 τ i j in (3.1:1), we have g jj g ii g j, i not summed. g jj Thus (4) σj i = g jj g ii τ j i, i,j not summed,

30 Sec EQUATIONS OF EQUILIBRIUM IN are the physical components of the tensor τj i, of uniform physical dimensions, representing the components of the stress vector T i resolved in the directions of the reciprocal base vectors. Note that, in general, σ ij σj i σj i, except for orthogonal coordinates. If the coordinates are orthogonal, g i and g i are in the same direction. When curvilinear coordinates are used, we like to retain the liberty of choosing coordinates without regard to dimensions. Thus, in cylindrical polar coordinates (r, θ, z), r and z have the dimensions of length and θ is an angle. The corresponding tensor components of a vector referred to polar coordinates will have different dimensions. For physical understanding it is desirable to employ the physical components, but for the convenience of analysis it is far more expedient to use the tensor components EQUATIONS OF EQUILIBRIUM IN CURVILINEAR COORDINATES In Sec. 3.4 we discussed the equations of equilibrium in terms of Cartesian tensors in rectangular Cartesian coordinates. To obtain these equations in any curvilinear coordinates, it is only necessary to observe that the equilibrium conditions must be expressed in a tensor equation. Thus the equations of equilibrium must be (1) () τ ij j + X i =0, in volume, τ ji i = T i, on surface. The truth is at once proved by observing that these are truly tensor equations and that they hold in the special case of rectangular Cartesian coordinates. Hence, they hold in any coordinates that can be derived from the Cartesian coordinates through admissible transformations. The practical application of tensor analysis in the derivation of the equations of equilibrium in particular curvilinear coordinates will be illustrated in Secs and 4.1. It will be seen that these lengthly equations can be obtained in a routine manner without too must effort. Because the manipulation is routine, chances of error are minimized. This practical application may be regarded as the first dividend to be paid for the long process of learning the tensor analysis. Problem 3.9. Let us recast the principal results obtained above into tensor equations in general coordinates of reference. In rectangular Cartesian coordinates, there is no difference in contravariant and covariant transformations. Hence, the Cartesian stress tensor may be written as τ ij, orτ ij,orτj i. In general

31 96 STRESS TENSOR Chap. 3 frames of reference, τ ij, τ ij, τ i j are different. Their components may have different values. They are different versions of the same physical entity. Now prove the following results in general coordinates. (a) The tensors τ ij, τ ij are symmetric if there is no body moment acting on the medium; i.e., τ ij = τ ji, τ ij = τ ji. (b) Principal planes are planes on which the stress vector T is parallel to the normal vector. If we use contravariant components, we have, on a principal plane, T j = σ j = σg ij i = τ ij i,whereσ is a scalar. If we use covariant components, we have correspondingly T j = σ j = σgj i i = τj i i. Show that σ must satisfy the characteristic determinantal equation τ i j σδ i j = 0 or its equivalent τ ij σg ij =0. (c) The first invariant of the stress tensor is τ i i,orτ ij g ij. However, τ ii and τ ii are in general not invariants. (d) The stress deviation tensor s i j is defined as s i j = τ i j 1 3 τ α α g i j. The first invariant of s i j zero. The second invariant has the convenient form J = 1 si ks k i. (e) The octahedral shearing stress has the same value in any coordinates system.