Elastic Wave Theory. LeRoy Dorman Room 436 Ritter Hall Tel: Based on notes by C. R. Bentley. Version 1.

Transcription

1 Elastic Wave Theory LeRoy Dorman Room 436 Ritter Hall Tel: Based on notes by C. R. Bentley. Version 1.1,

2 Chapter 1 Tensors Elastic wave theory is concerned with the propagation of small stresses and strains through deformable media. Stress and strain are both second-order tensor quantities, meaning that in 3-space, that 9 (i.e. 3 2 ) components of either stress or strain must be specified at a point in order that the stress or strain be fully specified at that point. Furthermore, these 9 quantities in one coordinate system must be related to the corresponding quantities in another coordinate system according to a certain transformation law involving the direction cosines relating the two coordinate systems. The transformation property is the formal definition of a tensor. We note by a ij the cosine of the angle between the i th axis of one right-handed Cartesian coordinate system and the j th axis of another such coordinate system, where i and j may have the values 1, 2, or 3. Then a tensor of order n (sometimes called rank n) is defined as a set of 3 n quantities such that if their values are given in one coordinate system, the values in any other coordinate system can be found by application of a certain transformation equation. To understand better the physical significance of a second-order tensor, however, we will not start with the formal definition but with a discussion of linear vector functions. Consider a vector v whose value is determined when the value of another vector y is known. We can then say that v is a function of y. v = f(y) (1.1a) If, when y = y 1 + y 2, v = f(y) = f(y 1 + y 2 ) = f(y 1 ) + f(y 2 ), then v is said to be a linear vector function of y. The simplest linear vector function is v = ky, where k is a constant; for example (Fig. 1), 2

3 3 F = kx (1.1b) where x is the extension of a long spring, k is the spring constant, and F is the restoring force. A more complicated example arises if we express y and v in terms of their components: y = y 1 i 1 + y 2 i 2 + y 3 i 3 and v = v 1 i 1 + v 2 i 2 + v 3 i 3 where i 1, i 2, and i 3, are unit vectors in the direction of the Cartesian coordinates x 1, x 2, and x 3 respectively, and let each component of v be a different linear multiple of the corresponding component of y. v 1 = k 1 y 1, v 2 = k 2 y 2, v 3 = k 3 y 3, v then differs from y in both magnitude and direction, and three quantities k 1, k 2, and k 3 are needed to specify v given y. For example, suppose that a mass is fixed to the ends of three long springs along the coordinate axes (Fig. 2). The spring constants are k 1, k 2, and k 3 respectively. Then the components of force, neglecting gravity, are F 1 = k 1 x 1, F 2 = k 2 x 2, F 3 = k 3 x 3, for a rest position at the origin,

4 4 CHAPTER 1. TENSORS and F = k 1 x 1 i 1 + k 2 x 2 i 2 + k 3 x 3 i 3. (1.1c) Another example of the specification of a linear vector function by three parameters is given by the expression for the linear velocity v of a point on a body rotating with angular velocity ω, at a distance x from the axis of rotation: v = ω x = (ω 2 x 3 ω 3 x 2 )i 1 + (ω 3 x 1 ω 1 x 3 )i 2 + (ω 1 x 2 ω 2 x 1 )i 3 Here indicates the vector (cross) product. In the most general case, each component of v is a linear function of all three components of y: v 1 = k 11 y 1 + k 12 y 2 + k 13 y 3 v 2 = k 21 y 1 + k 22 y 2 + k 23 y 3 (1.2) v 3 = k 31 y 1 + k 32 y 2 + k 33 y 3 Note that k ij is the coefficient relating v i to y j. These equations can be written v 1 = 3 k ij y j, i = 1, 2, 3 j=1 or, using the summation convention (see Appendix 1), v i = k ij y j (1.3) The nine coefficients k ij constitute a Cartesian tensor of the second order. An example of this most general form of a linear vector function is the case of the three perpendicular springs mentioned above, but no longer oriented along the coordinate axes. The restoring force due to each spring will now have components in all three coordinate directions, and each component of the vector force will depend on all three components of displacement and all three spring constants (Fig. 3).

5 5 Before showing this explicitly, it is first necessary to discuss the rotation of coordinate axes. Consider a second set of axes with origin coincident with that of the x 1, x 2, x 3 set, but oriented at an angle to them. Let us call this the primed system, and denote coordinates in this system by x 1, x 2, x 3, and unit vectors by i 1, i 2, i 3, (Fig. 4.). We can express each unit vector in the unprimed system by the sum of its components in the primed system. Designating the component of i i in the i j direction by a ij, we have i i = a ij i j (1.4a) By taking the dot product of each side with i j, we see that a ij = cos(i i, i j ) i i i j

6 6 CHAPTER 1. TENSORS The quantities a ij are thus known as the direction cosines. Note that a ij = a j i (1.4b) since the cosine of an angle does not depend on the sense in which it is defined (since the cosine function is an even function of its argument). It is clear that the direction cosines cannot all be independent since the angles between axes in each system are fixed. To find the relations between them, we write for i j, according to (1.4a), and rewrite (1.4a) in the equivalent form i j = a jk i k (1.5) i i = a ik i k (1.6) Let us define the Kronecker delta, δ ij, as 0 when i j, and 1 when i = j. Then, since the unit vectors i i and i j are normal to each other except when i = j, their scalar product clearly equals δ ij. Thus from (1.5) and (1.6), a ik a jk = δ ij (1.7a) The equations (1.7a) are known as the orthogonality relations among the direction cosines. Note that we can write also the equivalent expressions a k i a k j = δ ij and a ki a kj = δ i j Now let us write any vector x in terms of each system: (1.7b) (1.7c) x = x i i i and x = x j i j. Form the dot product of each equation with i j, use (1.4a) and equate: x j = a ij x i. (1.8a) Similarly, by forming the dot product with i i, we find x i = a ij x j (1.8b) These are the vector transformation equations; they express the components of a vector in one system in terms of its components in another system and the cosines of the angles between the corresponding axes. It is important to note the implicit assumption involved in the derivation of the transformation equations, namely that the vector itself, as viewed by a

7 7 detached viewer, does not change as the system of coordinates is rotated. If this were not so, the vector would have little use as the expression of any physical quantity. This transformational property is so fundamental to the nature of a vector that it may be used as the definition: a vector is a set of three quantities such that if their values in one coordinate system are x i, their values in any other coordinate system are given by (1.8a). Now let us rotate the coordinate axes in the case or three perpendicular springs. For this example we will write equations our rather than use the summation convention. This should serve the double purpose of clarifying the transformation relations and demonstrating the great convenience afforded by the summation convention. Writing out (1.4a), i 1 = a 11 i 1 + a 12 i 2 + a 13 i 3 i 2 = a 21 i 1 + a 22 i 2 + a 23 i 3 i 3 = a 31 i 1 + a 32 i 2 + a 33 i 3 and substituting into (1.1c), we find that the force in the new coordinate system is given by But from (1.8b), F =(k 1 x 1 a 11 + k 2 x 2 a 21 + k 3 x 3 a 31 )i 1 (k 1 x 1 a 12 + k 2 x 2 a 22 + k 3 x 3 a 32 )i 2 (1.9) (k 1 x 1 a 13 + k 2 x 2 a 23 + k 3 x 3 a 33 )i 3 x 1 = a 11 x 1 + a12 x 2 + a13 x 2 = a 21 x 1 + a22 x 2 + a23 x 3 = a 31 x 1 + a32 x 2 + a33, which, upon substitution in (1.9), shows that the 1 component of F is F 1 =(k 1 a k 2 a k 3 a31 2 )x 1 +(k 1 a11 a 12 + k 2 a 21 a 22 + k 3 a 31 a 32 x 2 +(k 1 a11 a 13 + k 2 a 21 a 23 + k 3 a 31 a 33 x 3 with similar expressions for F 2 and F 3. Thus each component of F depends on all three components of displacement x 1, x 2, and x 3. In general, if we have a linear vector function v i = k ij y j, we will wish to find the form of the corresponding linear function in any other coordinate orientation. We now return to the use of the summation convention, showing the great simplification in expression over the example we have written out in full. Suppose that the axes are rotated to a new primed orientation such that y j = a jl y l and v k = a k iv i. Then by substitution we have v k = a k i k ij y j = a k i k ij a jl y l = k k l y l,

8 8 CHAPTER 1. TENSORS where, using (1.4b), k k l = a k i a l j k ij. (1.10) Equation (1.10) is known as the tensor transformation equation. Exercise: Write (1.10) in matrix notation. Any second-order Cartesian tensor can be transformed to a rotated frame of reference by a transformation of this form. The transformation equations for different order tensors are analogous. For a first-order tensor, or vector, the transformation equation is that which we have already been using, namely y i = a i j y j. Third-order tensor transformations would involve products of three direction cosines. The transformation law for a scalar, or zero-order tensor, is that it be unchanged by any rotation of the coordinate axes. A second-order tensor may be thought of as a set of linear coefficients which relate one vector to another. The transformation equation is a consequence of our requirement that the physical significance of the vectors not be a function of the orientation of the coordinate axes. For example, in the case of the three springs, the force resulting from a given vector displacement measured relative to the system be unchanged, however the arbitrary coordinate system may be oriented. We may think of the vectors as fixed while the coordinate system is rotated. The transformation equation for vectors was derived by the simple geometrical projection of the vector components in one coordinate system onto the rotated coordinate axes. In concept, the tensor transformation is just as simple, and may be thought of as merely representing geometrical consistency. This is clearly necessary if a vector, or any other tensor, is to have any physical significance. A second-order tensor, then is a set of nine quantities defined for all coordinate orientations in such a way that it can have physical significance in relating one vector to another. Just as we may have vector properties of a material or space which vary from one point to point (vector point functions), such as the particle velocity in a fluid, so we may have tensor properties of a meterial or space; k ij will in general be a function of position. (We will henceforth mean second-order tensor when we write tensor, unless otherwise specified.) Any property of matter which linearly relates one vector to another is a tensor property. Examples, besides stress and strain (to be shown in later sections), include the dielectric constant (relating electric displacement and intensity vectors), permeabiity (relating magnetic induction and intensity vectors), and inertia (relating momentum and velocity in a rigid body). Because of symmetry or isotropy, some components may be zero, e.g., the dielectric constant and permeability are constants for isotropic media, but this does not change the basic tensor nature of the corresponding properties. The elements of a tensor k ij may be shown conveniently in matrix form: k 11 k 12 k 13 k 21 k 22 k 23 k 31 k 32 k 33.

9 1.1. SIMPLE TENSORS Simple tensors As illustrations we may consider the simpler forms of the linear vector equations. Thus the matrix of coefficients in (1.1b) is k k 0, 0 0 k and those in (1.1c) is k k k 3 If the tensor elements are such that k ij = k ji, the tensor is said to be symmetric; it then has only six different elements. In matrix form it may be written k 11 k 12 k 13 k 12 k 22 k 23. k 13 k 23 k 33 If the tensor elements are such that k ij = k ji, the tensor is said to be antisymmetric or skew. In this case there are only three independent elements, since it is clear that k ij = 0 when i = j. The matrix form is 0 k 12 k 13 k 12 0 k 23. k 13 k 23 0 An antisymmetric tensor is closely equivalent to a vector. To see this, let k 12 = ω 3, k 13 = ω 2, k 23 = ω 1. Let the position vector be x with components x i, and form the product We than have v i = k ij x j. v 1 = ω 2 x 3 ω 3 x 2 v 2 = ω 3 x 1 ω 1 x 3 v 3 = ω 1 x 2 ω 2 x 1, i.e., in vector notation, v = ω x. Thus multiplication by an antisymmetric tensor is equivalent to vector multiplication by an equivalent vector. From these definitions it follows that any tensor may be expressed as the sum of a symmetric and an antisymmetric tensor, since we may write k 12 = 1 2 (k ij + k ji ) (k ij k ji ). (1.11)

10 10 CHAPTER 1. TENSORS In general, a tensor equation is not commutative, i.e., k ij y j y j k ji, (1.12) unless k ij is symmetric. If k ij is antisymmetric, clearly k ij y j = y j k ji. 1.2 Visualization of tensors The stress and strain tensors which we will be concerned (as well as many other tensors of physical importance) are both symmetric. Let us therefore see if we can find a geometrical interpretation of a symmetric tensor which will aid in visualizing its properties, just as an arrow in space depicts the properties of a vector. Let us suppose that k ij is a symmetric tensor defined by the linear vector function as in (1.3), v i = k ij y i (1.13) If we think of k ij as a point function in some medium, i.e., as having a specified value at each point in the medium, then (1.3) gives us the value of the vector v i at each point when the vector y j is known at that point. Let us now form the scalar product of the two vectors: v i y i = k ij y i y j. Writing out the terms of this product gives, since k ij = k ji, k ij y i y j = k 11 y k 22 y k 33 y (k 12 y 1 y 2 + k 13 y 1 y 3 + k 23 y 2 y 3 ) C (1.14) where C is a scalar function of y i. It then follows that v i is just half the derivative of C with respect to y i. For, taking the partial derivative of C with respect to y 1, C y 1 = 2k 11 y 1 + 2k 12 y 2 + 2k 13 y 3 = 2k 1j y 1, and similarly for y 2 and y 3, whence C y i = 2k ij y j = 2v i (1.15) Thus v is one-half the gradient of C in y i coordinates, and is everywhere normal to the surface C = C 1, where C 1 is constant. C = C 1 is the equation of 2 nd degree called the tensor quadric centered around the point in question. It may be an ellipsoid (Fig. 5), a hyperboloid of one or two sheets, or a degenerate form, such as two parallel planes, depending on the signs and values of the tensor elements.

11 1.2. VISUALIZATION OF TENSORS 11 Since v is everywhere normal to the surface C = C 1 and, since v is a vector point function which is unchanged by a rotation of the coordinate axes, the quadric must also be unchanged by such a rotation. But for any quadric surface there exist three mutually perpendicular principal planes of mirror symmetry, the normals to which are the principal axes. Thus if the coordinate system is rotated to a new primed orientation along the principal axes, the equation of the quadric surface in the primed system must be an even function of y 1, y 2, and y 3, i.e., be of the form k 1 1 y k 2 2 y k 3 3 y 2 3 = C. (1.16a) However, since the rotation of the coordinate axes cannot change the value of the scalar product v i y i, i.e., v i y i must = v i y i, the equation of the quadric surface must still be v i y i = k i j y i y j = C, (1.16b) whence, comparing (1.16a) and (1.16b), it follows that k i j = 0 when i j. We have thus shown that for any symmetric tensor there exists some orientation of the coordinate axes such that all the non-diagonal elements of the tensor vanish. In matrix form rotated to that orientation the tensor k ij becomes k k 2 0, 0 0 k 3 where k 2, k 2, and k 3 are associated with the principal axes and are called the principal values or eigenvalues of the tensor. Then v 1 = k 1 y 1, v 2 = k 2 y 2, and v 3 = k 3 y 3. Note that if, for example, y 2 = y 3 = 0, then v = k 1 y. Thus for any symmetric tensor there always exists at least one set of three perpendicular

12 12 CHAPTER 1. TENSORS directions along which y and v are collinear. Expressing the quadric equation in the form k 1 y1 2 + k 2 y2 2 + k 3 y3 2 = C 1, (1.16c) we see that each principal value is equal to the inverse square of the corresponding semi-axis of the particular quadric C 1 = 1. Note that the largest principal value is associated with the shortest semi-axis, and vice versa. Let us summarize the physical picture which we have built up. At each point is space at which the symmetric tensor k ij is defined, we may imagine the quadric surface v i y i = constant, defined in terms of the coordinates y i, as expressing the relationship between y and v at that point in space. One should not think of the tensor quadric as extending away from the particular point in physical space any more than one does a vector, (unless y i are spatial coordinates). The shape of the tensor quadric at each point is a function of k ij only, and remains unchanged as y and v change, although the scale factor, the magnitude of C, depends on the magnitude of y. We may thus imagine a family of quadric surfaces, from which we choose that one for which a given y is a radius vector. The normal to the surface at the end of the radius vector, of magnitude equal to one-half the magnitude of the gradient of C, represents v (Fig. 5). 1.3 Some useful tensors A tensor whose components are the same for any orientation of coordinate axes is called an isotropic tensor. The corresponding quadric must also appear the same for any orientation of the coordinates, hence must clearly be a sphere. But any centered sphere has the equation y y y 2 3 = a 2, where a is the radius of the sphere. Comparing with (1.14), we see that We may then write where k 11 = k 22 = k 33, and k ij = 0 when i j. k ij = kδ ij, k = k 11 = k 22 = k 33. (1.17) Thus any second-order isotropic tensor can be written as a constant times δ ij, and any axes are principal axes. Another useful isotropic tensor is the third-order tensor ɛ ijk, defined as zero if any two subscripts are equal, 1 for the subscripts equal to 1, 2, 3 in that order or in a cyclic rearrangement, and -1 for the arrangements of 1, 2, and 3. Thus ɛ 123 = ɛ 231 = ɛ 312 = 1, ɛ 321 = ɛ 213 = ɛ 132 = 1,

13 1.4. TENSOR INVARIANTS 13 and all other ɛ 123 = 0. This tensor provides us with a compact way of writing the vector product (cross product) of two vectors in subscript notation. It can be seen by expansion that if w = y v, then w i = ɛ ijk y j v k. (1.18) The curl of a vector may also be simply represented with the aid of ɛ ijk. If v = y, then v i = ɛ ijk y k (1.19) 1.4 Tensor invariants We have previously pointed out that the vector and tensor transformation equations follow from our requirement that the value of a vector or tensor should not depend on the orientation of the coordinate system to which it is referred. To express this value, we would like to find some quantity or quantities related to the vector or tensor which do not change with rotation of the coordinate axes. Such quantities are called tensor invariants; it is clear that they must be scalars. The only invariant of a vector y is the sum of squares of its components, y i y i. In terms of the graphical representation of a vector as a directed line segment in space, the vector invariant is simply (the square of) the length of the line segment. A second-order tensor has three invariants. The first is the sum of the diagonal elements, called the trace, or spur of the tensor, which we express by the symbol Θ: Θ = k ii = k 11 + k 22 + k 33. (1.20) It is easy to show that Θ is invariant: let the trace in a rotated coordinate system be k j j, then k j j = a ij a kj k ik = δ ik k ik = k ii The second invariant,, involves products of pairs of terms: = k 11 k 22 + k 11 k 33 + k 22 k 33 k 12 k 21 k 13 k 31 k 23 k 32 = 1 2 (k iik jj k ij k ji ), (1.21)

14 14 CHAPTER 1. TENSORS and the third is the determinant of the tensor elements: k 11 k 12 k 13 D = k 21 k 22 k 23 k 31 k 32 k 33. (1.22) = 1 6 ɛ ijkɛ rst k ir k js k kt. The proof or the invariance of and D is left as an exercise. If the tensor is symmetric, we may relate the invariants to the tensor quadric. It is convenient first to express them in terms of the principal values of the tensor: Θ = k 1 + k 2 + k 3 (1.23) = k 1 k 2 + k 1 k 3 + k 2 k 3 (1.24) D = k 1 k 2 k 3. (1.25) If b 1, b 2, and b 3 are the lengths of the semiaxes of the tensor quadric C = 1, we have from (1.16c) that k = 1. Then b 2 i Θ = 1 b b b 2 3 (1.26) = 1 b b2 2 b b2 3 b 2 1 b2 3 D = 1 b 2 1 b2 2 b2 3 (1.27) (1.28) Thus Θ is related to the lengths of the semiaxes, to the areas of the rectangles formed by the pairs of semiaxes, and D to the volume of the rectangular solid formed by the semiaxes. The three equations (1.26), (1.27), and (1.28) determine b 1, b 2, and b 3 when Θ,, and D are known. They can be thought of as expressing the invariance of the shape of the tensor quadric. For an antisymmetric tensor, Θ = 0, D = 0, and = k k k For the example considered earlier, we see that the square of the associated vector. = ω ω ω 2 3,

15 Chapter 2 Stress The forces acting on a volume element within an elastic medium can be divided between external and internal forces. The former type, such as gravitational, centrifugal, and electromagnetic forces, normally depend upon the mass or charge density of the volume element, hence are called body forces. The sources of body forces are outside the elastic medium so that reactions on the sources can be ignored. The internal forces, which result from action and reaction between neighboring elements of the medium, comprise tractions, pressures, and surface tractions acting on the surfaces of the volume element. The limit of this elastic force per unit area acting on a surface as the area becomes very small (but still large compared to molecular dimensions so that the medium may be considered a continuum) is called the stress or traction on that surface. The stress will, in general, be a function not only of position but of orientation of the surface considered, and is a vector quantity. Let us consider the interaction between two adjacent rectangular volume elements with faces parallel to the coordinate planes (Fig. S-1). Let their common face be normal to the x 1 direction: designate the volume element to the left by M 1 and that to the right by M 2 ; M 1 and M 2 are shown separate for clarity, but be thought of as joined together. We define a positive x 1 surface as one whose outward normal is in the positive x 1 direction; the positive x 1 face of M 1 is common with the negative x 1 face of M 2. 15

16 16 CHAPTER 2. STRESS The vector stress p 1 exerted by M 2 on the x 1 face of M 1 may have an arbitrary orientation in space; we will call its components in the x 1, x 2, and x 3 directions p 11, p 12, and p 13 respectively. In general, we designate by p i the stress which acts on an x 1 surface, and by p ij the stress component which acts on an x 1 surface in the x j direction. We define p ij as positive when it acts in the positive x j direction on a positive x i surface. When i = j, we speak of a tensional stress or tension if p ij > 0, and a compressional stress or compression if p ij < 0. When i j, p ij is a shear stress. Now the stress on the negative x 1 face of M 2 is, by Newton s third law, equal and opposite to that on the positive x 1 face of M 1. Thus for consistency, since the stresses are acting on the same point in the body, we must define the stress components on the negative x 1 face of M 2 as positive when they act in the negative coordinate directions. Therefore, the component p ij is also positive when it acts in the negative x j direction on a negative x i surface. If we now consider the opposite faces of a single volume element, we see that a positive stress tends to distort the volume in the same direction regardless of the face on which it acts: positive p 11, for example, acting on either x 1 face, tends to stretch the volume element, and a positive p 12 acting in either face tends to shear the volume element in a consistent sense. 2.1 Continuity conditions The stress components are subject to a continuity condition which we may state as follows. The stress component p ij is everywhere continuous in the x i and x j directions. To prove this it will suffice to show that p ij is continuous in the x i direction since, as we will show shortly, p ij = p ji. Consider the volume element M 1 with sides x i, density ρ, volume V, and center point Q (Fig S-1.5). By Newton s second law, the sum of the components of elastic and body forces acting in the x j direction must equal the mass times the x j component of acceleration, a j. Let F j be the component of body force per unit mass in the

17 2.1. CONTINUITY CONDITIONS 17 x j direction so that ρ V F j is the body force on M 1. Let us denote the stress components p ij evaluated at Q by p ij Q. Then p 1j on the center of the positive x 1 face will be p 1j Q+ 1 2 x1, with similar expressions for the other faces. Since M 1 is arbitrarily small, we may consider the elastic forces to be equivalent to concentrated forces acting at the center of each face. Thus ( ρ V (a j F j ) = ( + + p 1j Q+ 1 2 x1 p 1j Q 1 p 2j Q+ 1 2 x2 p 2j Q 1 Dividing both sides by V = x 1 x 2 x 3, 2 x1 ) x 2 x 3 ) x 2 x2 1 x 3 ( ) p 3j Q+ 1 p 3j 2 x3 Q 1 x 2 x3 1 x 2 ρ(a j F j ) = p 1j Q+ 1 2 x1 p 1j Q 1 2 x1 x 1 + p 2j Q+ 1 2 x2 p 2j Q 1 2 x2 x 2 + p 3j Q+ 1 2 x3 p 3j Q 1 2 x3 x 3. (2.1) Now hold x 2 and x 3 fixed, and let x 1 0. Then, in order for a j to remain finite the numerator of the first term must approach zero. Therefore lim [p 1j Q+ 1 x x1] = lim x 1 0 [p 1j Q 1 2 x1] = p 1j Q. i.e. p 1j is continuous in the x 1 direction. Letting x 2 and x 3 also approach zero, we see that in general p ij is continuous in the x i direction. Note that the stress is not necessarily continuous in all directions. For instances, p 12 may be discontinuous in the x 3 direction, and p 11 in both the

18 18 CHAPTER 2. STRESS x 2 and x 3 directions. As an example, let us consider two rectangular volume elements, one of steel and the other of hard rubber, are welded together at a common x 1 face (Fig S-2). If we stretch the joint bar in the x 1 direction, most of the extension will take place in the rubber, the ratio of strains in the rubber and steel being such that the stress will be the same in each. On the other hand, if we stretch the bar in the x 2 direction in such a way that the extensions in the steel and rubber are the same, then p 22 in the steel will be much greater than in the rubber, and p 22 will be discontinuous in the x 1 direction. Since the stress is continuous in the sense defined, and is finite in any physical system, it is reasonable to assume that the derivatives of the stress are continuous and finite in the same sense. This allows us to express the stress components on the faces of M 1 in terms of a Taylor series around the point Q: p 1j Q± 1 = p 2 x1 1j ± p 1j Q x 1 x x 2 i ( ) 2 x1 ± Q 2 with similar equations for p 2j and p 3j. Because x i is small, each term in the series is small compared to the one of next lower order. Second and higher terms will be negligible for our purposes; we must retain the first derivative terms since, the differences between stresses on opposite faces will be important. Thus, to sufficient accuracy, p 1j Q± 1 2 x1 = p 1j ± p 1j x 1 Q x 1 2, p 2j Q± 1 2 x2 = p 2j ± p 2j x 2 Q x 2 2, and p 3j Q± 1 2 x3 = p 3j ± p 3j x 3 Q x 3 2. (2.2) We next wish to show that p ij is a symmetric tensor. To show symmetry we will prove that p 12 = p 21 ; similar arguments could be used to show that p 13 = p 31 and p 23 = p 32. Consider the moment about an axis through the center of M 1 parallel to the x 3 axis (Fig. S-1). Only the shear stress components exert a moment since the

19 2.1. CONTINUITY CONDITIONS 19 stress forces on the x 3 faces act as points on the axis, and the extensive stresses on the x 1 and x 2 faces have no moment arm. Taking moments positive when they tend to rotate M 1 from the x 1 axis toward the x 2 axis and using (2.2), we have for the moment due to the stress on an x 1 face: (p 12 ± p 12 x 1 x 1 2 )1 2 x 2 x 3 x 1, x 2 x 3 being the area of the face, 1 2 x 1 the moment arm, and the plus and minus signs referring to the positive and negative x 1 face, respectively. The sum of these moments is simply p 12 V. A similar analysis follows for the p 21 stresses; since they act in the opposite sense from the p 12 stresses, their resultant moment is p 21 V. (The derivative terms cancel because of symmetry in taking moments about a centered axis. For moments about any other axis, these terms will not disappear identically, but will still be negligible compared to the zeroorder terms. The moments thus will be the same about any axis parallel to the x 3 direction. Similar shear stress components acting on opposite sides of a small volume element are elements of a couple. Let the body force moment on M 1 acting around x 3 be ρf V L, where L is the moment arm. This force must act somewhere within M 1, since any forces acting outside are transmitted only by means of the stress. Thus L must be of the order of x 1 or smaller. The total moment is thus (p 12 p 21 + ρf L) V. We equate this to the 1 product of the moment of inertia, 12 ρ( x2 1 x 2 2), and the angular acceleration Ω, and divide both sides by V, giving p 12 p 21 + ρf L = 1 12 ρ( x1 1 + x 1 2) Ω As we let x i approach zero, and require Ω to remain finite, the right-hand side approaches zero, as does ρf L for finite body force. Hence p 12 = p 21, (2.3) completing the proof of symmetry. To demonstrate the tensor character of the stress, we will first cite the proof normally given (Fig. S-3). Consider a tetrahedron three sides of which coincide with the coordinate planes. Let the fourth face be arbitrarily oriented, and call the axis normal to it x 1. Designate the areas of the faces by S 1, S 2, and S 3, respectively. For any S i we may write S i = a 1 i S 1 (2.4) since each face in a coordinate plane has a side in common with the diagonal face, and the corresponding heights are in the ratio a 1 i : 1. We may consider any of the S i as the projection of S 1 onto the corresponding coordinate plane.

20 20 CHAPTER 2. STRESS The total elastic force on the tetrahedron in the x j direction, p 1 j S 1 p ij S i equals the mass of the tetrahedron times the x j component of acceleration, minus body force terms which are also proportional to the mass. As we let the size of the tetrahedron become very small, the mass will approach zero as the cube of the linear dimension, whereas the area of the faces will diminish only as its square. Hence for the acceleration to remain finite each force component must be identically zero, i.e. Using (2.4), this becomes p 1 j S 1 = p ij S i (2.5) p 1 j S 1 = p ij a 1 i S 1 whence p 1 j = a 1 ip ij. Since an analogous derivation holds for the components of p 2 write in general p m j = a m ip ij. and p 3, we may This equation expresses the components of the stress vector acting on the x m face in terms of the stress components on the other faces. Project this vector onto the x n direction: p m n = a n jp m j = a m ia n jp ij, (2.6) which we recognize as the tensor transformation law. This completes the proof. Let us now consider the stress tensor in terms of a linear vector function. (2.4), which was derived geometrically, simply expresses the vector character

21 2.1. CONTINUITY CONDITIONS 21 of the elemental area. For if we multiply both sides by a 1 i and perform the indicated summation, we have a 1 i = a 1 ia 1 i S 1 = S 1. But the 1 axis was arbitrarily oriented; the same proof would follow for the 2 and 3 axes, hence S k = a k i S i, (Why do the a 1 ia 1 i disappear?) which is the vector tranformation law. We may now think of the diagonal face of the tetrahedron as any element of arbitrarily oriented in the x i coordinate system. It is designated by the vector S directed along the normal to the surface, and has Cartesian components S i represented by the three right triangular faces of the tetranedron. Now we may look at the force equation in a new light. If we let the elastic force vector acting on S be denoted by f we may still write (2.5) in the form f j = p ij S i, (2.7) showing that f is a linear vector function of S. p ij is therefore a tensor. If we consider the diagonal face and the three right triangular faces as altenative surfaces on the same side of a volume element, (2.5) shows that the elastic force is the same in either case. More generally, the elastic force on any set of surfaces whose vector sum is S is the same regardless of the configuration. Since the stress is a symmetric tensor, it may be represented by the stress quadric C = f i S i = p ij S i S j = constant C 1. In this case, the radius vector to the quadric surface is S and the elastic force acting at the center of the quadric on S, given by f i = 1 C 2 S i, is depicted by the normal to the quadric at the point of intersection (Fig. S-4). The stress quadric does not have physical extension in the medium, but it is described in a space with coordinates S i.

22 22 CHAPTER 2. STRESS Principal stresses, P i exist along the principal axes of the stress quadric, hence any state of stress at a given point can be represented by three tensions or compressions alone. In general, however, the directions of the principal axes will vary throughout the medium. It is usually more convenient to work with shear stresses than to relate the stress to the principal axes. In the course of demonstrating the stress continuity condition, we derived (2.1) expressing the dynamic equilibrium of the volume element M 1. The requirement that the elastic forces, which appear on the right-hand side of (2.1), remain finite as M 1 becomes very small led to the continuity condition that there be no finite difference between stresses acting on surfaces an infinitesimal distance apart. There can, however, be infinitesimal stress differences leading to finite accelerations. To evaluate these, we substitute (2.2) into (2.1), obtaining ρ(a j F j ) = p 1j x 1 + p 2j x 2 + p 3j x 3 = p ij x i (2.8) Interchanging subscripts i and j, using the symmetry of p ij, and substituting a i = 2 u i t, where u 2 i is the displacement, we rewite (2.8) as ρ 2 u i t 2 = ρf i + p ij (2.9) which is the dynamic equilibrium equation, or equation of motion for any infinitesimal volume under the action of the combined elastic and body forces. If there is no acceleration, (2.9) reduces to the static equilibrium equation. ρf i + p ij = 0. (2.10)

23 2.1. CONTINUITY CONDITIONS 23 Multiplying (2.8) by V and using the symmetry of p ij, we see that the total elastic force on M 1 has components pij V. The force on a body of finite size will then be tbe sum of the forces on the constituent volume elements, i.e. p ij dv (2.11) V However, from (2.7), the elastic force on M 1 also has components p ij S j. Let the unit vector along S have components n j ; then S j = Sn j, where S is the scalar area. If we combine volume elements to produce a finite body, the internal stresses will produce no net force on the body, since the forces will be equal and opposite on opposite sides of each internal surface element. The net elastic force will thus be just the sum of those acting on the external surfaces of the body, i.e. p ij n j ds (2.12) Combining (2.11) and (2.12), we obtain p ij dv = V S S p ij n j ds, (2.13) which is an example of Gauss s, or the divergence theorem applied to a secondorder tensor.

24 Chapter 3 Strain In a rigid, incompressible body, the distance between two material particles cannot change. The general displacement of the body comprises a translation, in which all particles are displaced by the same distance, and a rotation about some axis, in which the distance of all points from the axis of rotation is unchanged. In a deformable medium, however, distances between particles are not fixed; the size and shape of any portion of the medium can change. Seismic waves consist of strains propagating through such a medium. In order to arrive at an analysis of strains, we must first consider the general displacement of a deformable body. Let us start at a time t = 0 when there is no deformation, so that every particle in the medium is in its equilibrium position. Let us consider a particular particle which at time t = 0 has coordinates b i, and let us say that at some other time t = t 1 this particle is at the point P with coordinates x i (Fig. D-1). The displacement of the particle from its equilibrium position, which we designate u i is simply u i = x i b i. (3.1) Our problem is to describe the variations of displacement in time and space. Note that a particular set of b i always refers to the same particle and a particular set of x i always refers to the same point in the coordinate system. Let the coordinates of a neighboring point Q be x i + x i, where x i is a infinitesimal quantity, and suppose that the particle found at Q at time t = t 1 had initial coordinates b i + b i. The difference in displacement between the particle at Q and the particle at P, u i is u i (Q) u i (P ) = u i = x i b i. (3.2) Figure D-2 shows this situation projected onto the x 1 x 2 plane. We do not wish to consider ruptures in the medium nor fluid flow, so we require u i to be continuous, hence u i and b i are infinitesimal quantities since x i is. Note that the symbol means the difference between the value of Q and P with time fixed. 24

25 25 Figure 3.1: D-1 Figure 3.2: D-2

26 26 CHAPTER 3. STRAIN Figure 3.3: D-3 Since we are also interested in temporal variation, we now advance time an infinitesimal amount δt. The particle formerly at P will move to a new position which we designate as P and a new particle, formerly at some point P will move to P (Fig. D-3). There are, therefore, two ways in which we can describe the change δu i at time δt. We may either follow the particle formerly at P, in which case b i is unchanged, or we may consider the changes at point P, in which case it is x i which is unchanged. These two points of view are the bases of the Lagrangian and Eulerian descriptions of the motion, respectively. In the Lagrangian description we consider b i and t to be the independent variables in terms of which we describe the position and displacement of any particle (or any other function). Partial derivatives with respect to time imply that b i is fixed, i.e. that rates of change in the neighborhood of a particular particle are being considered. and ( u i ) L = u i b j b j (3.3) (δu i ) b = ( u i t ) bδt, (3.4) where the derivatives are evaluated at the point P. In the Eulerian description it is the x i which is independent. Partial derivatives which respect to time now imply rates of change at a particular point in the coordinate system, occupied successively by various particles. The differential relations are ( u i ) E = u i x j (3.5)

27 27 and (δu i ) x = ( u i t ) xδt, (3.6) In the equations 3.4 and 3.6, the subscript indicates the variable which is held fixed. Note that we can hold either b i of x i fixed in a partial derivative, but not both. Hence ui b j implies only t fixed, not x i. Similarly with ui. u i refers to the difference between displacements at points P and Q in both Eulerian and Lagrangian coordinates, hence has the same value at a particular time in either description. On the other hand, δu i refers to points P and P in Lagrangian coordinates, but to P and P in Eulerian coordinates, hence has a different value in the two descriptions. It is also important to remember that Lagrangian and Eulerian coordinates refer to the same coordinate axes, whether this system of reference is fixed in space or moving with some general motion of the medium as a whole. Lagrangian and Eularian coordinates must not be confused with fixed and moving coordinate systems. It is often useful to be able to express one time derivative in terms of the other. Consider u i to be a function of s i and t, and x i in turn, to be a function of b i and t. Then ( u i t ) b = u i t ) x + ( t ) u i b This equation states that the rate of change of u i moving with a given particle, which we call the material rate of change equals the rate of change at a fixed point, called the local rate of change plus the scalar product of the gradient of u i and the velocity with which the particle moves through the gradient. It is common practice to write dui dt for the material rate of change instead of ( dui dt ) b, and du i instead of (δu i ) b for the corresponding differential. Mathematically, this notation is improper since u i is a function of the two independent variables b i and t, hence dui dt does not exist, and du i = (δu i ) b + u i. We will not, however, have occasion to employ the true total differential, so that we may simply define the operator d as indicating the change associated with a particular particle at time dt. Considerable notational simplification results, and confusion is avoided by keeping the significance of the symbols in mind. We then write du i dt = u i t + v u i j, (3.7) where v j = dx j (3.8) dt is the velocity of a particular material particle, and a partial differentiation with respect to t implies that x i is fixed. Also, rewiting (3.4), du i = du i δt. (3.9) dt The concepts we have developed hold not only for u i but for any other function f of position and time. The relevant differential relations are obtained by substituting f for u i in (3.3) through (3.7).

28 28 CHAPTER 3. STRAIN An important result which we will now prove is the interchangeability of the order of operations δ and. We will show this as applied to u i ; the proof follows similarly for any other function. At time t, the displacements at P and Q are u 1 and u i + u i, respectively. In Eulerian coordinates, the change in displacement in time δt at Q is, by (3.6) δ(u i + u i ) = t (u i + u i )δt = u i t δt + t ( u i)δt = δ(u i ) + δ( u i ). Thus at time t + δt, the new displacement of the particle formerly at Q, is u i + u i + δu i + δ( u i ). On the other hand, since the displacement at P at time t + δt is u i + δu i, the difference in displacement at Q at the same time is, by (3.5) (u i + δu i ) = t (u i + δu i ) x j = u i t x j + t (δu i) x j = (u i ) + (δu i ), so that the displacement at Q at time t + δt can also be written u i + δu i + u i + (δu i ) Equating these two expressions yields the desired result: δ( u i ) = (δu i ). (3.10) Similarly, in Lagrangian coordinates, d( u i ) = (du i ). (3.11) By combining (3.5),(3.6), and (3.10), it is easy to see that this is just the proof of the interchangeability of the order of partial differentiation with respect to space and time. A graphical representation of this proof is shown in figure D-4.

29 29 As we pointed out previously, the important property of a deformable body is that the distance between two particles can change. Let us now find an expression for the change in distance between particles now at P and Q from that in their initial states. The square of the initial distance s o is given by s 2 o = b i b i, and the square of the distance s at any other time by so that the difference is s 2 = x i x i, s 2 s 2 o = x i x i b i b i. (3.12) In Eulerian coordinates we may apply the differential form of (3.5) to the dependent variable b i so that whence From (3.1) so that b i b i = b k δb k = b k x i x i b k x j, s 2 s 2 o = (δ ij b k x i b k x j ) x i x j. s 2 s 2 o = b k = (x k u i ) = δ ij = u k, x i x i x i = [ δ ij (δ ik u k ( ui + u j u k u k x i x i x i )(δ jk u k ] ) x i x j. )

30 30 CHAPTER 3. STRAIN We now define then η ij 1 2 ( u i + u j x i u k x i u k ); (3.13) s 2 s 2 o = 2η ij x i x j. (3.14) The set of quantities η ij is called the Eulerian strain tensor. In an analogous manner, using (3.3), we find that in Lagrangian coordinates, s 2 s 2 o = 2e ij b i b j where (3.15) e ij 1 2 ( u i b j + u j b i + u k b i u k b j ) (3.16) is the Lagrangian strain tensor. Note that in either formulation s 2 s 2 o is the change in the square of the infinitesimal distance between P and Q occurring in any finite time interval. We may now write for the displacement difference between P and Q in Eulerian coordinates, using (3.5) and (3.13): u i = t i + r i, (3.17) where t i η ij x j, (3.18) and r i φ ij x j, (3.19) φ ij 1 2 ( u i u j x i + u k x i u k ) (3.20) When η ij = 0, there is, by (3.14) no change in the length of any fiber in the medium. Then u i = r i, whence it follows that r i is the displacement difference due to rotation alone. We thus call φ ij the Eulerian rotation tensor. Clearly t i is deformation involving no rotation (because addition of a rigid rotation to a state of strain cannot change s 2 s 2 o, hence cannot change η ij or e ij ). The proof or the tensor character of the strain and rotation is simply that they are the coefficients of the linear vector equations (3.18) and (3.19) In Lagrangian coordinates (3.17) still holds, but now where is the Lagrangian rotation tensor. t i = e ij b j and (3.21) r i = φ ij b j (3.22) φ ij 1 2 ( u i b j u j b i + u k b i u k b j ) (3.23)

31 31 We have shown thus far that for any neighboring points P and Q, the displacement at Q can be expressed in terms of that at P by the equation u i (Q) = u i (P ) + u i = u i (P ) + r i + t i Since u i (P ) is the same for any Q, it is a translation; we have already shown that r i is a rotation and t i is a deformation. The strain tensors are symmetric by definition, so by Chapter 1, there exist three principal axes, x i, and three principal strains, e i (or η i ), such that the components of the deformation vector, t i given by t 1 = e 1 x 1, t 2 = e 2 x 2, t 3 = e 3 x 3, (3.24) is directed along the corresponding principal axes, and is thus either an extension of a contraction along that axis. This concludes the proof of an important theroem due to Helmholtz: The most general motion of a sufficiently small element of a continuous, deformable body can be represented as the sum of a translation, a rotation, and an extension or contraction in three mutually orthogonal directions. Thus far in our discussion we have made no assumptions about the magnitude of the deformation. It happens, however, that elastic deformations are very small, that is, all strain components are 1. Furthermore, if we allow our coordinate axes to move with any rigid rotation of the body as a whole (e.g. fix them in the earth), the components of the are also very small. It follows that u to a very good degree of approximation we may assume all i and ui b j 1. The product terms in the strain and rotation tensors can be dropped, and these tensors become linear functions of the displacement derivatives. It further follows that we may drop the distinction between Lagrangian and Eulerian derivatives. To see this, we write, from (3.5), (3.3), (3.2) x j = b j b j = Rearranging terms and using (3.5): ( b j ) x j = ( x j u j ). b j u j = b j u j x k x k. Since this must be true for any x j, and since uj x k 1, = (3.25) b j Also from (3.7) du i dt = u i t (3.26)

32 32 CHAPTER 3. STRAIN so that the Eulerian and Lagrangian time derivatives of displacements are also approximately equal. Note the equivalence of the time derivatives applies only to the displacement (or any other function whose gradient is 1), whereas the equivalence of spatial derivatives applies to any properly defined function. One further result is useful. From (3.3) we have but also, from (3.11) and (3.3) ( ) ui d( u i ) = d b j = d b j ( ui b j d( u i ) = (du i ) = (du i) b j b j, both equations holding for any b j. Hence or, using (3.25) ( ) (ui ) d b j ( ) (ui ) d = ) b j (du i ), ) b j, = (du i ), (3.27) The greatly simplified theory which follows from these approximations is called infinitesimal strain theory. The strain tensor becomes e ij = 1 ( ui + u ) j (3.28) 2 x i and the rotation tensor is φ ij = 1 2 ( ui u ) j x i (3.29) and, from (3.17), (3.21), (3.22), a general deformation is expressed u i = (e ij + φ ij ) x j. (3.30) The rotation tensor is now clearly antisymmetric, whence if follows that there is an associated vector, which we will designate by φ, where φ 1 = φ 23, φ 2 = φ 31, and φ 3 = φ 12, such that Note that r = φ u 2φ = u, (3.31) thus in infinitesimal strain, the curl of the displacement vector is a measure of the amount of rotation involved in the displacement. The rotational displacement

33 33 of Q relative to P, r, is one half the vector product of the curl of u and the position of Q relative to P. Let us now consider the significance of the strain tensor. We first imagine a small rectangular volume with diagonally opposite corners at P and Q, and with edges of length x i parallel to the principal axes of strain. Its volume, V, is x 1 x 2 x 3. Suppose the medium undergoes strain with no rotation; The coordinates of then become, from (3.17), x i + t i, i.e. by (3.24), (1+e 1 ) x 1, (1 + e 2 ) x 2, and (1 + e 3 ) x 3 ; these quantities are also the new lengths of the sides of the volume element. If we now add a rotational displacement, the coordinates of Q will change, but the lengths of the edges of the volume element wil not, since, as we have shown, the rotation involves no change in length of any position vector. Thus, for any displacement we can say that the new volume of the volume element, V, is (1 + e 1 )(1 + e 2 )(1 + e 3 ) x 1 x 2 x 3. Thus V = V (1 + e 1 )(1 + e 2 )(1 + e 3 ) V (1 + e 1 + e 2 + e 3 ). We define the cubical dilatation as the specific increase in volume: V V V = e 1 + e 2 + e 3 which we recognize from chapter 1 as the trace of the strain tensor. Since this is the only trace we will use regularly, we will designate it by Θ, without a subscript. But we have shown that Θ is an invariant of the tensor hence, by eq. 1.20, Θ = e ii = u i. (3.32) We return now to the general form of the strain tensor, and consider the meaning of the individual components. Consider first a fiber in the medium parallel to the x 1 axis and with length x 1. After displacement, the length of the fiber, x 1, is x 1 = x 1 + u 1 = x 1 + (e 1j + φ 1j ) x j = x 1 + e 11 x 1 the last step following since φ 11 = 0 and, in this case, x 2 = x 2 = 0. This the specific change in length x 1 x 1 x 1 = e 11. Since the fiber could just as well be taken along the x 2 or x 3 axis we see that, for any deformation, the diagonal strains are the specific changes in length along the corresponding axes. To understand the significance of the non-diagonal terms, we consider a small rectangular solid with its corner at P and with edges x 1 parallel to