Infinitesimal Strains and Their Accumulations

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Infinitesimal Strains and Their Accumulations Chapter 2 Infinitesimal Strains and Their Accumulations In this chapter we will study infinitesimal deformation and the rate of deformation. Most geologic structures are the integral of infinitesimal deformations, so that it is important to understand how infinitesimal and finite deformations are related. 2.1 Infinitesimal strain The lifetime of a deformation zone or an orogenic belt can be over millions of years, during which time rock masses undergo large deformations. Tectonism is fairly slow and the incremantal defor- mation in our usual time-scale is very small. Therefore, the deformations detected by a geodetic survey are usually very small. Geology detects the result of long-term deformations. Geological measurement is much less accurate than geodetic one. Thus, it is usually difficult for geology to detect such small strains as geodesy detects. However, it is possible in some cases1 that geologic structures provide clues to quantify those strains. In addition, it is necessary to know the theories of infinitesimal strains to understand the following sections in which the dynamics of tectonic deformatoins is discussed. 2.1.1 Definition Infinitesimal strain is a very small strain. If a deformation gradient tensor is nearly equal to the iden- tity tensor (F ≈ I), it is called infinitesimal deformation. Let us first rewrite F with the displacement2, u = x − ξ. 1Note that strain is defined as the change of length compared to the original one. Given a tiny strain, the associated length change can be large enough for detection by geological measurement. 2The famous textbook by Turcotte and Schubert [245] defines the displacement u = ξ −x in order to deal with contraction and compression as possitive strain and stress, respectively. 31 32 CHAPTER 2. INFINITESIMAL STRAINS AND THEIR ACCUMULATIONS We assume that the deformation and associated displacement are very small. Namely, ξ ≈ x. Three components of ξ are independent of each other, consequently ∂ξi/∂ξj = δij. Therefore, we have the deformation gradient tensor ∂x ∂(ξ + u) ∂u ∂u F = = = I + ≈ I + . ∂ξ ∂ξ ∂ξ ∂x The last term is a tensor indicating the spatial gradient of displacement, so that we use the expression ∇u = ∂u/∂x for it. Taking the limit u → 0, we obtain F = I + ∇u, where the small departure from I is indicated by ∇u = ∂u/∂x. The equation F = I indicates no deformation. Let us write the difference as δF, with which we have the deformation gradient for infinitesimal deformation F = I + δF (δF ≈ O), (2.1) where δF = ∇u. In this case, 2 G + I = FT · F = (F + δF) T · (F + δF) ≈ I + δF + δFT, where the quadratic term δFT · δF was neglected. Geometrical linearity applies if this approxima- tion is satisfied. Rearranging the above equation, we have Green’s strain tensor 1 1 G = δF + δFT = ∇u + (∇u)T , (2.2) 2 2 the components of which are 1 ∂uj ∂ui 1 ∂uj ∂ui Gij = + = + . (2.3) 2 ∂ξi ∂ξj 2 ∂xi ∂xj Almansi’s strain tensor (Eq. (1.25)) is linearlized to be − −1 1 2 A + I = 2 I − F · FT = 2 I − I + δF · I + δFT −1 ≈ 2 I − I + δF + δFT ≈ I + δF + δFT. Therefore, Green’s and Almansi’s strain tensors are identical (G = A) for infinitesimal strains. We, accordingly, definethe infinitesimal strain tensor, 1 T 1 ∂uj ∂ui E = ∇u + (∇u) or Eij = + . (2.4) 2 2 ∂xi ∂xj 2.1. INFINITESIMAL STRAIN 33 This is often referred to as a strain tensor. Elongation and contraction are indicated by the pos- itive and negative components of this tensor (2.1.3). If we use the opposite sign convention, the infinitesimal strain tensor is written as 1 1 ∂u ∂uj e = − ∇u + (∇u) T or = − i + . 2 2 ∂xj ∂xi It is obvious that the infinitesimal strain tensors E and e satisfy the following equations, E = ET, e = eT, E = −e. (2.5) 2.1.2 Relationships among strain tensors for infinitesimal deformations We have seen that the infinitesimal strain tensor E was derived from G and A by geometrical lin- earization. Then, how is E related to other strain tensors? By the linealization, the right and left Cauchy-Green tensors become T 1 T 1 U = I + δF · I + δF 2 ≈ I + δF + δF 2 1 ≈ I + δF + δFT = I + G = I + E = V, (2.6) 2 indicating that both tensors coincide with each other for infinitesimal strains. The orthogonal tensor R, which is related to the deformation gradient F by the equation F = R · U, becomes a tensor approximately equal to the unit tensor: T −1 T − δF δF δF δF R = F · U 1 = I + δF · I + + ≈ I + δF · I − − 2 2 2 2 1 ≈ I + δF − δFT = I + X. (2.7) 2 The last term is called the rotation tensor defined by the equation 1 T 1 ∂ui ∂uj X ≡ ∇u − (∇u) or Ωij = − . (2.8) 2 2 ∂xj ∂xi E and X are the symmetric and antisymmetric parts of ∇u, respectively. Using Eqs. (2.6) and (2.7), we obtain U + R = 2 I + δF = I + F = 2 I + G + X. Thus F = U + R − I = V + R − I = I + G + X. (2.9) Since G approaches E for small strains, F = I + E + X. (2.10) For finite strains F = R · U = U · R, as matrix maltiplications are generally non-commutative. By contrast, matrix summations are commutative, so that Eq. (2.10) indicates that the order of rotation X and shape-change E are commutative for infinitesimal deformations. 34 CHAPTER 2. INFINITESIMAL STRAINS AND THEIR ACCUMULATIONS 2.1.3 Physical interpretation of infinitesimal strain and spin tensors To interpret the tensors E and X, we shall see how a small rectangle element deforms. Let us examine the diagonal components of E, first. If displacement is parallel to the coordinate axis O-1 (Fig. 2.1(a)), ∂ui u2 = u3 = 0, = 0(i = 1, 2, 3; j = 2, 3). ∂xj Therefore, all the components of E vanishes except for E11. In this case, neglecting higher-order terms, we have the length change ds ds0 ∂u1 ∂u1 ds − ds0 ≈ u1 + dx1 + dx1 − u1 − dx1 = dx1 = E11dx1. ∂x1 ∂x1 Therfore, we obtain (ds − ds0)/ds0 = E11. Accordingly, the diagonal terms of E represent the elongation along the coordinate axes. Using the angles θ and ψ shown in Figs. 2.1(b) and (c), we have ∂u1 ∂u2 2E12 = 2E21 = + = tan ψ + tan θ. ∂x2 ∂x1 For infinitesimal deformations, the last two terms of this equation become ψ + θ. These two com- ponents indicate the angular decrease of the corner at which two sides of the rectangle originally parallel to the axes O-1 and -2 meet. The indices of E12 and E21 correspond to the axes. Conse- quently, 2E12 and 2E21 represent the engineering shear strain of the corner. The term “shear strain” often refers to the diagonal components of E, which are half of the engineering shear strains. Com- paring Figs. 1.11 and 2.1(b), the parameter q in Eq. (1.13) is related to the displacement gradients as ∂u1/∂x2 = 2q and ∂u2/∂x1 = 0. Therefore, q equals shear strain for infinitesimal simple shears. The infinitesimal strain tensor is symmetric (Eq. (2.5)), so that it has three real eigenvalues, E1, E2, and E3, and three mutually perpendicular eigenvectors. Consider a rectangular parallelepiped whose edges are parallel to the eigenvectors. The length of the edge parallel to the ith eigenvector is elongated by (1 + Ei). Therefore the volume change is given by V − 1 = (E1 + 1)(E2 + 1)(E3 + 1) − 1 ≈ trace E = EI, (2.11) V0 where V0 and V stand for the volume before and after deformation. This quantity is called the dilatation, and is equal to the first basic invariant of the tensor E for infinitesimal strains. Figure 2.1(d) shows the counterclockwise rotation of a rectangle around the O-3 axis. Two sides at the corner were parallel to the coordinate axes before deformation. We are considering very small rotations associated with a infinitesimal deformation, so that the sides that were initially parallel to the O-1 and -2 axes are rotated by the angles ∂u2/∂x1 and −∂u1/∂x2, respectively. If it is a rigid-body rotation, the two quantities are equal. The average of these angles is 1 ∂u2 ∂u1 Ω12 = − . 2 ∂x1 ∂x2 2.1. INFINITESIMAL STRAIN 35 Figure 2.1: The gradient of u and the deformation of a rectangle (dotted line), the sides of which were originally parallel to the coordinate axes. Accordingly, the ijth component of X represents the mean rotation about the kth coordinate axis, where i = j = k. Generally an antisymmetric tensor has three independent components. The vector that is made up of these components is called the axial vector of the antisymmetric tensor. Rotation vector = T (1,2,3) is defined as the axial vector of the rotation tensor: ⎛ ⎞ 0 −3 2 ⎝ ⎠ X = 3 0 −1 . (2.12) −2 1 0 The above equation is alternatively written as Ωij = − ijkk. (2.13) Not only the correspondence of the components, but also the tensor and vector show an important property—for an arbitrary vector a the dot-product of the tensor and the cross-product of the vector 36 CHAPTER 2. INFINITESIMAL STRAINS AND THEIR ACCUMULATIONS are identical: X · a = × a. (2.14) This is a vector quantity, and the first component is Left-hand side = −3a2 + 2a3, Right-hand side = 2a3 − 3a2, therefore both sides are identical. The identity of Eq. (2.14) is readily demonstrated for other components. Combining Eqs. (2.8) and (2.12), we obtain 1 ∂u ∂u ∂u ∂u ∂u ∂u = 3 − 2 , 1 − 3 , 2 − 1 T, 2 ∂x2 ∂x3 ∂x3 ∂x1 ∂x1 ∂x2 which is the expression of the rotation vector by means of displacement.
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