Principal Stresses

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Principal Stresses Chapter 4 Principal Stresses Anisotropic stress causes tectonic deformations. Tectonic stress is defined in an early part of this chapter, and natural examples of tectonic stresses are also presented. 4.1 Principal stresses and principal stress axes Because of symmetry (Eq. (3.32)), the stress tensor r has real eigenvalues and mutually perpendicu- lar eigenvectors. The eigenvalues are called the principal stresses of the stress. The principal stresses are numbered conventionally in descending order of magnitude, σ1 ≥ σ2 ≥ σ3, so that they are the maximum, intermediate and minimum principal stresses. Note that these principal stresses indicate the magnitudes of compressional stress. On the other hand, the three quantities S1 ≥ S2 ≥ S3 are the principal stresses of S, so that the quantities indicate the magnitudes of tensile stress. The orientations defined by the eigenvectors are called the principal axes of stress or simply stress axes, and the orientation corresponding to the principal stress, e.g., σ1 is called the σ1-axis (Fig. 4.1). Principal planes of stress are the planes parallel to two of the stress axes, or perpendicular to one of the stress axes. Deformation is driven by the anisotropic state of stress with a large difference of the principal stresses. Accordingly, the differential stress ΔS = S1 − S3, Δσ = σ1 − σ3 is defined to indicate the difference1. Taking the axes of the rectangular Cartesian coordinates O-123 parallel to the stress axes, a stress tensor is expressed by a diagonal matrix. Therefore, Cauchy’s stress (Eq. (3.10)) formula is ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ t1(n) σ1 00 n1 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ t(n) = t2(n) = 0 σ2 0 n2 . t3(n) 00σ3 n3 1ΔS is sometimes called the stress difference [2]. However, we keep this term for another quantity (Eq. (11.19)). 101 102 CHAPTER 4. PRINCIPAL STRESSES Figure 4.1: Stress ellipsoid and principal stress axes. White lines indicate the principal planes of stress. ∴ t1(n) = σ1n1,t2(n) = σ2n2,t3(n) = σ3n3. (4.1) The planes perpendicular to the stress axes have unit normals, n = (100)T,(010)T,(001)T. Hence, the traction vectors upon the planes are ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ σ1 0 0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ t(n) = 0 , σ2 , 0 . 0 0 σ3 Therefore, the principal stresses equal the normal stresses working on the principal stress planes. Shear stress vanishes on the planes. The normal and shear stresses are written in this coordinates as 2 2 2 σN = t(n) · n = σ1n + σ2n + σ3n (4.2) 1 2 3 1/2 = | |2 −| |2 1/2 = 2 2 + 2 2 + 2 − 2 + 2 + 2 2 σS t(n) σN σ1n1 σ2n2 σ3n3 σ1n1 σ2n2 σ3n3 . (4.3) If there is a plane on which traction vanishes, the state of stress is said to be plane stress (§3.4). The stress tensor corresponding to this state has a null principal stress. If σ3 = 0, traction on the 2 + 2 + 2 = σ1σ2-plane vanishes. Combining Eq. (4.1) and n1 n2 n3 1, we have t (n) 2 t (n) 2 t (n) 2 1 + 2 + 3 = 1. (4.4) σ1 σ2 σ3 This is an ellipsoid equation whose principal axes coincide with the stress axes and radii equals |σ1|, |σ2| and |σ3| (Fig. 4.1). This is called Lame’s´ stress ellipsoid. Regarding t(n) as a position vector with its initial point at the coordinate origin, the end point of the vector is on the ellipsoid, provided that 0 <σ3. The ellipsoid becomes a sphere for the hydrostatic state of stress and its radius equals the hydrostatic stress. The stress ellipsoid is given uniquely from the stress tensor regardless of the orientation of coordinate axes, although Eq. (4.1) was derived for a specific orientation of the axes. The basic invariants (§C.6) of the stress tensor r are σI, σII and σIII. Those of the tensor S are SI, SII and SIII. These quantities designate the “absolute values” of the stress tensor, independent from the orientation of rectangular Cartesian coordinates. We can define variables with the same 4.2. TECTONIC STRESS 103 independence from the basic invariants. Among them is the stress ratio σ − σ Φ= 2 3 . (4.5) σ1 − σ3 It is seen that the ratio is in the range of 0 ≤ Φ ≤ 1. When the state of stress is hydrostatic, the stress ratio is not defined. For anisotropic stress states, the stress ratio designates the shape of the stress elliosoid. If Φ=0, σ3 = σ2 <σ1, then, the ellipsoid is prolate. Φ=1 corresponds to an oblate ellipsoid, σ3 <σ2 = σ1. The stress ellipsoid has axial symmetry for the extreme cases Φ=0 and 1. In contrast, intermediate Φ indicates triaxial stresses σ3 = σ2 = σ1. In this case, the stress ellipsoid and stress tensor have orthorhombic symmetry. 4.2 Tectonic stress The lithostatic state of stress causes isotropic contraction for a homogeneous material. Deviation from this state results in strain. Tectonic stress is generally defined as this deviation. The lithostatic state of stress is used as the reference. Other states of stress can also be taken as the reference (§7.2). The average σ + σ + σ trace r σ S + S + S trace S S σ ≡ 1 2 3 = = I or S ≡ 1 2 3 = = I (4.6) 0 3 3 3 0 3 3 3 is said to be mean stress. For lithostatic stresses, mean stress coincides with pressure. Accordingly, it is convenient to define deviatoric stress s = r − σ0I (4.7) as the deviation from the lithostatic state of stress. This is the deviatoric tensor introduced in Section C.6. Let us define the symbol T as T = S − S0I. (4.8) It is obvious that a deviatoric tensor is symmetric, so that the tensor has its principal orientations parallel to the stress axes (§C.5). If T1, T2 and T3 are the principal deviatoric stresses, they are related to the principal stresses as Ti = Si − S0. (4.9) Combining Eqs. (4.6), we have 2S − S − S 2S − S − S 2S − S − S T = 1 2 3 ,T= 2 3 1 ,T= 3 1 2 . (4.10) 1 3 2 3 3 3 Using the stress ratio, these are rewritten as 2 − Φ 2Φ − 1 Φ+1 T = ΔS, T = ΔS, T = − ΔS. (4.11) 1 3 2 3 3 3 104 CHAPTER 4. PRINCIPAL STRESSES Figure 4.2: Diagram for the explanation of the Hubbert-Rubey model. Hubbert-Rubey model In the previous chapter, we considered only the lithostatic state of stress. Our first application of the deviatoric stress tensor is the statics of a thrust sheet. There are thrust sheets that travel horizontally for more than 100 km. The Hubbert-Rubey model explains why such long-distance displacement is possible or what it implies. The very low friction of the fault surface is the key [86, 87]. Consider a thrust sheet that has a thickness of t and is advancing in the x direction (Fig. 4.2). For simplicity, the topographic surface coincides with the top of this sheet. We pay attention to a part of the sheet with length L in the x direction and unit length in the y direction. The tectonic forces pushing at the rear and front of this portion are τxxt in the +x direction and −(1 +Δ)τxxt in the −x direction. Therefore, the net force to push the portion of the sheet in the +x direction is −Δτxxt. Frictional resistance is the one significant force for the sheet. The resistance per unit basal area is σzx, and the law of friction gives σzx = μfσzz = μfρgt, where μf is the coefficient of friction and ρ is the mean density of the sheet. Therefore, a portion feels the resistance, −μfρgLt, in the +x direction. The motion of the sheet is so slow that inertia is negligible. Hence, all forces acting on the portion is balanced: −Δτxxt = −μfρgLt. Therefore, we obtain Δτxx = ρμfgL. In order to evaluate the coefficient of friction, we substitute Δτxx = 100 MPa as the representative magnitude of tectonic stresses and ρ = 2.75 × 103 kg m−3 in this equation. We consider a thrust sheet that travels hundreds of kilometers so that we assume L = 100 km. The result is that μf = 0.036. The friction at the base of those giant thrust sheets is very low compared to the coefficient of the friction of ordinary rocks. Pore fluid lowers the friction (§6.5). Octahedral shear stress The deviatoric stress tensor has the characteristic equation shown in Eq. (C.48), and its first basic invariant is TI = trace T = T1 + T2 + T3 = 0. (4.12) 4.2. TECTONIC STRESS 105 Figure 4.3: Basic invariants of deviatoric stress tensor versus stress ratio. The second basic invariant satisfies 1 1 T = T : T = T T II 2 2 ij ij i,j = 1 2 + 2 + 2 + 2 + 2 + 2 T11 T22 T33 T12 T23 T31 (4.13) 2 = 1 − 2 + − 2 + − 2 + 2 + 2 + 2 (S11 S22) (S22 S33) (S33 S11) S12 S23 S31 6 1 = (S − S )2 + (S − S )2 + (S − S )2 ≥ 0. 6 1 2 2 3 3 1 It is obvious in the last line that τII ≥ 0. TII is related to the principal deviatoric stresses as 1 1 T = T 2 + T 2 + T 2 = (T + T + T )2 − 2T T − 2T T − 2T T II 2 1 2 3 2 1 2 3 1 2 2 3 3 1 = −T1T2 − T2T3 − T3T1. (4.14) The transformation at the second equal sign is based on Eq. (4.12). Substituting Eq. (4.11), we obtain Φ2 − Φ+1 T = (ΔS)2. (4.15) II 3 1 Δ 2 Φ= 1 1 Δ 2 Φ= Therefore, TII ranges from 4 ( S) at 2 to 3 ( S) at 0 and 1.
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