Chapter 4

Principal Stresses

Anisotropic 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. 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 of the 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. The third basic invariant is TIII = T1T2T3. Combining this equation and Eq. (4.11), we have

(2 − Φ)(2Φ − 1)(Φ+1) T = − (ΔS)3. III 27

− 2 Δ 3 ≈− Δ 3 Φ= 2 Δ 3 ≈ Δ 3 Therefore, TIII ranges from 27 ( S) 0.074( S) at 1to 27 ( S) 0.074( S) at Φ= Φ= 1 = 0. When 2 ,wehaveTIII 0. Figure 4.3 shows the graphs of TII and TIII. 106 CHAPTER 4. PRINCIPAL STRESSES

Figure 4.4: An octahedral plane and its unit normal.

Octahedral shear stress √ Taking coordinate axes parallel to the stress axes, the eight unit vectors e⊥ = (±1, ±1, ±1)T/ 3 are the unit normals for the faces of a regular octahedron. The octahedral plane is the plane which makes equal angles with the principal stress axes (Fig. 4.4). The normal and shear stresses upon the planes are called the octahedral normal stress and octahedral shear stress. The former is given by the equation oct = ⊥ · · ⊥ = ⊥ = SN e S e S : e S0, (4.16)   where e⊥ ≡ e⊥e⊥ is the projector onto the vector e⊥. Equation (4.16) designates that the octahe- dral normal stress equals the mean stress. The octahedral shear stress is used in the theory of plasticity (§10.2), and is proportional to the second basic invariant of the deviatoric stress tensor        2 ⊥ 2 1 2 Soct = S : I − e  = (T − T )2 + (T − T )2 + (T − T )2 = T , S 9 1 2 2 3 3 1 3 II or, equivalently, (  2 1 Soct = T = (T − T )2 + (T − T )2 + (T − T )2. (4.17) S 3 II 3 1 2 2 3 3 1

oct For the convention that compression is positive stress, we use the symbol σS for the octahedral shear stress.

4.3 Mohr diagram

The Mohr diagram is useful to illustrate the state of stress and the combinations of normal and shear stresses for various directions of surface elements. These combinations are important when we consider friction on a fault plane. Traction on a surface element depends on the direction of its normal n, and the normal and shear components also changes with n. The admissible combinations 4.3. MOHR DIAGRAM 107

Figure 4.5: Mohr’s circles illustrate the admissible combinations of normal and shear stresses, σN and σS, for a given set of principal stresses σ1, σ2 and σ3 which are indicated by shaded  area. The σ2+σ3 σ1+σ3 sign σS indicate the sense of shear. The centers of the circles are located at 2 , 0 , 2 , 0  +  σ1 σ2 Δ and 2 , 0 . The maximum shear stress equals σ/2.

for a given state of stress are represented by areas bounded by three circles C1,C2 and C3 on O-σNσS plane (Fig. 4.5). They are called Mohr’s circles. The horizontal axis O-σN passes the centers of the circles of which the diameters are σ1 − σ3, σ1 − σ2 and σ2 − σ3. The first one equals the differential stress Δσ. The circles are symmetric with respect to the horizontal axis O-σN, so that the upper half of Mohr’s stress diagram, or Mohr diagram can illustrate the state of stress. No matter how great or small the magnitude of σ2 is, the maximum shear stress for a given state of stress is determined only by σ1 and σ3 as σ − σ Maximum shear stress = 1 3 . (4.18) 2

This is equal to the diameter of the outermost circle C2 (Fig. 4.5). Given the principal stresses and the unit normal of a surface element, the normal and shear stresses on the element are represented in Fig. 4.5 as the shaded area bounded by the Mohr’s circles. The relationship between the position of the point in the diagram and the direction of n = (cos φ, cos β, cos θ)T is established by taking Fig. 4.6 into consideration. These components are the direction cosines of n. Since a stress tensor has orthorhombic symmetry, Fig. 4.6(a) shows the first octant of a sphere on which a point Q represents the direction of n. The line segment QP is parallel to n, where P is the center of the sphere. It is shown (Exercise 4.1) that the octant is mapped onto the shaded area in Fig. 4.6(b), and that the points “A” through “H” on the octant correspond to the points “a” through “h” in the figure. The point “Q” that represents the direction of n corresponds 108 CHAPTER 4. PRINCIPAL STRESSES

Figure 4.6: Correspondence between the direction of n and the point in Mohr’s stress diagram. (a) The direction is represented by a point “Q” on an octant of a sphere. The points “K”, “Q” and “D” indicate the directions that meet the σ1 axis at the common angle φ. (b) The points “a”, “b” and “c” are plotted at the points (σ1, 0), (σ2, 0) and (σ3, 0), respectively. The points “A” through “H” and “Q” on the sphere correspond to the points “a” through “h” and “q” in the diagram. The small circle “KQD” corresponds to the circle “kqd”. to the point “q” in Mohr’s stress diagram. To find the point “q” in the diagram for a given n = (cos φ, cos β, cos θ)T, one has to plot

firstly the point “h” that is rotated clockwise along the circie C1 from the point “c” by the angle 2θ. Secondly, the point “d” is plotted on the circle C3. The point is rotated counterclockwise from the point “a” by the angle 2φ. Thirdly, one draws arcs “hf” and “dk” that are concentric with the circles

C3 and C1, respectively. The intersection of the arcs is the point “q”. The maximum shear stress is indicated by Eq. (4.18). Obviously, shear stress becomes the maximum at the point “m” in Fig. 4.6(b), where σN = (σ1 + σ3)/2 and σS = (σ1 − σ3)/2. The point “m” corresponds to the point “M” in Fig. 4.6(a), and ∠APM = ∠CPM = 45◦, i.e., shear stress is ◦ the maximum on the surface that is parallel to the σ2 axis and that makes an angle of 45 with the σ1 and σ3 axes.

4.3.1 Classes of stress

The state of stress at a point can be classified as shown in Fig. 4.7.

Lithostatic or hydrostatic pressure (σ3 = σ2 = σ1 = p): The normal stress across all planes is equal to pressure p, but there are no shearing stresses. In this case, the state of stress is symmetric, r = pI. The stress plots on the Mohr diagram degenerate into a point on the abscissa. 4.3. MOHR DIAGRAM 109

Uniaxial compression (0 = σ3 = σ2 <σ1): The only stress applied is a compressive stress in one direction. The state of stress has axial symmetry about the σ1 axis. The Mohr diagram for this case is a single circle tangent to the ordinate at the origin.

Uniaxial (σ3 <σ2 = σ1 = 0): The only stress applied is a tension in one direction. The state of stress has axial symmetry about the σ3 axis. The Mohr diagram for this case is a single circle tangent to the ordinate at the origin.

Axial compression (σ3 = σ2 <σ1): The state of stress has symmetric about σ1 axis. The Mohr diagram degenerates into a single circle.

Axial tension (σ3 <σ2 = σ1): Some authors avoid using this term, because “tension” seems to indicate negative principal stresses. The distinction should always be made clear [248]. The

state of stress has symmetry about the σ3 axis. The Mohr diagram degenerates into a single circle for this case.

Plane stress (0 = σ3 <σ2 ≤ σ1): If a principal stress is zero, plane stress is said to exist. This is synonymous with biaxial stress [91]. The state of stress has orthorhombic symmetry.

Pure shear stress (σ3 = −σ1,σ2 = 0): The normal stress on planes of maximum shear stress is zero. The state of stress has orthorhombic symmetry. The Mohr diagram for this case is symmetric with respect to the ordinate and to the origin.

Triaxial stress (σ3 <σ2 <σ1): All the principal stresses are different. The state of stress has orthorhombic symmetry. Three distinct circles compose the Mohr diagram.

If there are planes on which traction vanishes (σN = σS = 0), they are called free surfaces or free boundaries. Uniaxial and plane stresses have those planes. If O-3 axis is chosen to be perpendicular = = | | = 2 + 2 = to the surface, then σ33 σN 0 and the shear stress on the surface is σS σ31 σ32 0. Therefore σ31 = σ32 = 0. Consequently, the stress tensor for the stress state has the form ⎛ ⎞ ••0 r = ⎝••0⎠ , (4.19) 000 where the black dots represent the components that are not determined by the boundary condition.

4.3.2 Mohr diagram for two-dimensional problems Axial stresses can be treated as two-dimensional problems, where there are two principal stresses,

σ1 and σ2, accompanied by two principal axes. One of the axes is parallel to the axis of symmetry for the state of stress. Even for triaxial stresses, such a problem can be considered to estimate the normal and shear stresses to investigate faulting (Chapter 6), as σ2 has a limited effect on faulting (Chapter 10). 110 CHAPTER 4. PRINCIPAL STRESSES

Figure 4.7: Mohr diagrams for special states of stress.

The admissible combinations of normal and shear stresses for a given state of stress are illustrated by a Mohr diagram that consists of a single circle. To see this, suppose a stress tensor σ 0 r = 1 , 0 σ2 for which the rectangular Cartesian coordinates O-12 are taken as parallel to the principal stress axes (Fig. 4.8). Other coordinates O-12 are defined to have the same origin O with the former coordi- nate system. The coordinate axes meet at an angle of θ. The orthogonal tensor for this coordinate transformation is given by Eq. (C.8). Thus, stress components are transformed from O-12 to O-12 as σ σ θ θ σ θ − θ 11 12 = cos sin 1 0 cos sin σ σ − sin θ cos θ 0 σ sin θ cos θ 21 22 2 2 + 2 − − = σ1 cos θ σ2 sin θ (σ1 σ2) sin θ cos θ 2 2 . (4.20) −(σ1 − σ2) sin θ cos θσ1 cos θ + σ2 sin θ

  It is obvious in Fig. 4.8(a) that σN and σS are equivalent to σ11 and σ12, respectively. These stress 4.3. MOHR DIAGRAM 111

Figure 4.8: Mohr diagrams for two-dimensional problems. The axes O-12 and O-12 represent two rectangular Cartesian coordinate systems with a common origin O. The former axes are taken to be parallel to the principal stress axes. O-1 axis is taken to be parallel to N that is the outward normal to the surface element on which the traction is considered. This surface element has positive and negative directions for N and n, respectively, leading to the opposite sign of the shear stress, SS and σS. See Fig. 3.2 for the relationship between positive shear directions. (b, c) Mohr’s circles for each of the sign conventions. components are shown in Eq. (4.20). Namely,

=  = 2 + 2 σN σ11 σ1 cos θ σ2 sin θ, =  = − − σS σ12 (σ1 σ2) sin θ cos θ.

2 = − 2 = + = 1 Using the formulas, sin θ (1 cos 2θ)/2, cos θ (1 cos 2θ)/2 and sin θ cos θ 2 sin 2θ,we obtain

1 1 σ = (σ + σ ) + (σ − σ ) cos 2θ (4.21) N 2 1 2 2 1 2 1 σ = − (σ − σ ) sin 2θ. (4.22) S 2 1 2 112 CHAPTER 4. PRINCIPAL STRESSES

Figure 4.9: (a) Planes perpendicular to the O-3 axis are assumed to be the free surfaces. (b) Mohr diagram for the plane stress.

Plane stress Suppose a state of plane stress where the free surface is parallel to the O-12 plane (Fig. 4.8(a)). That is, the stress tensor has the components ⎛ ⎞ S11 S12 0 ⎝ ⎠ S = S21 S22 0 . 000

The Mohr’s circle for this stress is drawn as the following procedure. The center of the circle B in

Fig. 4.9(b) is located at SN = (S11 + S22)/2, because the trace of a tensor is invariant to coordinate rotations. The points A and D on the circle represent the traction on the right-hand and top faces, respectively, of the rectangular parallelepiped in Fig. 4.8(a). The line segment AD is the diameter of the circle, as the faces make a right angle. The phase angle in a Mohr diagram is twice as much as an angle in physical space. The line segment AB in Fig. 4.8(b) is the radius of the circle, and 

= − 2 + 2 AB (S11 S22)/2 S12.

4.4 Boundary conditions of stress

In the shallow part of the solid Earth, one of the stress axes is perpendicular to the surface. To see this, consider two materials I and II that are firmly fixed to each other at the O-12 plane (Fig. 4.10a). r(I) = (I) r(II) = (II) Let σij and σij be the stress tensors in materials I and II, respectively, where the coordinates O-123 are defined in Fig. 4.10. The traction t(I) is exerted on material I at the point P on the interface. Likewise, material II feels the traction t(II) at the same point. If all forces are balanced and the materials are at rest, we have

t(I) + t(II) = 0. (4.23) 4.4. BOUNDARY CONDITIONS OF STRESS 113

Figure 4.10: Schematic illustrations for the explanation of stress boundary coditions. (a, b) Materials I and II are firmly fixed at the O-12 plane. O-2 axis is perpendicular to the page. (c, d) Materials I and II with a lubricated interface.

Given the unit normal to the interface n = (0, 0, −1)T, the first traction vector is written as ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ (I) (I) (I) (I) σ11 σ12 σ13 0 σ13 (I) = r(I) · = ⎝ (I) (I) (I)⎠ ⎝ ⎠ = − ⎝ (I)⎠ . t n σ21 σ22 σ23 0 σ23 (I) (I) (I) −1 (I) σ31 σ32 σ33 σ33 − = T (II) = (II) (II) (II) T The unit normal for material II is n (0, 0, 1) , and we have t (σ13 ,σ23 ,σ33 ) . Substitut- ing these traction vectors into Eq. (4.23), we obtain     (I) (I) (I) T = (II) (II) (II) T σ13 ,σ23 ,σ33 σ13 ,σ23 ,σ33 . Since the stress tensors are symmetric, we obtain ⎛ ⎞ ⎛ ⎞ •• (I) ••(II) σ13 σ13 ⎝ •• (I)⎠ = ⎝ ••(II)⎠ σ23 σ23 (I) (I) (I) (II) (II) (II) σ13 σ23 σ33 σ13 σ23 σ33 as the boundary condition of the stress field at the interface2. The black dots in these matrices represent the stress components that are not constrained by the boundary condition. If the interface is not fixed but lubricated, the shear stress vanishes on the interface. That is, we have that the boundary condition for this case is ⎛ ⎞   •• 0  ⎝ ⎠ r  = •• 0 . (4.24) interface 00σ33

2This is a special case of the jump condition of physical properties at a surface. If the surface and the materials on its sides move with different velocities, the impluse by the differential motions should be taken into account [38]. 114 CHAPTER 4. PRINCIPAL STRESSES

Namely, only the normal stress σ33 is transmitted across the interface, but the shear stress is not. Values of the components σ11, σ22 and σ12 may jump across the interface. The σH axes are oriented NNW–SSW far from the San Andreas Fault. However, those are nearly perpendicular to the near the fault, suggesting a very low friction on the fault surface3 [277].

Vertical and horizontal stresses The stress tensor in Eq. (4.24) is diagonalized for the third column and row. Therefore, σ33 is a principal stress and the third coordinate axis O-3 is the stress axis corresponding to this principal stress. That is, one of the stress axes is perpendicular to a lubricated surface. The other stress axes lie on the surface. The solid Earth is covered by the atmosphere or seawater. These fluids have much less viscosity than rocks. We regard the surface of the Earth as a lubricated interface which is covered by a fluid. Therefore, Eq. (4.24) holds at the surface of the Earth. The solid Earth is exerted a shear stress by winds and ocean bottom currents, but the shear stress is much less than tectonic stress. For this reason, we neglect those drag forces in the discussion of . The result is that one of the stress axes is perpendicular to the surface at the shallow levels in the Earth. In addition, if we assume that the Earth has a level surface, the stress axis is largely vertical.

The component σ33 in Eq. (4.24) is, in these cases, the atmospheric or hydrostatic pressure at the surface of the solid Earth. The atmospheric pressure is about 0.1 MPa, which is negligible compared to tectonic stress. The pressure at the bottom of ocean is usually negligible. Therefore, we have the stress boundary condition at the surface, ⎛ ⎞   ••0  ⎝ ⎠ r  = ••0 . surface 000

This is, of course, applicable to the surface of airless planets and satellites. However, hydrostatic pressure at the bottom of ocean trenches may not be negligible, as the pressure reaches 100 MPa. If one of the stress axes is vertical, stress in the shallow part of the Earth is expressed as ⎛ ⎞ σH 00 ⎝ ⎠ r = 0 σh 0 , 00σv where σH and σh are the maximum and minimum horizontal stresses, and σv is the vertical stress that is usually approximated by the overburden stress (Eq. (3.29)). The σH- and σh-axes are the corresponding horizontal principal axes. Some researchers use the notation, σHmax and σHmin, for the horizontal axes.

3Some researchers argue against this interpretation [207]. 4.5. IN-SITU STRESS MEASUREMENTS 115

Figure 4.11: Deformed borehole determined by well logging at ∼3 km below the surface [180].

4.5 In-situ stress measurements

The present state of stress is estimated by several methods. Among them, two methods of in-situ stress measurement in boreholes are introduced in this section4. If a deep borehole is filled with a fluid that has a smaller density than the surrounding rocks, there is a great gap between the pressure in the fluid and in the rocks. The pressure difference sometimes deforms or even collapses the borehole. Figure 4.11a shows the geometry of a borehole no longer smooth or round. A zone along the wall of a well which has failed is called borehole breakout. There are a few types of breakouts reflecting the state of stress in the surrounding rocks. Detailed observation of the borehole geometry allows us to infer the principal orientations of stress perpendicular to the hole. Hydraulic stress measurement is an active measurement of in-situ stress. Hydraulic fracture or hydrofracturing is a process of breaking up the rocks immediately out of the wall of a borehole under pressure by pumping water into an interval of the hole of which both ends are sealed. Hydraulic fracture experiments are usually done to increase permeability in hydrocarbons and geothermal reservoirs. The pressure opens the pre-existing discontinuity surface including joints and bedding planes. The opened surfaces tend to be oriented parallel to the far field σH orientation. Therefore, the borehole geometry again indicates the stress orientations. The experiment not only determine the orientations but also the stress magnitudes.

4See [54] for further reading. 116 CHAPTER 4. PRINCIPAL STRESSES

Figure 4.12: A basaltic intruding Oligocene basaltic lavas and tuffs cropping out on the Japan Sea coast of Northeast Japan. Dotted lines outline the dike. Note that the dike expands its width at the level of the tuff layer. The dike looks like the head of a cobra. A close-up photograph of the lower left part of the cliff (indicated by angles) is shown in 4.13.

4.6 Dikes as indicators

Dikes, tabular intrusions of magma, are the results of the natural analog of . Magma pushed its way through the country via its pressure for intrusion. This is demonstrated by the dike in the outcrop shown in Fig. 4.12. The dike vertically penetrates basaltic volcanic rocks and tuffs, and the width of the dike increases abruptly at the horizon of a tuff. Figure 4.13 is a close- up photograph of the tuff to the left of the dike. The tuff and overlying lava are intruded by an old dike that is cut by a fault parallel to the bedding in the tuff, and the upper block is transported away from the young dike shown in Fig. 4.12. The offset of the old dike is as large as the expansion of the young dike. These observations indicate that the young dike pushed its country rocks when it intruded. 4.6. DIKES AS PALEOSTRESS INDICATORS 117

Figure 4.13: A dike (D) broken into three parts by bedding parallel faults (solid lines) in the tuff layer neighboring the dike shown in Fig. 4.12.

The Mohr diagram in Fig. 4.14(a) shows the admissible orientations of dike intrusion relative to the principal axes of stress in the country rock. Consider a rock mass in which magma is intruding with a pressure pm, which has pre-existing cracks with various attitudes. The hatched area is bounded by the vertical line σN = pm, and the normal stress on the crack is smaller than the magma pressure in the area. Magma pressure has to overcome the normal stress for intrusion. The state of stress in the country rock is illustrated by the Mohr’s circles in Fig. 4.14(a), and the dark gray portion of the Mohr diagram shows the condition under which the magma pressure overcomes the normal stress. This portion designates the traction on the surfaces whose orientations are shown by the dark gray portion on the sphere in Fig. 4.14(b). Magma can intrude cracks with orientations that are indicated by the dark gray part of the sphere. As the magma pressure increases, if pm is smaller than σ3, no magma intrusion occurs. When the pressure is slightly above the minimum principal stress (pm  σ3), magmas can only intrude cracks nearly perpendicular to the σ3-axis. As the pressure increases, the admissible orientations of dike intrusion expands. Finally, magma with pm >σ1 intrudes any crack, resulting in a network of intrusions. The above argument demonstrate that a dike does not necessarily indicate the orientation of stress 118 CHAPTER 4. PRINCIPAL STRESSES

Figure 4.14: (a) Mohr diagram showing the magnitude of normal stress on surfaces. (b) Diagram showing the attitude of the surfaces relative to the stress axes of country rocks.

axes when it intruded. If the magma pressure was appropriate, that is, slightly greater than σ3, dikes 5 are paleostress indicators perpendicular to the σ3-axis . If a dike swarm, parallel dikes, occur in a large regional area, they are evidence for paleo σ3 orientation. One good point of dike swarms in determining paleostress is that the age of the paleostress is known from the radiometric age of the dikes. However, it should be determined from other types of data whether a horizontally extensional or compressional stress field affected the area. Magma pressure may decrease with the distance of a magma chamber. Dikes accompanied by the Otoge cauldron shown in Fig. 4.15 illustrate this idea. The magmatism occurred at ∼15 Ma. NNE–SSW to N–S trending dikes penetrates a Mesozoic basement in this area, but there are ring dikes in the vicinity of the cauldron. The dike swarm is perpendicular to the present σH orientation, but is parallel to the Median Tectonic Line, which is the most significant fault in Southwest Japan. The line is parallel to the arc, and juxtaposes the high T/low P metamorphic belt and the low T/high P belt. There are cauldrons with the same age along the Median Tectonic Line, and one other cauldron in the central part of the arc has a dike swarm parallel to the line. The swarm was associated with normal faulting, therefore, the dikes indicate an arc perpendicular extensional stress field at ∼15 Ma [102, 108].

4.7 First-order regional stress field

The state of stress is measured through out the world using various methods, and the gross pattern of the present stress field in the shallow levels in the Earth has been understood [279]. Figure 4.16 shows the results published in 1992, indicating several important points.

5It was attempted to estimate magma pressure and intermediate stress from the variation of dike orientations in [14]. 4.7. FIRST-ORDER REGIONAL 119

Figure 4.15: (a) Dike swarm and (b) ring dikes accompanied by the Otoge cauldron in Central Japan [235]. The cauldron is the result of extensive volcanism in the Southwest Japan arc at ∼15 Ma. There are too many dikes in this area so that their density, i.e., number per a length of 1 km across the trend of the dikes, is shown. Note the difference of scale.

First of all, it was shown that there are vast regions where the principal orientations of stress are roughly uniform. This was doubted because there are a variety of stress sources in the including topographic and lithological variations. Lithosphere thickness is a reference to argue how the stress field is uniform. The thickness of the mechanically strong part in the lithosphere has spatial and temporal variations, but is several tens of kilometers. There are regions where σH orientation is uniform over the dimension 20–200 times greater than the thickness of the lithosphere beneath the regions. The central to northwestern North America is one of them, where σH axis is oriented ENE–WSW. In the Far East, the axis is oriented E–W between Japan and northern China. The first-order stress field in each tectonic plate is believed to be controlled by collisional forces at plate boundaries. If this is the case, the intraplate stress field is determined by the forces and the shape of the boundary. In the northern Indo-Australia plate, σH-axis is parallel to the velocity of the plate relative to the Eurasia plate. The Indo-Australia plate is bordered by a mid-oceanic ridge. The axis is oriented perpendicular to the boundary, due to the force. The second point is that the stress gradients within a plate are smaller than expected. The small gradients suggest that the viscous coupling of the lithosphere and the underlying mantle is weak. If the plate had been fixed upon the stationary or horizontally travelling deep mantle, plate boundary forces caused by differential movement of neighboring plates would have resulted in some gradients 120 CHAPTER 4. PRINCIPAL STRESSES

Figure 4.16: σH orientations in the shallow levels of the [279]. Altitude and ocean depth are indicated by a gray scale.

in the intra plate stress field. The observed first-order stress field is not consistent with absolute motion, which also demonstrates little coupling.

The third point is that plateaus coincide with the region of . The in the western United States is an example where the surface has an altitude of ∼2 km and the vigorous extensional tectonics has been going on since the late Tertiary. However, it should be noted that topographic height is not the only source of intra plate stress. This is demon- strated by the existence of continental such as the Suez, where rifting occurs near and under the sea level. 4.8. EXERCISES 121

Figure 4.17: Mohr diagram showing a triaxial stress. The points P, Q and R designate the centers of Mohr’s circles.

4.8 Exercises

4.1 Explain mathematically why the admissible combinations of normal and shear stresses are indicated by Mohr’s stress diagram. Show the correspondence of the points with the labels of upper- and lower-case letters in Fig. 4.6.

4.2 A stress state is designated by the Mohr diagram in Fig. 4.17. (a) What are the orientations of the planes corresponding to the closed circles A, B, C and D in the diagram? (2) Which is the point in the diagram corresponding to an octahedral plane?

4.3 Suppose a hot diapir with a diameter of L = 1 km ascending through the crust. The diapir loses its heat during the ascent, so that it has to rise rapidly to cause contact in the country rocks when it reaches a shallow level of the crust. Estimate the lower bound of the ascending velocity of the diapir for the metamorphism to occur. Use the thermal diffusivity κ = 10−6 m2s−1 and a distance of 10 km for the diapir to ascend.