principal stresses - terrapub · chapter 4 principal stresses anisotropic stress causes tectonic...

Chapter 4

Principal Stresses

Anisotropic stress causes tectonic deformations. Tectonic stress is defined inan early part of this chapter, and natural examples of tectonic stresses are alsopresented.

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 stressesare numbered conventionally in descending order of magnitude, σ1 ≥ σ2 ≥ σ3, so that they are themaximum, intermediate and minimum principal stresses. Note that these principal stresses indicatethe 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. Theorientations 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 ofthe stress axes.

Deformation is driven by the anisotropic state of stress with a large difference of the principalstresses. Accordingly, the differential stress

ΔS = S1 − S3, Δσ = σ1 − σ3

is defined to indicate the difference1.Taking the axes of the rectangular Cartesian coordinatesO-123 parallel to the stress axes, a stress

tensor is expressed by a diagonal matrix. Therefore, Cauchy’s stress (Eq. (3.10)) formula is

t(n) =

⎛⎝t1(n)t2(n)t3(n)

⎞⎠ =

⎛⎝σ1 0 00 σ2 00 0 σ3

⎞⎠⎛⎝n1

n2

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 ofstress.

∴ t1(n) = σ1n1, t2(n) = σ2n2, t3(n) = σ3n3. (4.1)

The planes perpendicular to the stress axes have unit normals, n = (1 0 0)T, (0 1 0)T, (0 0 1)T.Hence, the traction vectors upon the planes are

t(n) =

⎛⎝σ1

00

⎞⎠ ,

⎛⎝ 0σ2

0

⎞⎠ ,

⎛⎝ 00σ3

⎞⎠ .

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

σN = t(n) · n = σ1n21 + σ2n

22 + σ3n

23 (4.2)

σS =[|t(n)|2 − |σN|2

]1/2=[σ2

1n21 + σ

22n

22 + σ3n

23 −

(σ1n

21 + σ2n

22 + σ3n

23

)2]1/2

. (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 theσ1σ2-plane vanishes. Combining Eq. (4.1) and n2

1 + n22 + n

23 = 1, we have[

t1(n)σ1

]2

+

[t2(n)σ2

]2

+

[t3(n)σ3

]2

= 1. (4.4)

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 vectorwith its initial point at the coordinate origin, the end point of the vector is on the ellipsoid, providedthat 0 < σ3. The ellipsoid becomes a sphere for the hydrostatic state of stress and its radius equalsthe hydrostatic stress. The stress ellipsoid is given uniquely from the stress tensor regardless of theorientation 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 areSI, SII and SIII. These quantities designate the “absolute values” of the stress tensor, independentfrom 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

Φ =σ2 − σ3

σ1 − σ3. (4.5)

It is seen that the ratio is in the range of 0 ≤ Φ ≤ 1. When the state of stress is hydrostatic, the stressratio is not defined. For anisotropic stress states, the stress ratio designates the shape of the stresselliosoid. If Φ = 0, σ3 = σ2 < σ1, then, the ellipsoid is prolate. Φ = 1 corresponds to an oblateellipsoid, 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 ellipsoidand stress tensor have orthorhombic symmetry.

4.2 Tectonic stress

The lithostatic state of stress causes isotropic contraction for a homogeneous material. Deviationfrom this state results in strain. Tectonic stress is generally defined as this deviation. The lithostaticstate of stress is used as the reference. Other states of stress can also be taken as the reference (§7.2).The average

σ0 ≡ σ1 + σ2 + σ3

3=

trace r

3=σI

3or S0 ≡ S1 + S2 + S3

3=

trace S3

=SI

3(4.6)

is said to be mean stress.For lithostatic stresses, mean stress coincides with pressure. Accordingly, it is convenient to

define deviatoric stresss = r − σ0I (4.7)

as the deviation from the lithostatic state of stress. This is the deviatoric tensor introduced in SectionC.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 orientationsparallel to the stress axes (§C.5). If T1, T2 and T3 are the principal deviatoric stresses, they arerelated to the principal stresses as

Ti = Si − S0. (4.9)

Combining Eqs. (4.6), we have

T1 =2S1 − S2 − S3

3, T2 =

2S2 − S3 − S1

3, T3 =

2S3 − S1 − S2

3. (4.10)

Using the stress ratio, these are rewritten as

T1 =

(2 − Φ

3

)ΔS, T2 =

(2Φ − 1

3

)ΔS, T3 = −

(Φ + 1

3

)ΔS. (4.11)


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 thedeviatoric stress tensor is the statics of a thrust sheet. There are thrust sheets that travel horizontallyfor more than 100 km. The Hubbert-Rubey model explains why such long-distance displacement ispossible 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 apart of the sheet with length L in the x direction and unit length in the y direction. The tectonicforces pushing at the rear and front of this portion are τxxt in the +x direction and −(1 + Δ)τxxtin 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 basalarea is σzx, and the law of friction gives σzx = μfσzz = μfρgt, where μf is the coefficient of frictionand ρ is the mean density of the sheet. Therefore, a portion feels the resistance, −μfρgLt, in the +xdirection. The motion of the sheet is so slow that inertia is negligible. Hence, all forces acting on theportion is balanced: −Δτxxt = −μfρgLt. Therefore, we obtain Δτxx = ρμfgL. In order to evaluatethe coefficient of friction, we substitute Δτxx = 100 MPa as the representative magnitude of tectonicstresses and ρ = 2.75× 103 kg m−3 in this equation. We consider a thrust sheet that travels hundredsof kilometers so that we assume L = 100 km. The result is that μf = 0.036. The friction at the baseof 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 basicinvariant 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

TII =12

T : T =12

∑i,j

TijTij

=12

(T 2

11 + T222 + T

233

)+ T 2

12 + T223 + T

231 (4.13)

=16

[(S11 − S22)2 + (S22 − S33)2 + (S33 − S11)2

]+ S2

12 + S223 + S

231

=16

[(S1 − S2)2 + (S2 − S3)2 + (S3 − S1)2

]≥ 0.

It is obvious in the last line that τII ≥ 0. TII is related to the principal deviatoric stresses as

TII =12

(T 2

1 + T 22 + T 2

3

)=

12

[(T1 + T2 + T3)2 − 2T1T2 − 2T2T3 − 2T3T1

]= −T1T2 − T2T3 − T3T1. (4.14)

The transformation at the second equal sign is based on Eq. (4.12). Substituting Eq. (4.11), weobtain

TII =Φ2 − Φ + 1

3(ΔS)2. (4.15)

Therefore, TII ranges from 14 (ΔS)2 at Φ = 1

2 to 13 (ΔS)2 at Φ = 0 and 1.

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

TIII = − (2 − Φ)(2Φ − 1)(Φ + 1)27

(ΔS)3.

Therefore, TIII ranges from − 227 (ΔS)3 ≈ −0.074(ΔS)3 at Φ = 1 to 2

27 (ΔS)3 ≈ 0.074(ΔS)3 atΦ = 0. When Φ = 1

2 , we have TIII = 0. Figure 4.3 shows the graphs of TII and TIII.


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/√

3are the unit normals for the faces of a regular octahedron. The octahedral plane is the plane whichmakes equal angles with the principal stress axes (Fig. 4.4). The normal and shear stresses upon theplanes are called the octahedral normal stress and octahedral shear stress. The former is given bythe equation

SoctN = 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 thesecond basic invariant of the deviatoric stress tensor(

SoctS

)2=∣∣S :

(I − e⊥)∣∣2

=19

[(T1 − T2)2 + (T2 − T3)2 + (T3 − T1)2

]=

23TII,

or, equivalently,

SoctS =

√23TII =

13

√(T1 − T2)2 + (T2 − T3)2 + (T3 − T1)2. (4.17)

For the convention that compression is positive stress, we use the symbol σoctS 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 andshear stresses for various directions of surface elements. These combinations are important whenwe consider friction on a fault plane. Traction on a surface element depends on the direction of itsnormal 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. Thesign σS indicate the sense of shear. The centers of the circles are located at

( σ2+σ32 , 0

),( σ1+σ3

2 , 0)

and( σ1+σ2

2 , 0). The maximum shear stress equals Δσ/2.

for a given state of stress are represented by areas bounded by three circles C1, C2 and C3 onO-σNσS

plane (Fig. 4.5). They are called Mohr’s circles. The horizontal axis O-σN passes the centers of thecircles of which the diameters are σ1 − σ3, σ1 − σ2 and σ2 − σ3. The first one equals the differentialstress Δσ. The circles are symmetric with respect to the horizontal axis O-σN, so that the upper halfof Mohr’s stress diagram, or Mohr diagram can illustrate the state of stress. No matter how great orsmall the magnitude of σ2 is, the maximum shear stress for a given state of stress is determined onlyby σ1 and σ3 as

Maximum shear stress =σ1 − σ3

2. (4.18)

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’scircles. The relationship between the position of the point in the diagram and the direction ofn = (cosφ, cos β, cos θ)T is established by taking Fig. 4.6 into consideration. These componentsare the direction cosines of n. Since a stress tensor has orthorhombic symmetry, Fig. 4.6(a) showsthe first octant of a sphere on which a point Q represents the direction of n. The line segment QP isparallel to n, where P is the center of the sphere. It is shown (Exercise 4.1) that the octant is mappedonto the shaded area in Fig. 4.6(b), and that the points “A” through “H” on the octant correspond tothe points “a” through “h” in the figure. The point “Q” that represents the direction of n corresponds


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 = (cosφ, 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 thepoint “a” by the angle 2φ. Thirdly, one draws arcs “hf” and “dk” that are concentric with the circlesC3 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 themaximum 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 isthe 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 equalto 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.


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

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

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

Axial tension (σ3 < σ2 = σ1): Some authors avoid using this term, because “tension” seems toindicate negative principal stresses. The distinction should always be made clear [248]. Thestate of stress has symmetry about the σ3 axis. The Mohr diagram degenerates into a singlecircle for this case.

Plane stress (0 = σ3 < σ2 ≤ σ1): If a principal stress is zero, plane stress is said to exist. This issynonymous 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 stressis zero. The state of stress has orthorhombic symmetry. The Mohr diagram for this case issymmetric with respect to the ordinate and to the origin.

Triaxial stress (σ3 < σ2 < σ1): All the principal stresses are different. The state of stress hasorthorhombic 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 freeboundaries. Uniaxial and plane stresses have those planes. If O-3 axis is chosen to be perpendicular

to the surface, then σ33 = σN = 0 and the shear stress on the surface is |σS| =√σ2

31 + σ232 = 0.

Therefore σ31 = σ32 = 0. Consequently, the stress tensor for the stress state has the form

r =

⎛⎝• • 0• • 00 0 0

⎞⎠ , (4.19)

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 symmetryfor the state of stress. Even for triaxial stresses, such a problem can be considered to estimate thenormal and shear stresses to investigate faulting (Chapter 6), as σ2 has a limited effect on faulting(Chapter 10).


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 illustratedby a Mohr diagram that consists of a single circle. To see this, suppose a stress tensor

r =

(σ1 00 σ2

),

for which the rectangular Cartesian coordinates O-12 are taken as parallel to the principal stress axes(Fig. 4.8). Other coordinates O-1′2′ 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 coordinatetransformation is given by Eq. (C.8). Thus, stress components are transformed from O-12 to O-1′2′

as (σ′11 σ′12σ′21 σ′22

)=

(cos θ sin θ− sin θ cos θ

)(σ1 00 σ2

)(cos θ − sin θsin θ cos θ

)=

(σ1 cos2 θ + σ2 sin2 θ −(σ1 − σ2) sin θ cos θ−(σ1 − σ2) sin θ cos θ σ1 cos2 θ + σ2 sin2 θ

). (4.20)

It is obvious in Fig. 4.8(a) that σN and σS are equivalent to σ′11 and σ′12, respectively. These stress


Figure 4.8: Mohr diagrams for two-dimensional problems. The axes O-12 and O-1′2′ represent tworectangular Cartesian coordinate systems with a common origin O. The former axes are taken to beparallel to the principal stress axes. O-1′ axis is taken to be parallel to N that is the outward normalto the surface element on which the traction is considered. This surface element has positive andnegative directions for N and n, respectively, leading to the opposite sign of the shear stress, SS andσS. See Fig. 3.2 for the relationship between positive shear directions. (b, c) Mohr’s circles for eachof the sign conventions.

components are shown in Eq. (4.20). Namely,

σN = σ′11 = σ1 cos2 θ + σ2 sin2 θ,

σS = σ′12 = −(σ1 − σ2) sin θ cos θ.

Using the formulas, sin2 θ = (1 − cos 2θ)/2, cos2 θ = (1 + cos 2θ)/2 and sin θ cos θ = 12 sin 2θ, we

obtain

σN =12

(σ1 + σ2) +12

(σ1 − σ2) cos 2θ (4.21)

σS = −12

(σ1 − σ2) sin 2θ. (4.22)


Figure 4.9: (a) Planes perpendicular to the O-3 axis are assumed to be the free surfaces. (b) Mohrdiagram 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

S =

⎛⎝S11 S12 0S21 S22 00 0 0

⎞⎠ .

The Mohr’s circle for this stress is drawn as the following procedure. The center of the circle B inFig. 4.9(b) is located at SN = (S11 + S22)/2, because the trace of a tensor is invariant to coordinaterotations. 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 diameterof the circle, as the faces make a right angle. The phase angle in a Mohr diagram is twice as muchas an angle in physical space. The line segment AB in Fig. 4.8(b) is the radius of the circle, and

AB =√[

(S11 − S22)/2]2

+ S212.

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 seethis, consider two materials I and II that are firmly fixed to each other at the O-12 plane (Fig. 4.10a).

Let r(I) =(σ

(I)ij

)and r(II) =

(σ

(II)ij

)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 pointP on the interface. Likewise, material II feels the traction t(II) at the same point. If all forces arebalanced 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) MaterialsI and II are firmly fixed at the O-12 plane. O-2 axis is perpendicular to the page. (c, d) Materials Iand II with a lubricated interface.

Given the unit normal to the interface n = (0, 0, −1)T, the first traction vector is written as

t(I) = r(I) · n =

⎛⎝σ(I)11 σ

(I)12 σ

(I)13

σ(I)21 σ

(I)22 σ

(I)23

σ(I)31 σ

(I)32 σ

(I)33

⎞⎠⎛⎝ 00−1

⎞⎠ = −⎛⎝σ(I)

13

σ(I)23

σ(I)33

⎞⎠ .

The unit normal for material II is −n = (0, 0, 1)T, and we have t(II) = (σ(II)13 , σ

(II)23 , σ

(II)33 )T. Substitut-

ing these traction vectors into Eq. (4.23), we obtain(σ

(I)13 , σ

(I)23 , σ

(I)33

)T =

(σ

(II)13 , σ

(II)23 , σ

(II)33

)T.

Since the stress tensors are symmetric, we obtain⎛⎝ • • σ(I)13

• • σ(I)23

σ(I)13 σ

(I)23 σ

(I)33

⎞⎠ =

⎛⎝ • • σ(II)13

• • σ(II)23

σ(II)13 σ

(II)23 σ

(II)33

⎞⎠as the boundary condition of the stress field at the interface2. The black dots in these matricesrepresent 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, wehave that the boundary condition for this case is

r

∣∣∣∣interface

=

⎛⎝• • 0• • 00 0 σ33

⎞⎠ . (4.24)

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


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 orientedNNW–SSW far from the San Andreas Fault. However, those are nearly perpendicular to the faulttrace 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 thirdcolumn and row. Therefore, σ33 is a principal stress and the third coordinate axis O-3 is the stressaxis corresponding to this principal stress. That is, one of the stress axes is perpendicular to alubricated surface. The other stress axes lie on the surface.

The solid Earth is covered by the atmosphere or seawater. These fluids have much less viscositythan 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 stressby winds and ocean bottom currents, but the shear stress is much less than tectonic stress. For thisreason, we neglect those drag forces in the discussion of tectonics. The result is that one of the stressaxes is perpendicular to the surface at the shallow levels in the Earth. In addition, if we assume thatthe 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 thesurface of the solid Earth. The atmospheric pressure is about 0.1 MPa, which is negligible comparedto tectonic stress. The pressure at the bottom of ocean is usually negligible. Therefore, we have thestress boundary condition at the surface,

r

∣∣∣∣surface

=

⎛⎝• • 0• • 00 0 0

⎞⎠ .

This is, of course, applicable to the surface of airless planets and satellites. However, hydrostaticpressure 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

r =

⎛⎝σH 0 00 σh 00 0 σv

⎞⎠ ,

where σH and σh are the maximum and minimum horizontal stresses, and σv is the vertical stressthat is usually approximated by the overburden stress (Eq. (3.29)). The σH- and σh-axes are thecorresponding horizontal principal axes. Some researchers use the notation, σHmax and σHmin, for thehorizontal 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-situstress 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 differencesometimes deforms or even collapses the borehole. Figure 4.11a shows the geometry of a boreholeno longer smooth or round. A zone along the wall of a well which has failed is called boreholebreakout. 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 stressperpendicular to the hole.

Hydraulic fracture stress measurement is an active measurement of in-situ stress. Hydraulicfracture or hydrofracturing is a process of breaking up the rocks immediately out of the wall ofa borehole under pressure by pumping water into an interval of the hole of which both ends aresealed. Hydraulic fracture experiments are usually done to increase permeability in hydrocarbonsand geothermal reservoirs. The pressure opens the pre-existing discontinuity surface including jointsand 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 onlydetermine the orientations but also the stress magnitudes.

4See [54] for further reading.


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

4.6 Dikes as paleostress indicators

Dikes, tabular intrusions of magma, are the results of the natural analog of hydraulic fracturing.Magma pushed its way through the country rock via its pressure for intrusion. This is demonstratedby the dike in the outcrop shown in Fig. 4.12. The dike vertically penetrates basaltic volcanic rocksand 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 olddike that is cut by a fault parallel to the bedding in the tuff, and the upper block is transported awayfrom the young dike shown in Fig. 4.12. The offset of the old dike is as large as the expansion ofthe young dike. These observations indicate that the young dike pushed its country rocks when itintruded.

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 tufflayer neighboring the dike shown in Fig. 4.12.

The Mohr diagram in Fig. 4.14(a) shows the admissible orientations of dike intrusion relative tothe principal axes of stress in the country rock. Consider a rock mass in which magma is intrudingwith a pressure pm, which has pre-existing cracks with various attitudes. The hatched area is boundedby the vertical line σN = pm, and the normal stress on the crack is smaller than the magma pressurein the area. Magma pressure has to overcome the normal stress for intrusion. The state of stressin the country rock is illustrated by the Mohr’s circles in Fig. 4.14(a), and the dark gray portionof the Mohr diagram shows the condition under which the magma pressure overcomes the normalstress. This portion designates the traction on the surfaces whose orientations are shown by the darkgray portion on the sphere in Fig. 4.14(b). Magma can intrude cracks with orientations that areindicated by the dark gray part of the sphere. As the magma pressure increases, if pm is smaller thanσ3, 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 pressureincreases, 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


Figure 4.14: (a) Mohr diagram showing the magnitude of normal stress on surfaces. (b) Diagramshowing 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, dikesare paleostress indicators perpendicular to the σ3-axis5. If a dike swarm, parallel dikes, occur in alarge 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 isknown from the radiometric age of the dikes. However, it should be determined from other types ofdata whether a horizontally extensional or compressional stress field affected the area.

Magma pressure may decrease with the distance of a magma chamber. Dikes accompanied bythe 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 ringdikes 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 Pbelt. There are cauldrons with the same age along the Median Tectonic Line, and one other cauldronin the central part of the arc has a dike swarm parallel to the line. The swarm was associated withnormal 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 patternof the present stress field in the shallow levels in the Earth has been understood [279]. Figure 4.16shows 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 STRESS FIELD 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. Thereare too many dikes in this area so that their density, i.e., number per a length of 1 km across the trendof 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 areroughly uniform. This was doubted because there are a variety of stress sources in the lithosphereincluding topographic and lithological variations. Lithosphere thickness is a reference to argue howthe stress field is uniform. The thickness of the mechanically strong part in the lithosphere has spatialand temporal variations, but is several tens of kilometers. There are regions where σH orientationis uniform over the dimension 20–200 times greater than the thickness of the lithosphere beneaththe regions. The central to northwestern North America is one of them, where σH axis is orientedENE–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 forcesat plate boundaries. If this is the case, the intraplate stress field is determined by the forces and theshape of the boundary. In the northern Indo-Australia plate, σH-axis is parallel to the velocity of theplate relative to the Eurasia plate. The Indo-Australia plate is bordered by a mid-oceanic ridge. Theaxis is oriented perpendicular to the boundary, due to the ridge push force.

The second point is that the stress gradients within a plate are smaller than expected. The smallgradients suggest that the viscous coupling of the lithosphere and the underlying mantle is weak. Ifthe plate had been fixed upon the stationary or horizontally travelling deep mantle, plate boundaryforces caused by differential movement of neighboring plates would have resulted in some gradients


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

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

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

4.8. EXERCISES 121

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

4.8 Exercises

4.1 Explain mathematically why the admissible combinations of normal and shear stresses areindicated 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 ofthe planes corresponding to the closed circles A, B, C and D in the diagram? (2) Which is the pointin 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 diapirloses its heat during the ascent, so that it has to rise rapidly to cause contact metamorphism in thecountry rocks when it reaches a shallow level of the crust. Estimate the lower bound of the ascendingvelocity 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.

principal stresses - terrapub · chapter 4 principal stresses anisotropic stress causes tectonic...

Documents