Chapter 6

Faulting and Brittle Strength

In this chapter, faulting is firstly considered as brittle failure. It is shown that tectonicstress is limited in its magnitude by the brittle strength of faults in the shallow levels inthe lithosphere.

6.1 Primary fractures

Faulting generates earthquakes that cause serious natural disasters so that great efforts have beenmade to understand it. Faults appear as shear fracturing in intact rock. In addition, faulting occursalong a plane of pre-existing weakness such as fractures and bedding planes. It is known that theCoulomb-Naiver criterion of brittle failure provides a basis for simple models for tectonic faulting.The criterion is a yield condition for intact rocks.

The results of experimental rock mechanics with triaxial testing apparatus were synthesized topropose the criterion. If the orientation of a fracture plane relative to σ1 axis is measured with theaxial load and confining pressure when fracturing occurred, the normal and shear stresses exertedon the plane just before fracturing can be calculated. Axial stress σ1 > σ2 = σ3 is applied tothe test sample, therefore, Eqs. (4.21) and (4.22) hold. The axial load and confining pressure areknown variables and are equal to σ1 and σ2, respectively. Thus, a Mohr circle can be drawn foreach experiment. The fracture plane angle θ is the angle between the σ1-axis and the fracture plane.This angle is equal to the orientation perpendicular to the fracture plane from σ3 axis, so that thenormal and shear stresses are indicated by the point that makes an angle of 2θ from (σ2, 0) alongthe Mohr circle (Fig. 6.1). The point represents the critical condition for shear fracturing. A seriesof experiments conducted for a specific type of rock with various confining pressures provide aseries of such critical points in the Mohr diagram. Connecting those points, we obtain the failureenvelope or Mohr envelope for the sample material. The envelope is symmetric with respect tothe horizontal axis of the Mohr diagram, because the specimens and the applied stress have axialsymmetry. Accordingly, the envelope is expressed by a one-valued function σS = f (σN). Failuretakes place when the Mohr circle expands with increasing differential stress so that it is just tangent

131

132 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Figure 6.1: Schematic diagram showing the normal and shear stresses at which the cylindrical spec-imens of the same rock are fractured by axial stresses (σ3 = σ2 < σ1).

to the envelope. According to experiments on rock failure, the shear stress needed to produce failureincreases as the confining pressure increases. On the other hand, rocks cannot bear very large tensilestresses. Therefore, the envelope has a V-shape, as shown in Fig. 6.1.

The envelope divides the O-σNσS plane into two regions. The region on the right side of theenvelope represents the combinations of normal and shear stresses that the type of rocks, for whichthe envelope was drawn, can support the stresses and does not fail. The rocks cannot support thestresses designated by the other region. The envelope therefore represents the fracture strength ofthe rocks.

Experiments shows that the slope of the envelope decreases for large confining pressures, andeventually vanishes. Then, a rock yields to the applied stress. Figure 6.2 shows the brittle strengthof various types of rocks. Carbonate rocks and halite have lower yield strength than plutonic rocks.

Any solid material breaks under very large uniaxial tension (σ1 = σ2 = 0 and σ3 < 0). Hence,the Mohr envelopes above and under the σN-axis are connected on the horizontal axis on the right ofthe σS-axis. The normal stress designated by the intercept is called tensile strength, and is designated

6.1. PRIMARY FRACTURES 133

Figure 6.2: Brittle strength of various types of rocks at room temperature [125]. The strength pre-sented in Eq. (6.7) is also shown. Note that the vertical axes of the five panels have different scales.

by the symbol σT. When the uniaxial stress with σN < σT, tension fracture occurs1. The Mohr circleis tangent to the envelope at the intercept (6.1). That is, a specimen tends to be split by a surfaceperpendicular to the extensional orientation.

Longitudinal splitting tends to occur under conditions of uniaxial compression (σ1 > 0 and

1Note that tensional stress is indicated by σN < 0.

134 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Figure 6.3: Compressive and tensile strengths of intact rocks at room temperature [116, 125]. Bar,minimum and maximum values; Box, ±25% intervals about average.

σ2 = σ3 = 0). The compressive strength is the minimum σ1 to cause brittle failure for the conditionsand is designated by σC. Figure 6.3 shows the tensile and compressive strengths of various types ofrocks.

The longitudinal splitting causes expansion perpendicular to the compressive axis. Hence, posi-tive confining pressures suppress the splitting, resulting in the formation of shear fractures.

Rocks have numberless micro cracks with various orientations. The growth and coalescence ofthe cracks control macroscopic failure. Under conditions of uniaxial tension, the applied stress buildsup the stresses immediately adjacent to cracks (Fig. 6.4(a)), i.e., stress concentration occurs at thetips of those cracks which are grown by the concentrated stresses. The crack propagation enhancesthe concentration. This feedback cycle is formed most favorably from the cracks perpendicular tothe applied tensile stress, resulting in tension fracture.

Shear fracture is a more complicated phenomenon [118]. Frictional sliding on closed surfacesoblique to the σ1-axis gives rise to crack propagation at the tips to the σ1 orientation (Fig. 6.4(b)).Unlike the formation of tensile fractures, the crack propagation does not directly make a feedbackcycle but mitigates the stress concentration at the tips. Therefore, compressive strength is muchgreater than tensile stress.

The Griffith theory of fracture formulates the behavior of those micro cracks as the equation forfailure envelope:

σN =σ2

S

4|σT|− |σT|. (6.1)

This is a parabola with the apex at the point (σT, 0) and (0, ±2σT). The failure envelopes determinedby triaxial deformation tests2 are roughly similar to the parabola (Fig. 6.2).

2Triaxial testing apparatus applies axial stresses to specimens. The Griffith-Murrell failure criterion, (τoct)2 = |8σTσ0|, isproposed for conditions of triaxial stress [91].

6.2. COULOMB-NAVIER CRITERION 135

Figure 6.4: (a) Stress concentration by tensile stress. (b) Formation of shear fracture by confinedcompression.

6.2 Coulomb-Navier criterion

If σS = ±f (σN) represents the failure envelope, its linear approximation to this equation,

|σS| = τ0 + μσN, (6.2)

is often used, i.e., brittle strength is linearly related with normal stress (Fig. 6.5(a)). This is knownas the Coulomb-Navier criterion3. In relation to the coefficient of friction, μ is called the coefficientof internal friction. The angle φ that satisfies

μ = tanφ (6.3)

is known as the angle of internal friction and is designated by ∠CAB in Fig. 6.5(a). The interceptτ0 is known as cohesion. These are material constants depending on lithology and other factorsincluding temperature and strain rate.

This criterion is often used to link faulting to stress. As differential stress increases, the Mohrcircle between the least and greatest principal stresses expand. When the circle eventually touchesthe failure envelope, a shear fracture is formed, i.e., a fault is generated. Note that the magnitude ofthe intermediate principal stress has nothing to do with the failure criterion. However, the attitude ofthe fault plane predicted by this criterion is parallel to the σ2-axis, because the points on the Mohrcircle that links the σ1 and σ3 represent the surfaces that contain the σ2-axis (Fig. 6.5(b)). Amongthose surfaces, one in the pair represented by the points C and D in Fig. 6.5(a) is chosen to form afault. The two surfaces are called the conjugate shear planes or conjugate faults. The half line BCor BC makes an angle 2θ with the half line BE (Fig. 6.5(a)). The point E designates the normalstress σ3. Accordingly, the point C or D indicates the angle θ between the σ3-axis and the lineperpendicular to the fault surface in physical space. The latter equals the angle between the surfaceand the σ1-axis (Fig. 6.5(b)). The acute angle between the conjugate faults equals 2θ. The angle θis sometimes called the angle of shear.

3Various names are used for this criterion, such as the Coulomb, Navier-Coulomb, and the Coulomb-Mohr yield criterion.

136 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Figure 6.5: (a) Mohr diagram showing the Coulomb-Navier criterion. (b) Diagram showing therelationship among the stress axes, shear angle and the sense of sear. Note that the σ2-axis is per-pendicular to the page.

The point B is the center of the Mohr circle, so that ∠BCA = 90◦. Therefore, we have

2θ + φ = 90◦, (6.4)

where ∠ABC = 2θ. Rearranging this equation, we have μ = tanφ = tan(90◦ − 2θ) = cot 2θ. Thatis, we obtain the relationship between the coefficient of internal friction and the shear angle,

tan 2θ = 1/μ. (6.5)

This equation shows that μ is inversely correlated with the angle of shear. The angle has a minimumθ = 0 and a maximum θ = 45◦ for μ = ∞ and 0, respectively. Laboratory experiments show that θis roughly equal to 30◦ for various rock types.

The criterion is simple and, therefore, convenient. However, laboratory experiments using ap-paratus that produces true triaxial stresses σ3 < σ2 < σ1 demonstrated that the magnitude of σ2

affects the brittle strength of rocks. The strength predicted by the criterion has errors of up to ∼20%(§10.3). In addition, the attitude of fault surfaces coincide with the criterion only for limited cases(§11.7).

Strength of ice The Coulomb-Navier criterion is used to describe the brittle strength not only ofrocks but also of ice. Many moons in the outer planets have thick icy crusts, so that they are calledicy satellites. Large icy satellites had tectonic activities that left tectonic features on the surface.Ganymede is one of them (Fig. 3.6). Therefore, the strength of ice is important to understand theactivities. The surface temperature of the icy satellites is lower than ∼100 K. Laboratory experimentshave presented the values μ = 0.55 and τ0 = 1 MPa at low confining pressures σ3 < 500 MPa andat temperatures between 77 and 115 K and pressures [172]. The compressive and tensile strength of

6.3. ANDERSON’S THEORY OF FAULTING 137

Figure 6.6: Types of faulting predicted by Anderson’s theory of faulting.

ice are estimated at 3.4 and –1.2 MPa, respectively, by Eq. (D.22). Both are one orders of magnitudesmaller than those of rocks (Fig. 6.3). That is, ice is weaker than rocks.

6.3 Anderson’s theory of faulting

One of stress axes stands upright near the surface if the surface is largely horizontal (§4.4). Com-bining this and Coulomb-Navier criterion, we obtain an important relationship between the type offaulting and stress. This is known as Anderson’s theory of faulting [4].

The criterion predicts that fault planes are parallel to σ2-axis, and that the planes make angles12 tan−1(1/μ), which is less than 45◦ with the σ1 axis. Therefore, if the σ3-axis is vertical, thestress gives rise to reverse faulting with the dip angle of the faults less than 45◦ (Fig. 6.6(a)).Namely, thrusts are formed. If the σ1- or σ2-axis is vertical, normal or strike-slip faults are formed,respectively (Figs. 6.6(b) and (c)). The latter should have vertical fault planes.

Anderson’s theory of faulting is convenient for infering the state of stress from faults and viceverse so that it is often used if no other evidence is available. However, it is now known that thetheory is not always correct. It cannot account for the abundance of oblique-slip faults in the field[20]. In addition, the fault activity predicted by this theory results in a plane strain4. However, strainsshould be generally three-dimensional even if they are caused by brittle faulting (§11.7).

4See Exercise 11.1 and its answer.

138 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Anderson’s scheme is currently used for the classification of stress regimes in the lithosphere5,involving the normal fault regime (σv > σH), reverse fault regime (σH > σh > σv), and strike-slipregime (σH > σv > σh).

6.4 Frictional strength of fault

A rock mass has discontinuous surfaces such as joints, fault planes and bedding planes. Thosesurfaces can be activated as faults by applied forces. Frictional sliding occurs on the planes. Thefrictional resistance or frictional strength is represented by the shear stress on the surface, and isgiven by

τf = μfσN, (6.6)

where σN is the normal stress acting on the surface and μf is the coefficient of friction. In the caseof shear stresses that are less than this strength, the surface is locked. When the applied shear stressovercomes this, sliding occurs on the surface.

The frictional strength of rocks is important for the investigation of earthquakes and landslidesso that the coefficient of friction of various rock types are experimentally estimated in the range0.6–1.0. The strength roughly obeys the piecewise linear relationship6 [33],

μf =

{0.85σN (σN < 200 MPa)0.6σN + 50 MPa (0.2 < σN < 1700 MPa).

(6.7)

An overburden stress of 200 MPa corresponds to the depth at about 6 km. This relationship issometimes referred to as Byerlee’s law. The continuous function of σN,

τf = σ0.94N (σN < 1700 MPa), (6.8)

is sometimes convenient [125]. The coefficient of the friction of ice is estimated at μf = 0.69 [19].The coefficient of friction introduced above was determined by laboratory experiments. How-

ever, field observations suggest that great faults have low friction [197]. The σH orientations near theSan Andreas Fault is consistent with this view. Heat-flow measurements around this fault suggestsfriction as low as μf < 0.2 [115].

Phyllosillicate minerals such as clay, serpentine, and mica show lower coefficients of friction.Since fault surfaces have those minerals, their strengths are important. Montmorillonite, a claymineral stable at low temperatures, has μf ≈ 0.2 [158]. Fault gauges and clay sampled from the SanAndreas Fault have the values 0.15–0.55 [158]. It is a matter of debate how these materials lowerthe frictional strength of faults. It is argued that clay minerals generally show strain hardening, sothat highly deformed fault clay is as hard as ordinary rocks. In addition, frictional sliding has similar

5The regimes are also known as the normal faulting, reverse faulting, and strike-slip faulting regimes of stress.6It is reported that surface roughness affects the strength in the shallow crust where the confining pressure lower than

∼300 MPa.

6.5. EFFECT OF PORE FLUID PRESSURE 139

Figure 6.7: Force to move an air-cushion vehicle.

microscopic processes with brittle failure at high confining pressure. Therefore, clay minerals maynot lower the strength of the faults.

Frictional sliding and brittle failure have similar microscopic phenomena. Surfaces on whichfrictional sliding occurs have roughness so that brittle failure or ductile deformation of bumps onthe surfaces are needed for the sliding. In other cases, bumps on one side of the fault plane shouldovercome the normal stress to climb over the others on the other side of the plane. The normal stressbecomes large with increasing confining pressure so that the microscopic brittle failure and yieldingcontrol the frictional strength at high confining pressures.

6.5 Effect of pore fluid pressure

Rocks at depth have pore water to a greater or lesser extent. Pore fluids have strong control overthe brittle and frictional strengths of rocks so that they are important when considering landslides,earthquake generation, and the strength of the lithosphere.

Consider an air-cushion vehicle with the basal area of S and the weight M . The horizontal forceneeded to overcome the basal friction and to move the vehicle is (μfMg/S ′)S ′ = μfMg, where S ′ isthe area on which the vehicle rests on the ground (Fig. 6.7). The air pressure p in its skirt lowers thisforce to μf (Mg − pS), S is the basal area of the skirt. The pS is the force due to the pressure p tofloat the vehicle. When the pressure reaches Mg/S, the vehicle floats on the surface and eventuallymoves.

The air pressure is comparable with pore fluid pressure7 pf to lower the frictional and brittlestrength of rocks. Without pore fluid, normal stress in a rock should be supported by a framework ofsolid grains. Pore fluid pressure, if it exists, takes a part of the forces in the rock to lower the normalstress. The decreased normal stress reduces the brittle and frictional strengths. Pore pressure doesnot affect shear stress but lowers normal stresses so that this effect is described by the equation

re = r − pfI, (6.9)

where re is the effective stress. The frictional strength is reduced by the pore pressure as

τf = μf (σN − pf ), (6.10)

7This is simply called pore pressure.

140 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Figure 6.8: Diagram showing the effect of pore-fluid pressure on brittle strength.

where (σN − pf ) is the normal stress due to re. It was experimentally confirmed [71] that the brittlestrength predicted by the Coulomb-Navier criterion reduces to

|σS| = τ0 + μ(σN − pf ). (6.11)

Instead of pore fluid pressure, the pore fluid pressure ratio

λf = pf/pL (6.12)

is often used to discuss the strength of the lithosphere.Note that re is different from r only through the hydrostatic pressure pfI (Eq. (6.9)). Therefore,

the two tensors have common principal orientations (§C.5). The tensors have a common shear stressacting on a particular plane. Therefore, the difference is graphically expressed by the leftward shiftof the Mohr circles (Fig. 6.8). If r is constant, increased pore fluid pressure shifts the circlesto traverse the failure envelope to activate a shear fracture. Consequently, pore pressure reducesthe brittle strength of rocks. Increased ground-water pressure through heavy rain can weaken thefrictional resistance of a pre-existing slip surface to cause a landslide. High pore pressure is essentialfor the formation of an extensive decollement surface (§4.2).

There are several origins of high pore-fluid pressures. Dehydration associated with metamor-phism can build up the pressure. High pore pressure zones are often met in hydrocarbon explorationin sedimentary basins. As formations are deeply buried, increased overburden stress encourages de-watering of the formations. However, the formations are sealed by an overlying impermeable layersuch as a claystone bed. The formations are hindered from dewatering, resulting in overpressuredformations. Without this type of cap rock, increased compaction impede dewatering (Fig. 6.9(a)).There are other origins of overpressured zones [234]. Tectonic stress can control pore-fluid pressure[70, 251].

An overpressured zone is defined as the zone where the pore-fluid pressure is higher than thehydrostatic pressure of the same depth (Fig. 6.9(b)). The existence of overpressured formations iscrucial for petroleum industry, because pore-fluid pressure in a hydrocarbon reservoir affects efficientdrain-up of the resource. In addition, pore-pressure affects efficient borehole drilling, because high

6.6. REACTIVATION OF FAULT 141

Figure 6.9: Schematic pictures showing overpressured formation in a sedimentary basin. (a) Porefluid pressure pf help the framework of solid particles to support normal stresses. (b) Pore spacesbetween sedimentary particles are linked to each other in the shallow part of the basin, but thecontinuity is reduced by compaction. Therefore, overpressured formation appears at depths.

pore-pressure weakens the wall of a borehole. Therefore, pore-fluid pressure is routinely monitoredin hydrocarbon exploration. Overpressured formations appear at a few kilometers below the surfaceor deeper [54].

6.6 Reactivation of fault

The Coulomb-Navier criterion predicts that the formation of a fault in an intact rock needs a differ-ential stress as great as the cohesion τ0. By contrast, frictional sliding is much easier, as the slidinghas τ0 = 0. Shear stress needed for frictional sliding is smaller than that for forming a shear frac-ture in an intact rock. Therefore, pre-existing fractures are easily activated as faults, if they havenot been cemented. Accordingly, a fault can be repeatedly activated. Many geological faults havedisplacements much larger than the representative displacement of one earthquake faulting, whichis less than ∼10 m. Therefore, they demonstrate the reactivation, and active faults that moved in theLate Quaternary are dangerous.

Example

Old large faults are often reactivated in different stress regimes. Normal faults in a passive continen-tal margin are sometimes reactivated as reverse or thrust faults when the ocean basin in front of themargin is consumed. This is known as tectonic inversion or basin inversion. Figure 6.10 shows anexample. Old reverse faults are sometimes reactivated as normal faults.

The timing of inversion is constrained by geological structures and the ages of involved forma-tions. Triassic to Jurassic rifting in the North Sea created normal faults before the earliest Cretaceous

142 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Figure 6.10: Tectonic inversion in the North Sea [13]. The upper panel shows the seismic profile inthe basin. The middle panel shows the line drawing for the structures interpreted from the profile.The lower panel shows the inferred geological structure when the horizon A/C deposited. Significanthorizons are labeled as IUC = a reflecting horizon in the Upper Cretaceous, IC = a horizon inthe Upper Cretaceous Chalk Group, A/C = Aptian-Cenomanian (Lower and Upper Cretaceous)boundary, BC = base Cretaceous, TR = Triassic, Z = top Zechstein Supergroups (Upper Permian),R = top Rotliegend Group (Lower Permian). Note that normal faults that bounded wedge-shapedhalf grabens (lower panel) were reactivated as reverse faults (middle panel). Arrows indicate onlap,offlap or erosional truncation of sedimentary layers.

in this area, and some of these faults have been reactivated by compression since the Late Cretaceous.

The angular unconformity at the horizon B/C in the lower panel in Fig. 6.10 demonstrates thetermination of rifting before the middle of the Cretaceous. Arrows pointing to the horizon BC frombelow show that a sedimentary sequence older than the horizon is tilted and truncated by erosion. Theformation between Z and BC shows lateral variations in thickness. The formation thickens towardnormal faults, indicating syndepositional normal faulting. The thickness of the Lower Cretaceous,which lies between BC and A/C, is relatively constant, suggesting that the differential subsidenceaccompanied by normal faulting had ceased before the accumulation of the Lower Cretaceous.

6.6. REACTIVATION OF FAULT 143

Figure 6.11: Reactivation condition of faults. (a) Two stress states with the same differential stress.The fault orientations, allowed to move under the stresses, which are indicated by gray fans inthe Mohr circles, become narrower with increasing mean stress. (b) Since the vertical stress σv isapproximately equal to the overburden pressure, faults are more easily reactivated by the normalfaulting regime of stress than the reverse faulting one if the two regimes have the same differentialstress and the same overburden pressure.

The timing of inversion is known from the structure and age of the overlying strata. The UpperCretaceous formations make asymmetric folds with the same intervals as the normal faults. The for-mations onlap the anticlines (arrows with number 1) so that the formations are thick in the synclinesand thin in the anticlines. These observations demonstrate that the folds grew with the accumulationof the Late Cretaceous. Therefore, compressional tectonics began sometime between A/C and IUC.

The compressional tectonics had ceased at around the horizon IUC, because the thickness of thestrata younger than this horizon is hardly affected by the folds. The young formations accumulatedthrough the widespread uniform subsidence of this region.

Reactivation condition

There are conditions for the reactivation of a fault. Theis reactivation would be difficult for the faultsurfaces that have been firmly fixed by the lithification of fault materials8. In addition, the newstate of stress should have favorable principal orientations for the faults. For example, fault surfacesparallel to a principal plane of stress could not move because shear stress on the surface vanishes.

The state of stress controls the mobility of pre-existing fractures. Figure 6.11(a) shows that theorientations of faults that can be activated by the same differential stress expand with decreasing σ3,because the brittle and frictional strengths increase with σN. Therefore, the orientations are morelimited with increasing depth and pore pressure. Great faults appear to have low friction (§6.4). Ifso, the admissible orientations are widen, also. Accordingly, great faults are easily reactivated.

More importantly, the normal fault regime of stress enables the activation or reactivation of faults

8If the lithified material has elastic constants different from those of country rocks, stress concentration at the fault surfacewould encourage reactivation.

144 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

more easily than the reverse fault regime if the differential stress and the burial depth are the same.This difference results again from the positive gradient of the strengths, μ > 0 and μf > 0 (Fig.6.11(b)). Vertical stress obeys the relationship, σv ≈

∫ z0 ρ(z)gdz ≈ ρgz. In the last approximation,

the density is assumed to be constant, because common rocks9 have densities around 3×103 kg m−3.Therefore, σv is determined by depth. If a rock mass is in the normal faulting regime of stress, wehave σv = σ1 and σh = σ3 = σv −Δσ < σv. On the other hand, if it is in the reverse faulting regime,we have σv = σ3 = σ1 − Δσ < σH and σH = σ1. The Mohr circles of the former regime are nearerto the strength line than those of the latter (Fig. 6.11(b)). Therefore, faulting occurs in a normalfaulting regime more easily than in a reverse faulting regime.

6.7 Depth dependence of brittle strength

Rocks support differential stresses less than their strength. When differential stress reaches rockstrength, brittle failure occurs to mitigate the stress. Consequently, the brittle strength determinesthe maximum differential stress in the shallow part of the Earth, where brittle failure is the mostsignificant deformation mechanism. Let us derive the critical stress at depth z for normal and reversefault regimes from the friction law (Eq. (6.6)). Pre-existing fractures with various orientations areassumed to exist so that the law is employed here. Since we use the frictional strength (Eq. (6.6))that is independent from the magnitude of σ2, only the σ1 and σ3 orientations are considered. Thecoordinate system O-xz is defined as shown in Fig. 6.12(a).

In order to distinguish the regimes, we define Δ = σH − σv. The sign of this quantity indicatesone of the regimes (Figs. 6.12(b) and (c)). This is equivalent to differential stress for a reverse faultregime. However, −Δ equals differential stress for a normal fault regime.

Secondly, we have a uniform density ρ and a level surface, so that stress axes are parallel to thex- and z-axes. The principal stresses are

σz = ρgz and σx = σz + Δ. (6.13)

We investigate whether a pre-existing fracture surface dipping at an angle (π/2 − θ) is activatedas a fault (Fig. 6.12(a)). Since we are considering a two-dimensional problem, Eqs. (4.21) and(4.22) apply to our problem.

In the case of the reverse fault regime, σ1 and σ2 in Eqs. (4.21) and (4.22) should be replaced byσx and σz, i.e., we have the normal and shear stresses acting on the footwall

σN =12

(σx + σz) +12

(σx − σz) cos 2θ =σx + σz

2+

Δ2

cos 2θ (6.14)

σS =12

(σx − σz) sin 2θ =Δ2

sin 2θ. (6.15)

Note that the definition of θ in this problem is different from that in Fig. 4.8(a) in its sign. Therefore,the right-hand side of Eq. (6.15) lacks the negative sign that is attached to Eq. (4.22).

9Here, materials that are abundant and have low densities (e.g., water and rock salt) are neglected.

6.7. DEPTH DEPENDENCE OF BRITTLE STRENGTH 145

Figure 6.12: Schematic illustration to explain the activation condition of a pre-existing fracturesurface as a fault. (a) The dip of the surface is π/2 − θ. Unit vector n points inward to the footwall.(b, c) The relationship of the sign of Δ with stress regimes.

In the case of the normal fault regime (Fig. 6.12(c)), we have the correspondence σ1 = σz andσ2 = σx. Comparing Figs. 4.8(a) and 6.12(a), we replace θ by π/2 − θ in Eqs. (6.14) and (6.15).By means of formulas cos(π − 2θ) = − cos 2θ and sin(π − 2θ) = sin 2θ, we obtain

σN =12

(σz + σx) − 12

(σz − σx) cos 2θ =12

(σz + σx) +Δ2

cos 2θ

σS =12

(σz − σx) sin 2θ = −Δ2

sin 2θ.

Consequently, the results for the two cases are different only in the signs of the shear stress.Therefore, combining Eqs. (6.13), the normal and shear stresses are expressed by the equations

σN = ρgz +Δ2

(1 + cos 2θ), (6.16)

σS = ±Δ2

sin 2θ, (6.17)

where the upper and lower signs in the right-hand side of Eq. (6.17) indicate the reverse and normalfault regimes, respectively.

The critical condition for this surface is given by Eq. (6.10), i.e., σS = μf (σN − pf ). SubstitutingEqs. (6.16) and (6.17), we obtain

±Δ2

sin 2θ = μf

[ρgz +

Δ2

(1 + cos 2θ) − pf

],

∴ Δ2

[± sin 2θ − μf (cos 2θ + 1)

]= μf (ρgz − pf ).

Rearranging this equation, we obtain the critical condition

Δ =2μf (ρgz − pf )

± sin 2θ − μf (cos 2θ + 1). (6.18)

146 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Figure 6.13: Graphs of F (θ) and G(θ) for the reverse and normal fault regimes, respectively. Opencircles designate their peaks. The gray region indicates the domain of the function.

Hence, the critical stress increases with depth z, but pf can cancel this effect.The continental crust has experienced a long deformation history that has created the planes

of weakness such as old faults, joints, metamorphic foliations, and bedding planes. Therefore, weassume that there are pre-existing planes of weakness with various orientations. If so, the surfaceswith the most favorable orientations are activated as faults. The optimal orientation is given bythe minimum differential stress |Δ|. In order to determine the optimal orientation, we seek θ thatmaximizes the denominator in the left-hand side of Eq. (6.18).

In the case of the reverse fault regime, the function F (θ) = sin 2θ − μf (cos 2θ + 1) should bemaximized for the domain 0 ≤ θ ≤ π/2. By way of the formula sin a cos b+ cos a sin b = sin(a+b),we have

F (θ) = sin 2θ − μf cos 2θ − μf =√

1 + μ2f

⎛⎜⎝ 1√1 + μ2

f

sin 2θ − μf√1 + μ2

f

cos 2θ

⎞⎟⎠ − μf

=√

1 + μ2f sin(2θ + α) − μf , (6.19)

wherecos α =

1√1 + μ2

f

, sin α = − μf√1 + μ2

f

, tan α = −μf , (−π/4 ≤ α ≤ 0).

Since the coefficient of friction ranges between 0 and 1, the phase angle α is in the range −π/4 <α < 0. Equation (6.19) represents the sine curve with a period of π. Therefore, F (θ) have themaximum

F (θ) =√

1 + μ2f − μf

within the range 0 ≤ θ ≤ π/4 (Fig. 6.13).For the case of the normal fault regime, the function

G(θ) = sin 2θ + μf cos 2θ + μf =√

1 + μ2f sin(2θ + α) + μf

6.8. JOINT 147

must be maximized, where

cos α =1√

1 + μ2f

, sin α =μf√

1 + μ2f

, tan α = μf (0 ≤ α ≤ π/4).

The function G(θ) has the maximum

G(θ) =√

1 + μ2f + μf

within the range 0 ≤ θ ≤ π/4 (Fig. 6.13).Substituting the maximum denominator into Eq. (6.18), we obtain the critical condition

Δ =2μf (ρgz − pf )√

1 + μ2f ∓ μf

, (6.20)

where the upper and lower signs in the denominator of the right-hand side indicate the reverse andnormal fault regimes, respectively. Using the pore-fluid pressure ratio (Eq. (6.12)), this is rewrittenas (Fig. 6.14)

Δ = ±

⎡⎢⎣ 2μf√1 + μ2

f ∓ μf

⎤⎥⎦ (1 − λf ) ρgz. (6.21)

The double signs correspond to the reverse (upper sign) and normal (lower sign) fault regimes.Equation (6.21) designates that the critical differential stress is proportional to depth z.

Figure 6.14(a) shows a graph for the depth-brittle strength relationship given by Eq. (6.21).Withing this context, the absolute value of the critical Δ is the strength. The figure shows that thestrength is smaller in the normal fault regime than in the reverse fault regime. In addition, porepressure has a strong control over the brittle strength. This difference results from the fact illustratedin Fig. 6.11(b). The dependence of the strength on μf is indicated by the factor within the bracketsin Eq. (6.21). Figure 6.14(b) shows that the brittle strength has a larger μf dependence in the reversefault regime than in normal fault regime. It should be noted that the differential stresses shown inFig. 6.14(a) are the limits, and those may be much smaller than the critical values in tectonicallyquiet regions.

Large faults may have a small coefficient of friction. In-situ stress measurement is a clue toestimate the frictional resistance of large faults. For this purpose, stress states were measured atvarious depths in tectonically active regions where the stress state is at or near the critical condition,therefore, Eq. (6.21) is applicable. The results indicate that the effective coefficient of friction isabout the same as that determined in the laboratory [3, 21, 278].

6.8 Joint

Fractures of rocks that show little displacement parallel to their surfaces but with very small openingsnormal to them are called joints. They are basically extension fractures. The lack of displacement

148 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Figure 6.14: (a) Brittle strength (Δ) as a function of depth z. Parameters, ρ = 2.7 × 103 kg m−3,g = 10 ms−2 and μf = 0.75 are used. (b) Dependence of the brittle strength on the pore-fluidpressure ratio.

for a crack is verified by the feather-like pattern of subtle ridges and grooves on the fracture surfaceand the lack of fault striation (Fig. 11.2). Most ourcrops contain one or more sets of joints. Eachjoint set has a characteristic orientation and spacing. The photograph in Fig. 6.15 shows two jointsets in Lower Miocene sedimentary layers. One set has a spacing of several centimeters and theother has that of a few tens of centimeters. Joint sets in the same rock mass is called a joint system.The abundance of joints designates that there are numberless cracks in the shallow part of the crust,so that they are important for the transfer of formation fluids and for the stability of a rock masscontaining fractures.

The relative age of joint sets is determined at outcrops by the abutting relationship, i.e., an oldjoint surface acts as a free surface so that younger crack propagation is stopped at the surface10.

Joints have multiple origins including the contraction of rocks due to cooling, unloading of over-lying rocks by erosion, or tectonism11. Joints can form in sediments before they have consolidatedinto rock. For example, extension fracturing of unconsolidated sediment in the presense of highpore pressure can result in the formation of clastic dikes. As evidence, Fig. 6.16 shows a clasticdike which penetrates a tuff breccia. The dike is tapped from the thin pumice tuff layer beneath thebreccia. The crack is filled with the pumice tuff which covers the laminated siliceous tuff and thickfine tuff. The substrata of the breccia probably deposited in a swamp. The negative pressure pro-duced by the extension fracturing in the crack with respect to the overpressured substrata not only

10See [248, §3.4] for further reading.11See [54] for further reading about the mechanics of joint formation.

6.8. JOINT 149

Figure 6.15: Joint system in alternating beds of sand- and siltstones, the Lower Miocene Oya For-mation, northwestern Kyushu, Japan. Bedding planes are gently inclined to this side. There are twosets of joints here; fractures trending toward the sea with smaller spaces, and those with larger spacesperpendicular to the grip of the hammer.

sucked up the pumice tuff but also broke the laminated layer into slices and gathered them under thedike. Consequently, the slices made an antiformal stuck, a type of duplex structure. Formation ofclastic dikes are comparable with hydraulic fracturing. They were formed in the shallow levels ofsedimentary basins.

Sediments gradually gain stiffness during burial. Consider a horizontally extensive layer of sed-iments that is assumed to be an elastic plate. Lithification of the sediments may be modeled byincreasing the tensile strength ST = −σT. The incremental depth of a horizontally lying elastic slabcauses a normal fault regime of stress within the slab, i.e., horizontally extensional stress (§7.2).Therefore, clastic dikes are formed vertically. However, stress levels there are very small, so thatmultiple factors give rise to disturbance in the stress field. Consequently, clastic dikes are not soparallel as swarms of igneous dikes. By contrast, joints surprisingly make parallel sets.

Figure 6.17 shows the orientation of joints in the Upper Miocene formations in the Boso penin-sula of central Japan. The joints are nearly vertical and largely trend NE–SW. This orientation is

150 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Figure 6.16: Photographs showing the source of a clastic dike. Upper Oligocene Shioseno-misakiConglomerate, northwestern Honshu, Japan. (a) A hammer is placed at the root of the dike. (b)Close-up of the root. The slices of a thin layer of laminated siliceous tuff makes an antiformal stack,a type of duplex structure, at the base of the dike. The coin is about 2 cm in diameter.

6.9. EXERCISES 151

Figure 6.17: Orientation of joints in the Upper Miocene Amatsu and Anno Formations, Boso penin-sula, central Japan. Each solid square indicates the pole to a joint surface that the correction for thedip of strata has been carried out. The open circle designates the orientation of fold hinge lines inthis area. Arrows indicate the trend of maximum shortening perpendicular to the hinge line.

perpendicular to the hinge lines12 of folds in this area. The folds suggest Late Miocene to EarlyQuaternary shortening in NE-SW trend. Therefore, the majority of the joint were perpendicular tothe σh orientation at that time, suggesting the coeval formation of the joints and folds under NE–SWcompression. The peninsula is located near a triple-trench-type plate junction, and has experiencedpolyphase tectonics since the Tertiary. This joint set was probably reactivated as normal and oblique-normal faults in the mid Quaternary (§11.5).

6.9 Exercises

6.1 Eliminate σN, σS and τ0 from Eq. (6.2), and rewrite the equation to relate σ1, σ3, σT and σC.

6.2 Pore fluid pressure is usually limited by overburden stress pL. Therefore, intact rocks are stableif the differential stress Δσ is smaller than their tensile strength |σT| regardless of pore pressure (Fig.6.18). Common rocks have tensile strengths of the order of 100–101 MPa. Converting these valueswith a constant density of 2.8 × 103 kg m−3, we have a depth of 102–103 m. Therefore, intact rocksshallower than this can support their overburden. However, sediments before consolidation have sosmall effective values of tensile strength that they do not allow extension fracturing.

By means of the failure envelope predicted by the Griffith theory (Eq. (6.1)), determine themaximum depth to which cracks are allowed to propagate in unconsolidated sediments to form

12The hinge line of a fold is the line in the folded surface along which the curvature is a maximum.

152 CHAPTER 6. FAULTING AND BRITTLE STRENGTH

Figure 6.18: Stability of intact rocks. If pore fluid pressure pf equals overburden stress pL, the Mohrcircle of the effective stress passes the the origin O of this diagram.

joints. Estimate the order of magnitude for this depth using the values ρ = 2.3 × 103 kg m−3 and|σT| = 0.1 MPa = 1 bar.