structural geology lectures series 1.pdf

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Structural Geology

Lecture 1

Introduction:

The parameters of structural geology(constants, conversion factors, etc)

Structural geology boils down to a study of Newton's famous laws of motion asthey pertain to the deformation of rocks within the earth. Newton's first law of motion

states that "Every body persists in its state of rest or of uniform motion in a straight line

unless it is compelled to change that state by forces impressed on it". The very presenceof faults and folds in the crust suggests that rocks, seemingly at rest, were once subject to

forces that changed their original state by motion of one point relative to another.

Structural geology is the study of the deformation of rocks. In its simplest form this is adescription of present geometries. A study of the motion causing the geometries within

rocks is called kinematics. A study of the forces that cause the motion is calleddynamics. The mathematics of structural geology are designed to simplify the study of

kinematics and dynamics.

Structural geology is the study of the geometry, kinematics, and dynamics of rock

structures. Geometric analysis is the descriptive or qualitative portion of structuralgeology. This portion of structural geology is as the name implies: A study of the size,

shape, and orientation of structures. This portion was covered in classical structural

geology courses. However, in this set of lecture notes the study of geometry will bedelayed until a good mathematical base is established. In the meantime, many of the lab

exercises will be devoted to geometric analysis. One of the most useful tools in geometricanalysis is the stereonet which is a qualitative tool that serves the same purpose as vectors

within a coordinate system.

Kinematic analysis requires a mathematical base for a rigorous treatment.

Kinematics, as you learned when taking elementary physics, is a mathematical description

of the motion of objects. In the case of structural geology kinematics is the description of

the path that rocks took during deformation. It is also the mathematical description of therelative position of two infinitesimal points during the deformation of rocks. Two points

can change by translating together, rotating around each other, or changing in distance

relative to one another. We shall call such a mathematical description deformationmapping.

Dynamics is the study of the forces which caused the deformations studied duringkinematic analysis. In the case of structural geology dynamics includes the study of how

rocks react to stress. For every stress the rocks respond with a finite strain. In a sense

rock structures would not have formed, if rocks had not been subject to a stress. A studyof dynamics starts with the fourth lecture.



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Like any physical science, structural geology comes complete with its table of

numerical data and symbols for physical properties. Physics has its table of fundamental

and derived physical constants which includes the speed of light (c = 3 x 108 m/sec),

Avogadro's number ( N o = 6.02 x 1023 / mole), and universal gas constant (R = 8.23

joules/(mole)(K°). A comparable table involves numerical data which depend on

circumstances such as geographic location and are, therefore, not strictly physicalconstants. Another table consists of a list of common physical properties which are

represented by symbols including letters of the Greek alphabet. These numerical dataand material properties form an important component of the language of structural

geology.

Table of symbols for the structural geologist

Symbol Name Units

ρ density ML-3

σ stress ML-1T-2

τ shear stress ML-1T-2

σn normal stress ML-1T-2

ε strain dimensionless [LL-1]

E Young's Modulus ML-1T-2

ν Poisson's ratio dimensionless

γ Engineering shear strain dimensionless

P p pore pressure ML-1T-2

φ porosity dimensionless

T temperature C°

q heat flow JL-2T-1

κ thermal conductivity JL-1T-1C°-1

z depth L



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Table of numerical data for the structural geologist

Symbol Name Magnitude

g average gravity at sea level 9.8 m/sec2

ρm mean density of the mantle 4.5 x 103 kg/m2 (4.5 g/cm2)

ρquartz density of quartz 2.65 x 103 kg/m2 (2.65 g/cm2)

Table of conventions for the structural geologist

Name Convention

Principal stresses σ1 > σ2 > σ3

Stresses in the crust

Maximum horizontal stress SH

Minimum horizontal stress Sh

Vertical horizontal stress Sv

Compressional normal stress positive

Tensile normal stress negative

Table of conversion factors for the structural geologist

Stress and pressure

1 atm = 14.5 psi = 1 bar = 106 dynes/cm2 = 105 N/m2 = 105 Pascals (Pa)

1 MPa = 10 bars = 106 N/m2

pressure applies to a fluid

stress applies to a solid



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Table of definitions for the structural geologist

Name Definition

A Component of Principal Stress σii or σi

Any Component of Stress σij

Differential Stress σd = σ1 - σ3

Maximum Shear Stress τ max =σ 1 −σ 3

2

Lithostatic Stress SH = Sh = Sv

Hydrostatic Pressure P p = P p = P p

Mean Stress σ m =σ 1 + σ 2 + σ 3

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Deviatoric Stress (3 components) σm - σ1, σm - σ2, σm - σ3

Effective Stress σi - P p

Table of equations for the structural geologist

Stress and pressure

P p = ρH2Ogz Sv = ρrock gz

if ρrock = 2.5 x 103 kg/m3, g = 9.8 m/sec2, z = 103 m,

then Sv = 2.5 x 105 kg/m-sec2 = 25 MPa/km

Finally, the average geothermal gradient (dT/dz) within the crust of the earth is

about 20°C/km where T is temperature and z is depth. This gradient can vary from

10°C/km in a glaucophane-schist terrain to 40°C/km. The low geothermal gradients can

occur in the vicinity of crystalline overthrusts where cold crust is depressed. Highgradients occur in a region of magmatic intrusion. Heat flow (q) at the surface is an



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indication of geothermal gradient provided the thermal conductivity (K) of the crust is

known

q = K(dT/dz).



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Structural Geology

Lecture 2

The mathematics of structural geology(vectors, tensors, Einstein summation)

Scalar - A physical quantity expressed by a single number; a tensor of zero rank.

Gradient of a Scalar - The direction at which a quantity increases the fastest.

grad φ = ∇φ (2-1)

Vector - A physical quantity expressed by a three numbers; a tensor of rank one.Coordinate system for a vector: RIGHT HANDED - looking up the x3 axis from the

origin in positive direction, the x1

axis will rotate into the x2

axis in 90° in a clockwise

direction. The components of a vector, F = F , are:

f i[ ] = f

1, f

2, f

3 (2-2)

The mathematical tool necessary to model Newton's laws of motion is the vector.

Newton invented calculus to manipulate vectors. Needless to say, quantitative structuralgeology is based on the use of vector calculus. Remember that a vector is a quantityhaving both magnitude and direction whereas a scalar has just magnitude. Within this

text a vector is represented by variables printed in bold letters (e.g. a). A knowledge of

direction implies that a coordinate system is known. For structural geology a rectangular or Cartesian coordinate system represented by the direction of three positive unit vectors

i, j, and k will do (e.g. i = 1, etc) (Fig. 2-1). Any vector can then be represented withscalar components f 1, f 2, and f 3 which represent distances along the axes denoted by the



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three unit vectors. Every vector in space many be written as a linear combination of unit

vectors i, j, andk . The vector f is then

f = f 1i + f 2 j + f 3k (2-3)

where f 1, f 2, and f 3 are components of the vector f . These three components areindependent of the choice of the origin point of the coordinate system. Vectors such as

af , bf can be added by summing the individual components so that the resultant vector Fis:

F = af + bf = (af 1 + bf 1 )i + (af 2 + bf 2) j + (af 3 + bf 3)k

Magnitude of a vector is

F = f 1

2+ f

2

2+ f

3

2

(2-4)

F2 = f

1

2 + f 2

2 + f 3

2 = f if

i1

3

∑(2-5)

Vectors are commutative:

F + G = G + F (2-6)

Vectors obey the associative law:

F + G + H = F + G( ) + H = F + G + H( ) (2-7)

Vector times a scalar is a larger vector.

Force f is a vector commonly used in explaining the causes of geological structures.Force is defined as the cause of an acceleration. Newton's second law states that if f is

the net force acting on an object of mass m moving with a velocity v, then force (f )

equals mass (m) times acceleration (a). This fundamental equation of classicalmechanics is an example of a vector multiplied by a scalar to derive another vector:

f = m(dv/dt) = ma.

Two common operations on a vector include the dot and cross products, respectively

(Fig 2-2). The dot product of two vectors is the projection of vector A on vector B:

A*B = AB cos θ



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where θ is the angle between the two vectors. An example of the dot product is the work done by a force. Consider a rock on which a constant force f acts. Let the rock be given

a displacement d. Then the work W done during the displacement of the rock is defined

as the product of |d| and the component of |f | in the direction of d, that is,

W = |f ||d|cos α = f * d.

where α is the angle between f and d.

Scalar product or Dot product of a vector - magnitude of one vector times the

magnitude of projection of 2nd vector on the first vector. This number of a scalar.

F ⋅ G = F G cos θ (2-8)

F ⋅ G = f ig

i= f

1g

1+ f

2g

2+ f

3g

3 (2-9)

Repeated suffixes means summation on i. i = 1to 3.

If the vectors are orthogonal then

F ⋅ G = 0 (2-10)

The cross product of two vectors is the magnitude of A times B times the sine of the

angle between then. The direction (u) is perpendicular to the plane of A and B.

A x B = AB sin θ u

where θ is the angle between A and B. According to plate tectonics theory, ascontinental plates move over the mantle a moment of force is generated. The moment of

force is also known as a torque τ which is a vector quantity (Figure 2-3).



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In mechanics the magnitude of the moment τ of a force f about a point Q is defined as

the product τ = |f |.d where d is the perpendicular distance between Q and the line of

action L of f . If r is the vector from Q to any point A on L, then d = |r|sin γ and

τ = |r||p|.sin γ = |r X p|.

Vector product or Cross product - The magnitude of H is the area of the

parallelogram whose sides are F and G.

H = F × G = − F × G = F G sin θ (2-11)

The cross product is not associative

F × G( ) × H ≠ F × G × H( ) (2-12)

With each square matrix we associate a number called its determinant. If the matrix is A,

then we demote its determinant by det A.

A =

f 1f

2f

3

g1g

2g

3

h1h

2h

3

⎛ ⎜⎜⎝

⎞⎟⎟ ⎠

det A =

f 1f

2f

3

g1g

2g

3

h1h

2h

3 (2-13)

det A =h1

f 2

f 3g

2g

3+ h 2

f 1

f 3g

1g

3+ h 3

f 1

f 2g

1g

2 (2-14)

where

h3

f 1f

2

g1g

2

= h3

f 1g

2− f

2g

1( ), etc.

(2-15)



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Einstein Summation illustrated using heat flow:

Heat flow, q (cal cm-2

sec-1

), in homogeneous, isotropic rock is indicated by a thermal

gradient,

∂T∂x

j .

q = − κ ∂T∂x

j (2-16)

where κ is the thermal conductivity.

For Granite: κ = 8 x 10-3

cal cm-1

sec-1

°C-1

.

For Steel: κ = 180 x 10-3

cal cm-1

sec-1

°C-1

.

Heat flow is parallel to a thermal gradient so it can be expresses in a vector formwhere for an isotropic homogeneous rock.

q1

= − κ ∂T∂x

1 (2-17a)

q2

= − κ ∂T∂x

2 (2-17b)

q3

= − κ ∂T∂x

3 (2-17c)

If the rock is anisotropic then the thermal conductivity is not the same in each of

the three directions. If

a j

= ∂T∂x

j , then

q1

= κ11

a1

+ κ12

a2

+ κ13

a3 (2-18a)

q2

= κ21

a1

+ κ22

a2

+ κ23

a3 (2-18b)

q3

= κ31

a1

+ κ32

a2

+ κ33

a3 (2-18c)

where each q is linearly related to a. If we consider heat flow through a cube,

then the first subscript relates to the direction of the heat flow and the secondsubscript relates to the plane on which the heat flow operates. Remember that the

‘1” plane is defined by the 2- and 3- axes of the coordinae system defining the

planes



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If the temperature gradient is parallel to the x1 axis, then

∂T∂x

j

⎡⎢⎣

⎤⎥⎦

= ∂T∂x

1

, 0, 0⎡⎢⎣

⎤⎥⎦ (2-19)

but heat flow still occurs in three directions according to

q1

= κ11

∂T∂x

1

⎡⎢⎣

⎤⎥⎦ (2-20a)

q2

= κ21

∂T∂x

1

⎡⎢⎣

⎤⎥⎦ (2-20b)

q3

= κ31

∂T∂x

1

⎡⎢⎣

⎤⎥⎦ (2-20c)

Because of anisotropy of the rock, there is heat flow in all three directions of therock even though the gradient was applied parallel to the x1 axis.

κij is a tensor of second rank represented by the matrix

κij[ ] =

κ11

κ12

κ13

κ21

κ22

κ23

κ31

κ32

κ33

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ (2-21)

For a temperature gradient which is not parallel to a coordinate axis in ananisotropic material:

qi= κ

ij

∂T∂x

j

⎡⎢⎣

⎤⎥⎦ j = 1

3

∑for i = 1, 3 (2-22)

If a dummy suffix is repeated with respect to any one term, then that term

assumes summation over the term. A free suffix must occur once on each side of

the equation.

The expression for hear flow in an anisotropic material may be written in its full

form as

q1

= κ11

∂T∂x

1

⎡⎢⎣

⎤⎥⎦

+ κ12

∂T∂x

2

⎡⎢⎣

⎤⎥⎦

+ κ13

∂T∂x

3

⎡⎢⎣

⎤⎥⎦ (2-23a)



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q2

= κ21

∂T∂x

1

⎡⎢⎣

⎤⎥⎦

+ κ22

∂T∂x

2

⎡⎢⎣

⎤⎥⎦

+ κ23

∂T∂x

3

⎡⎢⎣

⎤⎥⎦ (2-23b)

q3

= κ31

∂T∂x

1

⎡⎢⎣

⎤⎥⎦

+ κ32

∂T∂x

2

⎡⎢⎣

⎤⎥⎦

+ κ33

∂T∂x

3

⎡⎢⎣

⎤⎥⎦ (2-23c)

Tensor transformation -- The physical quantity, such as thermal conductivity, κij, is

the same regardless of the set of axis chosen. This is the chief characteristic of a

tensor. Transformation from one coordinate system to another depends on thedirection cosine relating the axes of each of the coordinate systems. The

transformation process will be discussed in detail during a later lecture.

Structural geology requires an understanding of vector differentiation, the first

aspect of vector calculus. Vector differentiation is best illustrated by considering the

description of a curve in space. If a particular vector P is the position vector p(t) joining

the origin O of a coordinate system and any point (x,y,z,), then the vector function u(t)defines x,y, and z as functions of t

p(t) = x(t)i + y(t) j + z(t)k

As t changes, the terminal point of u describes a space curve having parametric

equations:

x = x(t), y = y(t), z = z (t).

Then

∆p/∆t = {[p(t + ∆t)] - p(t)}/∆t

is a vector in the direction of ∆p (see next figure).



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Then the limit

dp/dt as ∆t › 0

will be a vector in the direction of the tangent to the space curve at (x,y,z) and is given

by

dp dx dy dz

__ = __ i + __ j + __ k

dt dt dt dt

The derivative of the vector of position p with respect to time t is velocity v.

Let u(t) be the position vector of a moving block of rock P in the earth where t is

the time. The u(t) represents the path C of P. We know that the vector

v = dp(t)/dt

is tangent to C and, therefore points in the instantaneous direction of motion of P. The

derivative of the velocity vector is called the acceleration vector a; thus

a(t) = dv(t)/dt = d2p(t)/d2t

Acceleration is the second derivative of position with time. Newton's second law

includes a derivative of the velocity vector v with respect to time dv/dt. Acceleration a

is velocity changing with time. Second derivatives are must useful in structural geology.Another useful operation of calculus is the so-called chain rule of differentiation. If w is

a differentiable function of x and x is a differentiable function of t, then

dw/dt = dw/dx.dx/dt.

This chain rule can be generalized for a function of two variables such as w = f(x,y) and x

= x(t) and y = y(t) then

dw/dt = ∂w/∂x.dx/dt + ∂w/∂y

.dy/dt.

The operation ∂/∂x is the well known partial derative of a function. If x and y are

functions of position within a coordinate system and t is a function of time, then w varieswith position and time. If in contrast to the previous example x = x(p,v) and y =

y(p,v) then

∂w/∂ p = ∂w/∂x.∂x/∂ p + ∂w/∂y

.∂y/∂ p

and



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∂w/∂v = ∂w/∂x.∂x/∂v + ∂w/∂y

.∂y/∂v.

Partial differential equations are time consuming to write so a convention called Einstein

summation is used. To illustrate Einstein summation lets consider a property of a rock

that relates two vectors. A position within a rock x is related to a displacement u by

displacement gradients E. Position and displacment are both vectors with components x1,x2, x3 and u1, u2, u3, respectively. x and are related in the following way:

u1 = E11x1 + E12x2 + E13x3

u2 = E21x1 + E22x2 + E23x3

u3 = E31x1 + E32x2 + E33x3

Rather than writing out all of these equations we can shorten then in the following

manner:

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u1 = Σ E1 jx j j = 1

3

u2 = Σ E2 jx j j = 1

3

u3 = Σ E3 jx j

j = 1

or more compactly, as

3

ui = Σ Eijx j

i = 1

We now leave out the summation sign:

ui = Eijx j

and introduce the Einstein summation convention: When a letter suffix occurs twice in

the same term, summation with respect to that suffix is to be automaticaly understood. In



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the above example the suffixes can be any letters. i is the free suffix and j is the dummy

suffix. By this we mean that j can take any variable

ui = Eijx j = Eik xk .



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Structural Geology

Lecture 3

Mapping of Points and Displacements during

Deformation(introduction to kinematic analysis)

In this lecture techniques for kinematic analysis are introduced. The final product isa full accounting for rock strain which is at the heart of kinematic analysis. We can start

our analysis with a one-dimensional example of strain. Strain ( ε ) is a dimensionless

quantity defined as a change in length (∆l) per unit length (l).

ε = ∆l /l

Strain is actually defined near a point by the limiting process of differential calculus.

ε = lim ( ∆l /l)

l ⇒ 0

Now differential quantities can be introduced by letting δx = l and δu = ∆l

δu ∂u

ε = lim ___ = ___

δx ⇒ 0 δx ∂x

Strain in rocks is mapped by following two points separated by ∆x through a deformation(Fig 3-1). One dimensional strain may be illustrated by drawing a line out from the origin

(O) through two points such as P and Q.

∆x + ∆u = Q' - P'

(Fig. 3-1)

If the line is stretched the points P and Q go to P' and Q', respectively. The length from the

origin to P is denoted as x whereas the distance between P and Q is ∆x. The stretch from P



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to P' is u so that the total distance from P' to the origin is x + u. The stretch of ∆x is ∆u so

that the distance between P' and Q' is ∆x + ∆u. The strain of PQ is then

ε = ∆l/l = ∆u/∆x

If ∆u = 0.1 and ∆x = 1, then ε = .1/1 = 10%.

Deformation is conveniently separated into three components, of which two are

displayed in this one-dimensional analysis. In one dimension there can only be a rigid-

body translation and a stretch. The third component shows up in two and three dimensionswhere there is an additional deformation knows as a rigid rotation were the body spins

about an axis. These concepts will be more fully developed as we progress through the

elements of strain analysis.

Motion of a Point in a Deformed Body

Figure 3-1 demonstrates that for strain analysis we must keep track of the relative position of two points within a deforming body. However, before looking at the relative

motion of two points, it is instructive to consider the absolute motion of one point.

Deformation mapping involves the tracing of a material point within a body from itsundeformed position to its deformed position. The material point in its undeformed

position is specified by a vector X with components X1, X2, X3 or Xi. In the new or

deformed position the vector x has components x1, x2, x3 or xi (Fig 3-2). Note that in

Figure 3-2 the undeformed body is shown as a circle, a very common geological shape. Inthe deformed state that body becomes an ellipse. The point that we are watching is the

center of the circle.

NOTE: You will use capital letters for coordinates and coordinate axes in theunreformed state and lower-case letters for coordinates and coordinate axes in the

deformed state. This convention is used largely to conform with Means (1976).



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(Fig. 3-2: Until this figure is redrawn xi = X’i)

There two techniques for keeping track of what happened to the center of the

undeformed body (i.e., the circle). First, we can develop a series of equations relating thefinal resting spot of the center of the circle. Or, we can develop a series of equations

describing a vector pointing to the final resting spot from the initial position of the center of the circle.

Mapping of Points

One way of looking at the description of the 2-dimensional motion of a particle of

rock is through the following functional relationship which is described by the deformation

equations. These equations act to locate the final resting place of a point in a deformed body using the location of the point before deformation.

x1 = f(X1, X2)

x2 = f(X1, X2)

or

xi = f(X j)

In three dimensions this same relationship is

x1 = f(X1, X2, X3)

x2 = f(X1, X2, X3)

x3 = f(X1, X2, X3)

or xi = f(X j)

Written in their complete form, the deformation equations are as follows:

x1 = a0 + a1.X1 + a2

.X2 + a3.X3

x2 = b0 + b1.X1 + b2

.X2 + a3.X3

x3 = c0 + c1.X1 + c2

.X2 + c3.X3



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The constants, a0, b0, and c0 specify the rigid-body translation. If there is no rigid-body

translation, then the deformation equations for a homogeneous deformation may be writtenin matrix form:

a1 a2 a3

b1 b2 b3

c1 c2 c3

Later we will use a double subscripted variable to represent the nine components of the

deformation equations.

There are some special cases for deformation which are important in simplifying

geological analysis. In three dimensions one example of a homogeneous deformation can

be represented by the following equations:

x1 = a0 + a1.X1 + a2

.X2

x2 = b0 + b1.X1 + b2

.X2

x3 = X3

The reason that this is called a homogeneous deformation is that all points x' i are linearly

related to points xi. This is an also example of deformation in plane strain where all

motion is parallel to the plane normal to the x3 axis. (Plane strain is further discussed with

regard to faulting in a future lecture). The point (0,0,0) goes to the point (a0,b0,0) which

defines the rigid body translation. Points along the direction x1 where x2 = 0 go to the point

x1 = a0 + a1.X1

x2 = b0 + b1.X1

In contrast a non-homogeneous deformation the points x'i are related to xi in a nonlinear

manner. The following is one example of such a deformation in plane strain

x1 = a0 + a1.X1 + a2

.X2 + a3X12

x2 = b0 + b1.X1 + b2

.X2 + b3X12

x3 = X3



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We can introduce an example of mapping of a deformation using the following

matrix to specify our deformation equations:

1 −1 0

2 2 0

0 0 1

The three equations mapping the deformation might be the following provided it is

understood that the rigid body translation (a0 = 5) is not specified by the matrix givenabove (Fig. 3-3):

x1 = 5 + X1 - X2

x2 = 2X1 + 2X2

x3 = X3

Figure 3-3 is a mapping of three points of an undeformed body to its deformed shape. Inthis example you can think of the undeformed body as the outline of a shark’s tooth. You

will see that the deformation actually ‘sharpened’ the shark’s tooth. All of this was done

using the deformation equations as given above.

(Fig 3-3)

Displacement Vector



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The motion of the material point from Xi to xi is also described by a displacement

uo or (uo)i. uo ,which is a displacement vector, is nothing more than the final position, x,

minus the initial position, X. In Figure 3-3 the motion of the corner’s of the shark’s tooth

was shown by two vectors. The vectors may be expressed in vector components

(u0)1 = x1 - X1 and

(u0)2 = x2 - X2

or

(uo)i = xi - Xi

or u0 = x - X

These equations are known as the displacement equations. The displacement (uo)i of the

point Xi represents a major part of the motion of all points within a rock body. The motion

(uo)i is called rigid-body translation because it does not describe the motion of particles or

rock relative to each other but rather specifies that all particles follow the same path. In

this brief introduction we have only specified how individual points move during

deformation. We have not yet considered the relative motion of the points.The relative motion of the points describes the shape change of the shark’s tooth.

The change in shape of the rock is predicted by the displacement gradients which will be

developed in the next lecture.



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Structural Geology

Lecture 4

Displacement and Deformation Equations(kinematic analyses)

Kinematics, as you learned when taking elementary physics, is a mathematicaldescription of the motion of objects. In the case of structural geology kinematics is the

description of the path that rocks took during deformation. It is also the mathematical

description of the relative position of two infinitesimal points during the deformation of rocks. Two points can change by translating together, rotating around each other, or

changing in distance relative to one another. We shall call such a mathematical description

deformation mapping.

(Fig. 4-1: Until this figure is redrawn and xi = X’i)

The deformation gradient

If during rigid-body translation the particles of rock move relative to each other we

must devise other equations to account for their relative motion. To the undeformed state

we can attach a line segment dX whose components are dXi. The study of deformation is

concerned with the change in orientation and length of dX as the point at X is moved by

deformation to the point at x. We say that the vector dXi is both stretched and rotated to

become the new vector dxi. To account for the relative motion of particles within the rock,

we consider how the motion ui of any vector dX differs from the motion (uo)i of the vector

X (Fig. 4-1). If xi = f(X j)



23

then

xi + ∆xi = f( X j + ∆X j )

By Taylor's expansion

x( ) =

f n

a( )n !

n = 0

∑ x − a( )

n

so

∆xi = ( ∂xi/∂X j) ∆X j.

∂xi/∂X jare coefficients called the deformation gradient and are a function of the location

within the rock (Xi). Referring back to the deformation of the shark’s tooth as discussed in

lecture #3:

∂ x1

∂ x j =

1 −1 0

2 2 00 0 1

The displacement of the vector ui is the summation of (uo)i plus Eijdx where Eij (a

displacement gradient) is a function of position within the rock.

Eij = ∂ui/∂xj

The three equations mapping the deformation might be the following provided it is

understood that the rigid body translation (a0 = 5) is not specified by the deformation

gradients (Fig. 4-4):

x1 = 5 + X1 - X2

x2 = 2X1 + 2X2

x3 = X3

In geology the displacement equations prove to be more useful than the deformation

equations. However, for the purpose of an introduction to strain analysis a brief

introduction of the deformation equations is instructive. Now it is time to consider thecharacteristics of the displacement equations.

Strain is represented in the same manner as the deformation of shapes which was presented in the previous lecture. Lets examine the problem of representing strain in

three dimensions using the coordinate system x1,x2,x3. We define a point P(x1,x2,x3)



24

which has been displaced to the point P'(x1,x2,x3) (Fig. 4-2). The displacement vector

between P and P' is uP where

uP = ux1' i + ux2' j + ux3' k.

(Fig. 4-2)

As is the case for one dimension strain in rocks is mapped by following two

points, so we need an additional point Q which is displaced to Q'. Its displacement vector is then

uQ = ux1" i + ux2" j + ux3" k.

Now let A be the vector from P to Q and A' be the vector from P' to Q' (Fig. 4-3). Here

strain is the difference between A and A' which is represented asδA = A - A'.

(Fig. 4-3)



25

δA/δx is the rate of change of displacement with position within the rock. By vector

analysis we have

uP + A' = A + uQ

and

δA = uQ - uP

A function u is said to be analytic at a point P if it can be represented by a power series of

powers of P - a with a radius of convergence R > 0. Every analytic function can berepresented by a power series. Because u is an analytic function we can express uQ in

terms of uP by employing a Taylor series expansion of uQ about P. Remember that a

Taylor expansion takes the following form:

f(z) = f(a) + f'(a).(z - a) + (1/2!)f''(a).(z - a)2 + ........

This is a series centered at a. So the expansion of uQ about uP is

uQ = uP + (∂u/∂X1)P∆X1 + (∂u/∂X2)P∆X2 +

(∂u/∂X3)P∆X3 + ........

where ∆X1 = A1 (i.e. the components of A). Here higher order terms may be

neglected because of there small value.

Using the summation of indices as discussed in the appendix

uQ = uP + (∂u/∂X j)PA j j = 1,2,3

Now

δA = (∂u/∂X j)PA j

By convection

(δA)i = δAi

and(∂u/∂X j)i = (∂ui/∂X j).

So

δAi = (∂ui/∂X j).A j.

∂ui/∂X j is called the displacement gradient. These scalar quantities are the components

of the displacement equations which take the following form:

u1 =∂ u1

∂ X 1 X 1 +

∂ u1

∂ X 2 X 2 +

∂ u1

∂ X 3 X 3



26

u2 =∂ u2

∂ X 1 X 1 +

∂ u2

∂ X 2 X 2 +

∂ u2

∂ X 3 X 3

u3 =∂ u3

∂ X 1 X 1 +

∂ u3

∂ X 2 X 2 +

∂ u3

∂ X 3 X 3

As a quick aside the derivation of the displacement gradient can also be illustrated

using pages 94 and 95 of Nye (1957). In the three dimensional case displacement ui, are

linearly related to the independent variables, in this case vectors dx j, by simultaneous

equations

u1 = (u0)1 + E11dX1 + E12dX2 + E13dX3

u2 = (u0)2 + E21dX1 + E22dX2 + E23dX3

u3 = (u0)3 + E31dX1 + E32dX2 + E33dX3

or ui = (uo)i + EijdX j.

Eij are nine numbers in a 3 x 3 matrix that describe all motion within a rock that is not

rigid-body translation. The terms Eij are called displacement gradients. As pointed out in

the previous lecture a term with two suffixes relates two vectors. In this example the

vectors are the displacement u and the position vector dXi. The Einstein summation

convention as introduced in Lecture 2 applies to these equations for displacement.

Displacement and deformation gradients can be distinguished by the nature of the

information that they impart. The displacement equations indicate the direction of motion of particles during the deformation. The vector u (u1, u2, u3) points to the final

position of the particle from the initial position of the particle given the initial coordinate

system (X1, X2, X3). The displacement gradients (∂ui/∂xj) are a description of how the

movement of the initial point varies with position in the rock. In contrast, thedeformation equations map the final position of a particle of rock (x1, x2, x3 ) in terms of

the initial coordinate system (X1, X2, X3 ).

Deformation is conveniently separated into three components, of which two aredisplayed in this one-dimensional analysis. In one dimension there can only be a rigid-

body translation and a stretch. The third component shows up in two and three dimensionswhere there is an additional deformation knows as a rigid rotation were the body spins

about an axis. These concepts will be more fully developed as we progress through the

elements of strain analysis. In terms of the displacement gradients the stretch is denoted as

εij = 1/2 (Eij + Eij)



27

and the rigid-body rotation is denoted as

ωij = 1/2 (Eij - Eij).



28

Structural Geology

Lecture 5

Geological Deformation(a first look at ‘real’ strain)

Having introduced the nature of deformation and displacement equations in the

previous lectures it is time to consider three of the more famous examples of geological

deformation. In the following examples the problems are specified in terms of both thedisplacement and deformation equations.

Problem #1 The first deformation to occur in the history of a sedimentary rock isoverburden compaction. This is represented by flattening in the vertical direction

with no deformation in the horizontal directions. The following equations

represent overburden compaction as shown in Figure 5-1.

Displacement equations:

u1 = 0X1 + 0X2 + 0X3

u2 = 0X1 + 0X2 + 0X3

u3 = 0X1 + 0X2 - 0.5X3

Deformation equations:

x1

= 1X1

+ 0X2

+ 0X3

x2 = 0X1 + 1X2 + 0X3

x3 = 0X1 + 0X2 + 0.5X3

(Fig. 5-1)



29

Problem 2#: If the initial body were turned on its side this deformation could represent a

tectonic compaction. The Martinsburg shales near Harrisburg, Pennsylvania,have been isoclinally folded. Beds which were once flat-lying are now standing

on end. Assuming that the shales were deformed with no internal strain we wish

to represent this motion. The following equations represent the rigid rotation of a

block by 90° as shown in Figure 5-2.

Displacement equations (A vector from the old point to its new location):

u1 = 0X1 + 0X2 + 0X3

u2 = 0X1 - 1X2 - 1X3

u3 = 0X1 + 1X2 - 1X3

Deformation equations (A map of the new point given the coordinates of its old position):

x1 = 1X1 + 0X2 + 0X3

x2 = 0X1 + 0X2 - 1X3

x3 = 0X1 + 1X2 + 0X3

These equations are represented in the following figure:

(Fig. 5-2)

For an example in Problem #2, the initial point (1,1,1) is mapped to the new point

(1, -1, 1). The vector between these two points is u = (0, -2, 0).



30

Problem #3: As a final review of deformation equations, we shall take a look at the

behavior of a fault zone subject to simple shear. In higher grade terrains thesefault zones are simply called shear zones. Simple shear is illustrated by the

deformation of a unit cube in Fig. 5-3.

Displacement equations (A vector from the old point to its new location):

u1 = 0X1 + 0X2 + 0X3

u2 = 0X1 + 0X2 + 0.5X3

u3 = 0X1 + 0X2 + 0X3

Deformation equations (A map of the new point given the coordinates of its old

position):

x1 = 1X1 + 0X2 + 0X3

x2 = 0X1 + 1X2 + 0.5X3

x3 = 0X1 + 0X2 + 1X3

These equations are represented in the following figure:

Figure 5-3



31

Structural Geology

Lecture 6

A Second Derivation of the Displacement Gradient(another look at strain)

Three examples of deformation mapping were given in the previous lecture. Thedeformation can be represented in two manners. The displacement vector u was mapped

in one case leading to a displacement gradient. In the other case the new position of each

particle was mapped based on the initial position of the particle leading to a deformationgradient. In this lecture we will show that in 2 and 3 dimensions deformation consists of

rotations as well as stretches.

We start by taking another look at strain with some simple definitions such as a

change in length of line per unit length of line.

ε = ∆l/l

This is equivalent to a stretch. A formal definition of shear strain (γ) is the change in

angle (ψ) between two initially perpendicular lines (Fig. 6-1).

γ = tan ψ

(Fig. 6-1)

A second measure of shear strain is the tensor shear strain which is half the tangent of the

change in angle between initially perpendicular lines.

Tensor shear strain = γ/2.



32

γ is sometimes called the engineering shear strain. Note here that shear strain isrepresented by line rotations. This gives the first indication that the strain can be

separated into a rotational and irrotational component. We will deal in more detail with

these concepts in the next lecture.

The concept of strain in one dimension uses l0 to indicate the initial length of aline and l1 the final length of a line. We will start with the following definitions

ε = ∆l/l0 (elongation)

S = l1/l0 = (1 + ε) (stretch)

λ = (l ⁄ /l0)2 = (1 + ε)2 (quadratic elongation)

ε = dl/l0 (infinitesimal strain)

ε = ∆l/l0 (small increment of strain)

l1

ε = ∫ dl/l0 = ln (l1/l0) = ln(1 + ε) = 1/2lnλ

l0

(natural strain)

Now let's take another look at the displacement gradients by looking at how arectangular element at P is distorted as shown in Figure 6-2.

Figure 6-2



33

The meaning of u1 and u2 is as shown above. Stretch of PQ1 parallel to the 1 axis

is:

∆u1 =∂ u1

∂ x1

∆ x1 = e11∆ x1

Anticlockwise rotation of PQ1 is given as

∆u2 =∂ u2

∂ x1

∆ x1 = e21∆ x1

The angle through which PQ1 turns is:

tanθ =∆u2

∆ x1 + ∆u1

Because ∆u1 << ∆x1:

θ =∆u

2

∆ x1

= e21

For no distortion the following situation holds:

Figure 6-3



34

Structural Geology

Lecture 7

Irrotational vs. Rotational Strain(another look at strain)

The rotational strain tensor, eij, is invariant with respect to arbitrary, superposed

shifts of the point of origin of the coordinate system. We will discuss the implications of

this property in detail when looking at the stress tensor. For the time, we must accept thefact that any tensor property is independent of the position and orientation of the

coordinate system.

Nomenclature for the rotational strain tensor eij is as follows:

eij

=

e11

e12

e13

e21

e22

e23

e31

e32

e33

The first subscript, identifying the axis (i.e. direction) of a displacementcomponent ui, is permuted from 1 to 3 in each of the columns, the second, identifying the

face (i.e. the surface normal to an axis) whose center is displaced by the (u) j, is permuted

from 1 to 3 in each row.

We wish to make a distinction between infinitesimal strains and finite strains. Aninfinitesimal strain is one for which the final positions are very close to the initial

positions of adjacent particles. Infinitesimal strain theory is used in the theory of elasticity where strains are very small. While infinitesimal strains are independent of

actual path of displacement, this is not the case for finite strains. Strain evident in

outcrop is finite.

The rotational strain tensor, eij, applies to infinitesimal strains and is a general (or

asymmetric) second rank tensor and can be expressed as the sum to a symmetric and an

antisymmetric tensor

e ij = ε ij + ω ij

where

εij

= 12 e

ij+ e

ji( )

and



35

ωij

= 12 e

ij− e

ji( )

Here is εij symmetric and ωij is antisymmetric. The symmetric component of the

infinitesimal rotational strain tensor, eij, involves only dilatation (the change in volume)and distortion (the change in shape). Because of symmetry, the component, εij , consists

of only six independent variables. The component, εij , can be regarded as strain proper,

or irrotational strain. The antisymmetric tensor, ωij , has only three independent

components and involves only the rotational component of the strain tensor.

With these rather simple definitions the distinctions between simple shear and pure shear become clear. In a previous lecture we learned that the displacement

equations for a pure shear involving overburden compaction were as follows:

u1 = 0x1 + 0x2 + 0x3

u2 = 0x1 + 0x2 + 0x3

u3 = 0x1 + 0x2 - 0.005x3

Note that this time I have indicated a very small overburden compaction. This was done because I am now dealing with infinitesimal (i.e. very small strains). The deformation

gradient matrix for these equations is

∂ui

∂x j

=0 0 0

0 0 0

0 0 − 0. 005

The rotational strain tensor, eij, is

eij

=0 0 0

0 0 0

0 0 − 0. 005

For infinitesimal strain the rotational strain tensor and the deformation gradient are the

same. The irrotational strain tensor, εij, is

ε ij =

0 0 0

0 0 00 0 − 0. 005

The rotational component of strain is

ωij

=0 0 0

0 0 0

0 0 0



36

For pure shear there is a remarkable similarity between the rotational strain tensor and the

irrotational strain tensor,. This similarity disappears for the case of simple shear.



37

Again recall that displacement equations which were previously discussed for the

same of simple shear as might be encountered within a fault zone. For infinitesimal

strain these equations were

u1 = 0x1 + 0x2 + 0x3

u2 = 0x1 + 0x2 + 0.005x3

u3 = 0x1 + 0x2 + 0x3

The deformation gradient matrix and the rotational strain tensor for these equations is

∂ui

∂x j

= eij

=0 0 0

0 0 0. 005

0 0 0

The irrotational strain tensor, εij, is

εij

=0 0 0

0 0 0. 025

0 0. 025 0

The rotational component of strain is

ωij

=0 0 0

0 0 0. 025

0 − 0. 025 0

The distinction between pure shear and simple shear is further clarified by

considering the principal strain axes. If the directions of the principal axes of strain donot change as a result of displacement, then that deformation is termed irrotational strain.

In the case of simple shear the principal axes of strain always differ depending on the

amount of shear. The difference defines the rotational component of strain which isknown as rotational strain. It is important to note that although simple shear is a

rotational deformation, there has been no actual rotation in space within a fault zone. The

development of rotational strain does not necessarily imply that the body has to spin physically around some axis. Because of this lack of real rotation of within a fault zone,

some like to refer to the rotation as an internal rotation. This type of rotation is in

contrast to our example of external rotation where bedded sediments being turned on endas given by the displacement equations below.

u1 = 0X1 + 0X2 + 0X3

u2 = 0X1 - 1X2 - 1X3

u3 = 0X1 + 1X2 - 1X3



38

Note that in the case of external rotation we are dealing with finite strain where the

components of the deformation gradient matrix are large.



39

Structural Geology

Lecture 8

Strain Markers(Strain Analysis)

Strain in rocks is measured using objects of known initial shape. The initial shapes

can vary from round crinoid columnals to the irregular shape of breccia fragments. In twodimensions initially round objects include scolithus tubes, crinoid columnals, reduction

spots, vesicules, concretions, and oolites. Initially elliptical objects include congolmerate

pebbles. Fossils are usually irregular in shape but some such as leaves, brachiopods andtrilobites may have a bilateral symmetry. More complicated shapes include the spiral of

the ammonite, cavities of the coral, belemnites, and the branches of the graptolite. Other

markers are deposited with centers at uniform distances from their nearest neighbors. In

this lecture we are going to consider the analysis of four situations where rock strain can beinferred from the shape or position of deformed markers.

The least complicated strain analysis is the measure the elliptical shape of initiallycircular objects. This approached was used to map strain over 45,000 km2 of the

Appalachian Plateau. The Devonian Catskill Delta of the Appalaachian Plateau containsmany beds in which crinoid columnals parallel the bedding plane. Their elliptical shape on

pavement surfaces testifies to the deformation of the Appalachian Plateau. The state of

strain parallel to bedding in these outcrops is represented by three components: principal

strains, ε1 and ε2, and the orientation of the strain ellipse as measured using one of the

principal strain axes relative to north. In the case of the Plateau the strike of the long axes

of the ellipse relative to north is designated as θ. Strain on the plateau may have arotational component but this was impossible to measure.

The actual measurements consisted of two numbers: the axial ratio and theorientation of the long axes. The axial ratio is a measure of ellipticity (R):

R = (1 + ε1 )/(1 + ε2 )

In a paper published in Geology, Engelder and Engelder 1 (1977) presented the a map

showing the strike of θ and the value of R which varied between 1.1 and 1.2. The values of

ε1 and ε2 are not possible to measure directly nor can they be calculated from the above

equation because the strain on the Plateau is a volume loss strain.

The second type of deformed object that proves to be a useful strain marker is theellipse with an initial ellipticity R i. Upon deformation the shape of the final ellipse R f is a

function of the orientation and ratio of the initial ellipse ralative to the deformation (Figure8-1). In the deformed state the orientation of the long axis of the ellipse relative to some

marker is Φ. Data on R f and Φ can be graphed to form R f /Φ plots. These plots can then

1 - As a matter of general interest, Prof. Engelder was a research scientist at Columbia

university at the time this paper was written. His brother, Richard, was then a graduate

student at Penn State.



40

be compared with standard R f /Φ reference curves for different values fo initial ellipticity

R i and the strain ellipse R s. The R f /Φ plots have two shapes depending on whether R i >

R s or R i < R s. In the former case the data envelope is symmetric about the orientation of

the long axis of the strain ellipse and shows maximum and minimum R f values. In the

latter case the data envelope is closed and the data points shown a limited range of orientations defining the fluctuation F.

(Fig. 8-1)

The center-to-center technique allows the assessment of the bulk strain of a rock.

This is commonly known as the Fry technique. Sometimes rigid inclusions such as fossils

do not deform uniformly with the matrix of the rock. The best technique for determiningthe behavior of the matrix is the measure the distance between neighboring grains. Figure

8-2 shows an example where the distance and azimuth between various grains has been

measured. From the distance/azimuth plot the ratio of the strain ellipse and its orientationcan be determined. The Fry technique may also be used to take advantage of the

characteristic center-to-center distance of the deformed matrix of a rock.

The most beautiful and, hence, popular strain markers are the trilobites and brachiopods with lines of symmetry. If the line of symmetry lies parallel to a principal

strain direction, then the angular shear strain of the principal lines in the fossil are zero. In

this orientation the deformed fossil is still in a symmetrical form. If the initial lines of symmetry are not parallel to principal strain directions the lines of symmetry appear to

shear during deformation. Final shape of the brachiopod or trilobite is an oblique form.

The oblique forms can be either right or left handed depending upon the deflection of thesymmetry axis. The symmetrical forms can appear in a narrow or broad form depending

the initial orientation of the fossil relative to the shortening direction (Figure 8-3).



41

(Fig. 8-2)

The initial shape of the fossil can be characterized in terms of its length to breadth:

r 0 = l0/b0

The strain of the rock can be determined by using the ratio of the final length and breadth.

In the narrow form

r n = ln/bn = (l0R)/b0

and in the broad form

r b = l b/b b = (l0R)/b0

Even though the original shape ratio r‚ is unknown we can calculate the strain ellipse andr‚ from the following equations.

R = (r n/r b)1/2

r 0 = (r br n)1/2.



42

(Fig. 8-3)

The strain ellipsoid represents strain in three dimensions where the three axes of

finite strain are known as (1 + ε1) > (1 + ε2 ) > (1 + ε3). The longest extension is in the ε1

direction (Fig. 8-4). The principal strain ratios are defined as

R xy =(1 + ε

1 )/(1 + ε2)

and

R yz = (1 + ε1 )/(1 + ε3).

The plot of R xy versus R yz became known as the Flinn Graph (Fig. 18-4). Flinn suggested

the parameter k to describe the position of the strain ellipsoid in the Flinn Graph.

R xy - 1

k = ________

R yz - 1

If k is greater than one then the strain ellipsoid is in the form of a cigar whereas if k is less

than one the strain ellipsoid is in the form of a pancake.



43

(Fig. 8-4)



44

Structural Geology

Lecture 9

Properties of Tensors(Tensor Analysis)

Einstein summation convention -- Here the Einstein summation convention applies and

the components σ ij relate two vectors f i and l j in a linear manner.

The components σij form a 3 x 3 matrix:

σ11

σ12

σ13

σ21

σ22

σ23

σ31

σ32

σ33

In the study of vector analysis we have introduced three sorts of quantities: thescalar; the vector; and then a nine component quantity which relates two

three-component vectors. These three quantities are all called tensors

where:

Tensor of zero rank = Scalar -- This is a single number unrelated to any

axes of reference.Tensor of first rank = Vector -- This is specified by three numbers or

components, each of which is associated with one of the axes of

reference.Tensor of second rank = (e.g. stress) -- This is specificed by nine

numbers, or components, each of which is associated with a pair

of axes (taken in a particular order).

Recall that the 2 dimensional Mohr's diagram described stresses associated with a

pair of axes. In this sense the Mohr diagram is a visual representation of one-thirdof the second rank stress tensor. A tensor has the unique property that it can be

transformed without losing its value. By this we mean that a tensor has

conponents representing a physical quantity which retains its identity regardless

of which coordinate system (e.g. rectangular, cylindrical, etc.) is used or how theaxes of that coordinate system may be rotated. The physical quantity is the sameregardless of the direction from which it is viewed.

Tensor transformation -- The physical quantity, such as thermal conductivity, κij, is

the same regardless of the set of axis chosen. This is the chief characteristic of a

tensor. Transformation from one coordinate system to another depends on the

direction cosine relating the axes of each of the coordinate systems.



45

aij are the components of the transformation matrix where a12 is the cosine of the

angle between x1 of the old coordinate system and x2' of the new coordinate

system. The complete transformation matrix is

x1'

x2'

x3'

x 1 x 2 x 3

a11

a12

a13

a21

a22

a23

a31

a32

a33

(9-1)

The first sugscript in the a's refers to the 'new' axes and the second to the 'old'.There are only three independent direction cosines of the nine given in the

transformation matrix. The coordinates of a point (x1,x2,x3) can also be

transformed to the coordinates (x1',x2',x3') using direction cosines in the same

manner (Fig. 9-1)

x1' = a11x1 + a12x2 + a13x3

x2' = a21x1 + a22x2 + a23x3

x3' = a31x1 + a32x2 + a33x3. (9-2)

Figure 9-1

Likewise the equations for the transformation of a vector from one coordinate

system to another resemble the equations for transforming a point.



46

pi' = a

ij p

j (9-3)

where p'i are the components of the vector in the new coordinate system and p j are

the components of the vector in the old coordinate system. Here the free suffix isi and the dummy suffix, j, is together in the right-hand term. Written in its full

form equation 3-2 becomes

p1' = a

11 p

1+ a

12 p

2+ a

13 p

3

p2' = a

21 p

1+ a

22 p

2+ a

23 p

3

p3' = a

31 p

1+ a

32 p

2+ a

33 p

3 (9-4)

We can reverse the transformation process to find the old vector in terms of the

new vector

pi= a

ji p

j'

(9-5)

Here the dummy suffix, j, is separated by the free suffix, i.

Relations between direction cosines - Since each row of the array (3-1) representsthree direction cosines of a straight line with respect to a coordinate system, x1,

x2, x3, the following equations hold true

a112 + a

122 + a

132 = 1

a212 + a

222 + a

232 = 1

a 312

+ a 322

+ a 332

= 1 (9-6)

In a condensed version equation 3-5 is written

aik

a jk

= 1, if i = j(9-7)

Since eachpair of rows of the array 3-1 represents the direction cosines of two

lines at right angles, we have a situation comparable to a scalar product of twovectors at right angles.

a ik a jk = 0, if i ≠ j (9-8)

We can combine the orthogonality relations into one equation

aik

a jk

= δij

=1 i = j( )0 i ≠ j( )

⎧⎨⎩ (9-9)



47

where the symbol, δij, is called the Kronecker delta.

Second rank tensor -- In general, if a property T relates tow vectors P = [p1,p2,p3] and

Q = [q1,q2,q3] in such a way that

p1 = T 11q1 + T 12q 2 + T 13q 3

p2

= T21

q1

+ T22

q2

+ T23

q3

p3

= T31

q1

+ T32

q2

+ T33

q3 (9-10)

where the components, T11, T12, .....are constants, then T11, T12, .....are said to

form a second-rand tensor.

T11

T12

T13

T21

T22

T23

T 31 T 32 T 33

⎡⎢⎢

⎢⎣

⎤⎥⎥

⎥⎦ (9-11)

The values of the coefficients T11, T12, .....depend on the orientation of the

coordinate axes x1, x2, x3. Now suppose we choose a new set of coordinate axes

x1', x2', x3' related to the old axes by the direction cosines, aij. If so, the vectors P

and Q have new components pi' and qi'. Next we find P' in terms of Q' the series

of equations shown below (The arrow means "in terms of").

P' → P → Q → Q' (9-12)

The following equations permit the transformation

pi' = a

ik p

k (9-13)

pk

= Tkl

ql (9-14)

ql= a

jlq

j'

(9-15)

Substituting

p i' = a ik T kla jlq j' (9-16)

or

pi' = T

ij' q

j'

(9-17)

So



48

Tij' = a

ik a

jlT

kl (9-18)

This is the Transformation Law for a tensor of second rank. Written out in full

form each of its nine equations are rather long. The general equation is



49

Tij' = a

i1a

j1T

11+ a

i1a

j2T

12+ a

i1a

j3T

13+ a

i2a

j1T

21+

ai2

a j2

T22

+ ai 2

a j3

T23

+ ai3

a j1

T31

+ ai 3

a j2

T32

+ ai3

a j3

T33

(9-19)

Just as the components pi and qi transform to pi' and qi', when the axes are

changed, the nine coefficients Tij transform to the nine coefficients T'ij. In this

example i and j are free suffixes and k and l are dummy suffixes. A second rank

tensor must transform.

At this point it is important to appreciate the characteristics of a stress vector (f )

which is nothing more that the stress tensor (σij) expressed as a vector acting on a

surface whose normal is the vector l.

f i = σ ijl j

The principal axes of a tensor - The equation for the surface of a quadric (i.e., a seconddegree surface such as an ellipsoid or hyperboloid) is written

Sijx

ix

j= 1

(9-20)

where Sij are coefficients. Performing the summation with respect to i and j

equation is written in full by assuming the symmetric condition which is thatSij = S ji.

S11

x12 + S

22x

22 + S

33x

32 + 2S

12x

1x

2+ 2S

31x

3x

1+ 2S

23x

2x

3= 1

(9-21)

The coefficients of equation 3-17 transform like the components of a symmetrical2nd rank tensor.

Sij' = a

ik a

jlS

kl (9-22)

The theory of the transformation of a symmetrical 2nd rank tensor is thus

identical with the theory of th transformation of a quadric. Thus, the surfacerepresented by equation 3-17 is called the representation quadric of a tensor, Sij.

An important property of a quadric is the possession of principal axes. These are

three directions at right angles such that, when the general quadric is referred to

them as axes, its equation takes the simpler form



50

S11

x12 + S

22x

22 + S

33x

32 = 1

(9-23)

When we refer thje general tensor to principal axes, we have reduced the tensor

components to the matrix

S1

0 0

0 S2

0

0 0 S3

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ (9-24)

where S1, S2, S3 are called the principal components of the tensor [S ij]. The

equation for a standard ellipsoid is written

x2

a2 +

y2

b2 +

z2

c2 = 1

(9-25)

In this equation the semi-axes of the representation quadric are the lengths1

S1

,1

S2

,1

S3 .

We now define a property of a principal axis as the radius vector which parallels

the normal to the quadric surface. The radius of the quadric is specified by Sijx j.

The normal to the surface of the quadric is specified by the vector, xi. The

condition that the radius vector and the normal are parallel is satisfied provided

that the corresponding components are proportional. This condition is satisfied bythe equation

Sijx

j= λx

i (9-26)

where λ is a constant. Equation 3-23 is three homogeneous linear equations in thevariables xi.

S11

x1

+ S12

x2

+ S13

x3

= λx1

S21

x1

+ S22

x2

+ S23

x3

= λ x2

S31

x1

+ S32

x2

+ S33

x3

= λ x3 (9-27)

The trivial solution to these equations is that xi = 0. If so, these equations would

be worthless. For a non trivial solution the determinant of the components of

these three equations must be zero.



51

S11

− λ S12

S13

S21

S22

− λ S23

S31

S32

S33

− λ

= 0

(9-28)

or

S11

− λ( ) S22

− λ( ) S33

− λ( ) − S23

S32[ ] − S

12S

21S

33− λ( ) − S

23S

31[ ] +

S31

S21

S32

− S31

S22

− λ( )[ ] = 0

(9-29)

This cubic equation is called the secular equation. The three roots, λ', λ'', and λ''',

give the three possible values of λ that ensure that equations 3-24 have non-zero

solution. Those values of λ are also called eigenvalues or characteristic values of the matrix S . Each of the roots defines a direction in which the radius vector of

the quadric is parallel to the normal, that is, the direction of one of the principal

axes. These directions are called eigenvectors or characteristic vectors of S .

The direction of the principal axes are found by substituting, say λ', into equation3-24. These three equations may be solved for the ratios x1': x2': x3'. Knowing

the ratios we can solve for the direction cosines of the unit vector by

ui' =

xi'

x1' 2 + x

2' 2 + x

3' 2

(9-30)



52

Structural Geology

Lecture 10

An Introduction to Stress(Tensor Analysis)

Hooke's Law

When a body changes shape we say that the body is strained. Strain, a

deformation, is measured as the ratio of the change in shape of a body to its initial shape.Previous figures illustrates strain using a long rod with an initial length (L). When the

rod is stretched the additional length (i.e., displacement) of the rod is ∆L. Strain (ε) is ∆Ldivided by L. If the rod is deformed elastically, it will return to the length, L, when the

forces are removed from the end of the body.

Another aspect of elasticity which we will examine is the relationship betweenthe amount of force and the amount of strain. Figure 10-1 shows an unloaded spring and

the same spring loaded with one, two, and three weights, respectively. The spring getslonger in equal increments with the addition of comparable increments of weight.

Remember that a weight is a force. Relationship between change in force and change in

stain is characteristic of an elastic material.

ELASTICITY OF A SPRING

Figure 10-1

It is accurate to imagine that the atomic forces which govern the atomic distancesin a lattice act just like springs which return to their natural shape regardless of whether they were subject to a push or a pull. The law that governs the action of a spring is called

Hooke's Law. We can write the equation for the Hooke's Law of a spring as follows:

force = spring constant X displacement

(10-0)



53

Equation 10-0 says that the 'stretch' or displacement of a spring is proportional tothe force on the spring. This elastic law can be illustrated in a plot of strain

(displacement) versus stress (force) (Figure 10-2). Such a plot is a straight line and the

relationship between force and displacement is linear. The constant which specifies the

relationship between the force on the spring and its displacement is called the spring

constant. Often rock deformation is given in units of stress (force per unit area) andstrain (change in length divided by initial length). In this case Hookes Law is written as:

stress = Young's modulus X strain

(10-00)

HOOKE'S LAW

s t r a i n ( o r d i s p l a c e m e n

t )

stress (or force)

stress ( σ ) = forceunit area

ε

σ

Young's Modulus =stress

strain

Spring Constant =force

displacement

Figure 10-2

In Figure 10-2 stress (force) is plotted as the independent variable as it should be.

However, geologists are an odd bunch who by convention always plot stress (force) on



54

the vertical axis as if it were the dependent variable. Don't be fooled by this conventionin the following plots.

Measurements of strain of rocks usually involve very small displacements. This

is because most rocks are very stiff relative to a rubber band. In fact, rocks are so stiff

that strain is measured using the unit, microstrain. A strain of 10-1

is a 10% strain whichmeans that a body of one meter in length will shorten by 10 centimeters. A strain of 10-3

is a 0.1% strain which means that a body of one meter in length will shorten by 1

millimeter. A microstrain is 10-6 or a 0.0001% strain which means that a body of one

meter in length will shorten by one micron (one micrometer = 10-3 millimeter).

If we wish to measure very small strains we must use strain gauges hitched to a

strain indicator. Strain gauges are small wires wound like an accordion (Figure 10-3).

The strain gauge wires are glued to the surface of specimen so that when the specimenstretches, the strain gauge wires also stretch. If the wires are stretched the cross sectional

area of the wires gets smaller. The electrical resistance of the wire increases if the cross

sectional area decreases. The strain gauge operates on the principle that the change inresistance of the accordion-like wire is equivalent to the strain of the sample. A strain

indicator measures strain in the strain gauge wires by measuring the change in resistance

of the wire as the specimen stretches.

RESISTANCE OF A WIRE

long and

thin

short a

nd thick

higher resistance

lower resistance

change in resistance = change in strain

S T R AI N G AU G E

resistance wire

solder

tabs

Figure 10-3

Stress Vector -- Mohr's diagram is a convenient graphical representation of state of stress within the crust of the earth. It works well for the two dimensional



55

representation of stress but in three dimensions it becomes less convenient.However, some of the rules that apply in two dimensions hold for three

dimensions. One of the more important rules that was demonstrated in two

dimensions was that the sum of stresses was invariant or unchanged by the

rotation of coordinate axes. In this lecture we are going to examine the more

general case for rotation of coordinate axes with the stress transformation. Tointroduce the stress transformation we appeal to the concept of a stress vector.

Figure 10-4

To understand stress in three dimensions, we start with a force vector f acting on

the surface of a rock or acting across a boundary (Fig. 10-4). This vector is also astress vector where

σ = lim ∆f /∆S (10-1)

where the lim ∆S approaches 0. Take any small surface element of area δS

containing a point P within the stressed rock. A unit vector l can be drawn

perpendicular to the area δS at the point P. Then, the force transmitted across the

area is σδS. Note that stress multiplied by area is a force. The force is exerted on

the rock on the positive side of the area (defined by the orientation of l) is resisted

by stress within the rock on the negative side of the surface δS. While the force

on the positive side is defined by a first rank tensor (the three components of theforce vector), stress on the negative side of the surface is expressed using a

second rank tensor (the nine components of the stress tensor).

We now want to examine the variation of σδS as the orientation of l is changedwithin the stressed rock. In this exercise the surface will be maintained so that the

unit vector l will always pass through the point P on the surface δS. For thisexercise we need to assume that the stress is homogeneous, that there are no

body-forces, and that the body is in equilibrium. By equilibrium we mean to say

that the rock is not presently being deformed. The force σδS does not change

with orientation of l. Stress can be treated as a vector acting on a plane, P. Byexaming Fig. 10-1 it can be seen that the force vector, f , can be resolved (by using



56

the dot product) into a component parallel and a component normal to the surface

of the plane, P. Likewise, if f /δS is treated as a vector, it can be resolved into two

components: σn, a normal stress, and τ, a shear stress. While the magnitude and

direction of the stress vector does not change upon reorientation of the surface,

the magnitude and direction of the σn, a normal stress, and τ, a shear stress

changes. σ = f /δS is known as the stress vector acting on a certain plane.

Figure 10-5

Reaction of the solid to a stress vector -- Now we examine a solid piece of the rock in

the shape of a tetrahedron with corners ABCO (Fig. 10-5). Because the rock is inequilibrium no surface of the tetrahedron is moving relative to any other surface.

Let the surface defined by ABC be the surface element δS discussed above. The

force σ(f ) transmitted across δS is σ (f) times (area ABC). The forces on the three

faces at right angles may be each denoted by three components σij so that we

have nine components in all. Each face has two components parallel to the

surface and one component normal to the surface. Note that these components are parallel to the three components l1, l2, and, l3 of the unit vector l. Remember that

a stress multiplied by an area is a force. Resolving forces parallel to Ox we have

f 1 (ABC) = σ11(BOC) + σ12(AOC) + σ13(AOB) (10-2)

Dividing the area of each side (e.g. BOC) by the area of the face (ABC) gives the

component of the unit vector l in the direction normal to the side. Thus:

f 1 = σ11l1 + σ12l2 + σ13l3

f 2 = σ21l1 + σ22l2 + σ23l3



57

f 3 = σ31l1 + σ32l2 + σ33l3. (10-3)

Hence,

f i = σ ijl j (10-4)

where both f i and l j are three components of the force vector f and the unit

directional vector l.

Mohr circle construction - We start with the principal axes of a tensor

σij

=

σ1

0 0

0 σ2

0

0 0 σ3

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ (10-5)

and consider the transformation of the components of stress by a clockwise

rotation about the principal axis σ3. For this rotation the direction cosine matrix is

the following

aij

=

a11

a12

a13

a21

a22

a23

a31

a32

a33

⎛ ⎜⎜⎝

⎟⎟ =

cos θ - sin θ 0

sin θ cos θ 0

0 0 1

⎛ ⎜⎜⎝

⎟⎟(10-6)

We now transform σij

using the direction cosines to obtain

σij' =

σ11' σ

12' 0

σ12' σ

22' 0

0 0 σ3

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ (10-7)

According to the transformation we generate four components. For example,

σ11' = a

1k a

1lσ

kl= a

11a

11σ

1+ a

11a

12σ

2

where σ11 = σ1 and σ22 = σ2 so that

σ11' = σ

1cos 2 θ + σ

2sin 2 θ

σ22' = σ

1sin 2 θ + σ

2cos 2 θ

σ12' = + σ

1sin θ cos θ − σ

2sin θ cos θ

(10-8)



58

These equations may be rewritten to the classic equations (1-29a&b) for the Mohr circle stress analysis. From the Mohr's Circle analysis we note the property of

invariance.

12 σ

11' + σ

22'( ) = 1

2 σ11

+ σ22( ) (10-9)

Field tensors verses matter tensors - Stress are strain are examples of field tensors

whereas the tensors which measure crystal properties such as magneticsusceptibility are matter tensors. Matter tensors must conform to crystal

symmetry whereas stress is not a crystal property but is akin to a force impressed

on the crystal. Both field and matter tensors have similar special forms.



59

Field Tensor Symmetry of Matter Tensor

P p 0 0

0 P p 0

0 0 P p

⎡⎢⎢⎢

⎣

⎤⎥⎥⎥

⎦ Hydrostatic Stress

M 0 0

0 M 0

0 0 M

⎡⎢⎢⎢

⎣

⎤⎥⎥⎥

⎦ Isometric

σ1

0 0

0 0 0

0 0 0

⎡⎢⎢⎣

⎤⎥⎥⎦ Uniaxial Stress

σ1

0 0

0 σ2

0

0 0 0

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ Biaxial Stress

M1

0 0

0 M1

0

0 0 M3

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ Tetragonal

σ 1 0 0

0 σ2

0

0 0 σ3

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ Triaxial Stress

M1 0 0

0 M2

0

0 0 M3

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ Orthorhombic

σ1

0 0

0 − σ2

0

0 0 0

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ Pure Shear

M1

M12

0

M21

M2

0

0 0 M3

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦ Monoclinic

M11

M12

M13

M21

M22

M23

M31

M32

M33

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦Triclinic



60

Structural Geology

Lecture 11

Force-Balance and Mohr's Equations(Mohr Stress)

The Force-Balance Analysis

In high school physics we were taught that pressure is defined as force per unit area.

Stress may have the same definition. The distinction between pressure and stress instructural geology is based on the nature of the material on which the force is acting. The

distinction is made depending on whether the material in question has a shear strength.

Materials such as rock are said to have a shear strength because they maintain their shape

when placed unsupported on a table. If we are dealing with a rock that has a shear strength (fluids such as water and gases do not), then we say that it exerts a stress on its

surroundings. Materials which have a shear strength can exert different stresses indifferent directions. In contrast, water, without a shear strength would proceed to runover the table top seeking the lowest spot. If we wish to describe the force per unit area

that a liquid or gas is exerting on its container, we use the term pressure. Water in the

pores in rocks exerts a pressure on the grains surrounding the pore.

The most useful equations for teaching the concept of stress are the equations for normal

(σn) and shear (τ) stress in terms of principal stresses (σ1,σ3). The derivation of these

equations is based on a force-balance problem which assumes that a body subject to

forces is in equilibrium which means that all forces in any direction add to zero.

θ

sin θ

cos θ

sin θ

χοσ



61

(Fig. 11-1)

Consider the forces acting parallel to the front face of the triangular solid shown in Figure

11-1. These forces are acting in a plane so that this force-balance problem is two

dimensional. If the area of the hypothenuse face is unity (i.e. 1), then the area of the left

face is unity times sin θ whereas the area of the bottom face is unity times cos θ. Forcesacting on the three faces are stresses multiplied by the area of the face [Force =

(force/area) × (area)]. Summing the forces in the horizontal and vertical directions (theforces add to zero in both directions), we obtain

Σ Fh = 0 = Fx - S sinθ - N cosθ (11-1a)

Σ Fv = 0 = Fy - N sinθ + S cosθ (11-1b)

Rewriting the force terms using components of stress, we obtain the normal and

shear stresses acting on a plane an angle of θ to σ1 in terms of the principal stresses.

σ1 cos θ − σn cosθ − τ sinθ = 0 (11-2a)

σ3 sin θ − σn sinθ + τ cosθ = 0 (11-2b)

In many geological applications the plane of interest is a fault plane. Now we can solve

these two equations for σn and τ

σn = σ3 cos2θ + σ1 sin2θ (11-3a)

τ = (σ1 - σ3) cosθ sinθ (11-3b)

Remember that

sin2θ = 2 sinθ cosθ (11-4)

and also

cos2θ = 1/2(1 + cos2θ ) and sin2θ = 1/2(1 - cos2θ ) (11-5)

We derive

σn = 1/2(σ1 + σ3) + 1/2(σ1 - σ3)cos2θ (11-6a)

and

τ = 1/2(σ1 - σ3)sin2θ . (11-6b)

Note that in these equations θ is the angle between σ1 and the normal to the fault plane.

This convention leads to a positive sign in the middle of equation 11-6. Chapter 10 of

Marshak and Mitra (1988) uses the convention that θ is the angle between the normal to



62

the fault plane and σ1. If the θ is the angle between σ1 and the fault plane the sign

convention equation 1-29a becomes

σn = 1/2(σ1 + σ3) - 1/2(σ1 - σ3)cos2θ (11-7).

Unless each convention is memorized the reader will be confused when using theseequations in the future.

The general force-balance problem is shown in figure 11-2 where the back and side facesof the unit triangle are subject to both normal and shear stresses. In other words the

coordinate system x-y is not parallel to the principal stress directions. The general case is

given to illustrate a property of stress called invariance with respect to coordinate system.

In two dimensions we define stress as a force per unit of line length, in contrast to thethree dimensional situation where stress is a force per unit area (Fig. 11-1). Consider acoordinate system Oxy with an arbitrary line AB cutting the x and y axes such that the

normal to the line AB makes an angle θ with the x-axis. This gives a right triangle AOB

with sides OA (parallel to Ox) and OB (parallel to Oy) and a hypothenuse AB. Acrossthe line AB a stress vector p can be applied making an angle θ with the x-axis.

Remember that p = δf/δA when δA ⇒ 0, so a stress vector can represent stress at a point.

Otherwise stress is defined on a line (2-D) or a surface (3-D). The stress vector p can beresolved into components parallel to the x and y axes: p = px + py. Even though P is

called a vector it still has units of stress (force/length in two dimensions). Because the

triangle ABO is in equilibrium the sum of the force-vectors on all sides must balance. In

2-D stress multiplied by line length will give a force vector. So

pxAB = σxOB + τyxOA. (11-8)



63

(Fig. 11-2)

If length a = AB, then a×cosθ = OB and a×sinθ = OA. If we divide by length a, then

px = σx cosθ + τyx sinθ. (11-9a)

Likewise

py = σy sinθ + τxy cosθ (11-9b)

Now consider a normal stress σn and shear stress τ across AB in terms of the

components of the stress vector

σn = px cosθ + py sinθ (11-9a)

The equality holds for τ

τ = py cosθ - px sinθ. (11-9b)

Substituting for px and py and remembering that sin2θ = 2 sinθ cosθ and also

cos 2θ = cos2θ - sin2θ

σn = σx cos2θ + 2τxy sinθ cosθ + σy sin2θ (11-10a)

and

τ = 1/2(σy - σx) sin2θ + τxy cos2θ. (11-10b)

These are the general equations for a stress system in which the orientation of the

principal stresses are unknown. The following exercise illustrates the case for the

principal stresses being parallel to the coordinate axes.



64

(Fig. 11-3)

Now we consider the effect of rotating the coordinate system on the values of

stress (Fig. 11-3). To accomplish this coordinate rotation let σn and τ be σx' and τ x'y' where Ox'y' is rotated θ from Oxy. From a previous equation we have

σx' = σx cos2θ + 2τxy sinθ cosθ + σy sin2θ (11−11a)

Now we must find σy' in the new coordinate system. This is accomplished by replacing θ

by θ + 1/2π and we get

σy' = σx sin2θ - 2τ xy sinθ cosθ + σy cos2θ (11-11b)

If we add σx'

and σy'

and remember that sin2θ + cos2θ = 1, we get

σx' + σy' = σx + σy. (11-12)

This shows that the sum of the normal stresses is invariant or unchanged by rotation of

the coordinate system. Likewise

τ x'y' = 0.5(σy - σx) sin2θ + τ xy cos2θ. (11-13)

Principal stresses are found in planes containing no shear stress. If in these planes, τx"y"

= 0, then from the previous equation

2τxy

tan2Θ = ____ (11-14)

σx - σy

where Θ is the one angle between the coordinate system Ox"y" and Oxy where the shear

stresses vanish along the directions Ox" and Oy". In this coordinate system the only



65

stresses are the normal stresses σx" and σy". This coordinate system contains the principal

stress axes and the components, σx" and σy" are known as the principal stresses.

(Fig. 11-4)

Let us now call these principal stresses σ1 and σ2 (Fig. 10-4) and then we shall

choose the coordinate system such that the x and y axes are in the direction of the

principal stresses, σ1 and σ2. Now the normal σn and shear τ stresses across a line whose

normal is inclined at θ to σ1 is

σn = σ1.cos2θ + σ2.sin2θ (11-15)

This is the convention used in Chapter 10 of Marshak and Mitra (1988). Again

remember that

cos2θ = 1/2(1 + cos2θ ) and sin2θ = 1/2(1 - cos2θ )

we can rearrange the equation for the normal stress

σn = 1/2(σ1 + σ2) + 1/2(σ1 - σ2)cos2θ (repeat of 11-6a)

and shear stress

τ = 1/2(σ

1- σ

2)sin2θ . (repeat of 11-6b)

From this last equation we see that shear stress is greatest when θ = π/4 and 3π/4.

Using this same coordinate system where the axes are parallel to the principal

stresses σ1 and σ2, we can look at the stress vectors px and py. They become

px = σ1.cosθ and py = σ2.sin θ. (11-16)



66

(Fig. 11-5)

Substituting into the equation sin2θ + cos2θ = 1, we can generate the equation for anellipse where the px and py are on the ellipse called the stress ellipse.

px2 py

2

___ + ___ = 1

σ12 σ2

2 (11-17)

The semi axes of the ellipse are σ1 and σ2.The Mohr's circle is a graphical method of representing the state of stress of a

rock in two dimensions (Fig. 11-5). The equations used for the Mohr's circle

representation are those derived above for the coordinate system with axes parallel to the

principal stresses. The Mohr's circle may be used to derive the normal σn and shear τ

stresses on any plane whose normal is oriented at θ from σ1. The coordinate system for

the Mohr's circle representation is σn along the horizontal axes with increasing

compression to the right and τ along the vertical axes. Critical points along the σn-axes

are OP = σ1, OQ = σ2, and C = 1/2(σ1 + σ2). The angle measured as PCA counter

clockwise from OP is 2θ. Now we have

σn = OB = OC + CB = 1/2(σ1 + σ2) + 1/2(σ1 - σ2)cos2θ (11-18a)

and

τ = AB = 1/2(σ1 - σ2)sin2θ. (11-18b)



67

Structural Geology

Lecture 12

Mohr's Circle and Earth Stress(The Elastic Earth)

In the equations that we derived for Mohr’s circle, we measured the angle, θ, as

the angle between σ1 and the normal to the fault plane. On the Mohr diagram, θ is

represented by measuring 2θ counter clockwise from the maximum normal stress, σ1. In

general, θ will be approximately 60° so that the cos 2θ will be a negative number. Thesignificance of this is that the normal stress on active fault planes is less than the mean

stress (i.e., on the diagram below A < C). The components of Mohr's Circle are thefollowing:

A

B

Shear Stress

NormalStressC

D

A: σ n =σ 1 + σ 3

2+σ 1 − σ 3

2cos2θ

B: τ = σ 1 − σ 3

2sin2θ

C: The center of the Mohr's Circle -σ 1 + σ 3

2

2θ



68

D: The radius of the Mohr's Circle -σ 1 − σ 3

2

Elasticity and reference states for lithospheric stress

Discussion of two reference states is preceded by an introduction to elementary

elasticity. Such an introduction to elasticity is appropriate because many in-situ stress

measurement techniques capitalize on the elastic behavior of rocks. Furthermore, manynotions concerning state of stress in the lithosphere arise from the assumption that the

upper crust behaves as a linear elastic body.On scales ranging from granitoid plutons and salt domes (≈ 12-5 km) to the

thickness of lithospheric plates (≈ 100 km), the earth is approximately isotropic with

elastic properties independent of direction. If we assume that the lithosphere is subject tosmall strains, as is the case for elastic behavior, principal stress axes must coincide with

the principal strain axes. Then, elastic behavior of the lithosphere is represented by the

equations of linear elasticity which define the principal stresses2

of the three-dimensionalstress tensor as linear functions of the principal strains (Jaeger and Cook, 1969, section

5.2):

σ1 = (λ + 2ζ)ε1 + λε2 + λε3 (12-1)

σ2 = λε1 + (λ + 2ζ)ε2 + λε3 (12-2)

σ3 = λε1 + λε2 + (λ + 2ζ)ε3 (12-3)

where λ and ζ are elastic properties of the rock, known as the Lame's constants. ζ is

commonly known as the modulus of rigidity, the ratio of shear stress to simple shear strain. λ + 2ζ relates stress and strain in the same direction, and λ relates stress withstrain in two perpendicular directions.

If volumetric strain is defined as

∆ε = ε1 + ε2 + ε3, (12-4)

then then we may combine equations 12-1 to 12-3 as

σi = λ∆ε + 2ζεi. (12-5)

2 - This course treats compressive stresses as positive because a overwhelming majority of rocks withinthe earth are subjected to compressive stresses. If compression is positive, then contractional strain is

positive and extensional strain is negative. These sign conventions are confusing because the engineers

treat compressive stress as negative. Furthermore, structural geologists treat finite extensional strain as

positive largely because the very large elongation found in ductile shear zones is most convenientlymanaged as a positive natural logarithm (Ramsay and Huber, 1983) and, therefore, finite contractional

strain is negative. In dealing with elastic strains associated with earth stress the sign convention is of little

consequence because the strains in question are always very small.



69

The uniaxial stress state is of immediate interest in developing an understanding

of the elastic constants, Young's modulus (E) and Poisson's ratio (υ), both of which willappear several times during discussions of lithospheric stress in this book. Uniaxial

stress, a stress state for which only one component of principal stress is not zero, is rare

in the earth's crust except in the pillars of underground mines. For uniaxial stress, σ1 ≠

0, σ2 = σ3 = 0, and the three equations of elasticity are written as

σ1 = (λ + 2ζ)ε1 + λε2 + λε3, (12-6)

0 = λε1 + (λ + 2ζ)ε2 + λε3, (12-7)

0 = λε1 + λε2 + (λ + 2ζ)ε3. (12-8)

Using equations 12-7 and 12-8, strain parallel to the applied stress, ε1, is related to strain

in the directions of zero stress, ε2 and ε3,

ε2

= ε3

= − λ2 λ + ζ( )

⎛ ⎝

⎞ ⎠

ε1. (12-9)

To derive the ratio between stress and strain for uniaxial stress, values for ε2 and ε3 are

substituted back into equation 12-6. This ratio is Young's modulus,

E =σ

1

ε1

=ζ 3λ + 2ζ( )

λ + ζ. (12-10)

Equation 12-10 is a simplified form of Hooke's Law,

σij

= Cijkl

εkl (12-11)

where Cijkl is the stiffness tensor. Under uniaxial stress, an elastic rock will shorten

under a compressive stress in one direction while expanding in orthogonal directions.

The ratio of the lateral expansion to the longitudinal shortening is Poisson's ratio,

υ = −

ε2

ε 1 =λ

2 λ + ζ( )⎛ ⎝

⎞ ⎠ . (12-12)

υ for rocks is typically in the range of .15 to .30.

Two reference states of stress



70

While developing an understanding of lithospheric stress, it is convenient to start

with reference states which occur in a planet devoid of plate tectonics and, then describethe difference between these reference states and the actual state of stress. Stress

generated by plate tectonic processes make the difference. In the earth, geologically

appropriate reference states are those expected in 'young' rocks shortly after lithification.

Two general types of 'young' rocks are intrusive igneous rocks, shortly after solidificationwithin large plutons deep in the crust, and sedimentary rocks, shortly after the onset of

burial and diagenesis within large basins.

Lithostatic reference state The simplest reference state is that of lithostatic

stress found in a magma which has no shear strength and, therefore, behaves like a fluid

with

σ1 = σ2 = σ3 ⇒ SH = Sh = Sv ⇒ Pm (12-13a)

where Pm is the pressure within the magma. At the time enough crystals have solidified

from the magma to form a rigid skeleton and support earth stress, the 'young' rock, an

igneous intrusion, is subject to a lithostatic state of stress

σ1 = σ2 = σ3 ⇒ SH = Sh = Sv ⇒ Pc (12-13b)

where Pc is confining pressure. This, presumably, is the state of stress at the start of

polyaxial strength experiments in the laboratory. Strictly speaking, there are no principal

stresses in this case, because the lithostatic state of stress is isotropic. Looking back at

the Mohr diagram, the student will see that Mohr’s circle will plot as a point. Equation12-4 applies to calculate volumetric elastic strain accompanying the complete erosion of

an igneous intrusion assuming no temperature change. Adding equations 12-1 to 12-3

gives an expression for the response of rocks, including laboratory test specimens, to

changes in confining pressure,

∆P c =3λ + 2ζ

3⎛ ⎝

⎞ ⎠∆ε = κ∆ε =

1

β∆ε

(12-14)

where κ is the bulk modulus and its reciprocal, β, is the compressibility.

Lithostatic stress will develop if a rock has no long-term shear strength. Althoughsome rocks such as weak shales and halite have very little shear strength, experiments

suggest that all rocks support at least a small differential stress for very long periods

(Kirby, 1983). Indeed, heat will cause rocks to relax, but they never reach a lithostatic

stress state. Although metamorphic rocks retain a shear strength during deformation, themetamorphic petrologist commonly uses the term, pressure, when actually referring to a

state of stress which may approximate lithostatic stress (e.g., Philpotts, 1990). Astructural geologist reserves the term, pressure, for describing confined pore fluid or

other material with no shear strength. The terms signifying state of stress, of which

lithostatic stress in one, apply to rock and other materials that can support a shear stress.

Although the lithostatic state of stress is rare in the lithosphere, it is a convenientreference state.



71

Uniaxial-strain reference state A second reference state is based on the

postulated boundary condition that strain is constrained at zero across all fixed vertical planes (Terzaghi and Richart, 1952; Price, 1966, 1974; Savage et al., 1985). Such a

boundary condition leads to a stress state which approximates newly deposited sediments

in a sedimentary basin: the state of stress arising from uniaxial strain (Figure 12-1).

Uniaxial strain is also used as the model for the stress state describing the effect of overburden load on rocks at depth assuming these rocks develop fixed elastic properties

at some point after deposition. σ1 is the vertical stress arising from the weight of

overburden. If rocks were unconfined in the horizontal direction the response to an

addition of overburden weight would be a horizontal expansion. Because rocks areconfined at depth in the crust, Price (1966) suggests that horizontal expansion is restricted

by adjacent rock so that in the ideal case, ε2 = ε3 = 0. In practice, it is understood that

horizontal stress arising from a uniaxial strain state in a sedimentary basin is modified bychanges in elastic properties during diagenesis, creep relaxation, and slumping along

listric normal faults. Each of these processes bring the state of stress in sedimentary

basins closer to lithostatic. Mandl (1988) refers to any stresses arising from such

inelastic deformation mechanisms as prestress or a stress which is not explained bysimple incremental elastic deformation. As discussed in Chapter 10, remnant stress is a

version of prestress. Upon initiation of tectonic processes, horizontal stresses can varywidely from those calculated using the uniaxial reference state. During erosion and

removal of overburden weight, lack of contraction in the horizontal direction by uniaxial

strain behavior also leads to a large changes in horizontal stresses.

For the case of uniaxial strain, ε1 ≠ 0, ε2 = ε3 = 0, the equations of elasticity (12-1

to 12-3) are written (Jaeger and Cook, 1969, section 5.3):

σ1 = (λ + 2ζ)ε1 (12-15)

σ2 = σ3 = λε1. (12-16)

From equations 12-15 and 12-16, the relationship between vertical (Sv = ρgz = σ1) and

horizontal stresses (SH = Sh = σ2 = σ3) are given in terms of the Poisson's ratio

SH

= Sh

=υ

1 − υ( )⎡⎢⎣

⎤⎥⎦S v =

υ1 − υ( )

⎡⎢⎣⎤⎥⎦ρgz

(12-17)

where ρ is the integrated density of the overburden, g is the gravitational acceleration, andz is the depth within the earth. Major deviations from this reference state may signal that

the uniaxial-strain model is not a particularly effective model for state of stress in thelithosphere. Assuming a υ = 0.2, the Sh = 0.25 Sv. This stress state is illustrated by theMohr’s circle on page 62.

Tectonic stress

Stresses can vary from the reference state as a consequence of either natural or

man-made processes. Those components of the in-situ stress field which are a deviation



72

from a reference state as a consequence of natural processes include tectonic stresses,

residual stresses, and near-surface thermal stresses induced by diurnal and annualheating. Large deviations from a reference state of stress arising from mining, drilling,

excavation, and other societal activities are man-made and, thus, unrelated to stress in the

lithosphere. Traditionally, tectonic stresses are associated with stresses arising from the

largest-scale natural sources such as plate-boundary tractions (e.g., Zoback et al., 1989,Hickman, 1991) and local stresses arising from topographic loading, thermoelastic

loading, and unloading due to erosion are considered nontectonic. The term, contemporary tectonic stress, grew from Sbar and Sykes' (1973) paper to mean presentstress fields which are reasonably homogeneous on a regional or plate-wide scale.

However, the distinction between plate-wide and local stresses breaks down in certain

geological contexts. For example, the orientation of some late-formed joint sets iscontrolled by plate-wide stress fields and, yet, they propagate in response to stresses

developed during local unloading due to erosion (e.g., Hancock and Engelder, 1989).

Thus, the definition of tectonic stress is simplified if most constraints concerning scaleand source are removed from the definition. Tectonic stresses are usually horizontal

components of the in-situ stress field which are a deviation from a reference state as aconsequence of natural processes on all scales from plate-wide to local. By this

definition local stresses developed under topographic loading are considered tectonic.Unfortunately, this definition is still so loose that geoscientists will continue to quibble

about what qualifies as a tectonic stress.



73

Structural Geology

Lecture 13

Coulomb Failure Criterion(Mohr Envelope)

Structural geology is the study of deformed rocks. Deforamtion is the response of the rock to a state of stress. Because natural rock deformation is a slow process,

laboratory experiments must be used to "speed" time. One component of nature lost in

the laboratory is size. The deformation of small rock samples can be "mapped" in thelaboratory by curves showing the relation between the magnitude of stress applied to the

rock and the amount of strain or deformation the rock has suffered. These curves are

known as stress-strain curves. Strain (ε) is a measure of deformation when the rock

deforms uniformly. The simplest stress-strain curve (σ-ε curve) is one where

dσ/dε = E

where E is a constant. This differential equation is Hooke's Law for the linear-elastic

behavior of materials. In the case of Hooke's law the constant E is known as Young's

modulus. The reaction to a stress applied to an elastic body such as a rubber ball is a predictible and repeatable amount of strain. Elastic strain is recoverable. By this we

mean that once the stress is remove the body will return to its original shape. A meterial

is described as linear-elastic if Hooke's Law holds accurately. No material including rocks will sustain an indefinitely large amount of strain.

After a certain amount of elastic strain rocks will deform in some other manner. The

point of transition from elastic behavior to another type of behavior is called the yield

point. Yield connotes a non recoverable deformation. One common mode of rock behavior after yield is brittle fracture and another mode of rock behavior is ductile flow.

A rock that has failed by brittle fracture has usually broken into more than one peice.

In the earth each small volume of rock is surrounded by other rock. The effect of the surrounding rock is to confine the volume in question and apply a confining pressure

(sometimes called a lithostatic stress). In order for rock deformation to take place the

principal stress in one direction (σ1) must exceed the other two principal stresses (σ2, and

σ3) which are at right angles to σ1. This difference between σ1 and, say, σ3 is called the

differential stress to which the sample is subjected. In the analysis of rock deformation

σ3 is equivalent to the confining pressure.

Our knowledge of the behavior of rocks comes from experiments in thelaboratory. One common rock mechanics experiment uses a cylinder of rock placed in a

rock deformation machine. Such a cylinder is shown in cross section (Fig. 13-1). Pistonscontact the end of the cylinder and create the stresses necessary to deform the rock. The

cylinder is surrounded by a confining medium which is prevented from flowing into the

pores of the rock sample by an impermeable jacket. In Figure 13-1 the sample isstippled, the confining medium is represented by the inward pointing arrows, and the

loading pistons are shown as the darker objects on either end of the rock sample. The



74

application of differential pressure is shown by the dark, vertical arrows on either end of

the piston.

(Fig. 13-1)

A typical stress-strain curve is shown to the right of the rock deformaiton

experiments. As stress increases on the rock, the rock strains. Once the fracture strengthof the rock is reached the rock fails along one or more fracture planes. The failure is

denoted by a sudden stress drop in the stress-strain curve.

θθ

θ

(Fig. 13-2)



75

In the late 18th century a French naturalist, Coulomb, observed for rocks that

shear stress |τ| necessary to cause brittle failure across a plane is resisted by the cohesion

of the material So and by a constant *µ times the normal stress σn across that plane:

|τ| = So + *µσn.

The constant *µ is called the coefficient of internal friction. *µ is not to be mistaken for µ, the coefficient of sliding friction which will be discussed in detail in a future lecture.

Because a shear stress was parallel to the plane of failure, the brittle fracture is

commonly called a shear fracture. This mode of fracture should be distinguished from a

tensile crack that opens normal to the least principal stress (σ3). Recall that

σn = 1/2(σ1 + σ3) + 1/2(σ1 - σ3) cos2θ

and

τ = - 1/2(σ1 - σ3) sin2θ (see Fig. 13-2).

Substituting into the Coulomb relationship

τ - *µσn = 1/2(σ1 - σ3)[sin2θ - *µ.cos2θ] - 1/2*µ(σ1 + σ3)

For failure the difference between τ and σn must be maximum which occurs when

tan2θ = -1/*µ

2θ lies between 90° and 180°. Thus

sin2θ = (*µ2 + 1)-1/2,and

cos2θ = - *µ(*µ2 + 1)-1/2.

Again substituting

τ - *µσn = 1/2(σ1 - σ3)[*µ2 + 1]-1/2 - 1/2*µ(σ1 - σ3)

which gives the maximum value of τ - *µßn. Failure occurs when

2S0 = σ1{(*µ2 + 1)1/2 - *µ} - σ3{(*µ2 + 1)1/2 + *µ}.

This is the famous Coulomb criterion for brittle failure. The Coulomb failure envelope

can be plotted on a σ1 vs. σ3 graph (Fig 13-3). In fig. 13-3 the uniaxial compressive

strength is C0 where

C0 = 2S0[( *µ2 + 1)-1/2 + *µ]



76

(Fig. 13-3)

Another representation of the Coulomb criterion is obtained by introducing mean

stress σm and maximum shear stress τm where

σm = 1/2(σ1 + σ3) and τm = 1/2(σ1 - σ3).

Then the Coulomb criterion becomes

(*µ2 + 1)1/2τm - *µ.σm = S0

Now if *µ = tan φ where φ is the angle of internal friction

τm = σmsin φ + cos φ

which is a line in the σm -τm plane of inclination tan-1(sin φ) and intercept S0 cos φ on the

τm-axis (Fig 13-4).

This result is similar to the Mohr circle representation where the Coulomb

criterion is represented by a straight line of slope *µ = tan φ and intercept S‚ on the |τ|

axis. If the σ1 - σ3 circle touches the line of the Coulomb criterion then brittle failure.

Mohr's hypothesis was that when shear failure takes place across a plane, the

normal stress σn and shear stress τ are related by a function

|τ| = f(σn).

Failure takes place if the circle of center C on the σ1 - σ3 just touches AB. The failure

curve is obtained experimentally and is the envelop of many Mohr circles corresponding

to failure under a variety of confining pressure (σ3) conditions.



77

(Fig. 13-4)



78

Structural Geology

Lecture 14

Microcracks and Shear Fracturing(Rock Failure)

The Coulomb criterion is empirical requiring several rock deformation

experiments at different σ1 - σ3 conditions. From these experiments the failure of a

certain type of rock can then be predicted. The criterion contains no information on themechanism or the microscopic process involved in rock fracture. Interestingly, the

Coulomb criterion was formulated about 1770 and more than 150 years passed before the

actual process of rock failure was understood. About 1920 an engineer named Griffithfirst developed a mechanistic failure criterion which could be applied to rocks. Griffith's

theory was originally formulated to explain fracture in glass.

Griffith's contribution

Griffith's criterion was based on the observation that glass is not internallyhomogeneous but rather contain pores. The same can be said for rocks. Pores in rocks

can be original spaces between grains that did not completely fill with cement or they can

be microcracks that develop when the rock is stressed. These voids play an important

role in rock failure directly through weakening the rock and indirectly through hosting pore water which under pressure can also effect the rock strength. The effect of voids on

the elastic properties of a rock can include: 1.) The Young's modulus (E) of a rock

containing voids and cracks is less than the intrinsic E of a solid rock (an intrinsic property refers to that property in a body without flaws or cracks); 2.) The Poisson's

ratio (υ) of a rock containing cracks is less than its intrinsic υ; 3.) cracks give rocks adifferent modulus during loading and unloading (this difference in behavior is known as

hysteresis).

Griffith observed that prior to glass (i.e. rock) failure microcracks and pore

cavities within rocks grow in length or propagate. Many of these microcracks must

propagate and then intersect before the rock can fail along a shear zone. Griffith

examined the conditions under which individual microcracks propagated. First, heobserved that cracks must be pull apart to cause propagation. In other words, the

microcracks must fail in tension. Griffith's approach was to examine the amount of

energy available to cause the crack to propagate. Recall that work, a scalar, (W) is the

dot product of a force vector f and displacement vector d.

W = f (dot) d.

Griffith (1924) recognized that during crack propagation surface energy, dUS, is

necessary to create new surface area. As the crack propagates, a rock may undergo achange in strain energy, dUE which will contribute to the surface energy. Boundary



79

tractions may also contribute to the surface energy in the form of mechanical work, dWR

Both dWR and dUE are forms of mechanical energy. The total change in energy for crack

propagation is

dUT

= dUS

− dWR

+ dUE

. [14-1]

Under certain circumstances crack propagation may take place without changing the total

energy of the rock-crack system. This is known as the Griffith energy-balance conceptfor crack propagation where the standard equilibrium requirement is that for an increment

of crack extension dc,

dUT

dc= 0

. [14-2]

For equilibrium, the mechanical (-dWR + dUE) and surface energy (dUS) terms within therock-crack system must balance over a unit crack extension dc. The total energy of the

crack-rock system is at a maximum at equilibrium. Any perturbation of the system such

as an increase in tensile load will cause the crack to propagate spontaneously without

limit. During crack propagation the crack walls move outward to some new lower energyconfiguration upon removal of the restraining tractions across an increment of crack. In

effect, the motion of the crack walls represent a decrease in mechanical energy while

work is expended to remove the restraints across the crack increment. The work toremove the restraints is the surface energy for incremental crack propagation.

To evaluate the three energy terms relating to crack propagation, Griffith cited a

theorem of elasticity which states that, once an elastic body is subject to a load, its

boundaries will displace from the unloaded state to the equilibrium state so that

WR

= 2UE . [14-3]

For a body of rock containing an elliptical crack with major axis perpendicular to a

uniform tension (i.e., Sh < 0), Griffith calculated that

UE

=π 1 − ν 2( )

E

⎡⎢⎣

⎤⎥⎦c2 S

h( )2

, [14-4]

provided that the rock is thick enough for plane strain conditions to hold. According to

Jaeger and Cook (1977) the effect of a crack on the elastic properties of the rock shows

up on the π and crack length terms. For the surface energy of the crack, Griffith definedcrack length as 2c and recognized the crack propagation produces two crack faces.Therefore,



80

US

= 4c γ[14-5]

where γ is the free surface energy per unit area. Substituting Equations [14-3], [14-4],and [14-5] into [14-1] and then applying Equation [14-2], Griffith solved for the critical

stress for crack propagation

σ3

c = T0

=2Eγ

π 1 − ν 2( )c

⎡⎢⎣

⎤⎥⎦

1

2

[14-6]

where T0 is the uniaxial tensile strength of the rock. Here σ3

c= − S

h for a dry rock in

the near surface. Equation 3-22 is the famous Griffith crack propagation criterion.

This is the famous Griffith criterion for rock failure in tension (Figure 14-1).

(Fig 14-1)

Another way of illustrating Griffith's criterion is to consider the simple crack system shown in figure 14-2. The mechanical energy of this system is (-WL + UE)

where UE is the potential energy stored in the system and WL is the work due to

displacement of the outer boundary. The lower end is fixed rigidly and the upper end is

loaded with force F. The crack system will behave as an elastic spring by Hook's Law

u = λF

where λ is the elastic compliance of the system and u is the displacement of the top end

of the system. The strain energy of the system is then



81

UE = 1/2Fu = 1/2F2λ = 1/2u2/λ

provided we do not let the crack (2c) extend to δc. If the crack does extend by δc when

force F is applied, then the elastic compliance of the system increases as can be shown bydifferentiating

δu = λδF + Fδλ

For δc > 0, δF < 0 and δλ > 0. If the crack extends, the mechanical energy of the system

should decrease.

δWL = Fδu + uδF = Fδu

The total mechanical energy change is therefore

δ(-WL + UE) = -1/2F2δλ

(Fig. 14-2)

This leads back to consideration the distinction between failure in tension and

failure in compression. Griffith's criterion gave a convenient physical explanation for crack propatation when the crack is subject to tensile stresses. The crack propagates in

the plane normal to the tensile stress or least princial stress (T = -σ3). However, we have

learned that whole-rock stresses including σ3 increase with depth in the crust of the earth.

Yet, rock fracture in shear depends on microcrack propagation where the propagation is

caused by tensile stresses. Where are the tensile stresses generated when rock is deepwithin the crust of the earth? The answer is that on a microscopic scale grains press



82

against each other at sharp contacts. These contacts are called "stress risers". Under

these microscopic contacts very large tensile stresses can be generated even though thewhole-rock stress in highly compressive.

Figure 14-3 shows an example of a stress riser. The point contacts of each disk

serve to concentrate the stress which was distributed over the end plattens. The increasein stress is shown by the circular lines originating from the point contact. These lines

represent lines of equal stress difference. The numbers indicate that the stress difference

increases as the point contacts are approached.

(Fig. 14-3)

Another characteristic of point contacts is that a portion of the volume under the

point contact is in a state of tension. Such an volume is illustrated in figure 14-3. When

the rock is compressed under high σ1 - σ3 conditions, many of these microscopic contacts

are activated and microscopic crack propagation is common. As the population of

microcracks increases a zone of weakness develops near but not along a plane of

maximum shear stress (τmax). This zone of microcrack intersection eventually hosts the

through-going shear fracture.

Shear fractures do not form at 45° to σ1 and in the plane of †max but rather in a

plane whose normal is closer to σ3 than σ1. The reason for this behavior is found in a

closer examination of the Coulomb criterion where

τ = S0 + µ0σn



83

where σn and τ the normal and shear stress on the plane of failure respectively and S0 is

the cohesive strength of the rock (Fig. 14-4). µ0 is the angle of internal friction

µ0 = tan φ

e

e

(Fig. 14-4)

Tan φ can not be meassured directly but, rather, is derived from the slope of the Coulombfailure envelop. µ0 should be distinguished from the coefficient of sliding friction (µ)

which relates τ and σn during slip of a fault

µ = τ/σn.

(µ will be discussed in a future lecture) (Fig. 14-5).

µ0 predicts the angle of shear failure

µ0 = (τ - S0)/σn.

A graphical plot of τ, σn, and (τ - S0)/σn shows that shear failure occurs on the plane

where (τ - S0)/σn is maximized (Fig. 14-6).



84

(Fig.14-5)

(Fig. 14-6)

Shear fracturing requires a high differential stress. Such stresses are not common

within the crust. Places where σ1 - σ3 may be high include:



85

1.) Fault zones -- Along fault zones asperities which interlock can act as localstress concentrators. In fact, studies of earthquakes show that many faults lock and then

slip violently. Asperities cause the locking and more often than not the earthquake is a

manisfestation of the asperity being sheared off.

2.) Man-made structures -- Pillars in mines often have to bear loads many times

that of neighboring rock left undisturbed by mining. Mine shaft openings also act as

stress concentrators in much the same manner as microscopic point contracts within rock.The stresses around holes in rock can be many times as large as was found in undisturbed

rock.

3. Folds -- Rocks can act like elastic plates which bend. It is known from the

theory of elasticity that part of the bent plate is compressed. In this zone of compression

within folded rocks compressive stresses can become so large that the shear strength of the rock is exceeded.

There are several rock structures associated with shear fracturing. Cataclastic

rock is the product of concentrated microfracturing within a shear zone. Cataclasis is the process of mechanical reduction of grain size by brittle fracture and rigid grain rotation.

Microfractures develop across the shear fracture in the direction of σ1. Continued slip on

the shear fracture cause more microfracturing which leads to a wider zone of cataclasis.

Comminution is another term applied to the mechanical reduction of grain size. As the pressure on the rock increases, tension gashes form the the shear zone becomes

semiductile. Tension gashes are filled macrocracks or extension fractures. The origin of

extension fractures is the subject of a future lecture. Repeated shear fracturing causesdeformation bands. These are zones of cataclastic material most commonly observed in

sandstones. Deformation bands occur because each successive shear fracture or zone of cataclasis is permeable to cements which make the zone stronger than surrounding

sandstone. A successive shear fracture must then rupture intact rock or cemented fractureand is not focused by a previous plane of weakness.

The surface of a shear fracture can either be a slickensided surface or a fiber-coated surface. The latter has been erroneously called a slickensided surface.

Slickensided surfaces are the product of brittle polishing and commonly have carrot-

shaped wear grooves. Fiber-coated surfaces are the product of ductile slip where thedeformation mechanism is diffusion mass transfer. Pressure solution dissolves asperities

where are redeposited as fibers pointing in the direction of slip.



86

Structural Geology

Lecture 16

Stress Concentration(A process necessary for tensile failure)

Tensile cracking (jointing) occurs in the crust of the earth because of the

magnification of stress at the crack tip. Here we show that tectonic stress (SH) around a

circular borehole was magnified by a factor of three. In an infinite elastic plate containinga circular hole and subjected to a uniaxial far-field stress, local stresses are concentrated

about the edge of the hole (Kirsch, 1898). If rock containing a vertical cylindrical hole issubject to a uniform uniaxial horizontal stress at infinity, SH, then the stresses near the

hole expressed in polar coordinates at a point θ, and r are given by

σ rr =S

H

21 − R

2

r 2⎛ ⎜⎝

⎞⎟ ⎠ +

SH

21 + 3

R 4

r 4− 4

R 2

r 2⎛ ⎜⎝

⎞⎟ ⎠cos 2θ

[16-1]

σθ

=S

H

21 + R

2

r 2⎛ ⎜⎝

⎞⎟ ⎠ −

SH

21 + 3

R 4

r 4⎛ ⎜⎝

⎞⎟ ⎠cos 2θ

θ[16-2]

τr θ

= SH

21 − 3

R 4

r 4+ 2

R 2

r 2⎛ ⎜⎝

⎞⎟ ⎠sin 2θ [16-3]

where R is the radius of the borehole, r is the distance from the center of the hole, and θ is measured clockwise from the line parallel to the direction of the compressive stress SH

(the x-axis in Figure 16-1a). σθθ is the circumferential or hoop stress whereas σrr is the

radial stress. An analysis of the redistribution of stresses about a borehole shows that theuniaxial far-field stress, SH, is magnified by a factor of three at the point on the wall of

the hole where the stress is tangential to the hole (Figure 16-1b). At 0° and 180°, theuniaxial compression is reflected as a tension of the same magnitude as SH but opposite in

sign. Equations [16-1 to 16-3] are useful for calculating stresses at any point in thevicinity of the circular pore. For example the large compressive hoop stress at the wall of

the pore decreases rapidly with distance away from the pore wall. At a distance greater

than one radius the compressive normal stress is only slightly larger than the far-fieldstress, SH.



87

SH3

SH-

SH

compression

far-field

stress

tension

Kirsch's Solution

Pf

SHP

f=

θ r

borehole

x

y

R

a. b. c.

Figure 16-1

Joints form in tensile so we need to change the sign on the stress acting on the

hole in the rock and assume that Sh is tensile. At the tip of a crack stress Sh (it is nowtensile) is further magnified depending on the radius of curvature at the crack tip and the

aspect ratio of the crack. The aspect ratio is the aperature of the crack divided by the

length of the crack (i.e b/c in fig. 16-2).

cb

6

5

4

3

2

1

c

b _ = 1

S

S _

h

6

5

4

3

2

1

c

b _ = 3

S

S _

h

7

b

c

Pore Crack

hS hS

Figure 16-2

Conditions for initiation , propagation and arresting of a tension crack are as follows (Fig.

16-3). Conditions for propagation of a macroscopic crack can be described by specifying

only the crack edge intensity factor K. A crack will not propagate until a critical stress

intensity factor (K I) has been exceeded. For opening mode cracking

K I = (σyy).(π.c)1/2

where c is the initial crack length. Hence, K I > K is the criterion for crack propagation.

Note here that σyy is pulling the crack open and this is equivalent to a negative σ3.



88

CRACK-TIP STRESS CONCENTRATION

Sh = St (1 + 2c/b)

1:1

2:18:1 32:1

CRACK

RATIOS

3 5

17

65

STRESS

CONCENTRATIONS

Initial

Stress

Figure 16-3



89

Structural Geology

Lecture 17

Crack Propagation and Joint Formation(The behavior of tensile fractures in the crust)

The most pervasive mesoscopic structure of any outcrop will be cracks or fractures. Even in midcontinent areas where sedimentary rocks seems undisturbed by

any tectonic deforamtion cracks will be common in outcrop. These discontinuities will

be encountered in many engineering tasks involving national and local needs. Cracks inrocks carry ground water far more efficiently than the bulk rock with its relatively low

permeability. Disposal of both chemical and nuclear waste requires that the waste

repository be separated from any interconnected fracture network that might allow therapid migration of the waste into the biosphere. In contrast the recovery of geothermal

energy requires cracks in hot rock so that heat fluids pumped along the cracks can act as

the medium with which heated is transported to the surface. The recovery of petroleum is best accomplished when wells cut a fractured network containing large amounts of petroleum.

Rarely is there more than a meter or two between macroscopic cracks. The upper portion of the crust has a surprisingly high fracture density. Igneous bodies may be an

exception to the rule and here microcracks make up for the evident lack of large cracks.

Fractures are encountered in wells drilled to depths of ten kilometers or more within thecrust of the earth.

Mode IOpening

Mode IISliding

Mode IIITearing

MODES OF CRACK DISPLACEMENT

Figure 16-1



90

From an engineering point of view there are three modes of crack failure (Fig. 16-1). These are commonly called the opening mode (Mode I), the sliding mode (Mode II),

and the tearing mode (Mode III). Cracks within the crust may be either shear fractures of

the type discussed in the last lecture or tension fractures. Tension fractures are opening

mode cracks. Shear fractures are more complicated and often form as a mixed modecrack. Mixed mode cracking means that two or three of the above modes is active during

failure. Commonly opening mode failure is combined with one of the other two modes.The distinction between shear fracturing and tension fracturing is that tension cracks form

normal the the least principal stress σ3 and parallel to σ1, whereas the plane of a shear

fracture is inclined at some acute angle to the direction of σ1. Within the crust of theearth true tension is less common than a tensional situation set up by negative effective

stresses (e.g. σ3 - p p < 0). Fractures formed normal to σ3 in the presence of negativeeffective stress are called extension fractures or joints.

Joints are the most common fracture within the crust of the earth. This is so

because the differential stress (σ1 - σ3) in the crust rarely becomes large enough to causeshear fracturing. Hence, it is important to appreciate the stress state under which joints

form within the crust. This is accomplished by examining the Coulamb-Mohr failure

envelope.

Most Coulomb-Mohr failure envelopes are nonlinear particularly where the

failure envelope crosses into the field of tensile normal stresses (-σn). Because of itscurved nature the only place where the failure envelope can be intersected by the Mohr

Circle is at the point where shear stress (τ) is zero which means that failure occurs alonga principal stress plane.

Tensile cracking (jointing) occurs in the crust of the earth because of themagnification of stress at the crack tip. Here we show that tectonic stress (SH) around a

circular borehole was magnified by a factor of three. In an infinite elastic plate containing

a circular hole and subjected to a uniaxial far-field stress, local stresses are concentratedabout the edge of the hole (Kirsch, 1898).

Examples of either extension fractures or tension cracks include the following:

Mud cracks

Ice polygonsColumnar joints in basalts

Dykes and sills

Tension gashesCalcite and quartz filled veins

Joints in rocks



91

The surface morphology of extension fractures and tension gashes is unambigious. It is

best displayed on cracks in sandstones (See Chapter 5 of Davis and Reynolds). All suchdiscontinuities have initiation points which may be stress risers such as microcracks,

fossils, bedding plane roughnesses, or concretions. All cracks propagate a finite distance

before stopping at arrest lines. Just after the crack has started propagating the rupture

plane may be very smooth showing a mirror surface. As the speed of propagationincreases the surface becomes rougher in an area called mist. Finally the rupture front

may move so fast that the stress readjustments at the crack tip cause out-of-plane propagation which leaves hackle marks on the surface.

Hackle marks on the crack surface can be used to define the direction of propagation.

Their pattern is like a fan opening in the direction of rupture propatation. Arrest linesform normal to the direction of propagation. In some cases the direction of crack

propagation can be quite complicated. Hackle marks are believed to form when crack

propagation becomes unstable. Under this condition the stress field at the tip of the crack can not abjust to the advancing crack tip. The result is that crack propagation is

momentarily deflected out of the plane of propagation.



92

Structural Geology

Lecture 18

Effective Stress(Pore Pressure within the Earth)

During the previous lectures crack propagation was discussed in terms of stressrisers. The example of a stress riser was the point-to-point contact between grains.

Stress risers also occur around pores in rocks. An example of a stress riser around a pore

may be illustrated by examining the magnification of tectonic stress at the edge of acircular hole in an elastic medium (i.e. a rock). Figure 18-1 shows that the tectonic stress

σ1 is magnified three times at the edge of the hole whose tangent parallels the orientation

of the tectonic stress. At 90° around the hole from the stress concentration the tectonic

stress causes a tensile stress whose magnitude is equal to - σ1.

If this hole serves as a model for a pore in a rock, it is apparent that the position of

easiest cracking is at the point of greatest tensile stress. This stress distribution favors

crack propagation in the plane of the maximum compressive stress and normal the least

compressive stress ( σ3).

(Fig. 18-1)

So far we have discussed the concept of stress and one of its effects (e.g.microcracking and then brittle fracture in shear) on the deformation of rocks within the

upper crust. The effect of stress within the crust can be modified by several mechanismsand processes. One of the most profound "modifiers" of stress is pore water. As was

indicated in the discussion of the Griffith failure criterion rocks contain cracks and pores.

Most of these voids within rocks are interconnected and, hence, subject to penetration byfluids migrating through the crust. It is quite rare to find a completely dry rock within the

crust.



93

The fluids that penetrate pore space exert a "back pressure" on the pores. Thisinternal pressure tends to force the pores open, in contrast to the lithostatic stress of the

surrounding rock which acts to close the pores. Investigations of rock properties must

account for the effect of this "back pressure" which modifies the behavior expected for a

dry rock. The effect of this back pressure is shown in the fluid filled crack of Figure 18-2.

(Fig. 18-2)

Karl Terzaghi was a pioneer in the field of soil mechanics. Much of what he

proposed for the behavior of water saturated soils can be applied to the mechanics of rockswith fluid-filled pores. His most important concept is known as the effective stress

principle. The effective stress within a soil or rock is equal to the total stress minus the

pore pressure. The effective stress principle is as follows. Across any plane at element A

within a rock, there are acting a total stress σ and a pore water presssure p p. Total stress (

σ ) can be visualized as the weight of a water-saturated column rock. Two components of that weight are the rock with empty pores and the weight of the water that fills the pores.

We define effective stress as

σ = σ − P p

Distinguishing the total vertical stress σv and the total horizontal stress σh, we have

σ v = σ v − P p and



94

σh

= σh

− P p

I will define the coeffieient of lateral stress (Λ) based on effective stresses

Λ = σ h

σ v

Lets now consider the profile of various stresses and pressures with depth. To do

this we must define two constants: the density of water ρw and the total density of the rock

ρt which includes a component for water in the pore spaces. In the following equations z

is the depth of burial or depth below the surface of the earth. Total vertical stress is

σ v = ρ t gz = φρH 2O

+ 1 − φ( )ρQTZ⎡⎣ ⎤⎦

gz

Hydrostatic water pressure is

P p = ρH2 O

gz

Effective vertical stress is then

σ v = σ v − P p = ρt

− ρH2 O( )z

Now consider the horizontal stress generated from the vertical load. To do this we

assume plane strain. This means that imaginary vertical planes within the rock are not

allowed to move regardless of the vertical stress. This is like putting sand into a glass jar

and pushing on top of the sand. Even though we can gradually push harder, the walls of the glass jar do not expand. How is a vertical load transfered to the horizontal direction. If

we squeeze a rubber ball in a vertical direction it will want to expand in a horizontaldirection. However, if we do not allow the ball to expand in the horizontal direction by

clamping the ball, the clamps will have to sustain increasing larger horizontal stresses

with the additional vertical stresses. This response of an elastic body by an expansion perpendicular to a compression is known as the Poisson effect. The coeffieient of lateral

stress (Λ) as defined above is a measure of the magnitude of the Poisson effect. For our example where the only horizontal stress is due to a vertical stress then

σh

= Λσ v

Now the total horizontal stress is

σh

= σh

+ P p = Λσ v + P p = Λ σ v − P p( ) + P p

σh

= Λρ t z + 1 − Λ( )ρH2 O

z



95

This is an illustration of how pore water pressure effects horizontal stress within the crust.

The final concept to be introduced in this lecture is the effect of pores on strength of

rocks. A rock without pores is considerable stronger. This stands to reason because pore

space is unsupported within the rock. With the presence of unsupported space the intact

rock between the pores has to take larger loads to sustain an overall applied stress. Thisimportant effect also has a direct bearing on the effective stress law which is more properly

written

σ = σ − αP p

Terzaghi suggested that α = 1 based on experiments in soils. His interpretation was that

grain boundaries and grain contacts both have an effective porosity of one, which is to say,water at a pressure p p filled all space between grain contacts. Others including Geertsma

have suggested that

α = 1 - (K /K 0)

where K and K 0 are the effective and grain bulk moduli. Nur and Byerlee have a very

interesting derivation of α. They assume an isotropic aggregate of solid material withconnected pores. The outside of the solid is subject to confining pressure Pc and the pores

are subject to pressure p p. Both of these pressures have an effect on the volumetric strain

of the aggregate Θ. But what we will do is consider the effect of these pressures bydividing the application of the confining pressure into two steps. The first application of

pressure involves balancing p p and pc, so that p p' = pc. The second application of

pressure involves the remaining confining pressure pc'' = p

c- p

p. Volumetric strain Θ

of a dry aggregate caused by the addition of a confining pressure is

Θ = β. pc

where β = 1/K is the effective compressibility of the dry aggregate. Now the volumetric

strain of the aggregate Θ'' due soley to the confining pressue pc'' is

Θ'' = β. pc'' = β. pc - β. p p.

The volumetric strain Θ' caused by the equal addition of pressure around the rock body andinside the pores is of interest. The solution is to assume a solid in which the pores arefilled with a solid. Then an external confining pressure pc' causes an equal hydrostatic

compression pc' in the solid. The volumetric strain of the solid Θs is then

Θs = βs.pc'



96

where βs is the compressibility of the solid devoid of any cavities. In this analysis we can

remove the solid from the pore and replace it with a fluid at p p' = pc'. In doing so the

deformation of the solid remains unchanged. Consequently

Θ' = Θs = βs.pc'

Adding the equations to find the total strain

Θ = Θ' + Θ'' = β. pc - ( β - βs)p p.

This equation can be written in terms of the bulk moduli where K is the bulk modulus of

the dry aggregate and K s is the intrinsic bulk modulus of the solid itself

Θ = 1/K . pc - [1/K - 1/K s]. p p

Because the volmetric strain can be written

Θ = 1/K . pc''

the expression for the effective pressure can now be written

pc'' = pc - [1 - (K /K s)]. p p

and

α = 1 - (K /K s).

As discussed in the previous lecture, extension fractures are by far more common in

the crust of the earth. The explanation is simple: the state of stress within the crust of the

earth is such that the differential stress (σ1 - σ3) is rarely large enough to fracture intact

rock. Many smallish appearing faults are nothing more then reactivated extension

fractures. Reactivated means that slip occured parallel to the extension fracture after it

propagated normal to the least principal stress (σ3). Slip in shear requires that the stress

field be reoriented so that the extension fracture is no longer sitting in a principal stress plane.



97

(Fig. 18-3)

If differential stress within the crust of the earth is rarely large enough to cause

shear fractures, how are most fractures formed? This question is particularly interesting

because tensile stresses are required for the generation of extension fractures. Yet, how are

tensile stresses generated when lithostatic stress which is compressive become increasingly

large with depth in the crust. Remember that the vertical stress (σv) is proportional to the

density (ρ) of the rock and the depth of burial (h).

σv = ρ.g.h

where g is the acceleration due to gravity. In the absence of tectonic stresses the

horizontal stress (σh) increases with depth as a fraction of the vertical stress.

σh = υ/(1 - υ ). σv = υ /(1 - υ)

.ρ.g.h

Figure 18-3 shows extreme examples of the development of σh with depth of

burial in a delta environment. The difference depends on whether water can leak or drain

out of the rock during burial. Clay, the forerunner of shale has such a low permeability

that its leak rate is too slow to keep up with burial. Therefore we call a clay "undrained".For an undrained clay water supports the matrix. A material such as water with no shear

strength has a υ = 0.5. If a sandstone is allowed to drain, it has a υ = 0.25. There is a

dramatic difference in the overburden induced σh between these two situations.

However, there are four mechanisms by which these large compressive stresses can be

negated. These mechanisms either contribute large tensile stresses or act to create

negative effective stresses within the crust of the earth. These mechanisms include: 1.)increasing pore pressure (negative effective stress); 2.) thermal cooling (tensile stress);



98

3.) bending of rock (tensile stress); and 4.) Poisson contraction (tensile stress). Often

more than one mechanism is active at the same time within the crust of the earth.

PORE PRESSURE In a previous lecture we have already discussed the equations

involved in pore presssures. The effective stress concept was presented. A simple model

for effective stress involves pore water in a crack. The lithospheric stress which may beeither σv or σh depending on the orientation of the stress acting to close the crack. These

stresses are counterbalanced by water pressure within the crack which acts to open the

crack (Fig. 18-4). In a cross section through the rock if water has access to pores that fill

the cross-sectional area then a change in pore pressure ∆ p p is effective in balancing the

same change in lithospheric stress ∆σ. Experimental evidence suggests that where cracksare in point contact, water cannot penetrate the point contacts to push the crack open If

this is the case then the pore fluid in a cross-sectional area does not fill the area and a ∆ p p

does not act to offset ∆σ. Here, then, we would say that the effective stress changesless. Hence, the general effective stress law is

∆σe = ∆σ - ψ. ∆ p p

where ψ = 1.0 in the former case and ψ < 1.0 in the latter case. There is reason to

believe that ψ is proportional to the ratio of pore space to total cross-sectional area as isshown in Figure 10-4.

(Fig. 18-4)

Evidence for abnormal pore pressures come from oil wells through out the worldwhere the pore pressure is higher than expected for the weight of the water column at that

level within a well. Pore pressure is abnormal if



99

Pp > ρH2Ogh

Figure 10-5 shows data from the Gulf of Mexico where pore pressure is above

hydrostatic pressure at depths greater than 7500 feet. Rocks of ancient deltas contain

shales that are undercompacted. Undercompaction means that the pore space was notcrushed by the weight of the overburden. To prevent crushing the pore space must be

supported by the pore water at abnormal pressures. Joints have propagated in associated

with the undercompacted shales. These joints propagate only if the pore pressure in therock has exceeded the inward compression of the tectonic stresses or the weight of the

overburden.

Three mechanisms for generating abnormal pore pressures include compaction,

aquathermal pressuring, and clay dehydration. For pressuring by compaction the

permeability of the rock must be low. Captured pore water then prevents the pores from

collapsing under the weight of the overburden. The weight of the overburden is then

transferred from the rock matrix to the water which increases in pressure. For aquathermal pressuring the thermal expansivity of pore water is higher than the surrounding rock. If the

water is heated in closed pore space it will exert an ever increasing pressure against thewalls of the pores. At the temperature of the rock increases aquathermal pressuring

increases at the rate of a few bars per °C. A common clay in a sedimentary environment is

a montmorillonite. As the montmorillonite is heated it will dewater (lose water) during a phase transformation to illite. This volume of water released during the transformation to

illite equals about half the volume of the altered montmorillonite. This process is called

clay dehydration.

(Fig. 18-5)

THERMAL COOLING: Under most circumstances materials including rock expand when heated and contract when cooled. This property to expand when heated is



100

called thermal expansivity. The cooling of ice to develop water is one exception to the rule

that materials expand when heated and this example involves a phase transformation. A

measure of expansivity is called the coefficient of thermal expansion (α). Aquathermal

pressuring is another example of the effect of thermal expansivity. It denotes the change in

length per unit length per degree of temperature change (∆T). Recall that a change in

length per unit length is a strain. A metal called Invar has an α = 0 ppm/°C. Rock iscommonly α = 5 x 10-6/°C or 5 ppm/°C. Linear strain caused by a temperature change is

thus:

ε = dl/l = α.∆T

If the rock is constrained and unable to expand or contract, a stress is developed in

proportion to the rock's Young's modulus (E):

σ = E.ε = E.α.∆T

If ∆T = 100 °C and E = 0.5 x 106 b, the stress developed is:

σ = (0.5 x 106)( 5 x 10-6 )(102) = 2.5 x 102 b

If the change in temperature was a cooling, this effects stress in rock by lowering thehorizontal stress by 250 b.

(Fig. 10-6)

POISSON'S EFFECT: As defined earlier, Poissons ratio (υ) is the ratio of expansion in one direction of a rock caused by a contraction at right angles:



101

υ = ε2/ε1.

From this equation it is straight forward to calculate the change in horizontal stress (σ2)

accompanying the erosion of, say, 1 km of overburden. If the density (ρ) of rock in the top

km of the crust is 2.7, then the vertical stress (σ1) caused by one km of overburden is then

σ1 = ρ.g.h = (2.7 g/cm3)(980 cm/sec2)(105 cm)

σ1 ≈ 270 x 106 dynes/cm2 = 270 bars

From Poisson's ratio

υ = ε2/ε1 = Eσ2/Eσ1

If υ = 0.25, then

σ2 = υ. σ1 = 67.5 b.

This calculation again assumes that the rock is constrained from moving in a horizontal

direction (i.e. a plane strain model).

Figure 10-6 illustrated two paths for horizontal stress with depth of burial. Both

paths lead to negative (tensile) effective stresses. This stress condition is necessary for theformation of joints as described in the previous lecture. Aquathermal pressuring leads to

jointing by a hydraulic fracturing mechanism at considerable depth of burial. Contraction by thermal cooling and the Poisson's effect leads to joint formation in near surface rock.

FOLDING OF ROCKS (i.e. bending of elastic beams) Moments applied to the end

of a beam would cause the beam of flex or bend with a constant radiius of curvature. Rock

layers can be treated as beams and respond in the same fashion to bending moments.Engineers recognized that if an elastic beam were bent by moments applied at the ends of

the beam, a neutral fiber would develop. The neutral fiber is the plane within the beam

where neither tensile nor compressive stresses would develop. Below the neutral fiber the beam or rock layer is compressed. Above the neutral fiber the beam or rock layer is

extended, and may, in fact, extend far enough to overcome the horizontal stresses due to

overburden. Stresses in folds will be discussed in more detail in the lectures on folding.



102

Structural Geology

Lecture 19

Joint Patterns(The orientation and spacing of joints, veins, and dikes)

Two parameters influence fracture patterns: the orientation of the the fractures andtheir frequencies. Orientation of fractures is based on the state of stress within the rock --

both stress difference and orientation of the principal stresses. In contrast, the frequency

or spacing of fractures is based on the properties of the rocks in which the fractures haveformed (Fig. 19-1).

Fig. 19-1

The solution to the fracture spacing problem is found by recalling that the critical

stress intensity factor (K I) is

K I = (σ3)(πc)1/2

where c is the crack length. Here stress intensity is a function of crack length and, thus, is

independent of the lithology. However, in a bedded quartzite-limestone sequence, we predict that a quartzite will have more closely spaced joints if a given extensional strain is

responsible for generating larger tensile stress. The Young’s Modulus for quartzite is

greater than for limestone. Hence, by applying Hooke’s law we predict that tensile stressin a quartzite bed might be larger for a given unit of strain. If tensile stress increases as a

consequence of the thermal cooling of rock the following equation for the development of

thermal stresses might apply:



103

σ thermal = E α ∆T

1 −ν

where E is the Young’s Modulus, α is the coefficient of thermal expansivity and ∆T is the

change in temperature. This equation also suggests that when quartzite and an adjacentlimestone bed are subject to the same thermal change, tensile stress in the quartzite buildsup faster because Young’s Modulus and thermal expansivity for quartz are larger, and

hence,

σqtz > σls

So, the fracture intensity of the quartzite is larger and jointing will form on a closer

spacing.

Fig. 19-2

Fracture nomenclature is in part tied to the type of rock in which the fractures

have formed. Orientation also plays a role in fracture nomenclature. For example, Balk documented several types of fractures in granite bodies (Fig. 19-2). The type of fractures

are based on their relationship to the flow lines of the granite as well as the orientation of

the surface of the granite body. It is fair to assume that most joints in granites originate asextension fractures. Sheeting joints for parallel to the topographic surface and are

generally subhorizontal. Longitudianl joints are steeply dipping which parallel the flow

lines. Cross joints in granites form perpendicular to flow lines.Granite quarries are laid out based on the direction in which the granite breaks in

tension most easily. The quarrymen call this direction of easy breaking the rift. Flow

lines and the longitudinal joints are ofter parallel to the rift of a granite. The grain of a



104

granite is the direction of next easiest breaking. Sheeting joints may well parallel the

grain of the granite. In New England granites the orientation of the grain and rift mayappear in the opposite orientation as described (i.e. sometimes the rift parallels the

topography. The hardway of the granite is the direction in which the granite is most

difficult to break. The hardway is vertical in all New England granites.

In sedimentary rocks the joint nomenclature depends on the orientation of localstructures, most commonly the fold axis. Stearns pointed out the folded sediments may

have four fracture sets (Fig. 19-3).

(Fig. 19-3)

Each set may consist of up to three fractures: an extension fracture and conjugateshear fractures. Conjugate shears form so that the acute angle between the shears faces

σ1. Each of the sets has no particular name, however, the extension fractures may be

referred to as either cross joint (i.e. that group of joints perpendicular to the fold axis) or a

strike joint (i.e. that group of joints parallel to the fold axis). Here the term cross joint in asedimentary rock refers to a different type of joint than that found in a granite.

A different set of nomenclature was developed for joints in flat-lying sedimentary

rocks. Hodgson observed that in the absence of shear fractures in sedimentary rocks theremay be several joint patterns. The most common three patterns are shown in Figure 18-4.

Extension joints are not one long discontinuity but rather several joints that form

end-to-end in a joint zone. Some authors preferred to restrict the term joint to extension

fractures. This is the case with Hodgson. The line that an individual joint makes on theoutcrop surface is called the joint trace. Joint zones are often parallel to other zones

within the outcrop. These joint zones make up one joint set and are called systematic

joints. Often there are joints that form roughly perpendicular to the systematic joints. If these latter joints are themselves systematic joints they may be called cross joints.

Otherwise they are non-systematic joints. Here we see a third use of the term cross joint.



105

(Fig. 19-4)

(Fig. 19-5)

Conjugate shear fractures may be used to map σ1is the vicinity of a fault zone.

For the Bonita Fault, a normal fault in New Mexico, the σ1 is vertical as witnessed by the

orientation of shear fractures in the fault zone (Fig. 19-5). Yet, as the fault is approachedthe local rocks are seen to drag into the fault zone. Likewise, the conjugate shears rotate



106

into the main Bonita Fault suggesting that the stress field was not homogeneous near the

Bonita Fault. In the case of the Bonita Fault each of conjugate shear fracture is inself asmaller normal fault.

Shear fractures in and around anticlines and synclines may vary in orientation.

For some limestones in Morocco, de Sitter showed that the acute angle of the conjugate

shears faced parallel to the fold axes over anticlines yet faced perpendicular to the foldaxes over the synclines (Fig. 19-6).

(Fig. 19-6)

On the New York portion of the Appalachian Plateau Engelder has used extensionfractures to map the stress field during the Alleghanian Orogeny (Fig. 19-7). Here there

appear to be a conjugate set of shear fractures. Yet, surface morphology studies and butting relationships show that all fractures are extension in origin. These extension

fractures may be called cross-fold joints. The double pattern of cross-fold joints indicatesthat there were at least two phases of compression during the Alleghanian Orogeny.

Evidence that joint sets are an accurate indicator of the maximum principal stress

within the crust come from two areas. Volcanic dikes may be thought of as equivalent toextension fractures. In the Aleutian Peninsula of Alaska volcanic dikes parallel the

direction of subduction of the Pacific Plate under North America (Fig. 19-8). It is well

known that the orientation of maximum principal stress in the lithosphere is parallel tosubduction directions.

Dikes around the Spanish Peaks area of Colorado can also be used for stress

markers at the time of injection. Ordinarily the stress around a hole in an elastic platewould be radial. However, if there is a tectonic stress in the vicinity of the hole then the

stress trajectories is deflected away from this radial pattern. The Sangre de CristoMountains served as a ram forcing the ß ⁄ into the east-west direction. This can be seen by

the deflection of some of the dikes from the Spanish Peaks into the east-west orientation

(Fig. 19-9).



107

(Fig. 19-7)

(Fig. 19-8)



108

Fig. 19-9



109

Structural Geology

Lecture 20

Friction(The behavior of faults)

Rock friction is of interest because it controls many processes in the earth's crustincluding: flexural slip folding; earthquakes; landslides; subduction of continental crust;

strike-slip motion of the earth's major plate as they slide past each other. Ultimately it is

the frictional strength of the crust that controls the magnitude of differential stress withinthe earth's crust. Friction is an active mechanism of deformation above the brittle-ductile

transition within the earth's crust. It determines the level of shear stress (τ) required toinduce slip along any existing discontinuity.

The plate tectonics model suggests that large portions of the crust of the earth

(lithosphere) move over the mantle (asthenosphere). This process is believed to be driven

by the gravitational forces developed as thermal cooling of oceanic lithosphere takes place next to mid-oceanic ridges. Motion of the lithosphere is resisted by slip along

strike-slip plate boundaries and sbuduction zones. The collision of plates allows stress to

build up within the plates. The question of commonly asked, "What limits the buildup of differential stress in the lithosphere?".

In previous lectures we made the point that the earth's crust is pervaded by

discontinuities, most of which were mode I cracks (extension fractures, joints, tensioncracks, etc.). Because of these discontinuities a buildup in earth stress is controlled by

friction rather than the intact strength of rocks. In a sense, the generation of fractures,

particularly by shear failure, is a local phenomena occurring on relatively small scales.Measurement of in situ stress (i.e. earth stress) by techniques such as hydraulic fracturing

shows that the differential stress throughout much of the crust is less than or equal to thefrictional strength of local rocks.

The information that we have on the frictional strength of rocks comes from

laboratory tests such as those in Fig. 20-1. Samples A and B are cross sections of cylinders cut for triaxial testing. A is a sample sliding on a saw cut in a triaxial test. B is

a sample sliding on a previously induced shear fracure in a triaxial test. C is a

conventional shear test without confining pressure but with a normal stress indicated by

the vertical arrow. D is a double shear test with one plate sliding between two other

plates. E is a torsion test. In tests C and D the normal stress σn and the shear stress τ is

measured directly. In the triaxial tests σn and τ must be calculated from the principal

stresses σ1 and σ3. For the torsion test the τ must be integrated over the surface of thecylinder. The advantage of the torsion test over the other four is that infinite (very large)displacement may be achieved during a test. Large displacements permit the evolution of

steady state conditions along the sliding surface. In all cases friction, µ, for a single

experiment is a the ratio τ / σn.

µ = τ / σn. (20-1)



110

If σn and τ for several experiments are plotted the line connecting the data points may

converge at the origin. In this case µ is the slope of the line and friction has the samemeaning as for the single experiment. But if the line does not converge to the origin, it

will cut the τ axis at the value S0 which represents the cohesive strength of the fault zone.

Again µ is the slope of the line but

µ = (τ - S0)/ σn. (20-2)

Fig. 20-1

There is a distinction between sliding friction and internal friction that is best

illustrated in Mohr Space. Shear fractures do not form at 45° to σ1 and in the plane of

τmax but rather in a plane whose normal is closer to σ3 than σ1. The reason for this

behavior is found in a closer examination of the Coulomb criterion where

τ = S0 + µ0σn

where σn and τ the normal and shear stress on the plane of failure respectively and S0 is

the cohesive strength of the rock (Fig. 20-2). µ0 is the angle of internal friction

µ0 = tan φ



111

Tan φ can not be meassured directly but, rather, is derived from the slope of the Coulombfailure envelop. µ0 should be distinguished from the coefficient of sliding friction (µ)

which relates τ and σn during slip of a fault

µ = τ/σn.

(µ is the subject of this lecture).

µ0 predicts the angle of shear failure whereas the µ predicts the angle of frictional slip.

µ0 = (τ - S0)/σn.

e

e

(Fig. 20-2a)



112

(Fig.20-2b)

Using one of the five sample configurations shown in figure 20-1, force-

displacement curves are derived for various lithologies. Figure 20-3 illustrates the

various types of force-displacement curves. A shows the ideal curve for a sliding frictiontest. Here the force increases until the rock slips. Once slip is initiated the rock will

continue to slip without further increase in the force necessary to maintain slip. B

illustrates a more realistic force-displacement curve where there is a slight increase inforce necessary to maintain slip. Curves A and B are examples of a behavior called

stable sliding. C illustrates a situation where force builds and then suddenly drops. Thissame cycle is repeated many times throughout the experiment. D is the same behavior

measured at the sliding surface. This contrasts with C where displacment was measured

at some distance from the sliding surface. The displacement shown in C is a reflection of the elastic distortion of the load frame during force buildup. The rock slips

instantaneously with a major drop in force as shown in C. If displacment is measured at

the rock sample, it becomes obvious that the force drop accompanies the slip of the rock

sample. This behavior is known as stick-slip and is believed to be similar to earthquakesalong a major fault zone within the crust.



113

Fig. 20-3a

Fig. 20-3b

As we learned during Lecture 21, stick-slip occurs because unloading of the

elastic system which was responsible for the force buildup can not follow the force dropof the rock. Figure 20-3b illustrated this behavior. The straight line AB is the

uncontrolled unloading of the elastic load frame holding the sample (the earth on either

side of a fault zone). Here the earthquake (stick-slip event) starts at point A where the

shearing force along the fault zone drops faster than the unloading of the load frame.This creates excess potential energy which must be absorbed by the loading system. Slip

on the fault will stop when the available excess potential energy represented by theshaded area below the line AB is absorded by the extra work represented by the shaded

area above the line AB. At point B the load frame must reloaded (shearing stress

increased) to the point C before slip starts anew. Stick-slip starts again at point D.



114

Structural Geology

Lecture 21

Fault Rocks(The behavior of faults)

As we learned during Lecture 20, stick-slip occurs because unloading of theelastic system which was responsible for the force buildup can not follow the force drop

of the rock. Figure 20-3b illustrated this behavior. The straight line AB is the

uncontrolled unloading of the elastic load frame holding the sample (the earth on either side of a fault zone). Here the earthquake (stick-slip event) starts at point A where the

shearing force along the fault zone drops faster than the unloading of the load frame.

This creates excess potential energy which must be absorbed by the loading system. Slipon the fault will stop when the available excess potential energy represented by the

shaded area below the line AB is absorded by the extra work represented by the shaded

area above the line AB. At point B the load frame must reloaded (shearing stressincreased) to the point C before slip starts anew. Stick-slip starts again at point D.

Fig. 21-1

The simplest model for rock friction consists on an intact rock sliding on an

intact rock along a discontinuity which itself is clean. Because no surface is perfectlyflat, rough spots called asperities will catch on opposite surfaces (Fig. 21-1). These

asperities will dig into the opposite surface in the manner illustrated. The effect of this

digging is to slow the sliding process and maybe stop motion all together. Shear stress

then increases until the asperities are sheared off. Other asperities dig into the slidingsurface to slow and then stop slip. This is the physical model for stick-slip on a rock



115

surface. The same model has been applied to even the fault zones housing the largest

earthquakes.Brittle rock sliding against brittle rock can produce slickensided surfaces. Initial

wear is manifested by carrot-shaped wear grooves on the surface (Fig. 21-2). These wear

grooves are nothing more than the tool marks of asperities seen in plan view. The

grooves are most often seen when rocks of two different hardnesses slide against eachother. One of the best examples of this behavior is seen in the Jurassic chalks of southern

England where chert within the chalks acts as the asperities leaving carrot shaped wear grooves on fault surfaces of chalk. Mineralized fault zones in sandstones show the same

characteristic grooves where quartz grains are harder than the mineralization of the fault

zone.

Fig. 21-2

Once wear takes place clean rock surfaces become cluttered with fault gouge.

Such a fault gouge is shown in Figure 21-3. A brittle gouge will contain riedel shears

which point in the opposite direction of slip. If the wall rock on either side of the faultzone contains microcracks, they also will point in the direction opposite the slip.

Remember that microcracks form parallel with the maximum principal stress. Larger

fragments within the gouge will rotate in a direction compatible with slip.Brittle and ductile gouge zones should be distinguished. A ductile fault zone

shows structures with a sense of shear compatible with slip (Fig. 21-4). Ductile

structures include feaatures stretched in the direction of slip. Ductile fault zones mayoccur at deeper levels in the crust. However, salt is one lithology which will undoubtly

shown ductile behavior at very shallow levels within the crust.



116

Fig. 21-3

Fig. 21-4

Surfaces coated with fibrous calcite and quartz are not to be mistaken for

slickensided surfaces. The fibrous surfaces are commonly called slickelites and are the

product of very slow but stable sliding. The process of pressure solution occurs along thefault surfaces and dissolves asperities. The dissolved material is redeposited as long

fibers pointing in the direction of slip (Fig. 21-5).



117

Fig. 21-5

The nature of slip (i.e. stick-slip versus stable sliding) varies within the crust of

the earth depending on the temperature and pressure conditions. High pore pressures andhigh temperatures generally favor stable sliding over stick-slip (Fig. 21-6). This suggests

that earthquakes occur at shallow levels in the crust where there are no abnormal pore

pressures.

Fig. 21-6



118



119

STRUCTURAL GEOLOGY

Lecture 22

The Mechanics of Faulting(The strength of the Earth’s crust)

Faults and fault zones are mechanical systems which lend themselves to rigorous

analysis using both experimental and analytical approaches. Deformation textures andfault patterns observed within natural fault zones serve as precise guides to the selection of

appropriate constitutive equations for the prediction of the behavior of faults. These

constitutive relations are known through lab experiments and if properly scaled define thereaction of fault zones to various stress conditions.

The purpose of this lecture is to review for the structural geologist the analytical

approaches to understanding the mechanics of faulting. The observational andexperimental approaches were covered in the lectures on the Coulomb Failure Criterion,

The Griffith Failure Criterion, and Friction. The ingredients for an analysis of fault zones

using mechanics will include: (1) a statement concerning state of stress along fault zonesincluding evidence for magnitude and orientation along natural fault zones: (2) the yieldcriterion for slip on faults; (3) the mode of slip and nature of deformation mechanisms

accompanying and following yield; and (4) the energy expended during the faulting

process as inferred from both rock mechanics and seismology.Anderson (1905) recognized that principal stress orientations could vary among

geological provinces within the upper crust of the earth. Armed with this insight Anderson

was able to deduce the connection between three common fault types: normal, strike-slip,and thrust and the three principal stress systems arising as a consequence of the assumption

that one principal stress must be normal to the eart h's surface (Fig. 22-1).



120

(FIG. 22-1)

Anderson's classification of faulting shown in Figure 22-1shows principal stresses

written with subscripts 1,2, and 3. These subscripts immediately identify the relative

magnitude of the three principal stresses with the subscript 1 denoting the maximum principal stress. The principal stress may also be written in terms of a coordinate system

where the subscripts are x, y, and z with these subscripts representing the orientation of the principal stresses rather than their magniatude. The Anderson classification can be viewed

quantitatively assuming that the stresses in the x, y, and z directions are the principal

stresses. Note that in this coordinate system the vertical direction is indicated by y.

The state of stress within the crust of the earth may change according to localconditions. For example in a magma chamber where the rock is molten the the three

principal stresses are equal so that

σyy = σxx = σzz = Pm

where Pm is the pressure of the liquid magma. This state of stress is found in a liquid whereall three principal stresses increase with depth in the liquid. Immediately upon cooling

below the solidus the three principal stresses in the solid rock will also be equal. Any timethis situation occurs in solid rock the state of stress is said to be lithostatic. Such a stress

state is unusual in the crust of the earth.

Another unusual state of stress is one in which the horizontal components of stress

contain no contribution from a tectonic stress. In this case the horizontal components proportional to but less than the vertical component of stress. Using the theory of elasticity

this situation is represented by a uniaxial strain model (Fig. 22-2). This model is equivalent

to placing a rock in a box with four sides but open top and bottom. In such a model thelateral boundaries are assumed to be fixed so that they do not move regardless of the

magnitude of the push against them. The horizontal components of stress are themagntiude of the push against the fixed boundaries.



121

(Figure 22-2)

For the uniaxial strain model the overburden pressure is the vertical component of

the stress

σyy = ρgz

The vertical component of stress arises from the weight of the rocks above the point inquestion. Note that the natural tendency for a cube of rock without fixed boundaries

would be to expand in the horizontal directions when loaded top and bottom. This

expansion is known as the Poisson effect where the horizontal expansion is some fraction

of the vertical shortening. The following equation represents the Poisson effect:

υ = (horizontal expansion)/(vertical shortening) < 0.5

where υ is Poisson's ratio. In uniaxial strain the vertical load causes a horizontal pushwhich is represented by the equation

σxx = σyy = υ /(1- υ ) σzz = υ /(1- υ ) ρgz

For anaylsis of faulting a plane strain model is used where two sides of a cubic piece of earth are fixed (i.e. not allowed to move in or out) (Fig. 23-3). Plane strain differs

from uniaxial strain by the small number (2) of fixed sides. Plane strain is a reasonable

representation of faulting because two parallel sides of a cube will converge whereas the

second pair of sides remain fixed and the third pair of sides expand.

(Fig. 22-3)



122

For the plane strain model the first state of stress to consider is the stress developed

by overburden weight without any tectonic stress. The overburden pressure is the verticalcomponent of the stress

σzz = ρgz

Note that the tectonic stress in the vertical direction is always zero. Any deformation in thevertical direction is always taken up by a change in elevation to readjust so that only the

weight of overburden is pressing down. Horizontal stress in the y direction in the absence

of tectonic compression σt is

σyy = υρgz

where υ is the Poisson's ratio of the rock at the point in question. Note that the z-directionis normal to the fixed boundaries. This is the stress caused purely by the load due to

gravity. The stress in the x-direction is, of course, zero because there is no boundary

against which rock will push. Note here that the tectonic stress (σt ) is that stresscomponent which adds to the state of stress caused by a gravitational load. The following

analysis is a variation on an analysis presented in Turcotte and Schubert (1982).

For thrust faulting the horizontal compression in the x-direction must exceed the

vertical overburden pressure by an excess stress ∆σxx where

∆σxx > 0

(Fig. 23-4). For high horizontal stress

σxx

= υρgz + σt

= ρgz + ∆σxx

This analysis assumes a plane strain model where there is no strain in the y direction. So,

the excess stress in the y direction ∆σyy is

∆σyy = υ∆σxx

and the horizontal stess in the y direction is

σyy = ρgz + υ∆σxx



123

zz

zz

zz

yy

yy

yy

(Fig. 22-4)

Thus the horizontal stress in the y direction exceeds the overburden pressure but is less thanthe horizontal stress in the z direction by the value of the Poisson's ratio of the rock at the

point of consideration. Then thrust faults satisfy the sterss condition

σxx > σyy > σzz

For the case of normal faulting the overburden pressure is still

σzz = ρgy

and the excess stress in the horizontal direction is now negative (tensional) where

∆σxx < 0

and so the horizontal stress σxx is now smaller than the vertical stress σyy. Here

σxx = υρgy - σt = ρgy - ∆σxx

and

σyy = ρgy - υ∆σxx



124

Again using the plane strain assumption to calculate the σyy the relationship between the

three principal stresses is

σzz > σyy > σxx.

For strike-slip faulting the vertical overburden pressure remains the same as above but the excess stresses are positive (compressional) in one direction and negative

(tensional) in the other direction. Here two cases can occur

∆σyy > 0 and ∆σxx < 0

or

∆σyy < 0 and ∆σxx > 0

so that in the case of strike-slip faulting the overburden pressure is always the intermediatein value between the horizontal principal stresses.The Anderson theory of faulting was developed by combining the classification of

faulting as outlined above with the failure criterion for slip on a fault known as Amonton's

law for sliding friction. Amonton's law, a simple constitutive relationship for slip on faults,

states that the force parallel to a fault f τ necessary to initiate slip on that fault is

proportional to the normal force across the fault f n

f τ = µf n

where the proportionality constant µ is the coefficient of static friction. Dividing by the

area of contact we have

τ = µσn

where τ and σn are defined in the Mohr construction.

Using Amonton's law it is possible to calculate the dependence of the angle of dip β on the coefficient of friction µ for normal and thrust faults and to calculate the dependence

of deviatoric stress on the coefficient of friction for both normal and thrust faulting

(Turcottte and Schubert, 1982). For this analysis it is assumed that the vertical stress σzz is

the overburden pressure and the horizontal tectonic compression is the sum of the

overburden pressure and a deviatoric stress ∆σxx. σxx and σzz are related to τ and σn by

the equations for the Mohr construction

σn = 1/2( σxx + σzz) + 1/2(σxx - σzz)cos2γ

τ = -1/2(σxx - σzz) sin2γ



125

where γ is the angle of the fault with respect to the vertical. Upon substituting theequations for the vertical and horizontal stresses into the Mohr construction equations, the

normal and shear stresses on the faults are

σn = ρgy + 1/2(∆σxx)(1 + cos 2γ)

τ = -1/2(∆σxxsin 2γ)

In general faults contain ground water at hydrostatic or higher pressure. The pore water

pressure Pw affects the shear stress necessary to initiate slip by Amonton's law such that

τ = µ(σn - Pw)

Substituting into Amonton's law for the σn and τ for a stress system associated with either

normal or thrust faults

± 1/2∆σxxsin2γ = µ[ρgy + 1/2∆σxx(1 + cos 2γ)] - Pw

where the equation is positive for thrust faults and negative for normal faults. From above

comes the expression for the deviatoric stress int erms of the angle of fault with respect tothe vertical

∆σxx = [2µ(ρgy - Pw)]/[ ± sin2γ - µ(1 + cos2γ)]

It is assumed by many that the upper crust contains faults of many orientations so slip willoccur on that fault for which the ∆σxx is minimum. To calculate the angle of faulting γ

which minimizes the previous equation, we set

d∆σxx/dγ = 0

and find that

tan2γ = ± 1/µ

Figure 23-4 shows how the coefficient of friction and dip angle of both thrust and normalfaults aare related to minimize ∆σxx. The tectonic stress corresponding to these angles of

dip are found by substituting the previous equation into the equation for ∆σxx.

∆σxx = [ ±2µ(ρgy - Pw)]/(1 + µ2)1/2 ± µ

This relationship is also shown in Figure 23-5.



126

(Fig. 23-5)

If the behavior of the fault was dependent on just the constitutive relationship of thefault zone, then the fault would slip only when the yield criterion was exceeded. But,

because the fault zone is part of a system that includes a loading frame with an elasticity,

slip on the fault is controlled in part by the loading system. Within the earth the loadingsystem is taken to be the wall rocks next to the fault zone. As fault slip takes place the

loading frame relaxes; the rate of this relaxation will determine whether the fault will slip

continuously or by jerky steps. If slip is by jerky steps then the fault is known to haveearthquakes.



127

STRUCTURAL GEOLOGY

Lecture 23

The Geometry of Thrust Faulting(A look at Appalachian Tectonics)

Thrust faulting is characteristic of mountain belts marking the convergence of

lithospheric plates as they travel on the asthenosphere. The external portion of these beltsis called the foreland and the internal part is called the hinterland. Structures in

sedimentary cover characterize the foreland whereas structures in crystalline basement

characterize the hinterland.During the latter part of the semester we will draw heavily on the Geology of the

Appalachian Valley and Ridge and Appalachian Plateau to illustrate various types of

faults and folds. We start with a regional map of the Central and Southern Appalachians by noting that in the Central Appalachians the major structures are mapped as folds

whereas in the Southern Appalachians the major structures are mapped as thrust faults

(Figure 23-1).



128

Figure 23-1

Overthrust faulting is best introduced with a simple example, the Pine Mountain

thrust seen in the Southern Appalachians. This thrust sheet has developed as a single

detachment surface which cut up from a detachment in the Cambrian shales through to adetachment in the Carboniferous section. A cross section of the Pine Mountain Thrust is

shown in Figure 23-2. Terminology for this type of structure is given in Figure 23-3.



129

Figure 23-2: Pine Mountain Thrust system (after Davis and Reynolds, 1997)

Figure 23-3: Terminology for thrust fault systems (after Marshak and Mitra, 1985).

Multiple thrusts typically cut through a section as is seen in the Southern

Appalachian Mountains (Figure 23-1). These might stack in a ramp-flat geometryillustrated in Figure 23-4. The Valley and Ridge of the Southern Appalachians is an

example of this ramp-flat geometry after the upper portion of the section has been

removed by erosion.



130

Figure 23-4: Ramp-flat geometry of a typical regional thrust system (after Davis and

Reynolds, 1997).

The geometry of the Central Appalachians is characterized by a blind

autochthonous roof duplex (Figure 23-5). In effect, the duplex forms under asedimentary cover because the detachment fault has not cut to the surface.

Figure 23-5: In sequence thrusting under a cover (after Davis and Reynolds, 1997).



131

Figure 23-6: Two types of thrust systems: Top is an imbricate fan and bottom is aduplex (after McClay, 1992).



132

STRUCTURAL GEOLOGY

Lecture 24

The Geometry of Normal Faulting(A look at Rift Tectonics)

While thrust faulting is characteristic of convergent margins of continental crust,

normal faults are characteristic of divergent margins where rift basins have formed.These basins eventually grow to become filled with basalts and sheeted dike complexes

that are characteristic of oceanic crust. The passive margins of most continents are

bounded by rift basins of the type described in this lecture. Rifting is also found withincontinental crust that has grown warm and stretched without extending to the point that

oceanic crust has formed in the rifts.

A classic rift zone of relatively young age is the Red Sea rift (Figure 24-1). Across section through the Red Sea rift shows a series of listric normal faults dipping

toward the sea (Figure 24-2).

Figure 24-1: The Red Sea



133

Figure 24-2: Cross section through the Red Sea (after Lowell and Genick, 1972).

Once continents have rifted apart each continent is bounded by a series of half

grabens that dip oceanward. A classic example of this is found between the Texas Gulf

Coast and the Gulf of Mexico (Figure 24-3). Seismic sections show a very elaborate

network of normal faults. The master faults are listric in nature but secondary faults areantithetic with dips back toward the continent (Figure 24-4). The largest listric faults on

a continental margin may cut down to the Moho. In the Texas Gulf Coast many of the

listric faults cut down to a detachment at the level of a salt layer in the Jurassic (Figure24-5).

Figure 24-3:



134

Figure 24-4:

Figure 24-5:

The basin and range of the western United State is characterized by rift faulting

with the level of erosion reaching well into basement rocks. Core complexes are believed

to be a manifestation of listric faulting reaching mid- to lower crustal levels (Figure 24-6).



135

Figure 24-6

There have been a number of interesting models to explain the geometry of rifting

at the edge of continents. The most difficult problem involves providing an explanationfor the faulting that does not open large gaps in the crust. The listric fault geometry

accomplishes this purpose. Another model that seems to work in some cases is called the

domino model (Figure 24-7).



136

Figure 24-7

Very elegant clay models have been constructed that capture the shape of faulting

at continental margins. An example from Ken McClay is shown below.

Figure 24-8

Normal faults are three dimensional. The geometry of the tip zone of normalfaults is well displayed at places like Canyonlands National Park where the detachment

surface for extension of the local area is a Permian salt.



137

Figure 24-9.



138

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