17

THE CAUSES: MECHANICAL ASPECTS OF

DEFORMATION
Mechanics is the study of the effects of forces on bodies. A solid body subjected to external forces
tends to change its position or its displacement or its shape. During rigid body deformation, rocks
are translated and/or rotated while their original size and shapes are preserved.

If a body absorbs some or all the forces acting on it instead of being moved, the body becomes
stressed. The forces then cause particle displacements so that the body changes its shape: it
becomes deformed. Strain is the change in shape, or non-rigid body deformation, of a rock
caused by forces.

In the Earths crust, the most important of these forces are due to gravity and to the relative
motions of large rock masses in the crust and upper mantle. Other possible forces are usually small
or act only for short periods of time so that no significant strain results.
Movement-related forces act for relatively long times. Structural geology is concerned with the
permanent deformation (failure) that produces structures such as folds and faults in rocks. If a rock
fails by fracturing and loses cohesion, it is brittle. If the rock deforms without losing cohesion into
intricate shapes that are retained when forces stop acting, the rock has recorded a permanent strain
and has been ductile. The behaviour of the rocks, that is whether they deform permanently or not
and whether any deformation is predominantly by folding, by faulting, or by yet other modes, is
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influenced by the interplay of a number of physical and chemical factors. Thus, a thorough
understanding of the deformation process is important.

The main purpose of this lecture is to examine these factors in order to gain a physical insight into
the manner in which rocks deform in nature. Through these means, the propensity for rock
deformation may be estimated using easily measured material properties such as the flow stress as a
function of strain, strain rate and temperature, and applied to a large number of rocks. The
description of any deformation process involves specification of loads applied, which is the goal of
a dynamic analysis. This discussion defines the concepts of stress, strain, rheology, and equations
of motion. As geologists, we cannot see stresses directly; we can only infer them from the results of
deformation. We will begin with this topic, and introduce the vector calculus it involves.

Physical definitions
Continuous medium
Rocks are complex assemblages of crystals, grains, fluids etc. whose properties and physical
parameters vary continuously and have spatial derivatives. It is therefore necessary to consider
infinitely small volumes of material in which physical properties are the same everywhere. This is a
continuous medium that models real materials without considering their fine (e.g. atomic) structure.
The mechanical discussion that follows considers rocks as continuous media.

Newtons axioms: Laws of motion

The Newtons axioms state the moving conditions of a body in response to an external quantity, the
force, and a characteristic of the body: its mass (i.e. the amount of material in the body). In
dynamics, only the driving forces of the movements are considered. Since deformation is the
relative movement of points, Newton's three laws of motion are taken as fundamental axioms.
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Law 1, (inertia principle)
A body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to
change that state by forces impressed upon it.
The rate of such a free-moving body is constant in terms of magnitude and direction.
Law 2, (action principle)
The change of motion is proportional to the force impressed; and is in the same direction as the line
of the impressed force.
Law 3, (reaction principle)
To every action F there is always opposed an equal reaction F R = F ; or, the mutual actions of
two bodies on each other are always equal and directed to opposite parts.
For example, a falling rock pushes the earth as much as the earth pushes the rock.

Dimension / Quantity
Mechanical properties of a material are expressed in terms of the three independent, physical
dimensions (i.e. measurable parameters) length [L], mass [M], and time [T], [ ] meaning has the
quantity of. Other dimensions, such as electrical charge [Q] and temperature [], are derived
dimensions.
A quantity is the numerically scaled magnitude of a physical dimension. Quantities are
conventionally expressed in the Systme international d'units (SI units). These units are meter (m),
kilogram (kg), and second (s) for length, mass and time, respectively.

Force
A force is what influences, or tends to change, the motion of a body.

Mathematical expression

A force possesses both magnitude and direction. Therefore, a force F is a vector quantity that
follows the rules of vector algebra. Conventionally, an arrow in a given coordinate system
represents it.
- The length of the line specifies the amount of the force (e.g. how strong a push is).
- The orientation of the line specifies its direction of action (which way the push is).
- An arrow, pointing to the acceleration direction indicates the sense of direction.
The action principle states that a force F acting on a body of mass m will accelerate the body in the

direction of the force. The acceleration a is inversely proportional to the mass m and directly
proportional to the acting force:

a = F m F= m.a

A force possesses both magnitude and direction. Therefore, a force F is a vector quantity that
follows the rules of vector algebra. This relationship is written:

mv d mv
=
F =
t
dt

where mv is the product of mass and velocity, i.e. the momentum, and t is time.

( )

The most familiar force known to us is the weight, which is per definition the force experienced by
a mass (the product of volume and density) in the direction of gravitys acceleration and hence
normal to the Earths surface.
As any vector quantity, a force may be resolved into several components acting in different

directions, according to the parallelogram rule of vector analysis. For example, any force F can be
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resolved in three components labelled Fx , Fy and Fz , parallel to the coordinate axes x, y and z,
respectively. This is conveniently written in a column form:
Fx
F = Fy
Fz

Orientation
To define orientations one takes from the vector analysis that a plane can be defined by two vectors.
This first definition is extremely useful in 3D applications since the product of these two vectors is
a vector perpendicular to the other two vectors, i.e. perpendicular to the plane that contains them.
The vector product is also a unit vector (its magnitude is 1). The practical application is the standard

Cartesian coordinate system, for which i is the unit vector along the x axis (thus orthogonal to the

yz plane), j is the unit vector aligned with the y axis (perpendicular to the xz plane), and k is the
unit vector along the z axis (perpendicular to the xy plane). Expressing fully the force vector is
therefore:
F
1
0
0

F = Fy = Fx .i + Fy . j + Fz .k = Fx . 0 + Fy . 1 + Fz . 0
Fz
0
0
1

or

F
x

F = Fy = Fx .ex + Fy .e y + Fz .ez
F
z

with e i the unit vector along the i -axis.

When one writes the three vector components, i.e. how long each vector is along each of the 3 axes,

one simply omit the unit vectors and coordinates are simplified to the coefficients of the i , j and

k parts of the equation. The vector entity, defined by three numbers and a coordinate system, is
mathematically described as a first-order tensor.
Forces and force components are added as vectors.

Dimension

The unit and dimension of a force are defined from the second Newtons law: F = m.a . The
dimension has the form:

[ F] = M.L.T 2
The mass is a scalar quantity, i.e. it requires only one number to define it. Its unit is the kilogram
(1kg). Mathematically, a scalar is an entity also called a zero-order tensor.
Acceleration needs a coordinate system to be defined.
The mass is a scalar quantity, i.e. it requires only one number to define it. Its unit is the kilogram
(1kg). Mathematically, a scalar is an entity also called a zero-order tensor. Acceleration needs a
coordinate system to be defined.
The Newton and the dyne are the basic units of force (1 N = force required to impart an acceleration
of 1 m/s-2 to a body of 1 kg:
2
=
1N 1kg.1m.1s
=
105 dynes

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Surface - body forces

Body forces
Body forces result from action of distant, outside forces (gravity, electromagnetic field, etc) on
every particle of the body; for example, gravity acts on every atoms of a pen, producing its weight.
Body forces are consequently proportional to the mass, and hence to the volume of the body.

In purely mechanical systems the body forces are of two kinds, those due to gravity and those due
to inertia.
Surface forces
Surface forces (or applied forces) act on the external boundaries of a body. For example: the action
to push on its extremity and displace a pen. Surface forces, such as friction, are consequently
proportional to the size of the area upon which they act. Surface forces can act on imaginary or real
surfaces within the body, and can result from action of the body on itself, such as the tension in a
stretched rubber band. No concrete physical surface or visible material boundary is required.
Surface forces are usually generated beyond the considered body and are transmitted to it through
the whole mechanically continuous region that connects it with the place where the force is exerted.
For example, tectonic forces can be transmitted through the plate from its boundaries.
Ratio of the body forces to the surface forces
Since gravitational forces are proportional to mass, the weight of an overlying column of rocks
constitutes a significant force on rocks at depth in the crust.
In general, each element of mass is in a state of dynamic equilibrium, which means that the sum of
body forces is equal and opposite to the sum of surface forces. If d is the characteristic length of a
small element of the body, we can write a ratio of the body forces to the surface forces:

(d )
Bo d y fo rces
=K
S u rface fo rces
( d )2
3

which tends to zero as d tends to zero. In other words the magnitude of body forces diminishes
more rapidly than the surface forces. Consequently, if the element of volume is small, we may
neglect the body forces in equilibrium with themselves. The scaling between body and surface
forces has, for example, important consequences in biological engineering; the strength of a bone is
proportional to its cross-sectional area but the weight of body is proportional to its volume. The
bones of larger animals have, therefore, greater diameter/length ratios to hold relatively bigger
weights. Such scaling relations also control mechanical deformation processes. For example,
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George Gabriel Stokes derived in 1851 a solution for the velocity of a sphere falling (or rising) in a
viscous fluid with different density. This velocity increases with the square of the radius of the
sphere because the buoyancy forces increase relatively more than the friction forces acting on the
sphere surface against its motion. Similar relationships between buoyancy and friction forces have
important tectonic implications for the rise of diapirs or the sinking of lithospheric plates into the
mantle.

Directed forces
Directed forces act in particular directions. In geology:
- Compression is a pair of in-line forces that tends to compress bodies;
- Tension is a pair of in-line forces that tends to pull bodies apart;
- Shear refers to coupled forces acting in opposite directions in the same plane but not along the
same line;
- Torsion is a twisting force.

Normal and shear components

A force acting on a plane is generally oblique to the surface and may be resolved into vector
components acting perpendicular and parallel to the plane.

F = FN +FS

FN and FS are the normal and shear forces, respectively.

In two dimensions, F , FN and F S are coplanar; the two perpendicular components are defined
according to the right angle trigonometry as:
=
FN Fsin

=
FS Fcos
with the angle between the applied force and the normal to the considered plane (line in 2D). The
magnitude is obtained using the Pythagoras Theorem:
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2
= FN

+ FS2

These equations show that the key to find the component magnitudes is to know (i) the magnitude
of the applied force vector and (ii) the angle it makes with the plane.

Action / reaction; static equilibrium

Imagine a cube of rock within a large volume of rock assumed to be a continuous material. The six
faces of the imagined cube are pressed on by adjacent parts of the rock while there are
corresponding reactions from the material within the cube. Newtons reaction principle stating that
forces occur in pairs that are equal in magnitude but opposite in direction expresses this situation. In
addition, each atom within the cube is acted on by gravity, but each atom outside the cube also.
Therefore, the general body force which is equal everywhere can be, in a first approach, considered
to be absent.

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In static equilibrium, the considered cube of rock is not moving and not deforming. The system of
forces is closed and the sum of all forces in all directions equals zero. Static equilibrium is the
situation treated to understand natural geological forces.
In that case, forces on opposite faces cancel out and there is no net couple that would rotate the
cube. This again requires that forces on opposite faces be equal in magnitude and opposite in sense.
The shear forces on opposite faces (two on each face) must also be balanced. To simplify the
argument, we take the cube edges as the principal axes of a three-dimensional coordinate system.
Then the shear component is resolved into two shear-components parallel to the face edges.

Stress in a continuous medium

A stress is what tends to deform a body.

Definition
The magnitudes of the forces that act on the external faces of the cube depend on the areas of these
faces: the larger the cube, the larger the force required to produce a change of shape or a movement.
The situation is complicated by variations in magnitude and direction of force from point to point
over each cube face. Therefore, it is convenient to have a measure of the deforming forces that is
independent of the size of the cube considered. This freedom in calculation is procured by
imagining the cube to shrink to a cubic point whose infinitely small faces have area A = 1.
The significance of the area on which a force is applied is intuitively known to all of us. The feet
sink when walking on snow, less if one has snowshoes, and one can even slip on skis. The force
(weight of the person) acting on snow is the same but increasing the contact area reduces the stress
on snow. This shows that the stress, and not the force, controls the deformation of materials (here
snow). Therefore, one needs to work with stress in order to investigate the deformation of rocks.
Traction
The traction T is formally defined as a force (F) per unit area applied in a particular direction at
a given location on a body surface. We will begin with the imaginary unit cubic point within a
body:

t=F A

A more precise definition of the traction at a point is given by the limiting ratio of force F to the
area A as the face area is allowed to shrink and approaches zero (Cauchys principle).
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F dF
=
t lim
=

A 0 A dA

25

In this equation, F is a vector quantity defined by its magnitude and direction. The traction is also
a vector defined by three quantities:
- its magnitude;
- its orientation;

- the orientation of the plane on which it is applied, which is defined by the normal unit vector n .

F = t .A.n
This definition contains two directional components: one for force and one for plane orientation. It
indicates that traction is a bound vector that may vary from point to point on any given plane, and
be infinitely different on the infinite number of planes that pass through any given point. Traction is
therefore always expressed with reference to a particular plane.
Stress
A traction applied to the external surface of a body sets up internal tractions within the body, which
is then in a state of stress. The same equation as for external traction defines the internal traction at
a point within the body, the stress. Stress is applied on any point of this body like spring tension:
there are equal and opposite forces on the other (hidden) three faces of the cube, with a balanced
action and reaction between adjacent parts. The stress comprises both the action and the reaction. A
stress is therefore a pair of equal and opposite forces acting on the unit area. It is transmitted
through the material by the interatomic force field.

Dimension
Stress, as pressure, includes the physical dimensions of force and those of the area on which the
force is applied:
[M*LT-2]/[L2] = [Mass * Length-1* Time-2]
The unit is the Pascal (1 Pa = 1 Newton.m-2, remembering that 1N = 1 kilogram meter per square
second: 1 kg.m-1.s-2) and Bars with 1 Bar = 1b = 105 Pa ~ 1 Atmosphere. Geologists more
commonly use 1Megapascal = 106 Pa = 1 MPa. A useful number to remember for discussion with
metamorphic petrologists in particular, is 1 kb = 100 MPa. 100 MPa is approximately the lithostatic
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pressure at the bottom of a rock column of 4 km height and with a density of 2600 kg.m-3 (2600
kg.m-3 x 4000 m x 9.81 m.s-2 = 102 MPa, see section Terminology for state of stress).

Stress components
With an infinitely small cube, body forces are negligible compared to surface forces. Hence, body
forces are in equilibrium with themselves and one can consider the state of stress at a point (the
infinitely small cube) within the body. Since stress cannot be defined without specifying the plane
upon which the stress acts, both the direction of the force and the orientation of the faces of the
cube must be considered.
Forces (and traction vectors) on each of a cube faces are resolved into three orthogonal components,
one normal to the face (the normal force) and two parallel to the face (the shear forces). Like forces,
stresses acting on an infinitely small cube whose faces are unit areas can be resolved into three
normal stresses perpendicular to the faces and three times two shear stresses parallel to each face,
each shear stress parallel to each of the coordinate directions contained in the face plane.
- The normal stress, transmitted perpendicular to a surface, is given the symbol .
- The shear stresses, transmitted parallel to a surface, have the symbol but is common
notation in the literature.

Exercise; graphic representation to be done with Excel

* Draw a square ABCD and a diagonal surface on it.
* Draw a vertical force F1 that acts on this surface.
* Write equations that express the normal and shear components on this surface.
* Show that the highest shear stress is obtained for an angle = 45 between the
surface and F1 . * Represent graphically variations of the normal force FN and the
shear force FS on the surface as a function of the angle .
Stress at a point in a continuous medium
The state of stress at a point is three-dimensional. It is convenient to use the edges of the
infinitesimally small cube (the shape of the point) as a system of Cartesian coordinates
( x1, x 2 , x3 ) .

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Employing the symbol i j to denote the component of stress that acts on the pair of faces normal to
x i (thus identifying the plane orientation) and in the direction of x j (thus defining the direction of

traction), we resolve the stresses that act on the faces of the cube normal to x1 into:
11 the normal stress component, perpendicular to the faces normal to x1 (or x).
12 and 13 the two shear components within the paired faces normal to x1 , each acting parallel to
one of the other coordinates axes x 2 and x 3 (or y and z), respectively.
For each pair of faces there is one face for which the inward directed normal stress, taken here as
positive, is opposite to the normal stress acting on the other face. The same procedure applies for
the faces normal to x 2 and x 3 (or y and z), so that for the three pair of faces we obtain a total of
nine stress components:
Pair of faces normal to x1 :
Pair of face normal to x 2 :

11
22

12
21

13
23

Pair of face normal to x 3 :

33

31

32

or written as

xx

xy

xz

yy

yx

yz

zz

zx

zy

These are written so that components in a row are those acting on a plane and components in a
column are those acting in the same direction. Using the symbol instead of yields the
following ordered array:
11 12 13

21 22 23
31 32 33
This geometrical arrangement represents the original set of coefficients that form the stress matrix:
11 12

21 22
31 32

13
23
33

(1)

A matrix that has the same number of rows and columns is a square matrix. One may collectively
call this matrix of coefficients or ij , identifying its elements in a simple form as:
11 12

ij = 21 22
31 32

13
23
33

This grouping of the nine stress components is the stress tensor.

Reminder: Mathematical definitions; what are we talking about?
Scalar: a quantity with magnitude only (i.e. a real number, such as for mass,
temperature, time).
Vector: a geometrical object with magnitude and one direction (e.g. force, velocity,
acceleration).
Tensor: a mathematical structure with magnitude and two directions (two vectors), one
(a unit vector) specifying a plane of action (e.g. permeability, strain, stress).
The stress tensor, which represents all possible traction vectors at a point with no dependence on the
plane (unit normal vector) orientation, fully describes the state of stress at a point. More
specifically, it is a symmetric tensor since the six off-diagonal components are interchangeable; it is
a second order tensor since it is associated with two directions. Accordingly, stress components
have 2 subscripts, where indifferently and independently i = 1, 2, 3 and j = 1, 2, 3. The subscripts i
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and j refer to the row and column location of the element, respectively. The diagonal components
i= j are the normal stresses and the off-diagonal components i j are the shear stresses.
If the elemental cube does not rotate (i.e. postulating equilibrium condition and no body forces),
shear stresses on mutually perpendicular planes of the cube are equal: three of the shear
components counteract and balance the other three, i.e. the rotating moments about each of the
axes, the torques read across the diagonal of the square matrix, are zero:

12 = 21
23 = 32
31 = 13
Reminder: Torque is the product of the force vector and the perpendicular distance between the
center of mass and the force.
Since ij = ji (i.e. subscripts for shear stress magnitudes are commutative), the symmetrical stress
matrix is:
11
ij = 12
13

12
22
23

13
23
33

Its nine components reduce to six truly independent stress components acting on any arbitrary
infinitesimal element in a stressed body:

11
12

normal stresses
shear stresses

22
23

33
31

Therefore, for an arbitrarily chosen set of orthogonal axes x, y and z six independent quantities are
necessary to specify completely the state of stress at a point, i.e. for every surface element leading
through the point.
The cube representation helps emphasising an important difference between stress and forces. A
directed force may be acting in a certain direction (say towards the left) but this statement has no
sense when applied to internal stresses. A stress component acting upon one side of a surface
element exists only together with a component of equal intensity but opposite direction, acting on
the other side. This is true for normal as well as shear stresses. Hence a stress may exist in a vertical
direction, but not up- or downwards.

Principal stresses
Even if six independent stress magnitudes and unconstrained orientations simplify the stress tensor,
the formulation remains somewhat cumbersome to employ. Fortunately, this situation can be
considerably simplified. It is always possible, at any point in a homogeneous stress field, to find
three mutually orthogonal planes intersecting at the point and oriented such that the shear stresses
vanish to zero. Thus:

12 =
23 =
31 =
0
In this case there remain only the normal components of stress and:
11 12

21 22
31 32

13
23
33

becomes

0
11
0
22

0
0

0
0
33

These three no-shear-stress planes are the principal planes of stress and intersect in three mutually
perpendicular lines known as the principal axes of stress at the point considered. The stresses
acting along these three axes are the principal stresses 11 , 22 and 33 denoted 1 , 2 and 3
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to avoid repetitive subscripts, with the convention that 1 2 3, the maximum, intermediate,
and minimum principal stresses, respectively. In other words, the principal stresses are the normal
stresses that act on planes of zero shear stresses. They coincide with the principal axes of the stress
ellipsoid, which we will define below.
Attention! Sign convention: In physics and in engineering, tensile normal stress that
tends to pull material particles apart is considered positive; the compressive normal
stress, which tends to push the material particles together is negative. In geosciences, it
is customary to make compression positive and tension negative because natural
stresses are always compressional, even in areas in extension. For example, in a nontectonic environment, the stress at any depth within the Earth is generated by the
overburden. It is a compressive vertical stress that induces a compressive horizontal
stress. Even at the Earth surface, the maximum compressive stress is equal to the
atmospheric pressure. Shear stresses are positive anticlockwise.
If the magnitudes and orientations of the three principal stresses at any point are known, the
components of normal and shear stress on any plane through that point can be computed. The state
of stress at a point may, therefore, be completely characterised by giving the size of these three
principal stresses and their directions. The six independent stress components are needed only when
the faces of the reference cube are not parallel to the principal planes of stress.

Terminology for states of stress

Some special stress states are:
1 = 2 = 0 ;
3 < 0
2 = 3 = 0;
1 > 0

Uniaxial tension
Uniaxial compression
Biaxial (plane) stress
2 = 0
1 > 2 > 3
General, triaxial stress
1 = 2 = 3 = p
hydrostatic state of stress; all shear stresses are zero.
If p < 0 (tensile) the stress state is referred to as a hydrostatic tension. Hydrostatic stresses will
cause volume changes but not shape changes in a material.
In geology the lithostatic pressure is often used to describe the hydrostatic pressure generated at a
depth z below the ground surface due solely to the weight of rocks, of mean density , in the
column. Naturally, this is equal to gz where g is the acceleration due to gravity. Such a statement,
however, requires some qualification because it assumes that the stress state at depth z has become
truly hydrostatic due to relaxation of all shearing stresses by some creep process. If the stress state
has not been allowed to become hydrostatic, and we are talking about the stress state due solely to a
pile of rocks of height z, then this is usually taken to be:
z

1 g.dz
0

2 = 3 = (1 ) 1

where is Poisson's ratio.

Mean stress

The mean stress or hydrostatic stress component p (also called dynamic pressure) is the
arithmetic average of the principal stresses:
= p = ( 1 + 2 + 3 ) 3 = ii 3
This pressure is stress invariant and is independent of coordinate system; it is induced by the weight
of overburden (the confining pressure) and characterised by equal stress magnitudes in all
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directions. A typical value is ca. 30 MPa/km (ca 3kbar/10km). The mean stress thus specifies the
average level of normal stress acting on all potential fault planes, which governs the frictional
resistance on fault planes to slip. Otherwise, the mean stress may only produce a change in volume,
either to reduce it if the mean stress is compressive, or to expand it if it is tensile.

Deviatoric stress
It is difficult to measure changes in volume in rocks. Observable strain results from distortion.
Therefore, one defines the deviatoric stress by subtracting the pressure: Any state of stress can be
considered as the sum of the hydrostatic mean stress p and a deviatoric stress:
1 12 13 p 0 0 s1 12

23 0 p 0 + 21 s 2
21 2=
31 32 3 0 0 p 31 32

13
23
s3

Since p is the mean stress, s1 + s 2 + s3 =

0 . The second matrix on the right-hand side is the stress
deviator. Its components are the deviatoric stresses. The principal deviatoric stresses are the
amounts by which each of the principal stresses differs from the mean stress. They define the
effective shear stress, which measures the intensity of the deviator:
2

12

1 2
2
2
=
eff sij=
sij
s1 + s 22 + s32 + 223 + 31
+ 12

The decomposition into a deviatoric stress sij and a volumetric stress ij , utilizing the standard
Kronecker delta, is written:
ij = sij + ij
and the normal stress relative to the mean stress is then described by the deviatoric stress:
sij = ij ij
The Kronecker delta is a mathematical function defined as
0 for i j
ij
1 for i = j
That is,
11 = 22 = 33 = 1
12 = 13 = 21 = 23 = 31 = 32 = 0

under another form the matrix:

11 12

21 22
31 32

13
23
33

is the identity matrix

1 0 0
0 1 0

0 0 1

In simpler words, where 1 2 3 we can think of the rock being affected by two components:
- the mean stress p =

( 1 + 2 + 3 )

- and three deviatoric stresses :

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3.
s1 = 1 p
s2 = 2 p
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31

s3 = 3 p

All shear stresses are deviatoric.

The main deviatoric stress s1 is always positive and the smallest one s3 is negative (with
compression positive); the intermediate deviatoric stress is nearly equal to the mean stress. The
positive deviatoric stress tends to shorten the rock in the direction of its action, while relative
lengthening is easiest in the direction of the negative (tensional) deviator.
As a corollary, only the deviatoric stresses leave permanent deformation in rocks.

Differential stress

The differential stress d is the difference between the largest and the least principal stresses:
d = ( 1 - 3 )
Its value, along with the characteristics of the deviatoric stresses, influences the amount and type of
deformation experienced by a body. Note that differential stress is a scalar.

Stress acting on a given plane

In the following demonstration, it is important to remember that the value of a stress not only varies
with the orientation and magnitude of the imposed force but varies also with the orientation and size
of the area of action.
We have already seen that a force F that acts on a real or imaginary plane P can be resolved into
components normal ( FN ) and parallel ( FS ) to the plane P. The components have magnitudes:

=
FN Fcos

and

=
FS Fsin

(2)

respectively.
We further consider that the cubic point previously used belongs to the plane P.
F is oriented to act normally to one cube face, for convenience vertical F on the top face of the
cubic point. F is contained in the square, vertical section orthogonal to P through the cube. In this
section, faces with unit area A are reduced to unit segment lengths.
By definition, the stress is the concentration of force per unit area, which can be visualised as
intensity of force. The stress on the cube face has the magnitude:
Stress = Force / (Area of the cube face)
=F A
Cutting across the cube is another plane, P, whose normal is inclined at angle to F. The area of the
plane P is:
P(Area) = Cube face(Area) / cos
AP = A cos
(3)
Hence, the normal components of force and stress acting on plane P are:

FN= F cos = A cos = A P cos 2

and the shear components:

(4)

FS= Fsin = A sin = A P sin cos

The general trigonometry states that :

sin cos= ( sin2 ) 2

The cube is such that Ap = unit area = 1. Thus, the magnitudes of the normal and shear components
of stress across P are
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N = FN A P = ( F A ) cos 2 = cos 2
and

(5)

S = FS A P = ( F A ) sin cos = ( 2 ) sin 2

Comparison of equations (2) and (5) shows that stresses cannot be resolved by the use of vectors as
though they were forces.
Typically, any rock is under a triaxial state of stress; 1 , 2 and 3 are the principal stresses with
1 2 3.
Remember! The convention in geology takes all positive stresses as compressive. In the
non-geological literature stresses are considered positive in extension!
For practical purposes we can take the arbitrary plane P parallel to 2 , in turn parallel to the
horizontal x-axis of the Cartesian coordinates. The angle between the line normal to P and the
vertical 1 (parallel to the coordinate z-axis) is also the angle between the plane P and 3 . We
approach the problem by considering a two-dimensional state of stress. For this simplification, we
only consider the two-dimensional principal plane (1, 3 ) while ignoring 2 , orthogonal to this
slicing plane. This simplification is consistent with the statement that it is the difference between
1 , and 3 that rules deformation while 2 does little and can, as a first approximation, be
discarded. We also consider that all lines in the

( 1, 3 )

plane represent traces of planes

perpendicular to it, thus parallel to 2 .

From equations (5) that stress components due to 1 are:
1N = 1 cos2
1S = 1 sin cos

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3 is orthogonal to 1 . We can use the same trigonometric construction to resolve stress
components due to 3 as:
3N = 3 sin 2
3S = 3 sin cos

Where the principal stresses are 1 and 3 the equations for the normal and shear stresses across a
plane whose normal is inclined at to 1 are:
2
2
N = 1 cos + 3 sin

=
S sin cos ( 1 3 )

From the general trigonometry we know that:

cos 2 + 1
2
1

cos
2
sin 2 =
2

cos 2 =

which we can substitute in the previous equation to write the normal stress component:
cos 2 + 1
1 cos 2
N = 1
+ 3

2
2

and simplify to:

=
N

1 + 3 cos 2 ( 1 3 )
+
2
2

and we can write the shear stress component as:

=
S

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1
sin 2 3 sin 2
2
2

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=
S

1
sin 2 ( 1 3 )
2

Where the principal stresses are 1 and 3 the equations for the normal and shear stresses across a
plane whose normal is inclined at to 1 are:

=
N

( 1 + 3 ) + ( 1 3 ) cos 2
2

( ) sin 2
S = 1 3
2

(6)

respectively.
Note that (6) reduces to (5) when 3 is zero. These relations are extensively used in geological
studies because 1 and 3 are often close to the horizontal and vertical (lithostatic) tectonic
stresses.
These equations demonstrate that the value of S in (6) is maximum when sin2 = 1 i.e. 2 = 90 .
Thus the planes of maximum shear stress make an angle of 45 with 1 and 3 .
In all cases where 1 > 2 > 3 the planes of maximum shear stress are only two in number and
intersect along 2 . Indeed, it has been observed in triaxial tests ( 1 , 2 and 3 have non zero
magnitudes) that shear fractures form angles close to 45 to the principal stress axis 1 . The paired
faults, called conjugate faults, develop more or less synchronously in both of the equally favoured
orientations. Remember that conjugate faults intersect in a line parallel to the intermediate principal
stress axis 2 . Normal compressive stresses on these planes tend to inhibit sliding along this plane;
shear stresses on these planes tend to promote sliding.
In the special situation where 2 = 3 or 1 = 2 there are an infinite number of such planes
inclined at 45 to 1 or to 3 , respectively.
In all cases the maximum shear stress has the value ( 1 3 ) / 2 .

Equation (6) also implies that for any arbitrary state of stress with 1 and 3 , there are surfaces on
which no shear forces are exerted. We will use this property to define directions of principal
stresses.

Relationship between normal stress and shear stress: Mohr circle

The principal stresses are those that are orthogonal to the three mutually orthogonal planes on
which shear stresses vanish to zero. Between these special orientations, the normal and shear
stresses vary smoothly with respect to the rotation angle . How are the normal and shear stress
components associated with direction?

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Analytical demonstration

Rearranging N of equations (6) and squaring both equations we get:

N (1 2 ) ( 1 + 3=
) (1 2 ) ( 1 3 ) cos2 2
2

(7)

(1 2 ) ( 1 3 ) sin 2
2

=
S2

We can add both equations (7) to write

N (1 2 ) ( 1 + 3 ) =
+ S2 (1 2 ) ( 1 3 )
2

( cos2 2 + sin 2 2)

Since cos 2 + sin 2 =

1 , for any angle:

N (1 2 ) ( 1 + 3 ) =
+ S2 (1 2 ) ( 1 3 )
2

in which we recognise the form of the standard equation of a circle in the coordinate plane (x,y)
with centre at (h,k) and radius r :

r2
( x h )2 + ( y k )2 =
1 3
2

1 + 3
The centre of the stress-circle on the axis is at:
2

Cyclic interchange of the subscripts generates two other circles for the other two principal stress
differences, ( 2 3 ) and ( 1 2 ) .

The radius of the stress-circle is:

Graphical construction
We see that equations (6) describe a circular locus of paired values N and S (the normal and
shear stresses, respectively) that operate on planes of any orientation within a body subjected to
known values of 1 and 3 . This leads to the two-dimensional representation of stress equations
known as the Mohr diagram.

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Stress in two dimensions (plane stress)
The construction of the Mohr stress circle proceeds as follows:
For a two-dimensional stress, the normal and shear stresses can be plotted along two orthogonal,
scaled coordinate axes, with normal stresses along the abscissa (or horizontal x axis), and shear
stresses along the ordinates (or vertical y axis). These axes have no geographic orientation but
have positive and negative directions. By convention, the right half of the diagram is positive for
normal stresses. Shear stresses that have an anticlockwise sense (consistent with the trigonometric
sense) are considered positive and are plotted above the abscissa axis.

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Knowing the magnitude of the maximum principal stresses 1 and 3 of a given a state of stress
acting on a plane making an angle with the 3 direction, one plots both 1 and 3 along the
abscissa, at a distance from the origin equal to their numerical values. Through points 1 and 3 a
circle is constructed with diameter ( 1 - 3 ) and centre at C= ( 1 + 3 ) 2 .
The maximum principal stress 1 is at the right extremity of the circle, the least principal stress
value 3 is the left extremity of the circle. All the paired values of N and S exist at some point
on the perimeter of the circle. Then any point P on the circle has coordinates ( N , S ) where N
and S are given by (6) and 2 is the angle between the N axis and the line PC measured in the

anticlockwise (trigonometric) sense from the right-hand end of the N axis. The coordinates of any
point P on the circle give the normal stress N (read along the abscissa) and shear stress S (read
along the ordinates) across a plane whose normal (ATTENTION : not the plane itself) is inclined at
to 1 . For simple geometrical reasons, is also the angle between the fault plane and the least
stress 3 . This construction may also be used to find 1 , 3 , and given N and S on two
planes at right angles. Since angles are doubled in this graphical format, the point representing the
plane orthogonal to P is diametrically opposed to the P-point.
The Mohr diagram thus allows the magnitudes of stresses on variously oriented planes to be plotted
together. It neatly shows that the points on the circle (hence attitude of planes) along which shear
stress is greatest correspond to values of = 4 5 .
*-* Note that the maximum stress difference ( 1 - 3 ) determines the value of greatest
simply because the differential stress is the diameter of the Mohr circle, which is
twice the vertical radius, the maximum shear stress.

Stress in three dimensions

The Mohr construction may also be applied to three-dimensional stress states. In this case, the
diagram has three circles: two small circles ( 1, 2 ) and ( 2 , 3 ) are tangent at 2 and lay within

the larger ( 1, 3 ) circle.

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The three diagonal components 1 , 2 and 3 of the tensor are normal stresses plotted along the
horizontal axis; non-diagonal components are shear stresses plotted along the vertical axis. All
possible ( N , S ) points plot on the large ( 1, 3 ) Mohr circle or between this circle and the
and ( 2 , 3 ) Mohr circles. The diagram area between these three circles is the locus of
stresses on planes of all orientations in three dimensions.

( 1, 2 )

Exercise
Draw Mohr circles of the following states of stress:
Hydrostatic, uniaxial, triaxial
Mohr diagrams are used extensively in discussions of the fracturing of rock masses because they
graphically represent the variation of stress with direction and allow finding the stresses on known
weak planes.

Effects of pore fluid pressure

Fluid pressure
Fluid pressure refers to the pressure in, and exerted by fluids contained in the cracks and pores of
rocks. Rocks within depths of a few kilometres of the crust commonly have either intergranular or
fracture porosity along which a column of fluids exists up to the surface. Accordingly, the fluid
pressure is closely approximated by the equation:

p = (f ) gz
where (f) is the fluid density, g the acceleration of gravity and z the depth. This is the hydrostatic
pressure, which differs from the lithostatic pressure (weight of rocks at the same depth).

Exercise
Calculate the gradient of hydrostatic pressure due to a column of pure water and that
of the lithostatic pressure. Plot on a two-dimensional diagram and discuss.
However, fluid pressures are encountered, which are occasionally less than, but more often greater
than the normal hydrostatic pressure. Abnormal fluid pressures are attributed to one or more of
several mechanisms such as high water pressure due to compaction of sediments, dehydration of
minerals, artesian circulation. Tectonic stresses in active areas may also increase the interstitial
water pressure. Carbon dioxyde released from the mantle or other sources is a common abnormally
pressured fluid.

Effective stress
The total stress field in a porous solid can be specified in terms of normal and shear components
across plane surfaces. The solid and its interstitial fluid combined exert the total normal stresses and
the total shear components. The total stress field is expressed by a tensor of total stress as above:
11 12

21 22
31 32

13
23
33

in which the stresses due to the fluids have also nine components:

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p11 p12
p
21 p 22
p31 p32

p13
p 23
p33

in which:

pii = p jj = p
pij = p ji = 0
because normal pressures are equal in all directions and fluids do not support shear stresses.
Therefore, a diagonal, isotropic matrix (a square matrix that has non-zero elements only along the
main diagonal) represents the fluid pressure:
p 0 0
0 p 0

0 0 p

The effective stress is the difference between the total stress and the fluid pressure Pf = p :
tot = eff + Pf

Changes in confining pressure

If a material contains a fluid under hydrostatic pressure Pf , this pressure counteracts equally all the
principal stresses due to an applied load. The pressure Pf reduces the normal stress on any plane. In
rocks it corresponds to a change in confining pressure.
In a Mohr circle representation, changes in the total normal stress ( 1 + 3 ) shifts the circle along
the abscissa axis by an amount equal to the pore fluid pressure Pf , without changing its diameter,
i.e. the circle moves to the left, towards smaller values of normal stresses and keeps its size.

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The values of all shear stresses remain the same, indicating that they are independent of the
hydrostatic component.
The effective mean stress is the difference between the mean stress and the fluid pressure:
Seff = S p

Stress Ellipsoid
The stress ellipsoid is a graphical way to portray the six parameters of the stress tensor. It is defined
by three mutually perpendicular directions, called the principal directions, and the intensities of
the stresses in these directions, called the principal stresses.

Numerical Approach
The stress system is first restricted to a two-dimension plane.

Exercise (to be done with Excel or Matlab)

* Take a point O on an horizontal plane P.
A vertical stress of 100 MPa and a horizontal stress of 50 MPa act at O.
* Using a calculator and the figure below, determine the absolute values of stresses
for planes inclined at 5 intervals through O.
* Make separate calculations for the normal, shear and total stresses.
* Describe variation of stress magnitudes as a function of orientations.

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An ellipse is generated for the total stress, called the stress ellipse, a reduction of the stress
ellipsoid.
We can imagine the same exercise
- on a vertical plane where the vertical stress is 100 MPa and the horizontal stress has an
intermediate value of 75 MPa.
- On a horizontal plane where the intensity of the two perpendicular stresses are 75 and 50 MPa,
respectively.
The combination of these three stress ellipses around O generates the stress ellipsoid.

Analytical approach

We now consider a stress acting across a plane P, within the Cartesian coordinate directions Ox,
Oy and Oz, with Oz vertical.

The unit vector n defines the plane P to which it is normal. The direction of the line OP, thus the
direction of the plane P, can be expressed by the spherical coordinates of n, which are:
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n x =sin .cos

n y = sin .sin

z cos
n=

(8)

The spherical directions are equal to the direction cosines {cos ;cos ;cos } , directly obtained
from the angles between the line normal to the plane and the coordinate axes. Since n is a unit
vector, these components must satisfy the unit length condition:

n x 2 + ny 2 + nz 2 = 1

(9)

Now we consider the infinitely small tetrahedron bounded by the plane P and by the three other
triangular faces containing the coordinate axes. The plane cuts the axes Ox, Oy and Oz at points, X,
Y and Z, respectively. The triangle XOY is the projection parallel to Oz of the face P onto the plane
xOy. The area of the XYZ face is:
(1/2).(base XY * height ZH)
The area of the XOY face is:
(1/2).( base XY * height OH)
The ratio between these two faces is simply the ratio OH/ZH, the two sides of the same triangle
with a right angle in O. A geometrical construction in the plane ZOH shows that OH/ZH = cos =
n z . A similar reasoning for the projection of XYZ on to the other two coordinate planes shows that
the proportionality factors are (projection parallel to Ox) and (projection parallel to Oy).

We now consider equilibrium, which means balance of forces acting on the infinitely small
tetrahedron under consideration. Forces applied to each face are decomposed into one normal and
two shear forces.
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xx
yy
zz

xy
yx
zx

xz
yz
zy

We take XYZ as the unit area on which the force applied F/1 is also a stress vector T whose
components parallel to the coordinate axes are Tx, Ty and Tz. T is then determined by the simple
vectorial sum where the cosine directions weight the vector components:

=
T Tx cos + Ty cos + Tz cos

(10)

Areas of the coordinate faces with respect to the XYZ unit area were calculated above as cos ,
cos and cos . Per definition, force components ij are stress components ij and ij multiplied
by the areas on which they are applied. Then we can write all force components as:
xx = nxxx
yy = nyyy
zz = nzzz
The example (10) becomes:

xy = nxxy
yx = nyyx
zx = nzzx

xz = nxxz
yz = nyyz
zy = nzzy

Tx = n x xx + n y yx + n z zx

(11)

Absence of rotation implies that ij = ji and equation (11) becomes:

Tx = n x xx + n y xy + n z xz

(12)

With similar equilibrium arguments along the other directions, the coordinate axes of the stress
vector T relative to n are:
Tx = xx n x + xy n y + xz n z

Ty = yz n x + yy n y + yz n z

Tz = zx n x + zy n y + zz n z
These 3 linear equations are written in matrix notation:
Tx xx

yx
Ty =

T
z zx

xy
yy
zy

xz n x

yz n y

zz n z

(13)

The 3 X 3 stress matrix that transforms linearly every column vector as n into another column
vector as T defines the second-order stress tensor. It links any given plane with its associated stress
vector. It is written in a condensed fashion as the Cauchys formula:
i = ijn j
Remember: If A and B are two rectangular arrays of variables, their product C is defined as:
C = A.B
Where the elements c ij of C is obtained from the ith row of A and the jth column of B by
multiplying one by the other, element by element, and summing the product:
c ij = a ikb kj
k

AB is defined only if the width (number of column) of A is equal to the height (number of rows) of
B and, in general AB BA. Finally AB=0 does not imply that either A or B is a null matrix.
The parallelogram construction and equation (12) show that the normal stress across the plane
with the normal n is given by:
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T.n = Tx nx + Ty ny + Tz nz

(14)
=n x 2 x + n y 2 y + n z 2z + 2 n y n z yz + n x n z zx + n x n z xy
while the corresponding shear stress is:
2 = T2 2
We are looking for the geometric representation of the variation of stress with direction. The theory
can be followed most easily for a two-dimension stress system where the angle xOP = . Then
n x = cos , n y = sin and n z = 0 . Equation (9) reduces to:
2

2
= x cos + y sin + 2 xy sin cos
In the vertical plane xOz, taking parallel to the horizontal axis Ox, the only forces acting on the
tetrahedron (prism in two-dimension) in the Ox direction are x * (area of P) and 1 * (area of
xOz). These forces must balance for equilibrium of the tetrahedron, so that
x = 1 (area of xOz / area of P)
Now, since:
Area of xOz = n x * (area of P),
(areas are reduced to lines in this 2D projection) we have:
x = n x1
and, by similar arguments,
y = n y 2
z = n z 3
Substituting cosine directions from equation (9), it follows that:

( x 2

) (

) (

12 + y 2 22 + z 2 32 =1

(15)

Equation (15) is the equation of an ellipsoid centered at the origin with its axes parallel to the
coordinate axes.

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The ellipsoid semiaxes are in the same direction, and have the same magnitudes, as the principal
stresses. This stress ellipsoid is a commonly used graphical representation of stress. Its principal
axes are known as the principal axes of stress, which are mutually perpendicular directions of zero
shear stress. The direction and magnitude of a radius vector of the stress ellipsoid gives a complete
representation of the stress across the plane conjugate to that radius vector. The radius vector is:

s=

x 2 + y 2 + z 2

The ellipsoid is the loci of all s-extremities. Notice that, in general, the plane corresponding to a
given radius vector is not normal to the radius vector. Intuitively, from the known symmetries of the
ellipsoid, there are always 3 orthogonal directions (the principal axes) for which T and n have the
same direction. The normal stress of equation (14) across a plane whose normal has direction
cosines n x ; n y ; n z is now given by:

= n x 21 + n y2 2 + nz 2 3
The magnitude of the shear stress across this plane is:

2 =

( 1 2 )2 n x 2 n y2 + ( 2 3 )2 n y2 n z 2 + ( 3 1 )2 n z 2 n x 2

If the directions are taken as coordinate axes, all shear components are null. The stress tensor
(equation 13) is then simplified to:
0
1 0

=
T 0 2 0
0
0 3

Stress ellipsoids - stress tensor

The ellipse numerically generated portrays a section of the stress tensor of ellipsoidal form in a
specific principal plane. The stress tensor is not a single vector. It refers to the whole collection of
stresses acting on each and every plane of every conceivable orientation passing through a discrete
point in a body at a given instant of time.
To describe the stress tensor we need the orientation, size and shape of the stress ellipsoid. Thus, we
need determining the orientations and lengths of the three principle axes of the stress ellipsoid. If
we can define the stress tensors at each and every point within a body, we can fully describe a stress
field, which is the entire collection of stress tensors. This exercise is a fundamental approach for
evaluating the relationship of stress and strain

Stress field
When surface forces are applied to any body, the resulting stresses within the body in general vary
in direction and intensity from point to point. The stress field is the distribution of all the stresses at
all points throughout a body (i.e. all stress trajectories together). The stress field can be portrayed
either as a set of stress ellipsoids, or as their stress axes, or as stress trajectories. If both the normal
and shear components are the same at all points, in magnitude and orientation, the stress field is
homogeneous. Otherwise, and commonly in geology, it is heterogeneous. The relative uniformity
of stress orientation and relative magnitudes is striking and permits mapping of regional stress
fields.
See the World Stress Map: http://dc-app3-14.gfz-potsdam.de/
Two or more stress fields of different origin may be superimposed to give a combined stress field.
The sources of the stress are manifold and, consequently, stress is unevenly distributed within the
Earth's lithosphere. Its magnitudes are highest within, or next to, the regions where causative forces
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are exerted. Stress gradually diminishes away due to the elastic and creeping strain energy
consumed in deforming rocks. The stress gradient is the rate at which stress increases in a
particular direction, for instance depth with a normal hydrostatic gradient = 10 MPa/km and an
overburden gradient = 23 MPa/km. Curves of iso-stress-magnitude (stress contours) illustrate such
gradients. In the lithosphere, stresses result from forces that are transmitted from point to point.
Knowing the magnitude and the orientation of principal stresses at any point allows calculating the
normal and shear components on any plane passing through this point.

Stress trajectories
In two dimensions, on a given surface (e.g. in map view), the stress trajectories are virtual curves
that represent the directions of the principal stresses at all points, and linking the stress axes of the
same class. For example, one set of lines determines the direction of the maximum principal stress
and a second one that of the minimum principal stress. The two sets are everywhere orthogonal.
Individual trajectories may be curved but principal stresses must remain at right angles to each other
at each point in the curve. Therefore, stress trajectories portray continuous variation in principal
stress orientation from one point to another through the body.
Adjacent trajectories coming closer together indicate stress concentration.
Points where principal stresses are equal are isotropic points. Entwining stress trajectories bound
positive isotropic points; dissociating trajectories define negative isotropic points.
Points where principal stresses are all zero are singular points.

Slip lines
Knowing the principal stress trajectories, we can construct potential shear surfaces at any point of
the stress field, which are surfaces tangent to the direction of maximum shearing stress at these
points. Traces of potential shear surfaces associated with a stress field are lines called slip lines. In
two dimension, two sets of lines represent curves of dextral and sinistral senses of shear. They
converge towards isotropic points.

Stress Measurement
The motivation to measure stresses stems from geological hazards, engineering activity and
resource exploration. Stress measurement is made indirectly in mines and boreholes by measuring
the rock response to stresses. When they are excavated, the stresses that were previously supported
by the rock taken away are transferred to the surrounding rocks. In intact elastic rock, the resultant
stress concentration around the borehole or the mine is well understood from elastic theory. The
two principal methods using stress concentration around boreholes to measure in situ stresses are
near-surface overcoring and hydraulic fracturing.
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Elastic strain: Overcoring and breakout

Overcoring consists in installing a strain gauge on the bottom of a tubular bore hole. Then a coaxial
and annular hole is drilled around and deeper with an internal radius larger than the first hole. This
procedure releases stresses from the rock cylinder bearing the gauge, which has become isolated
from the regionally stressed surrounding. The subsequent elastic deformation of the circular inner
core into an elliptical one defines the orientation of horizontal stresses with the long axis of the
ellipse parallel to maximum horizontal principal stress (more relaxation in the most compressed
direction). This strain can be converted into a stress magnitude if the elastic properties of the rock
are known.

After drilling, a circular bore hole may become elliptical (breakout) in response to stresses in the
surrounding rock. The long axis of the ellipse is parallel to the minimum horizontal stress.

Hydraulic fracturing
Hydraulic fracturing consists in injecting fluids into a sealed off, short length of a drilled hole and
to pressurize it until a fracture is generated in the rock of the borehole wall and is filled by the
coring fluid. The fluid pressure required to induce tensile fracture at the wall of the borehole is the
breakdown pressure. Fracturing will occur if the breakdown pressure is equal to the tensile
strength T of the rock, which in turn is assumed to be equal to the magnitude of the minimum
horizontal effective stress *h,eff :

*h,eff = T
The concept presumes that one principal stress is vertical (near surface condition) and aligned with
the vertical well bore. The magnitude of the vertical stress v is evidently the weight of overlying

rocks. The goal is to find the magnitudes of the greater ( H ) and smaller ( h ) principal stresses in
the horizontal plane, and their orientation. Hydraulic fracturing assumes that cracks form
perpendicular to the minimum horizontal stress. Hence, measuring the orientation of created
hydraulic fractures and the breakdown pressure provides a clue to the stress tensor.
However this technique does not specify the direction of principal stresses. In fact, hydraulic
fracture initiation also depends on stress regimes and wellbore orientation.
Once the injection is ceased, the propped fracture becomes a passage for hydrocarbon gas or water
flow from the drilled reservoir to the well, thus allowing enhanced production. Hydraulic fracturing
is a practical stimulation technique for hydrocarbon recovery from low permeability reservoirs.

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Focal mechanisms
If an earthquake corresponds to slip on a fault with near optimal orientation for reactivation with
respect to the regional stress field, the P, B and T axes, which define the elastic strain released in
the earthquake, approximate the active principle stress directions s1 , s 2 and s 3 , respectively.

Present-day tectonic stress field

Stress determination is both incomplete and sparse. Results show that the stress state is
characteristically heterogeneous and unpredictable in space and time. Extrapolations from
individual measurements remain very limited simplifications of the real information because local
perturbations are ruled by local geological features. However, syntheses demonstrate the existence
of remarkably uniform Andersonian stress provinces, with two of the principal stresses horizontal
and the vertical stress either equal to 1 (extensional regime) 2 (strike-slip regime) or 3
(compression regime). The common occurrence of anthropogenic seismicity during reservoir filling
and of earthquake-triggered earthquakes suggests that the continental crust is globally in a state of
frictional equilibrium.

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Heidbach, O., Tingay, M., Barth, A., Reinecker, J., Kurfe, D., and Mller, B., The World Stress Map
database release 2008 doi:10.1594/GFZ.WSM.Rel2008, 2008

Applications to geological structures

Very little is known concerning the stress fields that exist in rocks during deformation, although it is
one of the prime goals of the subject to define these fields as closely as possible. The lack of
knowledge is, in part, due to the complexity of the stress fields that exist in deforming bodies but
mostly results from an overall lack of information concerning the mechanical properties of rocks.
The application of normal and shear stresses can be illustrated with reference to two simple
geological examples: the stress at fault plane and the stress at a bedding plane undergoing flexural
slip folding resulting from opposed compressive forces. Clearly, the sense of fault displacement and
bedding-plane slip can be predicted if the direction of the force is known and vice-versa.

Conclusion
The kinematic analysis identifies four components of deformation:
- Translation (change in position)
- Rotation (change in orientation)
- Dilation (change in size) and
- Distortion (change in shape).
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Stress is an instantaneous quantity defined as a force per unit area. The stresses at a point are the
vector components of the stress vectors on the 3 planes of reference.
The stress tensor ij is a mathematical structure that reflects the force per unit area on an oriented
area of the solid. The first subscript gives the direction of the force, while the second gives the face
of the cube on which it is acting.
It requires nine numbers and a coordinate system to be defined: it is a second order tensor. The 9
components of the stress tensor are the 9 components of stresses. Stresses cannot be summed using
vector addition.
The state of stress at a point is described by the magnitudes and orientations of 3 principal stresses,
or of the normal and shear stress on a plane of known orientation. It can be represented by an
ellipsoid with axial lengths determined by the three principal stresses.
The Mohr construction relates stress points to material planes. The stress state S on planes of
varying orientation describes a circle passing through 1 and 3 on the Mohr diagram, on which
the magnitudes and orientations of shear stress S as a function of the normal stress N can be
visualised.
Stresses in the lithosphere have both tectonic and non-tectonic, local origins. Stress controls
deformation. Therefore, understanding stresses is essential to describe, quantify and predict rock
deformation and tectonic processes. The regional uniformity of the natural stress fields observed
argues for dominantly tectonic origins.

Recommended literature
Jaeger, J. C. 1969. Elasticity, fracture and flow: with engineering and geological applications.
Methuen & Co LTD and Science Paperback, London. 268 p.
Jaeger, J. C. & Cook, N. G. W. 1979. Fundamentals of rock mechanics. Chapman and Hall,
London. 593 p.
Jaeger, J. C., Cook, N. G. W. & Zimmerman, R. W. 2007. Fundamentals of rock mechanics. Fourth
edition. Blackwell Publishing, Oxford. 475 p.
Means, W. D. 1976. Stress and strain. Basic concepts of continuum mechanics for geologists.
Springer Verlag, New York. 339 p.
Oertel, G. F. M. 1996. Stress and deformation: a handbook on tensors in geology. Oxford
University Press, New York. 292 p.

Mechanical aspects of deformation

jpb, 2014