Geomechanics for Petroleum Engineers - School of Earth, Atmospheric and Environmental Sciences

1. University of Manchester
Geomechanics
for Petroleum Engineers
EART20422
Course Notes 2013-2014
School of Earth, Atmospheric and
Environmental Sciences

2. 2
The Scope of Geomechanics
The theories of stress, strain and elasticity are the tools that we use to
understand the mechanical behaviour of rocks. In the context of petroleum
engineering, we are concerned about how rocks behave in reservoirs and around
boreholes. This course is about these aspects, but it is important to put reservoir
and borehole geomechanics into the wider context of the strength and mode of
failure of rocks over a wide range of physical conditions.
Stress
2 types of force can act on a body, BODY forces and SURFACE forces.
(a) BODY FORCES, e.g. gravitational, magnetic, inertial
No body contact required. Described as FORCE per UNIT VOLUME.
(b) SURFACE FORCES, action by physical contact with other bodies
(including fluids). Described using concept of STRESS or PRESSURE (=
FORCE per UNIT AREA)
Resolution of forces on a surface
Sign Convention:
normal stress, compression +ve
Shear stress, Right-handed +ve
Stress components in 3D
1st
subscript is the NORMAL to the
plane across which
the stress component acts.
2nd
subscript is the DIRECTION in
which the component
acts.
Using  and , it is sufficient to write xx as x
Alternatively, we can use only  to symbolize both normal and shear stress, in
which case BOTH subscripts are needed, but we can use the convenient shorthand
ij, where i, j = 1 to 3. In this case the 3 axes would be identified as x1 x2 x3.

3. 3
The array of terms can be written in matrix form ( 11 1213 )
         (212223 )
( 31 32 33 )
Stress is a multicomponent quantity that is an example of a TENSOR quantity.
The terms are defined by reference to a reference frame, If the reference frame is
rotated the values of the terms change. It is the particular way that the values of the
terms change as the reference frame is rotated that defines stress as a tensor
quantity.
There are other examples of tensor quantities,
e.g. Strain (stress and strain are examples of FIELD tensors)
Magnetic susceptibility, permeability, piezoelectricity, elasticity (these are examples
of PHYSICAL PROPERTY tensors).
Symmetry of the stress tensor
Provided a body is in equilibrium with respect
to ROTATION about any reference axis,
ij = -ji, i.e. the matrix representing
the tensor is symmetric about the leading
diagonal
Homogeneous and Heterogeneous Stress
In a homogeneous stress field the values of the stress components do not change
from place to place. In a heterogenous stress field they do. For example the
distribution of stress in a hook supporting a load is immensely variable, and within
folded sedimentary layers this is also true.
Principal Stresses
For any state of stress, there is always an orientation of the reference axes for which
the shear stress components all become zero. The values of the normal stress
components in this reference frame are called PRINCIPAL STRESSES and are
labelled 1, 2 and3 (in order of decreasing magnitude). It is common practice to
represent graphically the shape of a stress field by drawing an orthogonal grid that
shows the variation of orientation of principal stresses from place to place. Where
the grid lines converge, the values of that stress component increase, and vice
versa. The orthogonal grid is called the STRESS TRAJECTORY pattern.

4. 4
This figure illustrates the stress trajectory pattern of a heterogeneously stretched
bar. To a certain extent the expected pattern of stress trajectories can be deduced
from commonsense, bearing in mind the form of the BOUNDARY CONDITIONS.
Some problems of stress distribution can be solved ANALYTICALLY; more general
problems have to be solved NUMERICALLY.
Homogeneous stress in two dimensions
In two dimensions we have only 4 stress components ( only 3 are independent).
The two commonest tasks in dealing with homogeneous stress states are:
(i) Calculating the values of Principal Stresses in terms of components referred to
an arbitrarily inclined reference frame.
(ii) Calculating the values of normal and shear stresses acting on an inclined plane
in terms of the principal stresses.
Using the concept of resolution of forces across areas it is a simple matter to derive
equations to solve these problems. They are stated below without proof.
(i) Values of Principal Stresses in terms of components
referred to an arbitrary reference frame.
_______________________
Thus 1, 2 = ½(x + y) + ½ (x + y)2
- 4(xy - 2
)
_____________
= ½(x + y) + ½ (x - y)2
+ 42
Orientations of the Principal Stresses, 
tan 2 = 2 / (x – y)
(ii) Stresses in a plane in terms of the principal stresses
Here, we are interested in the values of normal and shear stress resolved along any

5. 5
arbitrarily inclined plane, at o
to the max principal stress. This is of particular
importance in geoscience, because this is how we calculate the normal and shear
stress acting on an inclined fault plane. It is key to understanding the stability (or
otherwise) of faults
n = ½(1 + 2) + ½(1 - 2) cos2
 = + ½ (1 - 2) sin2
        
The + signs refer to the fact that the equations
also give the values of n and with the plane
and its normal transposed. (recall ij = - ji)
Note that  is maximum when sin 2 = 1,
i.e. = 45o
and zero parallel to the principal stresses
(= 0 or 90o
)
Variation of and n with 

e.g. setting 2 = -½ 1
The Mohr Circle construction
Note that these equations take the same form as the parametric equations to a
circle:
n = ½(1 + 2) + ½(1 - 2) cos2
 = + ½ (1 - 2) sin2



6. 6

(What, you don‘t know the parametric equations to a circle? )
(You do now!)
The similarity in form means that points on the periphery of a circle can be used to
represent all the different combinations of shear and normal stresses on variously
inclined planes with respect to the principal stresses.
The normal stress components where the circle cuts the abscissa (when shear
stresses are zero) are the principal stresses (maximum and minimum values of the
normal stress components).
Coordinates A and B give and n resolved on 2 planes 90o
apart
Note : tan 2 = 2 /( x - y) , from Pythagoras.

7. 7
Elasticity
Elastic deformations are the small (<1%), strains that accompany stresses are too
low to cause permanent deformation. The strains are sufficiently small that a
particular variety of finite strain theory, called infinitesimal strain theory, can be
applied. Elastic strain is recoverable on removal of load and it is through elastic
strains that stresses are transmitted from place to place throughout a material.
Elastic strain is (to a good approximation) linearly proportional to applied stress.
This proportionality is called Hooke‘s Law (Ut tensio, sic vis - As is the extension,
so is the force). The constants of proportionality are called Elastic Moduli. For an
isotropic, linear elastic solid, we can usefully define 4 elastic moduli:
Young’s Modulus, E =  i / i
Shear Modulus, G = ij / ij
Bulk Modulus, K =  / v , where we define  = (x + y + z)/3
Poisson’s Ratio, v = x / y, for load applied in y direction
The diagram (right) defines the concept of
Poisson‘s ratio. As the ratio of two strains,
this is just a number. The other elastic moduli
all have the dimensions of stress, as they
take the form stress/strain.
Note that for isovolumetric deformation
Poisson‘s ratio is 0.5 . For deformations of
real crystalline solids it is less than 0.5 ,
normally about 0.3, implying that elastic
deformation of a crystal involves elastic volume reduction.
Note that for small strains (< 3%), volumetric strain v = dV / Vo
can usefully be approximated as  (x + y + z)/3 (but NOT for large strains)
Longitudinal and shear strains are defined in the same way as for finite strains.
Interdependence of elastic moduli
For an ISOTROPIC (make sure you understand what this word means) material,
only 2 elastic constants are independent:

8. 8
These are the interrelationships:
K = E / 3(1- 2) ; G = E / 2(1 + v)
From which the other relations between groups of 3 can be derived.
Elastic Stress-Strain Equations
For a triaxial state of stress (1  2  3)  0 we can write:
1 = 1 / E – v2 /E – v3 /E = {1 – v(2 + 3)}/E
Similarly 2 = {2 – v(1 + 3)}/E
and 3 = {3 – v(1 + 2)}/E

Elastic Strain Energy
This is the potential energy stored in an elastic solid by virtue of the work done on it
by the external forces to produce the strained state.
Work done per unit volume = Force x displacement
Volume
W = Force . l = . d
Area lo
But  = E , thus for a total strain ,
W = 

= ½   (the area under the stress/strain curve)
Using principal stresses and strains in 3D
W = ½ )
Using the stress strain equations to eliminate strains:
W = 1 {






2E

9. 9
Introduction to Strain
Homogeneous and heterogeneous deformation. Finite and infinitesimal strain.
Strain measures :
Changes in the length of lines: conventional strain e = (l1 - lo)/lo = l/lo ,
(1+e) = l1/lo
where l1 = current length and lo = original length.
Quadratic elongation,
 = (l1/lo)2
= (1+e)2
Natural or logarithmic strain,  = ln (1+e)
Changes in angles:
Shear strain,  = a/b = tan 
Homogeneous Strain
Straight lines remain straight,
lengths of lines and angles change :
The Mohr circle construction can
be applied to strain just as it can
be to stress, though we will not
explore that avenue here.
Homogeneous deformation of a circle to an ellipse
The deformed circle illustrates graphically the amount of deformation
and is called the finite Strain Ellipse. In 3 dimensions a circle would be deformed into an ellipsoid.
The maximum and minimum values of strain are the Principal Strains,
and correspond with the principal axes of the ellipse.
Principal axes are also lines of no finite shear strain.
Intersection of initial circle with strain ellipse are directions of no finite
longitudinal strain (NFLS). These divide the ellipse into areas where lines have
been shortened from areas where they have been stretched.

10. 10
Area change: In general strain involves volume change (or area change in 2D).
Some strain can take place at const volume (or area).
Area change is given by (1 2 )0.5
– 1, x100 to express in percent.
Note area change can be either increase or decrease.
Strain trajectories, These are orthogonal sets of lines (like stress trajectories) that
are everywhere parallel to the principal axes of strain. They are useful for
representing graphically variation of strain from place to place in a heterogeneously
strained body. Note how strain along a particular trajectory increases as the
trajectories converge. Ellipses of different eccentricities (do you understand this
word? [= axial ratios]) can be used to represent variations in amount of distortional
strain.

11. 11
Rock Deformation Mechanisms
Rocks are not continuous materials, like glasses, but are made of crystalline
grains. This granular heterogeneity of structure, coupled with the possibility of
intergranular porosity, profoundly affects how rocks can deform. The
crystalline internal structure of grains also affects deformability. Rock
deformation mechanisms can be classified into two main groups, and each
mechanism is dominant over particular ranges of pressure/temperature
conditions encountered within the Earth:
(a) cataclastic deformation, in which the rock distorts by brittle
fracturing at the grain scale or larger, and fragments are
displaced with respect to one-another.
(b) intracrystalline plasticity, in which the internal crystalline
structure of grains is distorted, effectively by the slippage of
atomic planes over one-another.
We will examine first the microstructural characteristics of these
different deformation mechanisms. Then we will consider the different
mechanical properties that are diagnostic of the different processes.
Cataclastic deformation
The rock distorts by brittle fracturing at the grain scale or larger, and
fragments are displaced with respect to one-another.
There is no internal distortion of the crystal or lithic fragments. This is
how rocks deform in the upper crust, and the deformation is commonly
localized into fault zones.
The necessary accompanying volumetric expansion causes the
resistance to cataclasis to increase dramatically with pressure (depth of
burial), but the process is not very sensitive to temperature variations.
[note, this last sentence is profoundly significant, and one way or another
WILL figure in examination questions].

12. 12
Typical Structures and Microstructures:
Formation of oriented crack arrays, aligned preferentially parallel to the
greatest principal stress direction and originating at impingement points
between grains.
These crack arrays are called axial cracks. In ancient rocks they can
become cemented, so they are only seen as oriented fluid inclusion
arrays.
Axial crack arrays tend to coalesce,
or localize, after a critical density of them
has formed, to produce a planar fault,
typically at about 30° to 1 (above, max
stress is vertical)
In low porosity rocks such as limestones,
cracking may be mesoscale and pervasive
(see upper right).
Pervasively mesocracked rocks can form important hydrocarbon
reservoirs in which oil is held in the cracks, e.g. in Northern Italy.

13. 13
Strength and microstructural evolution by cataclasis depends strongly on
amount and nature of cement, and how much of the original
intergranular porosity remains.
Fault Rocks
The rocks that are produced in fault zones (ranging from >1 cm to 10‘s
of metres wide) are variously called fault gouge or cataclasites, and
are characterised by fragmentation and granulation of the protolith.
Fault gouges may be free of clays or other phyllosilicates, and be
made up of only fragmented quartz, feldspar of limestone (according to
the nature of the protolith), with various amounts of cement. However,
breakdown of feldspars or micas or growth of authigenic clay can result
in a clay-bearing fault gouge. These various characteristics determine
whether a fault zone permits or impedes the flow of fluids such as water,
oil and gas. Faults that are sufficiently impermeable can form reservoir
seals or can cause compartmentalization of reservoirs (the latter with
attendant production problems).
Clay-bearing, foliated fault gouge Cataclasite in fault zone in limestone
Intracrystalline Plasticity
Internal crystalline structure of grains
is distorted, effectively by the slippage
of atomic planes over one-another.

14. 14
More symmetric crystal structures can have several differently oriented,
crystallographically controlled slip planes.
The deformation is a
simple shear, hence is
constant volume, which
makes the process
insensitive to pressure.
However, it is strongly
sensitive to temperature.
From the mid-crust and
downwards, this is probably
the commonest deformation
mechanism in the Earth. At
shallow, upper crustal
depths (top 10 km) rock salt deposits are mobilized by this process and
this leads to the formation of important hydrocarbon traps and reservoir
cap rocks.
Typical Microstructural Features at the Grain Scale
These are:
(a) Flattened grains and
(b) Wavy extinction due to internal distortion of crystal structure.
(c) Subgrain formation
(d) Dynamic recrystallization, starting around the edges of grains.
This image below of deformed basic granulite shows all of these
features:
(image width about 4mm, crossed polars)

15. 15
Similar features (below) in experimentally deformed marble (700°C, original
grain size 200 micron).
(below) Dynamic recrystallization by grain boundary migration
(especially at higher temperatures). This progressively replaces the old,
strained microstructure by a new, strain free microstructure, often of a
different mean grain size (image right, experimentally deformed marble
800°C, original grain size 200 micron).
(c) and (d) are known as recovery processes, that progressively restore
the microstructure to an unstrained state.
Crystallographic Preferred Orientation (CPO)
This only develops to any significant degree with flow by Intracrystalline
Plasticity.
Recrystallized grain
size / stress
relationship
Recrystallized grain size
can be used as a
paleopiezometer.
Deformation at higher
stress produces smaller
recrystallized grains,
thus in naturally
deformed rocks we can
use recrystallized grain
size to measure the flow stress at the time of recrystallization:
The graph above shows the experimentally-determined grain size vs stress
relation for marble.

16. 16
Describing the strength of rocks
We can usefully consider this topic under three headings:
(a) Brittle rocks of low porosity at low temperatures (<200o
C)
(b) Brittle rocks of high porosity at low temperatures (<200o
C)
(c) Rocks that deform by flow processes (at high temperatures,
300o
C for most rocks, but only 80o
C [or more] for rock salt)
The strength and mode of failure of rocks is generally investigated in the
laboratory using a triaxial testing machine.
This consists of a thick-walled pressure vessel containing a cylindrical
specimen between two loading pistons. Depth of burial is simulated by
pressurizing a hydraulic fluid (liquid or gas confining pressure). The pressure is
transmitted to the rock via a rubber or thin-walled metal jacket. The end-load is
increased above hydrostatic by advancing the loading piston into the vessel. The
amount of this excess load (differential load) at failure is a measure of strength.
High temperature can be achieved either by heating the whole pressure vessel
(external heating) or by means of a smaller furnace inside the pressure medium
and placed around the specimen. Pore fluids under pressure up to the value of
the confining pressure can be introduced via a hollow loading piston. We have 8
of these machines in the lab at Manchester.
Testing Machines

17. 17
(Above right: Iron-jacketed specimen assembly)
Specimens can be tested in compression
(above left), extension (lower left) or
torsion (right, photo David Olgaard)

18. 18
Modes of Rock Failure
This term refers to whether rocks fail
(a) by faulting (localized deformation) or
(b) by ductile flow (distributed deformation)
The images show examples of faulting in the lab and in nature :
(left) single shear fault (photo John Ramsay) (right) conjugate faults (source unknown).
(above) in lab. (below) in nature
Stress - Strain characteristics
The results of
mechanical tests
(typically taken to about
15% strain) are usually
plotted as stress/strain
curves. They are very
different for the cases
of faulting and flow:

19. 19
Deformation in the upper continental crust (topmost 15 km) is usually by
cataclastic faulting. That is, cracking that becomes localized into a brittle fault
(brittleness pertains to cracking). Cataclasis is the name given to deformation
processes involving cracking. Prior to permanent deformation, in the elastic
regime the slope of the stress/strain plot is the Young‘s modulus.
Deformation in the lower crust (lowermost 15 km) generally is by flow, leading
to folding in layered rocks or pervasive deformation in isotropic rocks like
granitoids.
Deformation involving cracking usually results in a small amount of volume
expansion (about 1 or 2%), as the fractured pieces no longer fit together. The
volume expansion is resisted by confining pressure which increases as depth
increases. Thus rocks deforming by cataclastic processes get stronger as
depth increases. [remember, you were warned about the importance of
understanding this point] On the other hand, increasing temperature with depth
has little effect on resistance to cataclastic deformation.
Deformation involving flow processes can be almost at constant volume, and
hence are insensitive to variation in confining pressure. On the other hand,
flow processes are very sensitive to temperature and to rate of deformation
(higher temperatures or slower rates lead to lower strength. Thus rocks get
weaker at greater depths [higher temperatures]). This is exactly opposite to
cataclastic deformation. Thus the variation of rock strength with crustal depth
looks like this (the details vary with rock type):
The flow curve shifts to the left (lesser depth) as the strain rate is decreased
and vice-versa.
The faulting to flow transition usually is near mid-crustal depths. Because
seismicity is associated with brittle deformation, the faulting to flow transition
also corresponds to the depth limit of seismicity in the crust.

20. 20
How quickly do temperature and pressure increase with depth?
The rate of temperature increase with depth is called the geothermal gradient.
It is very variable according to the thermotectonic regime. In oceanic trenches it
is low, about 12 °C/km; about 22 °C/km in stable continental crust; about 30
°C/km in orogenic (mountain building) regions and can be as high as 150 °C/km
in volcanic areas.
The rate of pressure increase (the vertical component of normal stress,z) with
depth is called the geobaric gradient. To a useful approximation it is given by :
z = m g z
in whichm is average crustal rock density, g is the gravitational acceleration
and z is depth. For typical crustal rocks dz / dz ~ 35 MPa / km, because the
density of compact crustal rocks is about 2700 kg/m3
. Thus the pressure at the
continental Moho is about 1000 MPa (= 1 GPa).
[1 MPa = 10 atmospheres = 145 psi]
However, if rock density is much more variable with depth, perhaps at shallow
crustal depths because progressive pore compaction and diagenetic mineral
transformations are taking place with depth, a more precise calculation of
pressure at a given depth is required. Assuming density varies with depth
according to (z), the pressure at depth z below the surface is given by
z
z = g  z) dz
0
(z) for sedimentary basins can be obtained from density measurements on
core samples.
The thermobaric regime of the crust has a profound effect on how rocks
deform and on the conditions for the formation of oil and gas.

21. 21
Describing brittle/cataclastic deformation
(i) - the Mohr-Coulomb failure criterion
This describes well the brittle failure of rocks of less than about 10% porosity
over the range of upper crustal P/T conditions. A linear relationship between
principal stresses at failure is assumed:
Consider a rock subjected to an initial stress state such that 1 = 2 =3z
(wherez is the normal stress in the vertical direction), i.e. stresses are
equal in all directions and principal stresses are horizontal and vertical.
This is called a hydrostatic stress state. E. M. Anderson, who was famous for
his quantitative work on stress and faulting in the first half of the 20th century,
called this a standard state.
Normal Faults Reverse Faults Wrench Faults
Now imagine 1 gradually increased until the rock fails. Failure is usually in
shear, with the formation of a fault inclined at  to 1.  is usually about 30°.
(1 - 2) at any instant is known as the differential stress, and its particular
value at failure is called the strength of the material.
Anderson suggested that different types of fault formed in response to
different stress regimes in the Earth. Thus when the vertical stress is1 (i.e.
the greatest stress), normal faults form. When 1 is horizontal and the least
stress (3) is vertical, thrust faults form. When the intermediate stress (2) is
vertical, wrench faults form.
Potentially, 2 faults may form in a given stress state, forming a conjugate
pair that intersect in the 2 (intermediate stress) direction. Commonly, the first
one to form relieves the load so the second fault fails to form.

22. 22
Extensional and Contractional faults
A more general way of describing faults is to relate the stresses to the local
layering, which make a marker for the sense of fault offset.
Thus faults that try to thicken or duplicate the sequence are contractional
(with respect to the layering) and those that excise (cut-out) part of the
sequence are extensional in character.
One can still talk about extensional or contractional tectonic regimes with
respect to the orientation of the Earth‘s surface, though in the general
case the principal stresses are not necessarily parallel and perpendicular to
the Earth‘s surface as assumed by Anderson for his standard state.
Returning to our 2D state of stress (1 and 2 only, also known as plane
stress), imagine increasing the value of the standard state stress (e.g.
increasing depth of burial). Now, at the greater depth, as we increase (1 - 2)
to produce failure, we find that the differential stress is greater than before.
This can be demonstrated in experiments, when the initial hydrostatic state is
called the confining pressure. As the confining pressure is increased the
rock becomes stronger. This is a
characteristic of all brittle rocks.
Experimental results can be
represented in 2 ways:
1) A plot of 1 against 2 at
failure (or (1 - 2) against 2)
(right) :
This is usually a straight line, in
which case it can be represented by
1 = a + b 2, where a and b are
material properties, and can be
extracted from the intercept and
slope of the line.
2) On a Mohr diagram (right), as a

23. 23
set of circles of increasing diameter. The circles are delimited by a straight
line, called the Mohr envelope, that defines the strength of the material.
The Mohr envelope can be described by  = o + n,
where o and  are material properties called the cohesive strength and
coefficient of internal friction, respectively. Because the above are simply
two alternative methods of reporting the same data, there exist linear
transformations between the material parameters, given by
a = 2ob and b = (1 + sin  )/( 1 - sin 
 = tan , where  is the slope in degrees of the envelope.  is called the
angle of internal friction (commonly about 30°). These two equations are
equivalent statements of the Mohr-Coulomb failure criterion.
The angle of friction is related to the angle of the fault that forms at failure by
 = 45° - /2
The diagram below shows the relationships between the Mohr circle world
and the real world :
The Role of Pore Fluid Pressure - the Law of Effective
Stress
Pore fluid pressure contributes to the overall state of stress in a rock.
Whereas the applied (far-field) stresses try to push the grains together, pore
fluid pressure tries to push them apart. A liquid cannot support a shear
stress, only a hydrostatic normal stress, thus pore pressure reduces the
effective applied normal stress components by an amount equal to the
hydrostatic pore pressure. This can be represented on a Mohr diagram.
The pore pressure shifts all points on the circle to the left by an equal amount.
The diameter of the circle (and hence shear stress components) remain
unchanged, but the effective confining pressure is reduced.

24. 24
Because the Mohr-Coulomb criterion has a positive slope, the action of raising
the pore pressure can result in failure. The effect on the Mohr-Coulomb
criterion is given by:
(1 - 2) = a + (b - 1) (2 - p)  =o +  (n - p)
where p is the pore fluid pressure.
The following diagrams illustrate how raising the pore pressure effectively
prises the two sides of a fault apart and reduces its ability to support frictional
stress (figures by Steve Hickman).
How does pore pressure vary in the Earth?
If the pores of a rock at depth z are connected via continous channelways to
the free surface, the ratio of pore pressure p to overburden pressure z is
equal to the ratio of density of the column of fluid to density of the column of
rock:
= fluidgz / rockgz = fluid /rock
The density of crustal rock of moderate porosity is about 2.5 times fluid

25. 25
(water) density, hence the ‘normal‘ value of is expected to be about 0.4.
Various processes can cause  to increase above 0.4 :
1. Compaction and cementation - reduces pore volume hence raises pore
pressure
2. Erosion - removes overburden pressure, so rises (provided fluid cannot leak
away).
3. Osmosis - Separation of aqueous fluid reservoirs of different salinities by a
semipermeable membrane (e.g. clay-rich) layer causes fluid diffusion from
the less saline fluid to the more saline fluid.
4. Production of excess fluid through diagenetic or metamorphic reactions
(dehydration of clays, evaporites and other hydrous minerals).
5. Thermal expansion on heating.
> 0.4 in a reservoir causes blowout when the reservoir is pierced by drilling,
unless it is prevented by appropriate engineering.
can rise to become equal to the least principal stress (= 1). Rocks are
weak in tension so any attempt for to exceed 1 results in hydraulic
fracture of the rock (the excess fluid pressure blows the rock apart against
the least principal stress). In the case of a reservoir this means failure of the
seal and leakage of fluid into adjacent, non-overpressured rocks.
Hydraulic fractures tend to propagate along the maximum principal stress
trajectory. In mid-crustal rocks, hydraulic fractures are often preserved as
veins filled with hydrothermal mineralization such as quartz. Tectonic
brecciation is also thought to be produced in zones of high fluid pressure,
because the reduction of effective confining pressure makes it easier to bring
about the dilatation of the rock mass necessary to accommodate fractures.
Examples of hydrothermal quartz veins filling cracks (source unknown):

26. 26
(left) quartz veins later deformed
Quartz solubility is high in
(a) high P/T (Pressure/Temperature)
water and
(b) when there is dissolved salt in the
water.
When a crack forms, the dilatation drops
the pore pressure and reduces quartz
solubility, causing precipitation. In metamorphic complexes quartz is dissolved
as water is evolved in dehydration reactions at high P and T at depth, and
precipitates as the solute fluid works its way up the metamorphic pile and is
depressurized. Early-formed veins are themselves later deformed as
deformation progresses. Metamorphic rocks can lose a lot of volume in this
way.
A simple criterion for hydraulic fracture is:
p - 3 = T, where T is the tensile strength of the rock. i.e. cracking occurs
when pore pressure exceeds 3 by the amount of the tensile strength (on the
order of 0.1 MPa)
Shape of the Mohr-Coulomb criterion when a rock has tensile strength

27. 27
Byerlee‘s rule for rock-on-rock sliding friction
Jim Byerlee, of USGS, proposed in 1977 (from laboratory measurements) that
to a good approximation the friction coefficient of rocks is nearly independent of
rock type. This has become known as Byerlee‘s rule, and is widely employed in
estimating the behaviour of the brittle upper crust.
The open symbols in this graph are some of the data assembled by Byerlee
(compared to in-situ stress measurements in the KTB borehole [a 9 km
deep hole drilled in southern Germany during the 1990s]). At shallow
depths the friction coefficient is about 0.85, decreasing to about 0.6 at
greater depths.
Description of the Failure of Porous Rocks in the Upper
Continental Crust - Critical State Mechanics
Rocks with porosities less than 10% behave according to the linear Mohr-
Coulomb description over most of the upper continental crust (top 15 km).
Mohr-Coulomb failure involves a small amount of dilatation (on the order of 1
or 2%) to allow for the formation of cracks, and this is why such rocks get
stronger at higher pressures.
More porous rocks, including many reservoir rocks, require a more complex
description of failure because volume loss often occurs during burial and
deformation.
How does volume loss (compaction) occur?
At shallow depths (down to about 3 or 4 km), compaction is Mechanical,
involving initially rearrangement of grains and, at higher pressures, grain
fracture and sliding between fragments.
At greater depths, there is a transition to Chemical compaction and
porosity loss, involving interpenetration of grains by solution at contact points

28. 28
and precipitation of cement in pore spaces.
Primary porosity of loosely compacted sand is about 42%.
Mechanical compaction typically will only reduce this by up to 10%.
Chemical compaction can reduce porosity (eventually) to near zero.
Generally, older rocks have progressively lower porosities, first because
they have had more time for the processes to operate, and second because
they have usually been more deeply buried. Elimination of porosity in
(siliciclastic) sediments usually takes up to about 50 Ma (limestones are
quicker).
Mechanical compaction is largely time-independent - a given effective
pressure produces a given amount of compaction.
Chemical compaction is generally time-dependent. A rock compacts more at
(a) a higher temperature and (b) over a longer time period. This is to be
expected as chemical processes such as reaction and diffusion are
temperature and time dependent according to the Arrhenius rate law:
rate  exp(-H/RT)
The above figures show data from an
experimental study carried out in this dept
a few years ago on the rate of chemical
compaction of a synthetic silty mudstone.
It illustrates the temperature and time-
dependence of compaction rate. This type
of study leads to a model from which the
amount of compaction at any given depth
and elapsed time can be predicted.

29. 29
Porosity loss by mechanical compaction
This is usually described by the
normal compaction (or consolidation)
curve:
The hydrostatic pressure here is
effective pressure, the total applied
pressure less the pore fluid pressure.
Thus if pore pressure equals applied
pressure, there is nothing to drive
the compaction. Rocks wholly or
partially overpressured in this way
remain underconsolidated (more
porous than they would normally be for a given depth of burial). This is good
for hydrocarbon reservoirs because porosity is not lost by compaction, but it
does mean that once fluid starts to be extracted, compaction will occur and
the rock column will subside (i.e. a production platform will subside - a very
expensive mistake, e.g. Ekofisk platform in 1979), unless the fluid removed is
replaced by pressurized water, for example.
Erosion leads to an overconsolidated state. That is, the porosity was
previously reduced by a greater amount than that which corresponds to the
effective pressure the rock is now subjected to. Knowing the consolidation
state of reservoir rocks is vital to the management of a reservoir.
To a good approximation, the value
of effective pressure necessary to
start compaction x the mean grain
size is independent of rock type.
The graph (right) illustrates this. It is
effectively the normal consolidation
curve in log/log coordinates, and
represents a very useful rule-of-
thumb for porous sandstones in
particular. The graph shown (right) is
based on Zhang et al. J.Geophys.
Res, 1990, 95, 341-352, but seems
to apply to all sandstones.
Yield in Porous Materials
Porous materials, then, can yield even under effective hydrostatic stress,
by the permanent collapse of pore spaces. Compaction can also be induced
by combined hydrostatic and deviatoric stresses, until a balance between

30. 30
pore compaction and dilatancy due to new crack formation occurs. This
balance is called the critical state. It corresponds to deformation at constant
porosity, at constant flow stress, at the peak on the yield surface.
This figure (right) illustrates a pressure-
sensitive yield surface at low mean
pressures that closes to produce a yield
point (P*) on the hydrostat. This is called
a capped yield surface.
(note: effective mean stress
= [(1 + 2 + 3)/3] - P )
To a good approximation, the critical
state line is independent of rock type.
Whilst the yield surface is expanding,
deformation is dilatant (this is equivalent
to Mohr-Coulomb behaviour). As it
contracts it is compactive. The contracting region (of negative slope) is called
the region of shear-enhanced compaction. This is because the rock
compacts, even though the effective pressure is less than P*, because the
non-hydrostatic part of the stress state can assist in bringing about
compaction.
Stress paths link the yield and critical state lines. As porosity progressively
collapses in the post-yield region, the rock hardens until the critical state line
is reached.
The above example of Penrith sandstone (initial porosity 25%) shows how
the capped yield surface diagram (below) relates to the corresponding

31. 31
stress/strain data:
The figure (right), shows
data for 3 sandstones of
different porosities
(Penrith 25%; Darley
Dale 13%; Tennessee
7%), showing the
capped yield surfaces,
the stress paths linking
yield and critical state,
and the critical state line
(independent of rock
type).
From Cuss, Rutter &
Holloway, 2003, Int. J.
Rock Mech Min. Sci.,
40, 847-862, also the
figures below.
The stress paths have a slope of 3, which is a property of the triaxial
deformation method, because increasing the differential stress by x increases
the mean stress by x/3). These stress paths stay on the same line as
differential load is increased because these are drained tests. That is, pore
pressure is kept constant by
allowing fluid to flow into or out of
the specimen according to whether
deformation is dilatant or
compactive. Undrained behaviour
is considered later.
Yield curves can be linked into a
continuous yield surface in a plot
of deviatoric stress (proportional to
differential stress) vs effective
mean stress vs porosity (x grain
size), to show how the yield
surface expands as porosity is
reduced and how the critical state
line separates regions of dilatant
and compactive deformation.
The yield surface (right)
incorporates the individual yield
curves for the 3 rocks above.

32. 32
The effects of porosity differences between different rocks can be taken out
by normalizing the plot of differential stress vs effective mean stress. We
can divide mean pressure and differential stress by P* for that rock. This
tends to collapse all yield surfaces onto a common curve:
This means that
one can
compute the
expected
behaviour of a
given sandstone
from knowledge
only of its
porosity and
grain size. This
seems likely to
prove a powerful
unifying
concept, and is
presently a very
active area of research. (Fig. above from Wong et al., J.Geophys Res. 1997)
This type of approach (called Critical State Theory) has proved particularly
valuable in the description of the behaviour of soils and weak rocks (including
partially molten rocks and rocks undergoing deformation with dehydration
reactions).
How do strain and porosity vary between yield and critical state?
These data for porous (dehydrated) serpentinite at 450°C illustrate what
happens.
Porosity change (progressive compaction) and strain accumulation contours
tend to track the yield curve.

33. 33
Above data from: Rutter, Llana-Funez & Brodie, 2008. J. Struct. Geol, 165,
639-649.
What about undrained behaviour?
If fluid is prevented from leaving or entering the pores of the rock during
deformation, it is forced to deform at constant volume. Thus if the pores try to
dilate the pore pressure drops, or it rises if the pores try to compact. This
causes stress paths to deviate from the simple linear paths characteristic of
drained loading.
In nature, for example,
undrained loading during
compaction causes pore
pressure to rise, so the
formation becomes
overpressured.
The figure (right) illustrates
how such variations in pore
pressure cause the stress
path to deviate.
It is vital for petroleum
geologists and engineers to
understand these concepts.
They underpin modern
approaches to reservoir
evaluation and
management.

34. 34
Describing high temperature plastic flow
The main mechanism of high temperature flow (flow from mid-crustal depths and
deeper) is by intracrystalline plasticity - involving internal distortion of grains by
intracrystalline slip.
Intracrystalline plastic flow is a constant volume process and hence is
pressure-insensitive. On the other hand the flow stress is reduced markedly by
increasing temperature or decreasing deformation rate, unlike cataclastic
deformation. Absolute temperature (T), flow stress () and deformation rate =
(d/dt ) are linked by the equation
d/dt = A f() exp (-H/RT)
in which A is an empirical constant, H is the activation enthalpy for flow and R is
the gas constant.
---------------------------------------------------------------------------------------------
A note about Strain rate, = d/dt : dimensions are sec-1
.
-2 -4 -6 -8 -10 -12 -14 -16
! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
--- lab. strain rates--- --- natural strain rates ---
log10 strain rate (sec-1
)
---------------------------------------------------------------------------------------------
---
f() = n
at low stresses (or high temperatures) and
f() = exp (B) at high stresses (or low temperatures),
where B is an empirical constant. Exponent n takes a value between 3 and 7,
depending on the material.
Taking logs to base e : at high temperature
ln d/dt = ln A + n ln  - H/RT
and at lower temperature :
ln d/dt = ln A + B  - H/RT
Thus at constant T, eq.1 implies a linear relation between ln d/dt and ln ,
with slope n and eq.2 a linear relation between ln d/dt and , of slope B.

35. 35
At constant stress the relation between
ln d/dt and 1/T is linear. This is called
an Arrhenius diagram, and the slope
is -H/R.
A collection of experimental data for flow stress might look like:
Typical n
values:
Halite 5.5 Calcite 7.6
Quartz 3.4 Olivine 3.0

36. 36
Relation between cataclasis and intracrystalline plasticity
At progressively lower confining pressures, as T is increased (or d/dt
increased), there is a mechanism transition from cataclasis to plasticity
(sometimes erroneously called the brittle-ductile transition).
Application to the Earth
In the Earth, confining pressure increases with depth, temperature increases
with depth, thus there is a mid-crustal transition from cataclasis to intracrystalline
plasticity, and a mid-crustal strength maximum.
Recall the diagram below-left.
The strength maximum may
correspond to the depth limit of crustal
earthquakes, which should be
restricted to the brittle part of the crust.
To a first approximation, the upper
crust is made of quartz/granite, the
lower crust of feldspar/mafic minerals
and the upper mantle of olivine. Thus
the lithosphere shows a coarse
lithological stratification, which should
be reflected in the strength profile with
depth.
The resultant strength profile (right)
is called Rheological Stratification
of the Lithosphere :
(Rheology is the study of flow)
In order that Plate Tectonics can
work, the stress distribution with
depth at a plate boundary must meet
the failure envelope everywhere
(Whole Lithosphere Failure). Before
this happens, strong intervals may
behave as elastic struts that hold the
lithosphere together.
These constitutive flow laws can be
used to extrapolate laboratory

37. 37
mechanical data to geological conditions, so that mathematical models of
geological deformation of the lithosphere can be made.
In summary, such extrapolations indicate that :
(a) at low temperatures and pressures (uppermost one-third of the continental
crust), rocks fail by brittle faulting which can be seismogenic. Strength increases
with depth because the process is temperature- and rate-insensitive.
(b) with increasing depth there is a transition to ductile, plastic flow. Because it is
pressure-insensitive, temperature dependence dominates and strength
decreases with depth.
If there are also variations in rock type with depth (e.g. across the Moho),
corresponding fluctuations in rock strength also occur. This leads to the concept
of rheological stratification of the lithosphere.
What gives rise to in-situ stresses in the Earth?
Knowledge of in-situ stress state can be vital for geo-engineering. However, it is
one of the most difficult things to
measure. In-situ stress can arise from:
(a) Forces arising from plate tectonics,
e.g. ridge push and trench pull. These
generate a maximum compressive
stress linking constructive to destructive
plate boundaries (e.g. world stress map
showing orientation of maximum
compressive stress - European part
shown on right http://www.world-stress-
map.org).
(b) Forces arising from lithospheric
bending due to surface or buried loads.
(c) Stresses arising from uplift, burial
and compaction.
(d) Local radial stresses due to features
like diapiric uplift (e.g. salt) and pluton
emplacement.
(e) Stresses induced through volume changes associated with mineral
transformations (e.g. in mantle transition zone, or serpentinization).

38. 38
Storage of stresses in rocks (residual stresses)
These arise from arrays of defects in crystals (e.g. dislocations), or from
cementation of a stressed array of crystals, e.g. during burial and cementation or
by crystallization of residual melt in a stressed matrix. The release of stored
stresses by cracking during uplift and erosion is responsible for jointing in rocks.
Item (c) above :
Imagine burial with no lateral strain, because of the constraint of rocks to either
side. In this case z = gz, and a horizontal stress is induced through the Poisson
ratio effect, given by x = z / (1 - ). Such a stress path during burial is called a
ko stress path, and is often invoked to describe (or infer) the stress state in a
sedimentary basin because a basin can be regarded as a thin layer unable to
slide on its contact with basement rocks underneath and prevented from lateral
expansion by the wide expanse of neighbouring sediment on either side.
z (= 1) is given by gh, hence x is given by gh /(1 - ), and will be smaller.
The deviatoric (proportional to differential) stresses i‘ (where i = 1, 2 or 3) at
depth are obtained by subtracting the mean stress, () and pore pressure if
present.
 = (1 + 2 + 3)/3, hence 1‘ = 1 -  = gh(1 +)/([3(1 - )]
and 2‘ = 2 -  = - gh(1 - 2)/[3(1 - )]
Thus the horizontal deviatoric stress is tensional, although the total stress is
compressive.
Thus compaction of sediments in nature is never likely to take place under
hydrostatic stress. Only for very weak
rocks might stress approach hydrostatic
during normal burial.
The ko path (right) is usually flatter than
the critical state line, suggesting that
compaction occurs in the shear-enhanced
compaction regime. The ratio of stresses
x/z in the zero lateral strain path is called
ko (or the coefficient of earth pressure at
rest), and its value is commonly about 0.4
in sands and higher (about 0.7) in clay rich
rocks (the critical state line has a slope
near 1).

39. 39
Since the vertical stress is largest the stress state would be expected to give
rise to normal faulting. Faulting as a mode of failure would only be possible by
‘trading‘ compaction for the lateral extension usually associated with normal
faulting.
Consider the ‘inverse‘ problem, ko uplift and erosion. A pluton crystallizing
at depth under hydrostatic stress has the overburden removed under conditions of
no lateral strain (see Turcotte & Schubert, Geodynamics (Cambridge) p.109).
Thus erosion will induce stresses in the pluton as it approaches the surface. The
change in vertical stress dz is -g dz, progressively restoring it to zero. Thus the
change in horizontal stress becomes :
dx = dz  /(1 - ) from its original value gz. The horizontal stress at
the surface becomes gz - /(1 - ) = (1 - 2)gz/(1 - ).
This can be a substantial horizontal compressive stress, that can cause surface-
parallel cracking called sheeting, and
lead to a characteristic erosion pattern of granitic terrains (see images below).
How high are in-situ differential
stresses?
The figure (left) shows how vertical
stressz (lithostatic gradient) and pore
with = 0.4)pressure (hydrostatic gradient
are expected to vary with depth in a
sedimentary basin.
The shaded area shows how pore pressure
might be enhanced by compaction
below a seal horizon. Excess pore pressure
is easily measured in a borehole,
gz, but thevertical stress is calculated from
principal stresses in the horizontal
plane are estimated from hydrofrac

40. 40
measurements.
(source: Engelder, Stress in the Lithosphere)
The graph left shows variations of
horizontal principal stress Shmax and Shmin
in the San Andreas fault pilot borehole,
about 1.8 km from the fault zone (from
Zoback, Reservoir Geomechanics).
The pore pressure gradient Pp is also
shown.
The stresses are consistent with values
expected in a strike-slip fault regime with a
friction coefficient of 0.8 (equivalent to
laboratory-determined values).
This graph compares stresses
measured in the KTB (German
continental deep borehole) with
laboratory friction values. The in-situ
stresses are closely comparable to
the laboratory-determined values
(coefficient of friction between 0.85
and 0.6), indicating that the upper
crust can sustain stresses over long
time periods just below those needed
to initiate slip on faults.
This map (right) shows max principal
stress orientations on either side of the
San Andreas fault (data from world
stress map). The orientations are at a
high angle to the fault zone, meaning
that resolved shear stresses along the
fault zone are very low, yet the fault
still moves.
There is also a lack of a heat flow
anomaly along the fault zone,
consistent with a weak fault.

41. 41
Finding the explanation for this paradox of a weak San Andreas fault in a strong
crust is the main reason behind the present deep drilling project. Are the fault
rocks anomalously weak, or is there high fluid pressure in the fault zone?
Stress Transfer around Faults
A slip event on a fault causes a redistribution of stress around a fault, leading to a
stress change that can transfer stress to nearby faults and possibly trigger an
earthquake a short time afterwards. This is called stress triggering of earthquakes.
It has been successfully applied in California and in Turkey. In the latter case, a
succession of earthquakes in recent years have propagated from east to west
along the North Anatolian fault. Stress transfer calculations place a likely short-
term future earthquake in the Sea of Marmara just south of Istanbul.
The figure (right) shows calculated stress
changes around a fault segment following an
earthquake. These changes affect the stress
on nearby faults, potentially destabilizing
them. Note: the stress changes are very
small.
(below) Stress changes arising from
earthquakes on faults around Landers (S.
California) eventually led to the triggering of a
large earthquake on the Landers fault
segment.
This triggering mechanism can result in
earthquake storms.
(Figures source: Mechanics of
Faulting, by C. Scholz)

42. 42
Displacements following earthquakes can now be
measured by satellite using Interferometric
Synthetic Aperture Radar (InSAR) with a
resolution on the order of 1 cm (above image).
This allows stress transfer calculations and
modelling of coseismic displacement fields to be
made with great accuracy.
Halokinesis and Salt Tectonics
Under most crustal conditions, halite (NaCl,
rocksalt) is much weaker (less resistant to
intracrystalline plastic flow) than any other
common rock type. Halokinesis refers to the movement of salt by flow.
Halite is also
much less
dense than
other rocks,
even than
porous
siliciclastic
sediments, at
about 2200
kg/m3
compared
to 2500 to 2700
kg/m3
.
Bedded salt is
also very
impermeable,
thus it (and other evaporite rocks such as gypsum) can form very effective
stratigraphic oil traps and seals.

43. 43
The weakness and low density of salt mean that it is possible for salt bodies to
rise through the overburden under the driving force of gravity, punching its way
through along fault planes or by upwarping and puncturing overlying sediments
(diapirism). Mud can also be mobilized in the same way to produce mud
volcanoes.
(above: hot oil globule rising
through cooler immiscible oil
in a ‘lava lamp’ http://www.
fotosearch.com/photos-
images/lava-lamp.html)
(Note: Diaprism is a
manifestation of gravity
tectonics, but the term gravity
tectonics also applies to the
emplacement of extensional
nappes in the late orogenic
collapse of mountain belts).
(above) NASA satellite image
of salt intrusions in SW Iran – look on Google Earth.
There are good examples of salt tectonics in the oilfields of the Gulf of Mexico, the
North sea and in N. Europe. In SW Iran the salt in many places has punched right
through to the surface, forming salt glaciers that can flow downhill at ca 1 m/day
when there is rainfall. Salt layers beneath the seafloor can also flow down slopes,
mobilizing the overlying sediment.

44. 44
(Above: Salt domes in northern Germany (http://my.opera.com/nielsol/blog/salt-
tectonics) and beloe, offshore Angola (http://en.wikipedia.org/wiki/Salt_tectonics). A
great deal of salt tectonics imagery is available by searching salt tectonics on
Google.
In the column of salt forming a diapir, the z (salt) = 2200 g z whereas in the
sediments on either size z (sed) = 2600 g z. The pressure difference is
approximately 400 g z. Thus the surrounding sediments act like a heavy plunger
pushing on the underlying bed of salt and pushing it into the lower pressure region
which is the diapir root. Thus the higher the diapiric column, the greater is the
resultant upward driving force.
Salt tectonics is important (and leads to gas/petroleum traps in many parts of the
world, e.g. Holland and N Germany, Iran, Gulf of Mexico. During the late Miocene
(Messinian stage, ~ 8 Ma), the whole of the Mediterranean basin dried out when the
Straits of Gibraltar were closed off, forming salt deposits on the ocean floor (the
Messinian salinity crisis). These are now buried under a thin sediment veneer.
Similar salt deposits now lying on the continental slope beneath the Red Sea are
able to slide downslope, deforming the sedimentary cover above.
As the salt rises higher it can reach a level of reduced or neutral buoyancy, as the
less compacted overlying sediments become less dense. It can then spread out
laterally with a range of intrusive geometries. Updoming of surrounding sediments
due to diapirism, and formation of sealing structures in faults by penetrating salt,
mean that evaporite structures can form a range of oil trap and seal geometries.
oooOOOooo

45. 45
Nature of Problems in Solid Mechanics
Problems may be classified as:
(a) Statically Determinate
Provided it can be usefully assumed that materials are wholly rigid, undeformable
elastically or plastically, so that the problem of stress can be solved entirely from the
equilibrium condition.
(b) Statically Indeterminate
Both stress AND strain components must be found, so solutions must satisfy the
equilibrium equations, the stress-strain equations and the strain compatibility
conditions.
We will consider each of these in turn.
Examples of Statically Determinate
Problems
(a) A picture suspended by an inextensible
string.
The problem is solved by writing down the equilibrium
condition:
  cosW
  W
cos

46. 46
       
Where T is the tensile force in the string.
(b) A thin walled pressure vessel under internal pressure,
P. We require the tensile stress T in the vessel wall. The thin
wall means we can ignore the variation of the stress across the
thickness of the shell:
Internal pressure x area = wall stress x circumference x
thickness

Pr 
rthence P rt

MORE ABOUT STRESS
Homogenous Stress in 3D and Stress Invariants
It can be shown that principal stress magnitudes are given as the 3 roots of a cubic
equation (i.e eigenvalues of the characteristic equation of the matrix of terms in the
stress tensor:
   3
– I12
+ I2 – I3 = 0
Where:
I1 = 1 + 2 + 3 = x + y + z
I2 = 12 + 2 + 13 = xy + yz + xz – xy
2
– yz
2
– zx
2
I3 = 12 = xyz + 2xyyzzx - xxy
2
– yzy
2
– zxz
2
These are called the STRESS INVARIANTS. They are properties of the stress state
that do not depend on the choice of reference frame. It is not surprizing the the
values of the principal stresses can be obtained from invariants, because they do
not depend on choice of reference frame. Strain energy too can be expressed in
terms of invariants.
Orientations of principal stresses are best obtained by the iterative method based on
the radius-normal property of the representation quadric.
The Principle of Superposition states that 2 stress fields (referred to the
SAME coordinate FRAME) can be summed up adding their corresponding

47. 47
components:
ij = Sij + ij
Thus a total stress tensor can be broken down into the sum of a MEAN STRESS :
(= 1/3 ij kk) (where ij is the Kroneker delta and summation is implied)
and a DEVIATOR stress, ij
‘
(ij
‘
  = ij - 1/3 ij kk)
Deviator = total stress - mean stress
We can also separate the elastic strain energy into Wtotal = Wvolume change + Wdistortion
thus Wvolume = I1
2
(1-2)/[12G(1+)]
Wtotal can be expressed as 3Wvolume - 3 I2/ 6G
Hence Wdistortion = (I1
2
- 3I2)/ 6G = I2‘/ 2G
where I2‘ is the second invariant of the deviatoric stress tensor.
Infinitesimal Strain in 2 Dimensions
Distortion of a rectangle. Note: all changes at length and angle are SMALL
(<1%), so that it can be assumed all products of derivatives and 2nd
and higher
derivatives can be neglected. u and v are displacements along x and y respectively.

48. 48
Line AB changes to A1B1, which is approximately equal to A1B1’ because  is
small. Similarly A1D1 A1D1’
A1B1’ = x + u . x : A1D1’ = y + v . y
x y
B1B1 = v , : D1D1’ = u
x y
The infinitesimal strains are:
x = A1B1 - AB = x (1 + u
/x) - x = u
________ ______________ __
AB x x
Similarly yv 
y
tan  =  = v and tan u
x y
Because angle DAB was originally 90°, =  =  + 
vu
xy
Note: this is the TOTAL shear strain. To form a symmetric tensor it is usual to
define
  + )
2
The components x y  defined above are not independent, because they
arise from gradients of only 2 displacements, u and v.
x = u : x = 2
u : 2
x = 3
u
x y xy y2
xy2
= P
y = v : y = 2
v : 2
y = 3
v
y x xy x2
x2
y = Q

49. 49
 = v + u :  = 2
v + 2
u : 2
 = 3
v + 3
u
x y x x2
xy xy x2
y y2
x = R
Note that P + Q + R, which provides an equation that relates x, y and 
2
 = 2
ex + 2
y
xy y2
x2
This is called the STRAIN COMPATIBILITY EQUATION. There is only one in 2D,
but six of them in 3D. These conditions must be satisfied for any stress distribution
in an elastic solid to be valid.
Variations in Stress through a Body
Let us assume that the stress components vary with distance x, y, z through a
body at the rates

xyzxyxyxy
x yz xyz
Considering only the forces in the x direction first, we can find the resultant force in
that direction. X is any body force along x (force/volume). Consider a block of
material of dimensions x, y, z:
Resultant force = stress x area - stress x area = Fx
Resultant force due to normal stresses:

50. 50
(x + x x) y z - x y z
x
stress area
Plus forces due to shear stresses and body forces:
+ (yx + yx . y) xz - yx xz
y
+ (zx+zx.z) xy - zxxy + X xyz = Fx
z
Force = mass x acceleration = a . xyz ( xyzmass )
= 2
u xyz
t2
u = displacement in x direction. Divide throughout by volume (xyz) of
body and collect terms :
x + yx + zx + X = 2
u . 
x y z t2
Similarly in y and z directions :
xy + y + zy + Y = 2
v . 
x y z t2
xz + yz + z + Z = 2
w . 
x y z t2
These are called the stress equations of motion.
If 2
u = 2
v = 2
w = 0,
t2
t2
t2
the body is at rest or in steady motion, and equations are called the STRESS
EQUILIBRIUM equations. They must be satisfied for all valid stress distributions. In
2D there are 2 equilibrium equations:

51. 51
x + yx = 0 : xy + y = 0
x yxy
Generalizing using summation convention in suffix notation, the equations can be
written
j = 0
x

Stress trajectories: In 2D these can be used graphically to illustrate variation in
stress through a body. Stress trajectories are everywhere parallel to the principal
stress axes. Consider stretching with necking of a ductile bar :
Trajectories are ORTHOGONAL

Total force must be constant, therefore stress trajectories CONVERGE towards
regions of increased stress and DIVERGE toward regions of lower stress.
Local orientation of 1 and 2 are given by tan 2 = 2
x - y
From the theory of stress functions,  can often be obtained as a function of x & y,
from which the local orientations of principal stresses can be plotted on a grid of
points, and the directon field sketched in.
A quick graphical method is the Method of Isoclines:
Writing the slope of the stress orientation as dy/dx, we may have dy = f(x,y)
dx

52. 52
Solutions to this D.E. are y = f (x), a family of orthogonal curves which are the
stress trajectories. If not analytically soluble, the ISOCLINES may be determined:
e.g. if dy = x2
, Set k = x2
hence y = x2
dx y y k
(this particular DE is This is a family of parabolas
easily soluble!) for different k values
The stress trajectories intersect the isocline at a constant angle to the x axis (given
by k), thus the trajectory pattern can be sketched :
Plastic Yield Criteria
The Mohr-Coulomb criterion is reasonably satisfactory for rocks failing by brittle
faulting at low temperatures, but for plastic yield (no mean pressure dependence) of
rocks and metals at low temperatures, other criteria are better. The two simplest
ones are the TRESCA and VON MISES
criteria.
Tresca This states that
yield occurs when the max
shear stress (1 - 2)/2
reaches a critical value k. If
2 = 0, we can plot
If we add the third dimension

53. 53
and all principal stresses are non-zero, the yield surface becomes a regular
hexagonal prism centred on the hydrostat1 = 2 = 3.
von Mises
He suggested yield occurred when the 2nd deviatoric stress invariant reached a
critical level C2
. The criterion can be written as:
6C2
= (1 - 2)2
+ (2 -3)2
+ (3 - 1)2
The significance of C can be seen from a consideration of yield under simple
uniaxial loading, 1 = u ; 2 = 3 = 0
6C2
= 2 u
2
, hence C = u / 3
The von Mises yield criterion effectively states that yield occurs at a critical value of
distortional strain energy, for Wdist = I2‘/2G
Octahedral Stresses
Imagine a plane inclined equally to the 3 principal stresses. This is the Octahedral
Plane.
The normal stress across it n oct = (1 + 2 + 3)/3 = I1/3
The max resolved shear stress across it
oct is
oct = (1/3) {(2 - 3)2
+ (3
- 1)2
+ (1 - 2)2
} = (2I2/3)
thus the von Mises criterion is equivalent
to the statement that yield occurs when
oct reaches a critical value.

54. 54
Rocks yield in a pressure-sensitive way at low pressures, but flow is pressure-
insensitive at higher pressures (right).
Relation between Tresca and von Mises criteria
In 2D the von Mises is an ellipse that circumscribes the Tresca hexagon. i.e. they
are equivalent at 6 points. In 3D a von Mises circular cylinder circumscribes the
Tresca hexagonal prism, thus deviations from the hydrostat in any direction
produces yield. Experiments show that von Mises describes data better. Tresca is a
more conservative criterion than von Mises.
There are other variants on the Tresca and von Mises criteria that we will not
consider further here.
Statically indeterminate problems
Here we require to find all the stress and strain components (total 6 in 2D, therefore
need 6 equations, these are the stress equilibrium, strain compatibility and elastic
stress strain equations).
If we first consider PLANE STRESS (y, xy, zy = 0).
The solid is assumed ISOTROPIC and ELASTIC and HOMOGENEOUS.
The stress/strain relations become:
x = (x - z)/E (1)
z = (z - x)/E
y = -(/E)(x + z)
(ie: plane stress does not give rise to plane strain unless x = - z)
zx = xz = xz / G : G = E
2(1 + )
The equilibrium equations without body forces are:
x +  = 0 :  + z = 0 (2a) (2b)
x z x z

55. 55
The compatibility condition is:
2
x + 2
z = 2
 (3)
z2
x2
xz
Substituting (1) into (3) we can eliminate strains:
2
(x - z) + 2
(z - x) = 2(1 + ) 2

 z2
x2
xz (4)
Differentiation of (2a) wrt x and (2b) wrt z gives:
2
x + 2
 = 0 : 2
 + 2
z = 0
x2
zx xz z2
Adding:
2 2
 = - 2
x - 2
z (5)
zx x2
z2
Substitute (5) into (4) to eliminate shear stress terms:
2
(x - z) + 2
(z - x) + (1 + ) { 2
x + 2
z } = 0
z2
x2
x2
z2
rearranging causes the  terms to cancel out:
2
x + 2
z -  (2
z + 2
x) + 2
x + 2
z +  (2
z + 2
x) = 0
x2
x2
z2
x2
z2
z2
z2
x2
2
(x + z) + 2
(x + z) = 0
x2
z2
OR ( 2
+ 2
) (x + z) = 0
x2
z2
OR 2
(x + z) = 0 (6)
2
is the Laplacian operator, and (6) is called the STRESS COMPATIBILITY
CONDITION. Stress states satisfying (6) also satisfy (1), (2) and (3).

56. 56
Note that: x + z = 1 + 2 = I1
Also  = hydrostatic part of ij = I1/2 = P
Therefore, for any hydrostatic pressure distribution

2
P = 0
This is Laplace’s equation. It also applies to ANY steady-state distribution of
potential (e.g. electrical potential, ionic concentration). Solutions take the form
P = f (x, z), i.e a contour map of the potential in space.
Laplace’s equation is the first of a family of partial differential equations that
collectively describe an enormous range of physical phenomena:

2
 = k is Poisson’s equation
 2
 = k  is the heatflow, or transport equation. Its solutions
t describe monotonic approaches of the potential  towards
equilibrium with time.
 2
 = k 
 Has solutions periodic in time and is the wave equation
t 
 2
 = ik  Has solutions periodic in time but with quantized energy
t levels (amplitudes of the waves). This is Schroedinger‘s
equation.
PLANE STRAIN
Here xy = zy = y = 0 (i.e. y = 0 or a constant)
y
Using y = 1/E (y – (x + z)) and setting y = 0 :
y = (x + z)
i.e. stress y must be applied to keep y = 0
The stress/strain equations become:

57. 57
x = 1/E {(1 – 2
) x – (1 + )z}
z = 1/E {(1 – 2
) z – (1 + )x}
xz = xz = 2(1 + ) xz
G E
The equilibrium and strain compatibility conditions are the same as for plane stress.
Combining the 3 sets of equations leads to the same stress compatibility equation,
 2
I1 = 0
AIRY STRESS FUNCTIONS
The Airy stress function, , is defined as:
x = 2
 : z = 2
 : xy = - 2

z2
x2
xz

diffn. w.r.t. x diffn. w.r.t. z
x = 3
  = - 3

x z2
x z xz2

Add
x +  = 0
x z
i.e. is defined in such a way that it satisfies the stress equilibrium conditions.
Substitute for x and z in the stress compatibility equation, 2
(x + z) = 0
(  +  ) ( 2
 + 2
 ) = 4
 + 24
 + 4
 = 0
x2
z2
x2
z2
x4
z2
x2
z4
OR 

( 2
) = 0

58. 58
OR 
4
= 0
This is the BIHARMONIC EQUATION. Any function (x,z) satisfying this is
automatically a valid stress distribution in PLANE STRESS or PLANE STRAIN, i.e.
a 2D stress distribution.
If 1 and 2 are solutions to 
4
 = 0, then1 +2 is also a solution. i.e. the
superposition principle applies.
Examples of simple solutions :
(A) Polynomial Functions. Any polynomial to 3rd degree automatically
satisfies biharmonic equation, because the 4th order derivatives are zero.
e.g. Let
 = c + ax + bx2
+ cy2
+ dxy + ex3
+ fy3
+ gx2
y + hxy2
linear quadratic cubic
terms terms terms
1 2 3
Linear terms = 1 When differentiated twice to get stresses, they are all zero.
Linear terms give rise to NO STRESSES.
Quadratic terms e.g. 2 = k1x2
+ k2 y2
Hence x = 2k2 : y = 2k1 : xy = 0
Therefore x and y are1 and 2
This represents:
x and y are constant through the material
The stress-strain equations can then be applied to find the strains:
x = 1 (2k2 - 2k1) = 2 (k2 - k1)
E E

59. 59
y = 2 (k1 - k2)= z = -2 (k1 + k2)
E E
z = 0 only when k1 = -k2
To find displacements
x = u u = x dx
x
= 2(k2 - k1)x + f (y, z) + C
E
Boundary conditions must be specified before displacements can be found.
The term in xy, set  = dxy, x = 0, y = 0, = d
i.e. the xy term gives rise to a homogeneous shear stress
Overall, the quadratic terms describe a HOMOGENEOUS, GENERAL STRESS
STATE
The Cubic Terms
(a) = ex3
x = 0
y = 6 e x
 = 0
y increases linearly with x alone
(b)  = fy3
, x = 0 : y = 6fy ;  = 0
i.e. x increases linearly with y.
This is the type of stress distribution
that occurs in a bending elastic
beam:
Stresses are proportional to distance
from neutral surface.

60. 60
(c) Terms in x2
y and y2
x
Let  = gx2
y x = 0 ; y = 2gy ; = 2gx
Thus y = Linear function of y
Similarly, for  = hyx2
x = 2hx ; y = 0 ;= 2hy
Thus the cubic terms give rise to a heterogeneous stress state where components
vary linearly with x and y. For polynomial terms of higher degree the stress
components vary non-linearly.
(B) Periodic Functions that satisfy the biharmonic
e.g.  = ey
cos x
x = 2
 = ey
cos x 4
 = ey
cos x
y2
y4
y = 2
 = - ey
cos x 4
 = ey
cos x
x2
x2
 = 2
 = + ey
sin x 2 4
 = - 2 ey
cos x
xy x2
y2
add = 0
hence this function satisfies the biharmonic eq.
As y increases the stress
components increase
exponentially. In the
region y=0 the
components look like:
Plotting of stress trajectories :
Use tan 2 = 2 = -22
xy
x - y (2
 /x2
) - (2
 / y2
)

61. 61
Here, tan 2 = 2ey
sin x / 2ey
cos x = tan x
2 = x,  = x/2
Using isoclines method,
dy/dx = tan x = tan2
When x = 90°,  = 45° 
Isoclines are k = tan x 
When k = 1, x = 45°,  = 22.5°
hence x = tan-1
k, a family of
straight lines parallel to y axis.
Stress trajectories are similar to those in a
buckled viscous layer.
Delineation of probable failure zones
Mohr Coulomb or von Mises etc. criteria may be used to map out 1/2 conditions
that may lead to failure:
e.g. for Mohr Coulomb
_________________
1, 2 = ½(x + y) + ½ (x - y)2
+ 42
for each point in the material, or may be expressed as
1, = f1 (x, y); 2, = f2 (x, y)
Thus 1 (x, y) = a + b 2 (x, y)
Geological Example of Cartesian stress functions
The best known ‘classical’ study is that of Hafner (1951) “Stress distributions and
faulting” BGSA 62, 373-398. Hafner used Anderson’s “standard state” concept to
simplify the problem:
1 = Anderson‹s Standard state (Hydrostatic stress)
2 = Deviator (Supplementary stress, causing deformation)
These can be superimposed by addition:1 + 2
Boundary conditions at Earth‘s surface, = 0, @ z = 0

62. 62
also z = 0 at z = 0. z = f(z) is assumed given by z = gz.
If there is NO LATERAL STRAIN, the horizontal stress induced by gravity is
x = z / (1 - )
x = z only if  = 0.5 (incompressible fluid). However, for the “standard
state”, we assume
x = z = gz
and  = 0
With pore fluid pressure contribution, z = gz - P(z)
Derivation of Supplementary stress states (2)
If we assume the supplementary stress state consists only of a horizontal stress, the
implications can be explored:
thus z = 0
i.e. 2
 /x2
= 0 for all values of z.
This can be integrated twice to find the range of possible supplementary stress
functions :
2 = z dxdx = af1(z) + b
hence 2 = axf1(z) + bx + cf2(z) + d
This must satisfy the biharmonic equation, therefore the 2nd order derivatives of
f1(z) and f2(z) must be zero, constants or linear functions of z.
The stress components therefore become :
x = 2
2 / z2
= ax 2
f1(z)/ z2
+ c 2f2(z)/ z2
: z = 2
2 / x2
= 0
 = -2
2 / xz = - (af1(z) + b) = -af1(z)/ z
By examining combinations of linear functions, constants or zero for f1(z) and f2(z)
Hafner found that 3 groups of equations can be found that satisfy the surface
boundary conditions,
z = 0 :  = 0 : f1(z)/z = 0

63. 63
The 3 groups are as follows :
x z 
cz + f 0 0 Group I
ax 0 -az Group II
axz 0 -az2
/2 Group III
Linear combinations of these solutions for supplementary stresses are also
possible. Hafner explored all of them, and concluded that superposition of solutions
I and II were geologically most interesting (note g z has to be added to the normal
stress components to get the total stresses) :
Superposition of solutions I and II
x = ax + cz + f + gz ; z = gz ;  = -az
These can be illustrated graphically as follows :
Orientations of principal stresses obtained from: tan 2 = 2  = -2az
x- z (ax+cz+f)
 = ½ tan-1
( -2az )
ax+cz+f
_____________
Values of 1, 2 computed from ½(x + z) + ½ (x + z)2
+ 42
________________
= ½(ax+cz+f+2gz) + ½ (ax+cz+f)2 -4a2
z2

64. 64
Boundaries of potential failure areas can be delineated by
1 (x, z) = A + B (2 (x, z))
Resultant Pattern :
Geological ‘realization’
A foreland fold and thrust belt! Hafner carried out sandbox experiments to illustrate
this geometry of upward-curving thrusts.
Stress Functions in Plane Polar Coordinates
Relation between Cartesian and polar
coordinates
x = r cos 
y = r sin 
Stress components therefore become , r and
r
The body of theory is comparable to the Cartesian case so the derivations will not

65. 65
be repeated.
The stress equilibrium equations become :
 r/ r +  /r + (r - )/r = 0
in the radial direction and
 /r + /r + 2/r = 0
in the tangential direction.
The infinitesimal strain-displacement equations are :
r = u /r ;  = u/r + v/r)  ;  = u/r  + v/r - v/r
The strain compatibility equation is
2
r / r = 2r / 2
+ r 2
r / r 2
- r r / r
The stress/strain equation have the same form as in Cartesians. The stress
compatibility equation also looks the same :

2
(r + ) = 0 and may be derived as before. Alternatively one may convert
the Laplacian operator into plane polars. The result is :
( 2
+ 1 2
+ 2
) (r + ) = 0
r2
r 2
2
r
The Airy Stress Function. 
This is defined in a corresponding way to the Cartesian case, such that it
automatically satisfies the equilibrium equations:
r = 1 12
 :  = 2

 rr r2
2
r2
11 - 1 2


66. 66
r r  r2
 r r 
Substitution of these definitions into the stress compatibility equation results in the
biharmonic equation as before.
The circular hole in an infinite plate
This is the most important result in the theory of stress functions in polar
coordinates.
Finding the solution to the stress distribution is an example of a Boundary Value
Problem. One way of solving problems of stress distribution is to make intelligent
guesses about the form of the stress function and then to find out what problems it
solves. This can be done with this problem.
The boundary conditions are:
At x = Sx = S1 (max principal stress in the far-field)
At y = Sy = S2
As xy the stress field is uniform and stress trajectories are horizontal and
vertical.
(r)r = ½ (Sx + Sy) + ½(Sx - Sy) cos 2
r  = + ½(Sx - Sy) sin 2
also, at r = a, r =  = 0
The form of the appropriate stress function (part of the general series solution of the
biharmonic eqn) is:

67. 67
 = A log r + Br2
+ (Cr2
+ Dr4
+ Er-2
+ F) cos 2
The first two terms on the right hand side do not depend on , but the rightmost
does. A, B, C, D, E, F are constants that must be determined using the boundary
conditions.
Differentiating to get the stress components:
r = 1 12
 = A/r2
+ 2B + (-2C - 6Er-4
- 4Fr-2
) cos 2
__ __ + __ ___ ___
 rr r2
2
 = 2
= -A/r2
+ 2B + (2C +12 Dr2
- 6Er-4
) cos2
r2
11 1 2
 = (2C + Dr2
- 6Er-4
- 2Fr-2
) sin 2 
_ ( _ __ ) = _ __ - _ ___
r r  r2
 r r 
These stress components can be set at the far-field boundary and around the
periphery of the hole. Coupled with the condition that the solution must be true for
all  and stresses must not become infinite in the far-field, 6 simultaneous
conditions to be satisfied are obtained:
D = 0: 2B = ½(Sx + Sy) : -2C = ½(Sx - Sy) : A/a2
+ 2B = 0
-2C - 6Ea-4
- 4 Fa-2
= 0 : and 2C - 6Ea-4
- 2Fa-2
= 0
These are solved to give the non-zero constants:
A = -a2
(Sx + Sy)/2 : B = (Sx + Sy)/4 : C = -(Sx - Sy)/4
E = -a4
(Sx - Sy)/4 : F = a2
(Sx - Sy)/2
And substituted in the general equations for the stress components to give the
stress components :
r = ½(Sx + Sy)(1 - a2
/r2
) + ½(Sx - Sy)(1+ 3a4
/r4
- 4a2
/r2
)cos2

= ½(Sx + Sy)(1 + a2
/r2
) - ½(Sx - Sy)(1+ 3a4
/r4
)cos2
= + ½(Sx - Sy)(1- 3a4
/r4
+ 2a2
/r2
)sin2

68. 68

Add +P a2
/r2
to normal stress terms if pore pressure P present - see below
where a is the radius of the hole. This is called the Kirsch solution. Note that the
stress components consist of a part dependent on r only and a  and r dependent
part.
For uniaxial stress (Sy = 0), the stress trajectory pattern looks like
Special Cases :
When Sx = Sy = P these reduce to
r = = P (1 - a2
/r2
) :  = 0
 = P (1 + a2
/r2
)
This is the case of the infinitely thick-walled pressure vessel, I.D. = 2a, subject to all-
round hydrostatic pressure. By subtracting P at all points, we obtain the case of the
hole in an infinite plate subject to internal hydrostatic pressure, P, and zero load at r
= infinity.
r = Pa2
/r2
:  = -Pa2
/r2
:  = 0
Note: the stresses die away as 1/r2
This corresponds to the volcanic centre under internal magma pressure, P, or a

69. 69
wellbore overpressured by the weight of drilling mud. Tangential stresses are
tensile at r = a, so radial dykes (cracks) can develop.
Returning to the external pressure case,  max occurs at the hole boundary, r = a,
hence
r = P(1 - 1) = 0 :  = P(1 + a2
/r2
) = 2P
thus  is twice the applied pressure. This is the simplest case of stress
intensification at a geometric imperfection.
Max shear stress is also at r = a,  = (r)/2
= (2P - 0)/2 = P.
This is in a plane oriented at 45° to the
boundary. The principal stress directions are
radial and tangential and are isoclines. The
max shear stress trajectories are a possible
slip line field, and define spiral surfaces. Slip
on these tends to close the hole, and plastic
borehole collapse (e.g. in salt) may occur in
this way. Here is an example quotation of
this behaviour from a drilling engineer:
“ When drilling in the southern North Sea Gas
Basin it was possible for a 5" drill hole to
close up sufficiently to jam the bottom hole
assembly (BHA) in the time it took to drill 31
feet in salt. This would occur with diesel based
drilling mud sufficiently weighted to prevent hole collapse.
>
This would occur at depths of a few thousand feet (6-7000ft) at
relatively low temperatures. Jarring would be required to free the
BHA, although a few were left down hole after.
>
So salt can move laterally into a free space 25-30mm in 15 minutes at
temperatures of approximately 50-70C ! “
Stabilizing effect of drilling mud
Drilling mud is heavier than water and, provided it is sufficiently viscous that it
does not penetrate into the wallrock pores, where it would add to pore pressure,
exerts an internal radial pressure on the wellbore (viscosity increasing agents can
be added to the drilling or hydrofrac fluid to compensate for the decrease of
viscosity with temperature at depth in the hole). This counteracts the  (hoop
stress) to a degree (makes it less compressive) and reduces the risk of failure that
would tend to collapse the wellbore. If the mud weight is too great, it puts the hoop
stress into tension and induces radial tensile cracks. Sometimes additives to the

70. 70
mud are used to reduce penetration of the mud into the wallrock.
When one principal stress is zero (Sy = 0) :
Max tangential stress occurs when cos 2 = -1, i.e. at  = /2 or 3/2 & r = a.
 = 3Sx (compressive)
Min tangential stress occurs when cos 2 = 1, i.e. at  = 0 or  and r = a.
= - Sx, i.e. a TENSILE stress.
2) Non-hydrostatic loading - Borehole breakouts and drilling-induced tension
crack formation
When one principal stress is zero (Sy = 0) :
Max tangential stress occurs when cos 2 = -1, i.e. at  = /2 or 3/2 & r = a.
 = 3Sx (compressive)
Min tangential stress occurs when cos 2 = 1, i.e. at  = 0 or  and r = a.
 = - Sx, i.e. a TENSILE stress.
Thus the following possibilities for
failure around the hole are as
illustrated in figure right:
Failure of the borehole wall 90° to the
compression direction leads to the
formation of borehole breakouts. The
failure can either involve tensile
fracturing, causing fragments to spall
into the hole, or shear failure, with the
same result.
In a porous sandstone, if failure
spreads around the hole (e.g. if the
differential stress is small) the hole
diameter increases (forming a
washout), this leads to sand
production. Extensive borehole wall failure can
result in as much sand as oil being transported up
the hole.
A tension crack running into the wall parallel to Sx is
stable, and does not penetrate far from the borehole
wall.
The orientation of the breakouts can be sensed with
caliper gauges or borehole televiewer images (e.g.
right, the horizontal axis is angular distance around

71. 71
the hole, the vertical axis is depth range (1 metre), the arrows point to the
breakout features: source Zoback, Reservoir Geomechanics, Cambridge 2007, and
leads to an estimate of the orientation of the maximum and minimum principal
stresses in a plane normal to the borehole. This is the main source of stress
orientation evidence used for the construction of the world stress map.
The occurrence of breakouts and drilling-induced tensile cracking in the televiewer
images indicates relatively high differential stress is present.
(Photo above) Optical thin section of an experimentally produced breakout in
sandstone. Breakout features form parallel to the minimum stress. Cuss, Rutter &
Holloway, 2003, Int. J. Rock Mech Min. Sci., 40, 847-862
Hydraulic fracture in borehole walls.
A single tension crack running parallel to the maximum compressive stress can be
induced by pumping up the borehole fluid pressure until the crack forms when the
fluid pressure exceeds the least principal stress in the borehole wall by the amount
of the tensile strength. As fluid continues to be pumped into the hole, the crack
extends farther into the wallrock, and the injection pressure becomes equal to the
least principal stress. This is the basis of the hydrofrac method, which has two
uses (see Zoback, Reservoir Geomechanics).
(Do not confuse hydrofrac with drilling-induced tension cracking).
(a) for stimulating enhanced hydrocarbon recovery by increasing the effective
surface area of the hole (a widely used technique in secondary oil recovery),
(b) to measure the magnitude of in-situ stress. The tension crack forms when the
fluid pressure exceeds the minimum stress by the amount of the tensile strength
(which is always small). Together with other data, the form of the Kirsch solution can
also be used to estimate the maximum principal stress.

72. 72
To carry out a hydrofracture test, the bottom of the hole is packed off with an
inflatable packer (inflated with compressed air), through which the overpressured
fluid (water) can be pumped. To hydrofrac a section of the hole above the bottom, a
double packer (straddle packer) is used.
Hydrofrac assembly with straddle packer (above),
and single packer details (left) (source of figures,
Engelder, Stress in the Lithosphere).

73. 73
Graph (above) showing pressure cycles applied during a hydraulic fracture test
(also known as a Leak-off test [LOT]). Formation of the fracture stops the pressure
rise as fluid is forced into the expanding crack. Fracture occurs when the fluid
pressure exceeds the minimum stress in the plane normal to the borehole by the
amount of the tensile strength of the rock (usually about 1 MPa). In this way the
minimum stress SHmin is obtained.
Unwanted hydrofrac causes loss of circulation.
Note: further transient stress state modification can be caused by the cooling effect
of the drilling mud. This usually helps borehole stability.
Borehole breakout width and the problem of finding SHmax
Variation of stress components
around (vertical) borehole wall:
If the unconfined compressive
strength is known, the breakout
width can be predicted. Conversely,
breakout width can help constrain
SHmax. At higher SHmax, the
breakout gets wider for a given rock
strength (see figure below).
Whether a tension crack is induced in
the SHmax direction depends on the
relative values of Shmin and SHmax. If
Shmin is zero this is most favourable
for a tension crack because a tensile
stress = -SHmax is induced at the
bore wall. A tension crack appears
on the televiewer image as a
thin line halfway between the
breakout orientations. If
effective Shmin is compressive
the  sinusoid is lifted above
the zero line and a drilling-
induced tension crack cannot
form.
The mean value of Sv (= Sz) is
given by the overburden
pressure, but it varies
sinusoidally around the
borehole wall (right) owing to

74. 74
the Poisson ratio effect and the SHmax variation.
SHmax from hydrofrac experiments
For relatively shallow (<2km) boreholes a ‘classical‘ method (Haimson & Fairhurst
1970) was developed for estimating SHmax from hydrofrac experiments.
Leaving aside the effects of mud weight and thermally induced stresses, knowing
starting pore pressure Pp and tensile strength of the formation T, it is possible to
show that
SHmax = 3 Shmin - Pb - Pp + T
in which Pb is called the breakdown pressure and is obtained from the hydrofrac
test. After an initial hydrofrac has formed, Pb is the pressure that has to be re-
applied to start the fracture running unstably.
For deeper holes methods based on the observation of wall failure are more
reliable.
Figure (right) shows
stress measurements
in the vertical San
Andreas fault pilot
borehole (you have
seen this figure
before). SHmax and
Shmin are shown as
squares and circles.
Sv is the vertical stress
profile calculated from
Sv = g z.
Note that Shmin is
approximately equal to
Sv.
SHmax is the greatest stress, acting in the horizontal plane. The relative values of
SHmax, Shmin and Sv indicated the type of faulting implied by the Anderson theory
of faulting. In this case either strike- slip or thrust (reverse) faulting is equally likely.

75. 75
Pp shows pore fluid pressure profile expected for a  value of 0.4 (normal
hydrostatic gradient).
The dashed line shows expected value of SHmax assuming friction coefficient of
0.8.
The value of SHmax may in this case be limited by frictional sliding along favourably
oriented small faults, and the data show consistency with respect to laboratory
friction measurements.
Images of Experimental Hydraulic Fracturing
(a) Shale
(b) In transparent perspex (PMMA)
Here, pink wax has been used as
the hydrofracture fluid. Because it
is more viscous than water, the
crack grows more slowly. Two
hydrofracs have been produced
from opposite sides of the central
hole. The crack tip extends
beyond the limit of fluid
penetration. the separation is
called the fluid lag.

76. 76
Final note: Obtaining stress magnitudes from borehole failures and hydrofrac tests
is generally a complex business that requires careful interpretation of each situation
according to borehole orientation and rock properties. This is even more true for
deviated and horizontal boreholes. Refer to Zoback‘s book on Reservoir
Geomechanics and original references therein for a more thorough treatment of
these issues.
oooOOOooo
Obtaining information about in-situ stresses from sonic logs
A sonic logging tool carries an array of ultrasonic transducers that can send P and S
wave pulses along a section of a borehole. Measuring speed of sound in the
borehole wall carries information about the elastic properties of the wallrock, and
can for example be used to detect when a different rock type or a zone of pore fluid
overpressure is entered.
Microcracking generally results in a reduction of sonic velocity and a reduction of
the Vp/Vs ratio. If the microcracks have a preferred orientation, the rock is rendered
seismically anisotropic.
Microcracks produced during
progressive loading are generally
strongly oriented parallel to the

77. 77
direction of applied max stress (perpendicular to the min stress).
Figure (above) shows volumetric expansion in Merrivale Granite resulting from
formation of axially-oriented microcracks.
Crack formation affects elastic properties hence acoustic velocities e.g. for Vp (data
of Colin Sayers, Schlumberger).
Here, microcracking is
axisymmetric along the maximum
loading axis (33), so in the 22 and
11 directions waves propagate
across the cracks, hence are
slowed down. At a small stress,
the pre-existing (intrinsic) rock
anisotropy is reversed as a result
of the formation of stress-induced
microcracks.
Shear Wave Birefringence
An array of planar-oriented microcracks
causes shear wave anisotropy and shear
wave splitting (birefringence). The fast wave is
that vibrating parallel to the crack array, that is
generally parallel to the max stress direction
in the horizontal plane (or the plane
perpendicular to the borehole).
Thus a log of fast Vs azimuth indicates
orientation of max local stress:
(sources: Schlumberger)
Shear wave splitting is caused by the intrinsic

78. 78
rock properties (bedding fabric,
crystallographic preferred orientation, pre-
exisiting microcracks). Pre-existing cracks
tend to close normal to max stress and open
normal to min stress, so can carry information
about the stressed state of the rock.
Relative slowness of the split shear waves is frequency dependent (dispersion,
velocity is slower at high frequencies), but when the birefringence is wholly or
partially stress-induced, there is a crossover effect produced, the magnitude of
which is proportional to the differential stress.
This effect has in recent years been used to develop patented methods
for inverting in-situ stresses from sonic logs.
The theoretical aspects are complex, and interpretation of the results
must be done with care and take into account all information available.

79. 79
An excellent description of anisotropic elasticity is given by J.F. Nye,
Physical properties of crystals: their representation by tensors and
matrices. Cambridge Univ. Press. 1957. Super bedtime reading.
Further developments from the circular hole in the infinite
plate - the elliptical hole.
It is convenient to change
the coordinate system to
elliptical cordinates, of
which plane polars is a
special case (when C = 0).
The system is of a series
of confocal ellipses and
hyperbolae:
The transformation from
Cartesian to elliptical
coordinates is:
x = C sinh  sin 
y = C cosh  cos 
where 2C is the distance between the foci.  labels a sequence of confocal
ellipses and  labels a series of confocal hyperbolae.
If a particular ellipse o corresponds to the boundary of an elliptical hole, long axis
inclined to the y axis at angle , and subject to far-field principal stresses P1 parallel
to y and P2 parallel to x, by the methods of conformal transformations it can be
shown that the variation of tangential stress t around the hole boundary are given
by:
t = (P1 + P2) sinh 2o +(P1 - P2)(exp 2o cos2() -cos 2)/(cos 2o - cos 2)
o describes hole shape, and  describes position around the boundary.
The relations between o and the major and minor semi-axes a and b of the ellipse
are:
b/a = sinh o / cosho = tanh o

80. 80
Griffith (1924) supposed that the weakness of brittle solids was due to crack growth
fro pre-existing flaws (Griffith cracks), such that when the most favourably oriented
one became critically stressed, it would grow and cause tensile failure of the
material.
There are two ‘most critical‘ orientations:
(a) with the crack normal to max tensile stress - crack grows in-plane.
(b) a crack at 30% to max compression, crack grows out of plane towards max
compressive stress. This results in axial cracking.
A failure criterion can be derived, that predicts when the most unfavourably oriented
flaw will start to grow:
(a) when the least principal stress is zero or tensile, 1 < -3T, where T is the tensile
strength of the material, and is in principle measurable.
(b) When 1 > -3T, in which case the failure envelope becomes
(1 - 3)2
/(1 + 3) = - 8T
This is a parabolic failure criterion. Cohesive strength is predicted to be - 2T, and
uniaxial compressive strength is - 8T. Remember, this criterion predicts the stress
conditions at which cracks start to grow. In tension, one growing crack can cause