xx xy  0.1 0 
ij  
 

0.2
yx yy   0
y
(0,11)
(9,12)
(0,10)
(10,10)
(9,0)
(10,0)
X
 xx  yy  0.1 0.2  0.1
This is the trace of the strain tensor. In general the trace of the strain
tensor gives area change in 2-D and volume change in 3-D

The principal axes are directions along which the starting vector and
ending vector are parallel
Pure shear = principal axes do not rotate with time
Principal Axes
1
2
3
= maximum stretch direction
Intermediate stretch direction
Minimum stretch direction
(or most contractional)
The principal axes are all mutually orthogonal to
one another
xx xy  0.1 0.1
ij  
 

yx yy  0.1 0.1
x  0.1x  0.1y y  0.1x  0.1y
(10 ,0)becomes
(11,1)
(10,-10) remains fixed, as does (-10, 10)

(0, 10) becomes (1,11)
(10,10) becomes (12,12) etc...
y
(12,12)
(1,11)
(10,10)
(0,10)
(11,1)
(10,0)
X
In principal axis coordinate system
this tensor can be written:
xx xy  0.2 0.0
ij  
 

yx yy  0.0 0.0
Simple Shear
• In Simple shear the principal axes rotate
with increasing shear
• Simple shear applies only to finite strain
Rotational
strain
Marker
This part of marker
not disformed
Stress = Force/Area
Force is measured in units of
mass*acceleration
1 N (Newton) = 1
kg * m * s-2
another common unit for force is
the pound
Pressure is a number. It corresponds to a special kind
of stress.
Stress is a tensor, but it has the same units as
pressure (Pa)
1000 Pa = 1 kPa
1,000,000 Pa = 1 MPa (about 10 bars)
Traction is a Vector
• Tractions are vectors = force/area
• Traction can be resolved into two
components
Normal component to plane = normal
stress
Tangential component = shear stress
The stress tensor
•
•
•
•
The stress tensor is symmetric
The stress tensor has 3 principal axes
The principal axes are mutually orthogonal
principal axis = direction in which the
traction vector is parallel to normal to plane
=> no shear stress resolved on that plane
1
= maximum compressive principal stress
2
= intermediate compressive principal stress
3
= minimum compressive principal stress
Normal Stress and Shear Stress
 n = Normal Stress resolved on plane

= shear stress resolved on plane
Anderson Faulting Theory
• If 1 is vertical then a new fault will be a
normal fault (extensional)
• If 1 is horizontal and 3 is vertical then
reverse (thrust) fault (contractional faulting)
• If 1 and 3 are both horizontal then strikeslip (transcurrent) fault
Fault Angles and Principal
Stresses
 2 in the plane of the fault
 1 20°-40° from the plane of the fault
 3 50°-70° from the plane of the fault
3


1
1
3
n = (1+3)/2 - [(1-3)/2] cos 2
 = [(1-3)/2] sin 2
THESE ARE ALSO THE EQUATIONS FOR A CIRCLE
WITH A RADIUS OF (1-3)/2
AND A CENTER (1+3)/2 TO THE RIGHT OF WHERE
THE AXES CROSS!!!!

pe
o
l
ve
n
e
d
el
i
y
al
n
o
i
t
c
i
fr
=
atan 
n
fric
tion
al
yie
ld e
nve
lop
e


3
1
3
1
n
Let’s Look at internal friction
angles, coefficients of friction,
and theta
• If =10° (so =tan=0.18), then 2=80°,
so =40° and 1 axis is 40° from the
fault plane.
• If =20° (so =tan=0.36), then 2=70°,
so =35° and 1 axis is 35° from the
fault plane.
• If =30° (so =tan=0.58), then 2=60°,
so =30° and 1 axis is 30° from the
fault plane.
• If =40° (so =tan=0.84), then 2=50°,
so =25° and 1 axis is 25° from the
fault plane.
Cohesion
• Cohesion = shear strength that remains even
when normal tractions are zero
• Byerlee’s law with cohesion
• The cohesion represents the intercept value
Pre-existing faults
• If there are pre-existing faults, then figure in
previous slide predicts a range of
orientations of faults, with respect to
maximum principal stress direction that can
slip
• If there are no pre-existing faults, then only
one orientation is possible
Role of Fluid Pressure or Pore
Pressure
• Hydrostatic Pressure: Phydrostatic = water g z
• Lithostatic pressure is when entire weight of
the overlying rock (density rock) is being
supported
• Plithostatic = rock g z
Fluid Pressures and Tractions
• Fluid Pressures can support normal tractions
but not shear tractions!
• Elevated fluid pressures make the Mohr
circle move to the left
Effective Stress
• Effective Stress = total stress minus the
fluid Pressure
 1' = 1 - Pf
 2' = 2 - Pf
 3' = 3 - Pf
• Shear Tractions are not affected!

Pf
3
1
3
1
n
conjugate tensile
fractures (joints)

conjugate faulting
tensile crack or joint (with
a single orientation normal
to minimum stress axis)
n
Joints
• The