Geology Reference
In-Depth Information
traction vector
T
from a direction vector
n
.In
general, the components of this tensor change
with the position within the body. However, its
symmetry guarantees that we can always find
three directions
n
i
such that the shear stress is
zero along surface elements orthogonal to
n
i
.In
these conditions,
n
i
and
T
(
n
i
) are parallel each
other, so that by Cauchy's theorem (
7.3
)wehave:
T
n
By convention, the three stresses ¢
i
are sorted
in such a way that:
j
¢
1
j
>
j
¢
2
j
>
j
¢
3
j
.When¢
1
D
¢
2
D
¢
3
we say that the stress field is
hydrostatic
.
In liquids, the stress tensor is
always
diagonal and
¢
i
-
P
,where
P
is the
pressure
. An important
feature of the principal stresses ¢
i
at a point
r
is
that they allow to predict the planes of maximum
shear stress at that point. In geology, these are
the planes along which the probability of rupture
and faulting, for a given stress field, reaches a
maximum. Now we are going to show that there
are just two orientations of
n
such that the shear
component of
T
(
n
) is maximum, and ¢
1
and ¢
3
form angles of 45
ı
with each of these planes. Let
us consider an arbitrary plane with normal versor
n
in the principal axes coordinate system. In this
instance, by Cauchy's theorem the normal stress
along this plane is given by:
i
D
œ
i
n
i
i
D
£n
I
i
D
1;2;3 (7.5)
where the quantities œ
i
are scalars. To find a ver-
sor
n
i
that satisfies (
7.5
), we solve the following
eigenvalue equation:
i
.
£
Iœ
i
/ n
D
0
(7.6)
where
I
is the identity matrix of order three. The
parameter œ
i
that satisfies this equation is the
eigenvalue
of the equation, while
n
i
is the
eigen-
vector
. Equation
7.6
represents a homogeneous
system of three linear equations in the unknown
eigenvector components
n
j
, which has non-trivial
solutions only when:
T
N
.n/
D
T .n/
n
D
¢
i
n
i
(7.10)
Therefore, the squared shear stress component
will be given by:
T
S
.n/
D
T
2
.n/
T
N
.n/
D
¢
i
n
i
¢
i
n
i
2
(7.11)
det.£
Iœ
i
/
D
0
(7.7)
This is a cubic equation, which has three real
solutions because of the stress tensor symmetry.
To find the eigenvector corresponding to a given
eigenvalue œ
i
, we insert this value in (
7.6
)and
solve for the eigenvector components
n
j
,taking
into account that only two of the three compo-
nents are independent. The three eigenvectors
n
i
are orthogonal each other and define a new
local
reference frame that is called the system of the
principal axes
of stress. In this local coordinate
system, the stress tensor is represented by a
diagonal matrix ¢ given by:
To
find
the directions of
maximum shear
stress,
we
must
find
the
solutions
of
the
equations:
@T
S
@n
j
D
0
(7.12)
with the constraint that:
n
i
n
i
D
1
(7.13)
This is a classic problem of finding a con-
ditional maximum, which can be solved using
Lagrange's multipliers. In this instance, we will
search the solutions of the equations:
2
3
¢
1
00
0¢
2
0
00¢
3
T
£N
D
4
5
¢
D
N
(7.8)
@T
S
@n
j
D
œn
j
I
j
D
1;2;3
(7.14)
where
N
is a matrix formed with the components
of the eigenvectors
n
i
:
where œ is the Lagrange multiplier. Using (
7.11
),
these equations can be rewritten as follows:
N
ij
D
n
i
(7.9)
