Geology Reference
In-Depth Information
2n
j
h
¢
j
2
¢
i
n
i
¢
j
i
D
œn
j
I
j
D
1;2;3
(7.15)
Solving these equations with the constraint
(
7.13
) gives nine solutions, three of which are the
base versors
n
i
(for which
T
S
D
0). The remaining
six represent the planes where
T
S
has local or
global maxima:
1
p
2
;
˙
p
2
;0
1
p
2
;0;
˙
1
1
p
2
I
I
0;
1
1
p
2
p
2
;
˙
It is easy to
v
erify that
T
S
attains its maximum
for n
D
Fig. 7.5
Cauchy'
s s
tress surface. V
ersor
m
is parallel to
T
(
n
)and
¡
1=
p
2
n
D 1/
p
T
N
is the distance PQ
3
. For these directions,
it results that the shear and normal components of
stress are given by:
1
˙
n
We obtain the following direct formula for ¢
1
and ¢
2
:
8
<
:
¢
1
¢
3
2
I
T
N
D
¢
1
C
¢
3
2
T
S
D
(7.16)
r
£
12
C
1
2
.
£
11
C
£
22
/
C
1
4
.
£
11
£
22
/
2
¢
1
D
r
£
12
C
In the case of two-dimensional problems, for
example in the study of the state of stress along
vertical faults, it is easy to determine direct for-
mulae for the normal and shear components of
stress, and for the principal stresses. If
s
is the
strike of a vertical fault (clockwise angle from
the North) and the coordinate axes
x
1
and
x
2
are
directed, respectively, northward and eastward,
then:
8
<
1
2
.£
11
C
£
22
/
1
4
.£
11
£
22
/
2
(7.19)
¢
2
D
Now we are going to describe the variability
of the normal stress
T
N
(
n
) at a point
P
,which
is determined by (
7.10
) in the reference frame of
the principal axes, as a function of
n
. This can be
done through an elegant geometrical method due
to Cauchy. For any versor
n
at
P
, let us consider
the point
Q
along the direction of
n
, at distance:
1
2
.
£
11
C
£
22
/
C
1
2
.
£
11
£
22
/ cos 2s
T
N
.s/
D
C
£
12
sin 2s
T
S
.s/
D
:
1
2
.£
22
£
11
/sin 2s
C
12
cos2s
(7.17)
1
p
T
N
.n/
¡.n/
D
(7.20)
The set of points
Q
, whose distance from
P
is
¡(
n
)forvariable
n
, forms a surface
S
that traces
the variations of normal stress at
P
. This surface
is known as the
Cauchy stress surface
and has
the important property that its normal at
Q
has
the same direction of
T
(
n
)at
P
, as illustrated in
Fig.
7.5
.
To prove this assertion, we first note that if
q
i
are the Cartesian coordinates of
Q
in the principal
axes reference frame, which has origin at
P
,then
Then, we see that
T
S
D
0for:
2
tan
1
2£
12
1
s
D
s
1
D
(7.18)
£
11
£
22
One principal axis has direction
n
1
D
(cos
s
1
,
sin
s
1
), the other one has versor
n
2
D
(cos
s
2
,sin
s
2
)
D
(sin
s
,cos
s
), where
s
2
D
s
1
C
/2. The
principal stresses are obtained substituting
s
1
and
s
2
into the expression (
7.17
) for of
T
N
.
