Finding stresses P1 and P2 in the x1 and x2
directions on a plane of known orientation
x2
l1 and l2
are cosines of the angles between the
x1 and x2 axes and the pole to the
plane
x2
σ12
α
P2
α
Al1
P1
σ11
Plane P
α
α
σ21
x1
x1
Al2
cos(α) = l1
cos(90 − α) = sin(α) = l2
σ22
divide by A to get
x2
P1 = σ11 l1 + σ21 l2
P2
σ11 σ12
A
α
P2 = σ12 l1 + σ22 l2
P1
α
σ21
x1
σ22
P1 A = σ11 Al1 + σ21 Al2
force balance in the x 1 direction
P2 A = σ12 Al1 + σ22 Al2
force balance in the x 2 direction
!
we can write the above as a
matrix equation
" !
P1
σ11
=
P2
σ21
σ12
σ22
"!
or more compactly
P = Σl
l1
l2
"
Cauchy’s formula works in 3D too...



P1
σ11
 P2  =  σ21
P3
σ31
σ12
σ22
σ32

 
Suppose we solved for P1 and P2 and found
that shear stress on plane P was zero

σ13
l1


σ23
l2 
σ33
l3

l1
is the vector of direction cosines - aka a unit
 l2  vector in x1, x2, x3 coordinates, perpendicular
to plane P
l3


P1
 P2 is the stress vector in x1, x2, x3 coordinates,
acting on plane P
P3
From Cauchy’s formula...
P1 = σl1 = σ11 l1 + σ12 l2
P2 = σl2 = σ21 l1 + σ22 l2
Rearrange to form these equations:
(σ11 − σ)l1 + σ12 l2 = 0
σ21 l1 + (σ22 − σ)l2 = 0
x2
σ12
P2 = σl2
P2 σ
α
P1
P1 = σl1
σ11
α
x1
σ21
Conditions for
which this is true?
σ22
In matrix form these equations are
!
σ11 − σ
σ21
σ12
σ22 − σ
(Σ − σI)l = 0
This has a solution if
!
! σ11 − σ
!
! σ21
"!
l1
l2
Σl = σl
!
σ12 !!
=0
σ22 − σ !
"
=0
σ1 and σ2 are principal stresses
(σ11 − σ)(σ22 − σ) − σ12 σ21 = 0
σ − (σ11 + σ22 )σ + (σ11 σ22 − σ
2
2
12 )
=0
There are two solutions for σ, σ1 and σ2
Remember from algebra
ax2 + bx + c
√
−b ± b2 − 4ac
x=
2a
Sum of roots is -b/a and product of roots is c/a,
these are invariants (do not depend on x)
Eigenvectors come from
"
!
"!
σ11 − σ
σ12
l1
=0
σ21
σ22 − σ
l2
I = σ1 + σ2 = σ11 + σ22
2
II = σ1 σ2 = σ11 σ22 − σ12
Invariants
σ1 and σ2 are also called eigenvalues of the
stress matrix. The two pairs of
direction cosines associated with
each are eigenvectors.
We want l1and l2 so we can tell the orientations
of the principal stresses
My modeling problem - weak zones around
faults in the Mojave got weaker during the
Hector Mine earthquake, so they deformed.
and the fact that the length of each
eigenvector is 1
l12 + l22 = 1
Let’s do an example.
Fialko, Science, 2002
Modeling stresses on active faults in the
Mojave Desert
~100 MPa
N 20 E
Maximum horizontal stress is oriented N 20 E
and the horizontal principal stresses are 50 to 100 MPa.
-
According to my source in So Cal:
50 20
20 90
"
this is the state of stress in the Mojave Desert
uppermost crust
we ignore stresses acting in the vertical direction
units are MPa (N / m^2)
x1 direction is east
x2 direction is north
However, my finite
element model has
east, north coordinates.
I can only impose shear
and normal stresses in
east, north directions.
!
Now let!s make sure that this is consistent with the principal
stress directions - maximum stress oriented N 20 E.
North
I will drop the sign in front of the stress tensor - all components
have the same sign so this is okay for now
East
!
50 − σ
20
20
90 − σ
"!
l1
l2
"
=
!
0
0
"
(1) Find eigenvalues of the stress matrix by finding the
determinant of the above matrix and solving for σ
(50 − σ)(90 − σ) − 400 = 0
σ 2 − 140σ + 4100 = 0
σ1 = 41.72
σ2 = 98.28
(2) Find eigenvectors for each σ. These are the principal
stress directions (unit vector in the principal stress
vector!s orientation)
!
50 − 98.28
20
20
90 − 98.28
−48.28l1 + 20l2 = 0
20l1 − 8.28l2 = 0
Both equations give: l2
l=
!
1
2.41
"
= 2.41l1
"!
l1
l2
"
=
!
0
0
"
l=
!
1
2.41
"
l12 + l22 = 1
make the eigenvector length 1
x2
are cosines of the angles
between the x1 and x2 axes
and the pole to the plane
direction
of 98.28 MPa
principal stress
α
(2.41b)2 + (b)2 = 1
b2 = 0.14792
b = 0.3846
!
"
.3846
l=
.9269
l1 and l2
Al1
A
α
direction of
41.72 MPa
principal stress
Al2
This is the unit vector in the
direction the principal stress of
98 MPa acts. (direction cosines)
!
"
.3846
l=
.9269
α = 67.7 °
!
"
.9269
l=
.3846
α = 22.4°
Plane P
x1
cos(α) = l1
cos(90 − α) = sin(α) = l2
How do these weak zones deform in response to the
regional stress field? Can I match the inSAR data?
98 MPa
N 22 E
x1*
42 MPa
N 112 E
x2
x*2
-
[ 098
North
0
42
]
in the * coordinate system is
equivalent to
x1
!
-
East
50 20
20 90
"
in the original coordinate system.
Upward
displacements
Horizontal
displacements
What if we want to know what the stress
tensor is in a rotated coordinate system?
It turns out that if we rotate from x, y, z to x’,
y’, z’ coordinates we have 9 direction cosines
in 3D and 4 in 2D.
Each is for the angle between one of the new x’, y’,
or z’ axes and one of the old x, y, or z axes.
!
σij
= lip ljq σpq
!
σij
= lip ljq σpq
Calculate one stress component:
!
σ11
= l1p l1q σpq
in 2D this is
!
σ11
= l11 l11 σ11 + l12 l11 σ21 + l11 l12 σ12 + l22 l12 σ12
all 9 terms of the 3D version are on page 148 of the text