lecture12 strain examples
TRANSCRIPT
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Lecture 12: Strain Examples
GEOS 655 Tectonic GeodesyJeff Freymueller
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A Worked Example Consider this case of
pure shear deformation,
and two vectors dx 1 anddx 2. How do they rotate? Well look at vector 1 first,
and go through eachcomponent of ! .
"i
= e ijk # km + $ km( )d
x j d
xm
" =
0 # 0
# 0 0
0 0 0
$
%
& & &
'
(
) ) )
* = (0,0,0)
(1) dx = (1,0,0) = d x 1(2) dx = (0,1,0) = d x 2
dx (1)
dx (2)
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A Worked Example First for i = 1
Rules for e 1jk If j or k =1, e 1jk = 0 If j = k =2 or 3, e 1jk = 0 This leaves j=2, k=3 and
j=3, k=2 Both of these terms will
result in zero because j=2,k=3: " 3m = 0 j=3,k=2: dx 3 = 0
True for both vectors
"1
= e1 jk # km + $ km( )d x j d xm
" =
0 # 0
# 0 0
0 0 0
$
%
& & &
'
(
) ) )
* = (0,0,0)
(1) dx = (1,0,0) = d x 1(2) dx = (0,1,0) = d x 2
dx (1)
dx (2)
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A Worked Example Now for i = 2
Rules for e 2jk If j or k =2, e 2jk = 0 If j = k =1 or 3, e 2jk = 0 This leaves j=1, k=3 and
j=3, k=1 Both of these terms will
result in zero because j=1,k=3: " 3m = 0 j=3,k=1: dx 3 = 0
True for both vectors
"2
= e 2 jk # km + $ km( )d x j d xm
" =
0 # 0
# 0 0
0 0 0
$
%
& & &
'
(
) ) )
* = (0,0,0)
(1) dx = (1,0,0) = d x 1(2) dx = (0,1,0) = d x 2
dx (1)
dx (2)
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A Worked Example Now for i = 3
Rules for e 3jk Only j=1, k=2 and j=2, k=1
are non-zero
Vector 1:
Vector 2:
"3
= e 3 jk # km + $ km( )d x j d xm
" =
0 # 0
# 0 0
0 0 0
$
%
& & &
'
(
) ) )
* = (0,0,0)
(1) dx = (1,0,0) = d x 1(2) dx = (0,1,0) = d x 2
dx (1)
dx (2) ( j = 1, k = 2) e312 dx1" 2 mdx m= 1 #1 # $ #1 + 0 #0 + 0 #0( )= $
( j = 2, k = 1) e321
dx2
" 1m
dxm
= %1 #0 # 0 #1 + $ #0 + 0 #0( )= 0
( j = 1, k = 2) e312 dx1" 2 m dx m= 1 # 0 # $ #0 + 0 #0 + 0 #0( )= 0
( j = 2, k = 1) e321 dx 2" 1mdx m= %1 #1 # 0 #0 + $ #1 + 0 #0( )= %$
- #
#
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Rotation of a Line Segment There is a general expression for the rotation
of a line segment. Ill outline how it is derivedwithout going into all of the details.
First, the strain part
"i
= e ijk # km + $ km( )d
x j d
xm
"i( strain )
= e ijk # km d x j d xm = e ijk d x j # km d xm( )"
i( strain )
= e ijk d x j # $ d x( )k " ( strain ) = d x % # $ d x
( )
If the strain changes theorientation of the line, thenthere is a rotation.
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Rotation of a Line Segment Here is the full equation then:
This is used for cases when you have angle
or orientation change data, or when you wantto predict orientation changes from a knownstrain and rotation.
"i
=
e ijk d
x j #
kmd
xm+ $
i % $
j d
x j ( )d
x i" = d
x & # ' d
x( )+ $ % $ ' d x( )d x
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Vertical Axis Rotation Special case of common use in tectonics
Well use a local east-north-up coordinate system All motion is horizontal (u 3 = 0)
All rotation is about a vertical axis (only % 3 is non-zero) All sites are in horizontal plane (dx 3 = 0) The expression for % gets a lot simpler
"i
# " j dx j dx i $ " i " % dx = 0( )
" 3 = # 12 e 3 ij & ij "
3= # 1
2 e312 & 12 + e321 & 21 + 0[ ]"
3= # 1
2 1 %& 12 # 1 % #& 12( )[ ]
"3
= # & 12
The vertical axisrotation is directlyrelated to the rotationtensor term
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Strain from 3 GPS Sites There is a simple,
general way to calculate
average strain+rotationfrom 3 GPS sites If you have more than 3
sites, divide networkinto triangles For example, Delaunay
triangulation asimplemented in GMT
A
B
C
u AB
u AC
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Strain from 3 GPS Sites We have seen the equations for strain
from a single baseline before
C A
B u AB
u AC
" u 1 AB
" u 2 AB#
$ % &
' ( = ) 11 " x1
AB
+ ) 12 " x 2 AB
+ * 12 " x 2 AB
) 12 " x1 AB
+ ) 22 " x 2 AB + * 12 " x1
AB#
$ % &
' (
" u 1 AB
" u 2 AB
#
$
% &
'
( =" x1
AB " x 2 AB 0 " x 2
AB
0 " x1 AB " x 2
AB +" x1 AB
#
$
% &
'
( ,
) 11
) 12
) 22
* 12
#
$
% % % %
&
'
( ( ( (
Observations = (known Model)*(unknowns)
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Strain from 3 GPS Sites Including both sites we have 4 equations
in 4 unknowns:
C A
B u AB
u AC
" u 1 AB
" u 2 AB
" u 1 BC
" u 2 BC
#
$
% % % %
&
'
( ( ( (
=
" x1 AB
" x 2 AB
0 " x 2 AB
0 " x1 AB " x 2
AB )" x1 AB
" x1 BC " x 2
BC 0 " x 2 BC
0 " x1 BC " x 2
BC )" x1 BC
#
$
% % % %
&
'
( ( ( ( *
+ 11
+ 12
+ 22
, 12
#
$
% % % %
&
'
( ( ( (
+ 11
+ 12
+ 22
, 12
#
$
% % % %
&
'
( ( ( (
=
" x1 AB
" x 2 AB
0 " x 2 AB
0 " x1 AB " x 2
AB )" x1 AB
" x1 BC " x 2
BC 0 " x 2 BC
0 " x1 BC " x 2
BC )" x1 BC
#
$
% % % %
&
'
( ( ( (
) 1
*" u 1
AB
" u 2 AB
" u 1 BC
" u 2 BC
#
$
% % % %
&
'
( ( ( (
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Strain varies with space In general, strain varies with space, so you cant necessarilyassume uniform strain over a large area For a strike slip fault, the strain varies with distance from the fault x
(slip rate V, and locking depth D)
You can ignore variations and use a uniform approximation (see noevil)
Fit a mathematical model for ( x) and ' (x)
You can try sub-regions, as we did in Tibet in the Chen et al. (2004)paper
Map strain by looking at each triangle Or you can map out variations in strain in a more continuous fashion
This is a more powerful tool in general
" 12 =VD
2 # x 2 + D
2( )$ 1
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Southern California
Feigl et al. (1993, JGR)
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Strain Rates
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Rotations
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Model strain rates as continuousfunctions using a modified least-
squares method.
Uniqueness of the method: Requires no assumptions of stationary of
deformation field and uniform variance ofthe data that many other methods do;
Implements the degree of smoothingbased on in situ data strength. Method developed by Z. Shen, UCLA
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!!!!!!!!!
"
#
$$$$$$$$$
%
&
n
n
Vy
Vx
VyVx
Vy
Vx
...
2
2
1
1
=
!!!!!!!
!!
"
#
$$$$$$$
$$
%
&
'('''''
'('' '''
'('''''
nnn
nnn
x y x
y y x
x y x y y x
x y x
y y x
010
001
..................
010
001
010
001
222
222
111
111
Ux
Uy"
xx
" xy
" yy
#
$
%
& & & & & & &
'
(
) ) ) ) ) ) )
+
e x1
e y1
e x
2
e y 2
...
e xn
e yn
"
#
$ $ $ $ $ $ $ $ $
%
&
' ' ' ' ' ' ' ' '
Ux, Uy : on spot velocity components
xx, ! xy, ! yy : strain rate components
" : rotation rate
V i V j
Vk
xi x j
xk R
At each location point R , assuming a uniform strain rate field, the strainrates and the geodetic data can be linked by a linear relationship:
d = A m + e
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d = A m + e
Bd = BAm + Be
where B is a diagonal matrix whose i -th diagonal termis exp (-! x i 2 /D2 ) and e ~ N(0, E), that is, the errors are
Assumed to be normally distributed. D is a smoothingdistance.
reconstitute the inverse problem with a weighting matrix B:
m = (A t B E -1 B A) -1 At B E -1 B d
This result comes from standard least squares estimationmethods.
The question is how to make a proper assignment of D?
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W = " i exp (-! x i 2 /D 2 )
Trade-off between total weight W and strain rate uncertainty (
small D Average oversmall area
large D Average overlarge area
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Tradeoff of resolution anduncertainty
If you average over a small area, you willimprove your spatial resolution for variations instrain But your uncertainty in strain estimate will be higher
because you use less data = more noise
If you average over a large area, you will reducethe uncertainty in your estimates by using more
data But your ability to resolve spatial variations in strain
will be reduced = possible over-smoothing
Need to find a balance that reflects the actualvariations in strain.
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Example: Strain rate estimation from SCEC CMM3 (Post-Landers)
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Post-Landers Principal Strain Rate and Dilatation Rate
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Post-Landers Rotation Rate
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Post-Landers Maximum Strain Rate and Earthquakes of M>5.0 1950-2000
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Global Strain Rate Map
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Large-Scale Strain Maps The Global Strain Rate Map makes use of some
important properties of strain as a function of space: Compatibility equations
The strain tensors at two nearby points have to be related toeach other if there are no gaps or overlaps in the material
Equations relate spatial derivatives of various strain tensorcomponents
If we know strain everywhere, we can determine rotationfrom variations in shear strain (because of compatibility eqs)
Relationship between strain and variations in angularvelocity, so that you can represent all motion in terms ofspatially-variable angular velocity
Kostrov summation of earthquake moment tensors or faultslip rate estimates
Allows the combination of geodetic and geologic/seismic data
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Geodetic Velocities Used
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Second Invariant of Strain
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Invariants of Strain Tensor The components of the strain tensor depend
on the coordinate system For example, tensor is diagonal when principal
axes are used to define coordinates, not diagonalotherwise
There are combinations of tensorcomponents that are invariant to coordinaterotations Correspond to physical things that do not change
when coordinates are changed Dilatation is first invariant (volume change does
not depend on orientation of coordinate axes)
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Three Invariants (Symmetric) First Invariant dilatation
Trace of strain tensor is invariant I1 = & = trace( " ) = " ii = " 11 + " 22 + " 33
Second Invariant Magnitude
Third Invariant determinant
Determinent does not depend on coordinates I3 = det( " )
There is also a mathematical relationshipbetween the three invariants
I 2
= 12
trace " ( )2 # trace (" $" )[ ] I
2= "
11$"
22+ "
22$"
33+ "
11$"
33 # "
122 # "
232 # "
132
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