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(1)

Geodesy and Geodynamics

By

Christophe Vigny

National Center for scientific Research (CNRS)

&

Ecole Normale Supérieure (ENS) Paris, France

http://www.geologie.ens.fr/~vigny

(2)

DEFORMATION PATTERN IN ELASTIC CRUST

Stress and force in 2D

Strain : normal and shear

Elastic medium equations

Vertical fault in elastic medium => arctangent

General elastic dislocation (Okada’s formulas)

Fault examples

(3)

Stress () in 2D

- Force =

x

surface

- no rotation =>

xy

=

- only 3 independent

yx

….components :

….. 

xx

,

yy

,

xy

- Normal stress =  ii

- Shear stress =  ij

(4)

Applied forces

Normal forces on x axis





xx(x)

.

y



xx(x+

x)

.

y



y



xx(x)

.  

xx(x+

x)



y

d

xx/

d

x .

x (1) Shear forces on x axis





yx(y)

.

x



yx(y+

y)

.

x



x

d

yx/

d

y .

y (2) Total on x axis



d

xx/

d

x +

d

yx/

d

y



x

y

(5)

Forces Equilibrium

Total on x axis =

d

xx/

d

x +

d

yx/

d

y



x

y

Total on y axis =

d

yy/

d

y +

d

yx/

d

x



y

x

d

yy/

d

y +

d

yx/

d

x

 

d

xx/

d

x +

d

yx/

d

y

 

Equilibrium =>

(6)

Normal strain : change length (not angles)

- Change of length proportional to length

- 

xx

,

yy

,

zz are normal component of strain

+

+

(7)

Shear strain : change angles

xy = -1/2

(

1 +

2

)

= 1/2

(

dy

dx + dx

dy

)

xy =

yx (obvious)

(8)

Solid elastic deformation (1)

• Stresses are proportional to strains

• No preferred orientations

xx = (

G)

xx +



yy +



zz

yy =

xx + (

G)

yy +



zz

zz =

xx +



yy + (

G)

zz

•  andG are Lamé parameters

The material properties are such that a principal strain component

produces a stress (  G ) in the same direction

and stresses   in mutually perpendicular directions

(9)

Solid elastic deformation (2)

Inversing stresses and strains give :

xx = 1

/

E

xx -

/

E

yy -

/

E

zz

yy = -

/

E

xx + 1

/

E

yy -

/

E

zz

zz = -

/

E

xx -

/

E

yy + 1

/

E

zz

• E and  are Young’s modulus and Poisson ’s ratio

a principal stress component produces

a strain

1

/

E

in the same direction and

strains / in mutually perpendicular directions

(10)

III/ Elastic deformation at plate boundaries

What is the shape of the accumulated deformation ?

Two plates (red and green) are separated by a vertical strike-slip fault.

If the fault was slick then the two plates would slide freely along each other with no deformation. But because its surface is rough and there is some friction, the fault is locked. So because the plates keep moving far away from the fault, and don’t move on the fault, they have to deform.

(11)

Mathematical formulation

The plates are made of elastic crust above a viscous mantle.

The viscous flow localizes the deformation in a narrow band just beneath the elastic layer

We can compute the deformation of the elastic layer using the elastic equations detailed before

(12)

Mathematical formulation

•Symetry => all derivative with y = 0

yy

= 0

•No gravity => 

zz

= 0

•What is the displacement field U in the elastic layer ?

(13)

Mathematical formulation

•Elastic equations :

(3)

+ 

zz = 0 =>

 

xx +



zz = -2

G

zz

and

(2) =>

yy =

xx +



zz = -2

G

zz





xx = (

G)

xx +



zz

(2)

yy =

xx +



zz

(3)

zz =

xx +

(

 G)  

zz

xy =

G

xy

xz =

G

xz

yz =

G

yz

=>

xx = - (2

G

+



zz

and

(1) =>

xx =

[

- (2G)2

/ + ] 

zz

_________________________

(14)

Mathematical formulation

• Force equilibrium along the 3 axis

(x)

d

xx

/ d x

+

d

yx

/ d

y +

d

xz

/ d z

= 0

_________________________

(y)

d

xy

/ d x

+

d

yy

/ d

y +

d

yz

/ d z

= 0

(z)

d

xz

/ d x

+

d

yz

/ d

y +

d

zz

/ d z

= 0

x x

x x

• Derivation of eq. 1 with x and eq. 3 give :

d 2

xx

/ d x 2

= 0

• equation 2 becomes :

d

xy

/ d x

+

d

yz

/ d z

= 0

(15)

Mathematical formulation

relations between stress (

) and displacement vector (U)

xy = 2

G

xy = 2

G [ d U

x

/ d

y +

d U

y

/ d x ]

.1

/

2

_________________________

yz = 2

G

yz = 2

G [ d U

z

/ d

y +

d U

y

/ d z ]

.1

/

2

x

d / d x [ d U

x

/ d

y +

d U

y

/ d x ]

+

d / d z [ d U

z

/ d

y +

d U

y

/ d z ]

= 0

d 2 U

y

/ d x 2

+

d 2 U

y

/ d z 2

= 0

Using

d

xy

/ d x

+

d

yz

/ d z

= 0 we obtain :

x

(16)

Mathematical formulation

What is

U

y, function of x and z, solution of this equation ?

d 2 U

y

/ d x 2

+

d 2 U

y

/ d z 2

= 0

Guess :

U

y = K arctang (x

/

z) works fine !

Nb.

d

atan(

)

/ d  

1

/

(1+2 )

d U

y

/ d

x

=

K

/

z(1+x2/z2)

=> d

2

U

y

/ d

x2= -2Kxz

/

(z2+x2)

d U

y

/ d

z

=

-Kx

/

z2 (1+x2/z2)

=> d

2

U

y

/ d

z2= 2Kxz

/

(x2+z2)

(17)

Mathematical formulation

Boundary condition at the base of the crust (z=0)

U

y = K arctang (x

/

z)

U

y = K . /2 if x > 0 = K . – /2 if x < 0

=> K = 2.

V0

/

And also :

U

y = +V0 if x > 0 = –V0 if x < 0

(18)

Mathematical formulation

at the surface (z=h)

U

y = K arctang (x

/

z)

U

y =

2.

V0

/

arctang (x

/

h)

The expected profile of deformation across a strike slip fault we should see at the surface of the earth (if the crust is elastic) is shape like an arctangant function. The exact shape depends on the thickness of the elastic crust, also

(19)

Arctang profiles U

y =

2.

V0

/

arctang (x

/

h)

(20)

Sagaing Fault, Myanmar

GPS measurement on the Sagaing fault fit well the arctang profile

but with an offset of 10- 15 km

(21)

Palu Fault, Sulawesi

Part of the GPS data on Palu fault fits well an arctang profile. But wee need a second fault to explain all the

(22)

Altyn Tagh Fault, China

(23)

Altyn Tagh Fault, China (INSAR)

Interferogram Nov. 1995/ Nov. 1999

Fault-parallel velocity :

v = v0/ atan(x/D)

Slip rate V0 = 1.4 cm/yr Locking depth D = 15 km

(24)

San Andreas Fault, USA (INSAR)

(25)

Elastic dislocation (Okada, 1985)

The displacement field ui(x1,x2,x3) due to a dislocation uj (1,2,3) across a surface  in an isotropic medium is given by :

Where jk is the Kronecker delta,  and  are Lamé’s parameters, k is the direction cosine of the normal to the surface element d.

uij is the ith component of the

displacement at (x1,x2,x3) due to the jth direction point force of magnitude

Surface deformation due to shear and tensile faults in a half space, BSSA vol75, n°4, 1135-1154, 1985.

(26)

Elastic dislocation (Okada, 1985)

(1) displacements

For strike-slip For dip-slip For tensile fault

(27)

Elastic dislocation (Okada, 1985)

Where :

(28)

Subduction modeling

Velocity component // to convergence direction In the case of a subduction (dippping fault with downward slip) we use Okada’s formulas.

We find a very large deformation area (> 500 km) because the dipping angle is only 22°

With oblique slip we predict the surface vector will start to rotate above the end-tip of the subduction plane The profile of the velocity component // to the convergence shows a

“plateau” at this location

(29)

Subduction elastic curves

Normalized horizontal and vertical curves : . The applied dislocation on the subduction plane is equivalent to 1 cm/yr accumulation

(30)

GPS measurements in Chile and S-America

“fiducial” stations on the stable South American craton (Brazil, Guyana, Argentina, Urugay,) allow to define the:

South-American reference frame.

(31)

GPS measurements in Chile

Deformation (elastic def. induced by coupling on the subduction) is visible in Chile (Constitucion)

And

reaches far inland:

TUCU (Tucuman) and CFAG (Coronel

Fontana) in

Argentina show deformation more than 400 km away from the trench

(32)

Velocity profiles

As expected from elastic coupling, velocities decrease Eastward (from 35- 45 mm/yr along the coast to 10-15 mm/yr at the cordillera) and vector directions rotate from a direction // to plate convergence to East-West trending.

(33)

Subduction parameter adjustments

(34)

Elastic curves fit

(35)

Along strike comparison

Conception 37°S Coquimbo 30°S

(36)

Along strike comparison

(37)

Patial coupling model

(38)

potential for an Eq potential for an Eq . . Mw ~8.5 every Mw ~8.5 every plane: dip angle 15°

plane: dip angle 15°

. . locked at locked at 50km50km

slip: 2 – 2.5 cm/an slip: 2 – 2.5 cm/an

.. N 30°-35° N 30°-35°

Oblique subduction + strike slip faulting

at Arakan trench and Sagain fault in Myanmar, SE-Asia

(39)

END OF CHAPTER

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