KARAKTERISTIK DEFORMASI Strain dan Stres

KARAKTERISTIK DEFORMASI
Strain dan Stress
HERI ANDREAS
Mahasiswa Program Doktor
Prodi Geodesi dan Geomatika ITB
E-mail : heri@gd.itb.ac.id
September 2007

1. Pengertian Deformasi
Deformasi adalah perubahan bentuk, dimensi dan posisi dari
suatu materi baik merupakan bagian dari alam ataupun buatan
manusia dalam skala waktu dan ruang

2. Penyebab Deformasi
Bila dikenai Gaya (Force) maka benda/ materi akan terdeformasi

3. Objek dari Deformasi
rotasi bumi

Fenomena lain

abrasi
proses geologi lokal

longsoran
Alam

ocean loading

tsunami
tektonik lempeng

Manusia

pelapukan
erosi

subsidence

pasut
atmosferik

proses hidrologi
D. Sarsito, 2006

3. Objek dari Deformasi

S A MP N o r t h

15

He ight (m )
36.560
36.520
36.480
36.440
36.400
36.360
36.320
36.280
36.240
36.200

36.160
36.120
36.080
36.040
36.000
35.960
35.920
35.880
35.840
35.800
35.760
35.720
35.680
35.640
35.600
35.560
35.520
35.480
35.440
35.400

35.360
35.320
35.280
35.240
35.200
35.160
35.120
35.080
35.040
35.000

10

5

0

-5

-10

-15

-20

-25

0 3/06
16
6 9 3006 -12/1
6 6
0 3/ 06
6 -45/1
1 6/ 061 -48/1
03
6 -1
1 026/06
0 1
4 15 6/06
0 1

0 /0061-7
0 -1
6 60 1536/16
8 9
3 146/06
16/1-1023/00
4 05 6/06
7 08 6/06
7 8
126/06
1 /016 2-4
0 /006 -4
1 /0162-4
0 /0062-4
/ 01
/ 061 1
/9/
/ 060 -1
22 /9/
8/1

-10
2/113/1
-4
/0
240/9/-12
2370/9/-1206 /9/
-13 /9/ 06 -13/1
9/10 /1
/ 106/11-1
/06 -19/1
/ 5/1
06 -16/1
06 -1
/ 06 -16/1
6
/06/1
1 /113-1
1-16/1
3-16/1
1-16/1

2-16/1
/111-4
/102-4
/11 -4
/10 2-4
/10 3-4
/06 -1
/06/11-1
/11 1-4
/06/1
/06/1
/0/1 -1/0 1-19/1
/06/12-1
/06 -16
/06/12-1
/06/12-1
/06/12-1
/06/1

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

Tgl/Bln/Thn-Jam

4. Jenis dari Deformasi
Deformasi dapat dibagi menjadi 2 jenis yaitu Deformasi Statik dan
Deformasi sesaat
Deformasi statik bersifat permanen
Deformasi sesaat bersifat sementara / dinamis

5. Parameter Deformasi
Deformasi dari suatu benda/ materi dapat digambarkan secara
penuh dalam bentuk tiga dimensi apabila diketahui 6 parameter
regangan (normal-shear) dan 3 parameter komponen rotasi
Parameter deformasi ini dapat dihitung apabila diketahui fungsi
pergeseran dari benda tersebut persatuan waktu

Normal strain

Shear strain

6. Model dan pengamatan Deformasi
Secara praktis survey deformasi akan terpaut pada titik-titik yang
bersifat diskrit, dengan demikian deformasi dari benda harus
didekati dengan model.
Fungsi dari deformasi dinyatakan oleh persamaan dalam bentuk
matrik :
d=Bc
Dimana :
B, adalah matrik deformasi yang elemennya merupakan fungsi dari
posisi dari titik yang diamati, serta waktu
C, vektor yang koefisiennya akan diketahui

6. Model dan pengamatan Deformasi
Survey deformasi: penentuan perubahan posisi, jarak, sudut,
regangan : teknik geodetik, geofisika, dan lain-lain

7. Analisis Deformasi
Analisis Geometrik :
Bila kita hanya tertarik pada status geometrik (ukuran dan
dimensi) dari benda yang terdeformasi
Analisis Fisis :
Bila kita bermaksud untuk menentukan status fisis dari benda yang
terdeformasi, regangan, dan hubungan antara gaya dengan
deformasi yang terjadi

8. Analisis Deformasi aspek fisis
Dalam analisis fisis deformasi, hubungan antara gaya dan
deformasi dapat dimodelkan dengan menggunakan metoda
empiris (statistik), yaitu melalui korelasi antara pengamatan
deformasi dan pengamatan gaya
Metoda lain dalam analisis fisis yaitu metoda deterministik, yang
memanfaatkan informasi dari gaya, jenis material dari benda, dan
hubungan fisis antara regangan (strain) dan tegangan (stress)
pada benda

9. Normal strain :perubahan panjang

- Change of length proportional to length

-

xx, yy, zz are normal component of strain

nb : If deformation is small, change of volume is

xx + yy + zz (neglecting quadratic terms)
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10. Shear Strain : perubahan sudut

xy = -1/2 (1 + 2) = 1/2 (d dx + d dy )
xy = yx (obvious)
y

x

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11. Stress dalam 2 Dimensi
- Force =  x surface
- no rotation =>

xy = yx

- only 3 independent
….components :

…..xx , yy , xy

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12. Applied Forces
Normal forces on x axis



xx(x). y  xx(x+x). y
y xx(x).  xx(x+x)
 y dxx/dx . x (1)
Shear forces on x axis 
yx(y). x  yx(y+y). x
 x dyx/dy . y (2)
Total on x axis







d

xx/

dx +

dyx/



dy x y
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13. Forces Equilibrium

d
d
Total on y axis = 
Total on x axis =

Equilibrium =>

xx/

dx +

yy/

dy +

d
d


/ xy x
d

dyx/

dy x y

dyx

yy/

dy +

xx/

dx +


/ y
d

dyx/
dyx

dx 

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14. Solid elastic deformation
• 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)
and stresses 





in the same direction

in mutually perpendicular directions
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14. Solid elastic deformation
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
a strain 1/E
strains



 produces

in the same direction

and

/E  in mutually perpendicular directions
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15. Elastic deformation across a locked fault

What is the shape of the accumulated deformation ?
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Formula matematis

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Formula matematis

•Symetry =>

all derivative with y = 0

•No gravity =>

yy = 0
zz = 0

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

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Formula matematis
•Elastic equations :

(2)
(3)
(3)

xx = (G) xx + zz
xy = G xy xz = G xz
yy =  xx + zz
zz =  _________________________
xx + (G) zz yz = G yz
+ zz = 0 =>  xx + zz = -2G zz
and (2) => yy =  xx + zz = -2 G zz
=>

and (1) =>

xx = - (2G +   zz
2

(
2G)
/ +  ] zz
xx = [-

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Formula matematis

• Force equilibrium along the 3 axis
(x)
(y)
(z)

x
x
x

dxx / dx + dyx / dy + dxz / dz
dxy / dx + dyy / dy + dyz / dz
dxz / d_________________________
x + dyz / dy + dzz / dz

x

= 0
= 0
= 0

2xx / dx2
d
• Derivation of eq. 1 with x and eq. 3 give :
dxy / dx + dyz / dz
• equation 2 becomes :

= 0
= 0

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Formula matematis
relations between

stress () and displacement vector (U)

xy = 2G xy = 2G [dUx / dy + dUy / dx] .1/2
yz = 2G yz = 2G [dUz / dy + dUy / dz] .1/2
_________________________

dxy / dx + dyz / dz = 0 we obtain :
d/dx[dUx/dy + dUy/dx] + d/dz[dUz/dy + dUy/dz] = 0
Using

x

d2Uy / dx2 + d2Uy / dz2

x

= 0
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Formula matematis

d2Uy / dx2+ d2Uy / dz2

= 0

What is Uy, function of x and z, solution of this equation ?
Guess : Uy = K arctang (x/z) works fine !

datan()/

Nb.

dUy/dx=

1/

(1+2 )

d

/

=>
dU /dz=-Kx/z (1+x /z ) =>
K

y

z(1+x2/z2)
2

2

2

d2Uy/dx2= -2Kxz/(z2+x2)
d2Uy/dz2= 2Kxz/(x2+z2)
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Formula matematis
Uy = K arctang (x/z)

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

Uy = K . /2 if x > 0

= K . – /2 if x < 0

And also :

Uy = +V0 if x > 0

=>

= –V0

if x < 0

K = 2.V0 / 

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Formula matematis

Uy = K arctang (x/z)

at the surface (z=h)

Uy = 2.V0 / arctang (x/h)

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16. Arctang Profiles

Uy = 2.V0 / arctang (x/h)

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17. Deeping Fault

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18. Elastic Dislocation (Okada, 1985)
Surface deformation due to shear and tensile faults in a half space, BSSA vol75, n°4, 1135-1154, 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
F at (1,2,3)
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18. Elastic Dislocation (Okada, 1985)

(1) displacements
For strike-slip

For dip-slip

For tensile fault

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18. Elastic Dislocation (Okada, 1985)

Where :

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19. Case : Sagaing Fault Nyanmar

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20. Case : Palu Koro Fault

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20. Case : Palu Koro Fault (more complex)

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21. Case : Sumatra subduction zone

Natawijaya, 2007

Segmen Mentawai-Pagai belum sepenuhnya terpatahkan ??? !!!

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22. Strain rate and rotation rate tensors
To asses plate deformation :
1.

Look at station velocity residuals

2.

Compute strain rate and rotation rate tensors

Strain =

Velocity
_______

=

Distance

mm/yr
_____
km

= % / yr
d(Vx) / d(x)

d(Vx) / d(y)

d(Vy) / d(x)

d(Vy) / d(y)

Matrix tensor notation : Sij = d(Vi) / d(xj) =

Theory says : [S] =

[E]

+

[W]

Symetrical Antisymetrical
Strain rate rotation rate

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22. Strain rate and rotation rate tensors
E11
[E] = ½ ([S] + [S]T)

E12

=

[W] = ½ ([S] - [S]T)
E12

E22

0

W

-W

0

=

[E] has 2 Eigen values : 1, 2

1 and 2 are extension/compression along principal direction defined
by angle defined as angle between 2 direction and north

1 = E11 cos2 + E22 sin2 – 2 E12 sin cos
2 = E11 sin2 + E22 cos2 – 2 E12 sin cos
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22. Strain rate and rotation rate tensors
Minimum requirement to compute strain and rotation rates is :
3 velocities (to allow to determine 3 values 1,

2, and W)

Therefore we can compute strain rate and rotation rate within any
polygon, the minimum polygon being a triangle

No deformation

compression

rotation

Strain and rotations are unsensitive to reference frame
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23. Case : Strain & Rotation on GEODYSSEA network

Strains :

Rotations :

extension/compression/strike-slip

Anti-clockwise/clockwise
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24. Case : intensity of strain in GEODYSSEA network

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25. Case : Strain & Rotation in Nyanmar

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