Strain dan Stress

KARAKTERISTIK DEFORMASI

Strain dan Stress

HERI ANDREASMahasiswa Program Doktor

Prodi Geodesi dan Geomatika ITBE-mail : [email protected]

September 2007

1. Pengertian Deformasi

Deformasi adalah perubahan bentuk, dimensi dan posisi darisuatu materi baik merupakan bagian dari alam ataupun buatanmanusia dalam skala waktu dan ruang

2. Penyebab Deformasi

Bila dikenai Gaya (Force) maka benda/ materi akan terdeformasi

3. Objek dari Deformasi

tektonik lempeng

pasut

atmosferik

proses hidrologi

ocean loading

proses geologi lokal

rotasi bumi

Alam

Manusiapelapukan

erosi

abrasi

subsidence

longsoran

tsunami

Fenomena lain

D. Sarsito, 2006

3. Objek dari Deformasi

S AMP No r t h

-25

-20

-15

-10

-5

0

5

10

15

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

35.000

35.040

35.080

35.120

35.160

35.200

35.240

35.280

35.320

35.360

35.400

35.440

35.480

35.520

35.560

35.600

35.640

35.680

35.720

35.760

35.800

35.840

35.880

35.920

35.960

36.000

36.040

36.080

36.120

36.160

36.200

36.240

36.280

36.320

36.360

36.400

36.440

36.480

36.520

36.560

22 /9/ 06 -124 /9/ 06 -1326 /9/ 06 -127 /9/ 06 -1329 /9/ 06 -130 /9/ 06 -132/10 / 06 -13/10 / 06 -135/10 / 06 -16/10 /06 -138/10 /06 -19/10 / 06 -1311 /10 /06 -112 /10 /06 -1313 /10 /06 -1915 /10 /06 -716 /10 /06 -1618 /10 /06 -419 /10 /06 -1621 /10 /06 -422 /10 /06 -1624 /10 /06 -425 /10 /06 -1627 /10 /06 -428 /10 /06 -1630 /10 /06 -431 /10 /06 -162/11 / 06 -43/11 / 06 -165/11 / 06 -46/11 /06 -168/11 / 06 -49/11 / 06 -1611 /11 /06 -412 /11 /06 -1614 /11 /06 -415 /11 /06 -1617 /11 /06 -418 /11 /06 -1620 /11 /06 -421 /11 /06 -1623 /11 /06 -424 /11 /06 -16

Tgl/Bln/Thn-Jam

Height (m)

4. Jenis dari Deformasi

Deformasi dapat dibagi menjadi 2 jenis yaitu Deformasi Statik danDeformasi sesaat

Deformasi statik bersifat permanen

Deformasi sesaat bersifat sementara / dinamis

5. Parameter Deformasi

Deformasi dari suatu benda/ materi dapat digambarkan secarapenuh dalam bentuk tiga dimensi apabila diketahui 6 parameterregangan (normal-shear) dan 3 parameter komponen rotasi

Parameter deformasi ini dapat dihitung apabila diketahui fungsipergeseran dari benda tersebut persatuan waktu

Normal strain Shear strain

6. Model dan pengamatan Deformasi

Secara praktis survey deformasi akan terpaut pada titik-titik yangbersifat diskrit, dengan demikian deformasi dari benda harusdidekati dengan model.

Fungsi dari deformasi dinyatakan oleh persamaan dalam bentukmatrik :

d = B c

Dimana :B, adalah matrik deformasi yang elemennya merupakan fungsi dariposisi 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 dandimensi) dari benda yang terdeformasi

Analisis Fisis :

Bila kita bermaksud untuk menentukan status fisis dari benda yangterdeformasi, regangan, dan hubungan antara gaya dengandeformasi yang terjadi

8. Analisis Deformasi aspek fisis

Dalam analisis fisis deformasi, hubungan antara gaya dandeformasi dapat dimodelkan dengan menggunakan metodaempiris (statistik), yaitu melalui korelasi antara pengamatandeformasi dan pengamatan gaya

Metoda lain dalam analisis fisis yaitu metoda deterministik, yangmemanfaatkan informasi dari gaya, jenis material dari benda, danhubungan 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)

SEAMERGES GPS COURSE, 2005

10. Shear Strain : perubahan sudut

xy = -1/2 (1 + 2) = 1/2 (dydx + dxdy )xy = yx (obvious)


11. Stress dalam 2 Dimensi

- Force = x surface

- no rotation =>xy =yx

- only 3 independent

….components :

…..xx ,yy ,xy


12. Applied Forces

Normal forces on x axis xx(x). y xx(x+x). yy xx(x). xx(x+x) y dxx/dx . x (1)

Shear forces on x axis yx(y). x yx(y+y). x

x dyx/dy . y (2)

Total on x axis dxx/dx + dyx/dyx ySEAMERGES GPS COURSE, 2005

13. Forces Equilibrium

Total on x axis = dxx/dx + dyx/dyx y

Total on y axis = dyy/dy + dyx/dxy x

dyy/dy + dyx/dxdxx/dx + dyx/dyEquilibrium =>


14. Solid elastic deformation

• Stresses are proportional to strains

• No preferred orientationsxx = (G) xx + yy + zzyy = xx + (G) yy + zzzz = 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


Inversing stresses and strains give :

xx = 1/Exx -

/Eyy -

/Ezzyy = -

/Exx + 1/Eyy -

/Ezzzz = -

/Exx -

/Eyy + 1/Ezz

• 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

/E in mutually perpendicular directions


14. Solid elastic deformation

15. Elastic deformation across a locked fault

What is the shape of the accumulated deformation ?


Formula matematis


Formula matematis

•Symetry => all derivative with y = 0

yy = 0

•No gravity => zz = 0

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


•Elastic equations :

(3) +zz = 0 => xx + zz = -2G 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 xzyz = G yz

=> xx = - (2G + zz

and (1) => xx = [- (2G)2/ + ] zz

_________________________

Formula matematis


• Force equilibrium along the 3 axis

(x) dxx /dx + dyx /dy + dxz /dz = 0

_________________________

(y) dxy /dx + dyy /dy + dyz /dz = 0

(z) dxz /dx + dyz /dy + dzz /dz = 0

xxx x

• Derivation of eq. 1 with x and eq. 3 give : d2xx /dx2 = 0

• equation 2 becomes : dxy /dx +dyz /dz = 0

Formula matematis


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

xd/dx[dUx/dy +dUy/dx] +d/dz[dUz/dy +dUy/dz] = 0

d2Uy /dx2 +d2Uy /dz2 = 0

Using dxy /dx +dyz /dz = 0 we obtain :

x


Formula matematis

What is Uy, function of x and z, solution of this equation ?

d2Uy /dx2+d2Uy /dz2 = 0

Guess : Uy = K arctang (x/z) works fine !

Nb. datan()/d1/(1+2 )

dUy/dx=K/z(1+x2/z2) => d2Uy/dx2

= -2Kxz/(z2+x2)

dUy/dz=-Kx/z2 (1+x2/z2) => d2Uy/dz2

= 2Kxz/(x2+z2)


Formula matematis

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

Uy = K arctang (x/z)

Uy = K . /2 if x > 0 = K . – /2 if x < 0

=> K = 2.V0 /

And also :

Uy = +V0 if x > 0 = –V0 if x < 0


Formula matematis

at the surface (z=h)

Uy = K arctang (x/z)

Uy = 2.V0 /arctang (x/h)


16. Arctang Profiles Uy = 2.V0 /arctang (x/h)


17. Deeping Fault


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 isotropicmedium is given by :

Where jk is the Kronecker delta, and are Lamé’s parameters, k isthe 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)

18. Elastic Dislocation (Okada, 1985)

(1) displacements

For strike-slip For dip-slip For tensile fault


18. Elastic Dislocation (Okada, 1985)

Where :


19. Case : Sagaing Fault Nyanmar


20. Case : Palu Koro Fault


20. Case : Palu Koro Fault (more complex)


21. Case : Sumatra subduction zone

Triyoso, 2005

Natawijaya, 2007

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

22. Strain rate and rotation rate tensors

2. Compute strain rate and rotation rate tensors

1. Look at station velocity residuals

Velocity mm/yrStrain =

_______=

_____= % / yr

Distance km

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

d(Vx) / d(x) d(Vx) / d(y)

d(Vy) / d(x) d(Vy) / d(y)

Theory says : [S] = [E] + [W]

Symetrical Antisymetrical

Strain rate rotation rate

To asses plate deformation :



[E] has 2 Eigen values : 1, 21 and 2 are extension/compression along principal direction defined

by angle defined as angle between 2 direction and north

[E] = ½ ([S] + [S]T) =

E11 E12

E12 E22

[W] = ½ ([S] - [S]T) =

0 W

-W 0

1 = E11 cos2 + E22 sin2 – 2 E12 sin cos2 = E11 sin2 + E22 cos2 – 2 E12 sin cos



Therefore we can compute strain rate and rotation rate within any

polygon, the minimum polygon being a triangle

Minimum requirement to compute strain and rotation rates is :

3 velocities (to allow to determine 3 values 1, 2, and W)

No deformation compression rotation

Strain and rotations are unsensitive to reference frame


23. Case : Strain & Rotation on GEODYSSEA network

Strains :

extension/compression/strike-slip

Rotations :

Anti-clockwise/clockwise


24. Case : intensity of strain in GEODYSSEA network


25. Case : Strain & Rotation in Nyanmar


Strain dan Stress

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