KARAKTERISTIK DEFORMASI
Strain dan Stress
HERI ANDREAS
Mahasiswa Program Doktor
Prodi Geodesi dan Geomatika ITB
E-mail :
[email protected]
1. Pengertian Deformasi
2. Penyebab Deformasi
4. Jenis dari Deformasi
Deformasi dapat dibagi menjadi 2 jenis yaitu Deformasi Statik dan
Deformasi sesaat
Deformasi statik bersifat permanen
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
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 = B c
Dimana :
B, adalah matrik deformasi yang elemennya merupakan fungsi dari
posisi dari titik yang diamati, serta waktu
6. Model dan pengamatan Deformasi
7. Analisis Deformasi
Analisis Geometrik :
Bila kita hanya tertarik pada status geometrik (ukuran dan
dimensi) dari benda yang terdeformasi
Analisis Fisis :
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
9. Normal strain :perubahan panjang
- Change of length proportional to length
-
xx,
yy,
zz are normal component of strainnb : If deformation is small, change of volume is
xx +
yy +
zz (neglecting quadratic terms)10. Shear Strain : perubahan sudut
xy = -1/2(
1 +
2)
= 1/2(
d
y
d
x +d
x
d
y)
xy =
yx (obvious)11. Stress dalam 2 Dimensi
-
Force =
xsurface
-
no rotation =>
xy=
yx-
only 3 independent
….
components :
…..
xx,
yy,
xy12. 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
13. Forces Equilibrium
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
)
in the same direction
and stresses
in mutually perpendicular directions
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
/
E
in mutually perpendicular directions
SEAMERGES GPS COURSE, 2005
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 :
• Force equilibrium along the 3 axis
Formula matematis
relations between
stress (
) and displacement vector (U)
xy = 2G
xy = 2G
[
d
U
x/
d
y +d
U
y/
d
x
]
.1/
2Formula matematis
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
= 0Formula matematis
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 < 0Formula matematis
at the surface (z=h)
U
y = K arctang (x/
z)U
y =2.
V0/
arctang (x/
h)16. Arctang Profiles
U
y =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.
SEAMERGES GPS COURSE, 2005
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
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
22. Strain rate and rotation rate tensors
2. Compute strain rate and rotation rate tensors 1. Look at station velocity residuals
Velocity mm/yr Strain rate rotation rate To asses plate deformation :
22. Strain rate and rotation rate tensors
[E] has 2 Eigen values :
1,
2
1 and
2 are extension/compression along principal direction defined by angle
defined as angle between
2direction and north
[E] =
½
(
[S] + [S]T) =
22. Strain rate and rotation rate tensors
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