tectonic stress modelling with the finite element method · 1.2 fem for viscoelastic deformation...

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Numerical modelling with the finite element method: Some

principles and programs

Jason Zhao

2002

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Contents

1. Finite element method and programs

2. Some tests

3. Mantle flow after crustal faulting

4. References

5. Appendix

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1. Finite element method and programs

Although there are some commercial software packages on the market, most of them were not

designed for geodynamic modelling purposes. In addition, the vendors usually do not supply source

codes. In practice, it is quite common to modify the source codes of a program in order to develop

some new procedures for modelling some targeted problems. Moreover, most FEM programs cannot

be directly used for inverse modelling. Therefore, it is necessary to develop some new programs.

Three numerical models corresponding to elastic, viscoelastic, and viscous rheologies have been

constructed for stress analysis with FEM during 1999 - 2000 by assembling programs available in

published books and on the internet (e.g. Kirshnamoorthy, 1987; Zienkiewicz & Taylor, 1988; Smith

& Griffith, 1997). In the following, we present briefly the basic formulas associated with the FEM

models, and then introduce the main functions of the programs used in this study.

1.1 FEM for elastic deformation and stress analysis

Basic formulas (Zienkiewicz and Taylor, 1988):

ε = L u ………….. (strain → displacement) (1.1)

L1’ σ + p = 0 ………(in V) (1.2)

L2’ σ = q …………. (on S) (1.3)

σ = D ε …………….(stress → strain) (1.4)

where

LT = [b1 b2 b3],b1 = [∂/∂x 0 0 0 ∂/∂z ∂/∂y],b2 = [0 ∂/∂y 0 ∂/∂z 0 ∂/∂x],b3 = [0 0 ∂/∂z ∂/∂y ∂/∂x 0],

ε is the strain tensor, u is the displacement vector, L1 (relating to nodal stress to body forces) and L2

(relating nodal stress to traction) are the operators, σ is the stress tensor, p is the boundary force, q is

the surface traction, V is the volume of the 3D body, and S is the boundary surface of V. Elements

used in the FEM programs are shown in Figure 1.1. The advantage of using different types of

elements is that an object with complex geometry can be well approximated. The infinite element is

useful when investigating a local problem (Zienkiewicz and Taylor, 1988).

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8-node-brick 6-node-prism 4-node-tetrahedron

infinite element fault-slip node

Fig. 1.1 Elements used in the FEM programs

Input/output:

Input: geometry, forces, material coefficients, displacements (strains), weak zones, fault

slip values, temperature changes and others;

Output: deformation and strain/stress.

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1.2 FEM for viscoelastic deformation and stress analysis

The constitutive equations for a non-Newtonian viscoelastic material can be expressed as

(Melosh & Raefsky, 1980):

ε′ = f (σ0, σ′,σn-1, η) (1.5)

where σ′ and ε′ are the stress and strain rates, respectively, σ0 is the stress tensor, σ is the

fourth stress tensor, and η is the viscosity. For a linear Newtonian model used in this

study, n = 1, and the constitutive equations for a two-dimensional case are expressed as:

ε′xx = (1 + ν)/E [(1-ν) σ′xx - ν σ′yy] + σn-1 /(4η) [σxx - σyy] (1.6)

ε′yy = (1 + ν)/E [(1-ν) σ′yy - ν σ′xx] + σn-1 /(4η) [σyy - σxx] (1.7)

ε′xy = (1 + ν)/E σ′xy + σn-1 /(2η) σxy (1.8)

where E is the Young’s modulus, ν is the Poisson’s ratio, σ′xx, σ′xy, σ′yy and ε′xx, ε′xy,

ε′yy are the stress and strain rates, respectively, σxx, σxy, and σyy are the stress tensors.

Input/output:

Input: geometry, forces, material coefficients (e.g. viscosity values), velocities (strains),

weak zones, fault slip rates, and others;

Output: (flow) velocity field and strain/stress (rates).

1.3 FEM for viscous deformation and stress analysis

Basic formulas (Newtonian flow) (Bremaecker & Becker, 1978):

ε′ = L v ……(strain rate → velocity) (1.9)

∂vi/∂xi = 0 ..……………. (incompressible) (1.10)

∂τ ij/∂xi + pi = 0 ..………..(in V) (1.11)

τij nj = qi ………….…….. (on S) (1.12)

τij=τji ….………….. (Euler stress/symmetric) (1.13)

τij = - pδij + 2η vij ……….(Newtonian flow) (1.14)

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Where

LT = [b1 b2 b3],b1 = [∂/∂x1 0 0 0 ∂/∂x3 ∂/∂x2],b2 = [0 ∂/∂x2 0 ∂/∂x3 0 ∂/∂x1],b3 = [0 0 ∂/∂x3 ∂/∂x2 ∂/∂x1 0],

τ is the stress tensor, v is the (flow) velocity field, η is the viscosity, ε′ is the strain rate, pi

is the body force, qi is the traction, p is the pressure, V is the volume of the 3D body, and

S is the boundary surface of V. The method used to solve the equations is given in

Appendix A.

Input/output:

Input: geometry, forces, viscosity values, velocities, fault slip (rate), and others;

Output: velocity (flow) and strain/stress rates.

1.4 FEM programs

In the following, we briefly introduce main functions of the numerical models used for

stress analysis.

Table 2.1 FEM models for stress analysis

Model Functions Language

VISKX 3D viscoelastic stress analysis FORTRAN

FLOW 3D viscous stress analysis FORTRAN

(1) VISKX - a 3D viscoelastic FEM program for simulating deformation and stress change

generated by fault interaction and any internal and external loads. Applications of the

program include: modelling deformation due to post-seismic relaxation processes,

simulating deformation due to post-glacial rebound, and investigating viscoelastic

responses of the earth to loading or unloading.

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(2) FLOW - a 3D viscous FEM program for simulating crustal/lithospheric deformation

and stress evolution associated with plate motion and fault-slip on geological time scales.

The reactivation of a fault can be simulated by input of the strain rates of the fault points.

The programs are developed mainly based on a 3D code from the FEM book by

Krishnamoorthy (1987). The program structures and functions of most subroutines used

here have been described in the book, though some new routines have been added. It is

recommended to go through this book to fully understand the code structure.

2. Some tests

Test A:

A1. Model and data

A block contains 24 nodes and 5 brick elements.

The elastic modulus: E = 2.0 x 106

Poisson ratio = 0.00

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Nodal forces: F6(z) = -4 x 104, and F17(z) = -4 x 104.

Nodes of 1, 7, 13 and 19 are fixed.

Nodal data:

1,01,01,01,01,1,1, 0.000, 0.000, 30.000, 0,0,0, 2,00,00,00,01,1,1, 0.000, 40.000, 30.000, 0,0,0, 3,00,00,00,01,1,1, 0.000, 80.000, 30.000, 0,0,0, 4,00,00,00,01,1,1, 0.000, 120.000, 30.000, 0,0,0, 5,00,00,00,01,1,1, 0.000, 160.000, 30.000, 0,0,0, 6,00,00,00,01,1,1, 0.000, 200.000, 30.000, 0,0,0, 7,01,01,01,01,1,1, 20.000, 0.000, 30.000, 0,0,0, 8,00,00,00,01,1,1, 20.000, 40.000, 30.000, 0,0,0, 9,00,00,00,01,1,1, 20.000, 80.000, 30.000, 0,0,0, 10,00,00,00,01,1,1, 20.000, 120.000, 30.000, 0,0,0, 11,00,00,00,01,1,1, 20.000, 160.000, 30.000, 0,0,0, 12,00,00,00,01,1,1, 20.000, 200.000, 30.000, 0,0,0, 13,01,01,01,01,1,1, 0.000, 0.000, 0.000, 0,0,0, 14,00,00,00,01,1,1, 0.000, 40.000, 0.000, 0,0,0, 15,00,00,00,01,1,1, 0.000, 80.000, 0.000, 0,0,0, 16,00,00,00,01,1,1, 0.000, 120.000, 0.000, 0,0,0, 17,00,00,00,01,1,1, 0.000, 160.000, 0.000, 0,0,0, 18,00,00,00,01,1,1, 0.000, 200.000, 0.000, 0,0,0, 19,01,01,01,01,1,1, 20.000, 0.000, 0.000, 0,0,0, 20,00,00,00,01,1,1, 20.000, 40.000, 0.000, 0,0,0, 21,00,00,00,01,1,1, 20.000, 80.000, 0.000, 0,0,0, 22,00,00,00,01,1,1, 20.000, 120.000, 0.000, 0,0,0, 23,00,00,00,01,1,1, 20.000, 160.000, 0.000, 0,0,0, 24,00,00,00,01,1,1, 20.000, 200.000, 0.000, 0,0,0,

Element data:

1 1 19 20 14 13 7 8 2 1 0 0 0 0 0 02 1 20 21 15 14 8 9 3 2 0 0 0 0 0 03 1 21 22 16 15 9 10 4 3 0 0 0 0 0 04 1 22 23 17 16 10 11 5 4 0 0 0 0 0 05 1 23 24 18 17 11 12 6 5 0 0 0 0 0 0

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A2. Output

Nodal displacements (cm)

Node X-displacements Y-displacements Z-displacements

***** FROM T1= 0.000 yr. TO T2= 1.000 yr. ***** 1 0.0000000E+00 0.0000000E+00 0.0000000E+00 2 -0.2988061E-15 0.1270588E-01 -0.1827504E-01 3 0.1467739E-15 0.2258824E-01 -0.6666403E-01 4 0.1464052E-14 0.2964706E-01 -0.1376597E+00 5 0.3928661E-14 0.3388235E-01 -0.2236262E+00 6 0.6879913E-14 0.3529412E-01 -0.3175418E+00 7 0.0000000E+00 0.0000000E+00 0.0000000E+00 8 -0.2983724E-15 0.1270588E-01 -0.1827504E-01 9 0.1613835E-15 0.2258824E-01 -0.6666403E-01 10 0.1435592E-14 0.2964706E-01 -0.1376597E+00 11 0.3993333E-14 0.3388235E-01 -0.2236262E+00 12 0.6862566E-14 0.3529412E-01 -0.3175418E+00 13 0.0000000E+00 0.0000000E+00 0.0000000E+00 14 0.1044913E-14 -0.1270588E-01 -0.1827398E-01 15 0.2976528E-14 -0.2258824E-01 -0.6666930E-01 16 0.5986653E-14 -0.2964706E-01 -0.1376345E+00 17 0.9740520E-14 -0.3388235E-01 -0.2237464E+00 18 0.1413192E-13 -0.3529412E-01 -0.3169680E+00 19 0.0000000E+00 0.0000000E+00 0.0000000E+00 20 0.1042569E-14 -0.1270588E-01 -0.1827398E-01 21 0.2971364E-14 -0.2258824E-01 -0.6666930E-01 22 0.6013419E-14 -0.2964706E-01 -0.1376345E+00 23 0.9708814E-14 -0.3388235E-01 -0.2237464E+00 24 0.1412759E-13 -0.3529412E-01 -0.3169680E+00 -----------------------------------------

A3. Comparing with analytical results

Uz (cm)Distance fromthe end alongthe Y-axis (cm) Analytical FEM

0.00 (node 12) 0.29629 0.3175440.00 (node 11) 0.20859 0.2236280.00 (node 10) 0.12800 0.13766120.00 (node 9) 0.06163 0.06667160.00 (node 8) 0.01659 0.01828200.00 (node 7) 0.00000 0.00000

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Test B

Prism elements

B1: Nodal data: Same as Test A.

Element Data

1 1 19 20 13 7 8 1 0 0 0 0 0 0 2 1 13 20 14 1 8 2 0 0 0 0 0 0 3 1 20 21 14 8 9 2 0 0 0 0 0 0 4 1 14 21 15 2 9 3 0 0 0 0 0 0 5 1 21 22 15 9 10 3 0 0 0 0 0 0 6 1 15 22 16 3 10 4 0 0 0 0 0 0 7 1 22 23 16 10 11 4 0 0 0 0 0 0 8 1 16 23 17 4 11 5 0 0 0 0 0 0 9 1 23 24 17 11 12 5 0 0 0 0 0 0 10 1 17 24 18 5 12 6 0 0 0 0 0 0

B2: Results:

Nodal displacements (cm)

Node X-displacements Y-displacements Z-displacements

***** FROM T1= 0.000 yr. TO T2= 1.000 yr. ***** 1 0.0000000E+00 0.0000000E+00 0.0000000E+00 2 -0.5927777E-02 0.1641801E-01 -0.2312350E-01 3 -0.9426253E-02 0.2874495E-01 -0.8415672E-01 4 -0.1181673E-01 0.3746523E-01 -0.1732706E+00 5 -0.1306780E-01 0.4259793E-01 -0.2806678E+00 6 -0.1347275E-01 0.4395588E-01 -0.3975436E+00 7 0.0000000E+00 0.0000000E+00 0.0000000E+00 8 -0.2001348E-02 0.1428098E-01 -0.1819162E-01 9 -0.5728269E-02 0.2697011E-01 -0.7387230E-01 10 -0.9437412E-02 0.3626300E-01 -0.1589458E+00 11 -0.1184736E-01 0.4193661E-01 -0.2640486E+00 12 -0.1289858E-01 0.4427209E-01 -0.3797556E+00 13 0.0000000E+00 0.0000000E+00 0.0000000E+00 14 0.5927777E-02 -0.1641801E-01 -0.2312402E-01 15 0.9426253E-02 -0.2874495E-01 -0.8415859E-01 16 0.1181673E-01 -0.3746523E-01 -0.1732411E+00 17 0.1306780E-01 -0.4259793E-01 -0.2808431E+00 18 0.1347275E-01 -0.4395588E-01 -0.3970097E+00 19 0.0000000E+00 0.0000000E+00 0.0000000E+00 20 0.2001348E-02 -0.1428098E-01 -0.1819192E-01 21 0.5728269E-02 -0.2697011E-01 -0.7386952E-01 22 0.9437412E-02 -0.3626300E-01 -0.1589574E+00 23 0.1184736E-01 -0.4193661E-01 -0.2640368E+00 24 0.1289858E-01 -0.4427209E-01 -0.3795231E+00 -----------------------------------------

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B3. Comparing with analytical results

Uz (cm)Distance fromthe end alongthe Y-axis (cm) Analytical FEM

0.00 (node 12) 0.29629 0.3797640.00 (node 11) 0.20859 0.2640580.00 (node 10) 0.12800 0.15895120.00 (node 9) 0.06163 0.07387160.00 (node 8) 0.01659 0.01819200.00 (node 7) 0.00000 0.00000

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3. Mantle flow after crustal faulting

Objectives:

1. mapping viscoelastic flow after a rupture;2. analyzing stress distribution associated with mantle flow;3. discussing implications for interaction of multi-faults

3.1 Model setup

The volume of the block is 160 x 160 x 75km (Fig.1). The crust is taken to be 30km thick,and it is divided into four layers: 7.5km, 7.5km, 7.5km, 7.5km. The mantle is taken to be45km, and it divided into two layers: 15km, and 30km. The fault is 30km long (alongstrike) and its down-dip length is 32km. The dip-angle is 68.2 (deg) and the thrust value is1.132m.

Fig. 3.1

0

50

100

1500

2040

6080

100120

140160

0

10

20

30

40

50

60

70

Y(km)X(km)

Z(k

m) upper crust

lower crust

mantle

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Table 3.1 Rheological parameters

LayerElastic modulusE (Pa)

Poisson ratioν

Viscosityη (Pa s) notes

upper crust(0-15km)

3x1010 Pa 0.25 - elastic

lower crust(15-30km)

3x1010 Pa 0.25 1.0 x 1018 viscoelastic

mantle(30km-bottom)

5x1010 Pa 0.25 2.0x1020 viscoelastic

Boundary conditions: the bottom face is taken as fixed, and side faces are free-slip(only the vertical motion is allowed at the side points). Note that the two triangles in Fig. 1marked the profile along which the vertical deformation at four depth levels are shownbelow.

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3.2 Horizontal deformation

Coseismic horizontal deformation at four depth levels

Fig. 3.2 Coseismic horizontal deformation

50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

1.0m

50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

1.0m

50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

1.0m

50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

0.2m

depth = 0km

Horizontal deformation (t=0 yr)

depth = 15km

depth = 30km depth = 45km

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Postseismic velocity at t = 1yr (one year after the rupture)

Fig. 3.2 Postseismic horizontal deformation (t = 1yr)

50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

5cm

50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

5cm

50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

5mm

50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

5mm

Horizontal deformation (t = 1yr)


depth = 30km depth =45 km

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Postseismic velocity at t = 2yr (two years after the rupture)


50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

5mm

50 60 70 80 90 100 11050

60

70

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100

110

120

X(km)

Y(k

m)

10mm

50 60 70 80 90 100 11050

60

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110

120

X(km)

Y(k

m)

5mm

50 60 70 80 90 100 11050

60

70

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90

100

110

120

X(km)

Y(k

m)

0.5mm




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Postseismic velocity at t = 3yr (three years after the rupture)


50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

5mm

50 60 70 80 90 100 11050

60

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110

120

X(km)

Y(k

m)

10mm

50 60 70 80 90 100 11050

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120

X(km)

Y(k

m)

3mm

50 60 70 80 90 100 11050

60

70

80

90

100

110

120

X(km)

Y(k

m)

0.5mm




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3.3 Vertical deformation

Coseismic vertical displacements (t = 0 yr)

Fig. 3.6 Coseismic vertical displacement along a profile (marked triangle, see Fig.1)perpendicular to the fault strike at four depth levels.

0 50 100 150-50

0

50

dh(c

m)

0 50 100 150-50

0

50

dh(c

m)

0 50 100 150-50

0

50

dh(c

m)

0 50 100 150-50

0

50

dh(c

m)

Vertical deformation (t = 0yr)



Y(km)

Y(km)

Y(km)

Y(km)

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Postseismic vertical velocity at t = 1yr (one year after the rupture)

Fig. 3.7 Postseismic vertical displacement along a profile perpendicular to the fault strikeat four depth levels (t =1yr).

0 50 100 150-15

-10

-5

0

5

10

15

dh(c

m)

0 50 100 150-15

-10

-5

0

5

10

15

dh(c

m)

0 50 100 150-15

-10

-5

0

5

10

15

dh(c

m)

0 50 100 150-15

-10

-5

0

5

10

15

dh(c

m)




Y(km)Y(km)

Y(km)Y(km)

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Postseismic vertical velocity at t = 2yr (two year after the rupture)

Fig. 3.8 Postseismic vertical displacement along a profile perpendicular to the fault strikeat four depth levels (t = 2yr).

0 50 100 150-1

0

1

2

3

dh(c

m)

0 50 100 150-1

0

1

2

3

dh(c

m)

0 50 100 150-1

0

1

2

3

dh(c

m)

0 50 100 150-1

0

1

2

3

dh(c

m)



Y(km)Y(km)

Y(km)Y(km)


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Postseismic vertical velocity at t = 3yr (three years after the rupture)

Fig. 3.9 Postseismic vertical displacement along a profile perpendicular to the fault strikeat four depth levels (t = 3yr).

0 50 100 150-1

-0.5

0

0.5

1

1.5

2

dh(c

m)

0 50 100 150-1

-0.5

0

0.5

1

1.5

2

dh(c

m)

0 50 100 150-1

-0.5

0

0.5

1

1.5

2

dh(c

m)

0 50 100 150-1

-0.5

0

0.5

1

1.5

2

dh(c

m)



Y(km)Y(km)

Y(km)Y(km)


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4. References

Bremaecker J. C. and E.B. Becker, finite element models of folding, Tectonophysics, 50,349-367, 1978.

Chandrupatla, T.R. and A.D. Belegundu, Introduction to Finite Elements inEngineering, Prentice Hall, Englewood Cliffs, New Jersey, 414pp, 1991.

Cook, R.D., Concepts and Applications of Finite Element Analysis, John Wiley &Son, New York, 2nd ed. 537pp, 1981.

Krishnamoorthy, C.S., Finite Element Analysis: Theory and Programming, TataMcGraw-Hill Publishing Com. Limited, New Delhi, 551pp, 1987.

Melosh H.J. and A. Raefsky, The dynamical origin of subduction zone topography.

Melosh H.J. and Raefsky, A simple and efficient method for introducing faultsinto finite element computations, Bull. Seism. Soc. Am., 71, 1391-1400, 1981.

Smith I.M. and D.V. Griffiths, Programming the Finite Element Method, JohnWiley & Sons, Brisbane, 3rd ed., 534pp, 1997.

Zienkiewicz, O.C. and R.L. Taylor, The Finite Element Method, Vol.1, BasicFormulation and Linear Problems, McGraw-Hill Book Com., London, 4th ed., 648pp,1988.

Zienkiewicz O.C., C. Emson, and P. Bettess, A novel boundary infinite element,Int. J. Num. Meth. Engng., 19, 393-404, 1983.

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5. Appendix A

Some considerations for solving large matrix equations associated with

viscous models

When the dimension of the linear equation system in the viscous stress modelling is large,

some measures have to be taken in order to save memory and storage in a computer. For a

viscous modelling problem, we usually have to solve the following matrix equations:

A X + B Y – R = 0 (A.1)

B1 X = 0 (A.2)

Where X is the nodal velocity vector, Y is the pressure vector, and, A, B, R, and B1 are

coefficient matrices. The rigorous solutions of the equation can be expressed as:

Y = (B1 A-1 B)-1 B1 A-1 R (A.3)

X = A-1 (R-BY) (A.4)

However, this requires to directly obtain the inverse of the matrix A. In practice, the

dimension of the matrix A is very high, so this method is seldom used. In practice, the

penalty function method is often used to solve the equation approximately (c.f.

Chandrupatla and Belegundu, 1991). By introducing a penalty factor ‘a’, the equations can

be transferred into

(A + a A1) X – R =0

where a is the penalty factor, and A1 is a matrix related to B. The downside of the penalty

function method is that (1) the selection of the penalty factor is somewhat arbitrary; (2)

iteration is required in the computation, which therefore takes a longer time to solve the

equations; and (3) the solutions obtained are approximate.

We consider the following steps to directly solve the equations:

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(a) Since A is a symmetric matrix, we may store only its non-zero elements into a matrix

M. Usually, in such a way, more than 60% storage space can be saved

(Krishnamoorthy, 1987). The size of the matrix M is set to equal that of (B1 AB) so

that it can be used to store the elements of (B1 R2 B2) later, which has a dimension

equal to (B1 AB). Eliminate A from AX + B Y –R = 0 ( Note that A was stored in M),

we obtain:

X + B2 Y – R2 = 0 (now we no longer need A).

(b) Insert X into B1 X = 0, we have B1 (R2-B2 Y) = 0, and store B1 R2 B2 into the matrix

M.

(c) We now have M Y – B1 R2 = 0, and therefore can eliminate M to get Y.

(d) Get X from X + B2 Y – R2 = 0.

(e) Insert the solutions into the original equations to check the results which should meet:

‘0’ = ‘0’ or in a given error range.

This method requires a matrix M to store most of the elements, and the dimension of the

matrix is (np x np), where np is the total number of nodal points. So, the limit of a

machine/computer may be exceeded when np reaches a certain limit (it depends on the

type of computers).

tectonic stress modelling with the finite element method · 1.2 fem for viscoelastic deformation...

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