Please copy and paste this embed script to where you want to embed

Civil Engineering Analysis and Modelling (CIVL3140)

Dr. Zhenhe (Song) Song [email protected] GHD Pty Ltd

1

Part 1 Geomechanics (Plaxis) Dr. Zhenhe Song [email protected] Part 2 Hydraulics (Fluent) A/Prof. Tongming Zhou (Unit coordinator) [email protected] Part 3 Structures (Multiframe) Mr. Philip Christensen [email protected] 2

Yusuke Suzuki

[email protected]

Wensu Chen

[email protected]

Wen Gao

[email protected]

3

All the students to set up PLAXIS Version 9 software before tutorial. If you get your laptop this year, you may have PLAXIS 2010, you need to reinstall Plaxis V9 Please try to run PLAXIS in your laptop and make sure it works well. Please ask help from the IT support if you have any problems to open PLAXIS. IT Support: Keith Russell [email protected]

4

2x2hrs sessions per week First 2hrs: Lecture (Theory) Second 2hrs: Tutorials (Practice) 4 weeks in total 6% Weekly Practice; 14% Assignment 40% Exam (combined) 5

This note has incorporated the note from previous teaching by Prof. Yuxia Hu

The development of tutorial questions by Dr. Long Yu

6

Finite element analysis in geotechnical engineering: theory, David M. Potts, Lidija Zdravkovi Finite element analysis in geotechnical engineering: application, David M. Potts, Lidija Zdravkovi Guidelines for the use of advanced numerical analysis, David Potts, Kennet Axelsson, Lars Grande, Helmut Schweiger and Michael Long

7

Modelling and FEM in Geotechnical Engineering

8

Stability Loading on Structure Movement

Footing; Retaining Wall and Deep Excavation; Piles and Bridge Abutment; Embankment, Dams and Seawalls; Tunnel; Stockpile; Dynamic (Seismic Analysis)

10

Soils are neither elastic, nor homogeneous. Soils around the world vary. Same soil with different saturations and consolidations behaves differently. Soil properties are difficult to measure. In situ vs laboratory testing …

11

New civil engineer: 14/04/2005

13

Geotechnical engineering is complex. It is not because you’re using the FEM that it becomes simpler; The quality of a tool is important, yet the quality of a result (mainly) depends on the user’s understanding of both the problem and the tool; The design process involves considerably more than analysis.

14

Traditional methods of analysis often use techniques that based on assumptions that over simplify the problem at hand. These methods lack the ability to account for all of the factors and variables the design engineer faces and may severely limit the accuracy of the solution.

15

Equilibrium (stress) Compatibility (strain) Constitutive Relationship (stressstrain) Boundary Condition

Solution of Geotechnical Problems

“Exact” or Closed Form

Finite/ Boundary Element

Limit Equilibrium

Boundary Element

Numerical

Finite Difference

Limit Analysis

Finite Element

Empirical, Based on Experience

Discrete Element

17

Method of Analysis

Solution Requirements Stress Equilibrium

Compatibility

Limit equilibrium

(P)

X

Slip-line method

(P)

X

Limit Analysis -Lower Bound -Upper Bound

Constitutive behaviour Rigid plastic Rigid plastic

Design Information Stability

Displacements X X

Perfectly plastic

X

X X

X

Displacement finite element

Any P– partially satisfied 18

Receive Design Prescriptions (from a client) Obtain Soil Properties (Site investigations and lab testing) Model Geotechnical Problem Verification Detailed Design Report 19

Numerical modelling Geotechnical model

Silo

http://www.cofs.uwa.edu.au/Researh/centrifugeprojects.html

Physical modelling

http://www.pbase.com/image/41209293

20

Plain strain or axisymmetric

21

CL Footing (B/2)

22

Discretisation (mesh):Divide the model field (soil and/or structure) into parts (nodes and elements) Displacement Approximation: Over each part (element), displacement is expressed as function of nodal values Element Equation: Use an approximate variational principle (e.g. minimum potential energy) to derive an element equation KUE=PE Global Equation: Then assemble the parts of element equation to form a global equation KU=P Boundary Condition: Formulate boundary conditions and modify global equations. Loads affect P, displacement affect U Solutions: Solve displacement values at nodes and then stress and strain can be evaluated 23

CL Footing (B/2)

Element x x

x

Node x

Gauss point (integration point)

24

Element Type

Degree of Freedom per Element

Plane Strain

Axisymmetric

Integration rule Gauss point

Constraints per Element

Ratio Degrees of Freedom Constraints

Suitable

Integration rule Gauss point

Constraints per Element

Ratio of Degrees of Freedom Constraints

Suitable

3-noded constant Strain triangle

1

1-point

1

1

Y

3-point

3

1/3

N

6-noded linear Strain triangle

4

3-point

3

4/3

Y

6-point

6

2/3

N

10-noded quadratic Strain triangle

9

6-point

6

3/2

Y

12-point

10

9/10

N

15-noded cubic Strain triangle

16

12-point

10

8/5

Y

16-point

15

16/15

Y

4-noded quadrilateral

2

2x2

3

2/3

N

3x3

5

2/5

N

8-noded quadrilateral

6

3x3

6

1

Y

3x3

9

2/3

N

12-noded quadrilateral

10

4x4

10

1

Y

4x4

13

10/13

N

17-noded quadrilateral

16

5x5

14

8/7

Y

5x5

19

16/19

N 25

Sloan, S. W. and Randolph, M. F. (1982) “Numerical prediction of collapse loads using finite element analysis”, Int. J. Num. Ana. Meth. Geo.

(x3, y3) u 3, v 3

3

y x

Function: u(x,y) = a1 + a2x + a3y v(x,y) = b1 + b2x + b3y

v u 1 (x1, y1) u 1, v 1

u=?

2 (x2, y2) u 2, v 2

u1 = u(x1, y1) = a1 + a2x1 + a3y1 u2 = u(x2, y2) = a1 + a2x2 + a3y2 u3 = u(x3, y3) = a1 + a2x3 + a3y3

u1 u2 u3

1 x1 1 x2 1 x3

y1 y2 y3

a1 a2 a3

Solve for a 1, a 2, a 3 26

(x 2 y 3 Function of (x,y) N

N1 N2 N3

(x 3 y1 (x1y 2

x 3 y 2 ) x(y 2 y 3 ) y(x 3 x 2 ) 2A x1y 3 ) x(y 3 y1 ) y(x1 x 3 ) 2A x 2 y1 ) x(y1 y 2 ) y(x 2 x1 ) 2A

u1

Function of (x,y)

U

v1 u2

u

N1

0

N2

0

N3

0

v

0

N1

0

N2

0

N3 v2 u3 v3

27

(x6, y6) u 6, v 6 6

x

(x1, y1) u 1, v 1

u1 u2 u3 u4 u5 u6

(x5, y5) 5 u ,v 5 5

v u

1

Function: u(x,y) = a1 + a2x + a3y + a4x2 + a5xy + a6y2 v(x,y) = b1 + b2x + b3y + b4x2 + b5xy + b6y2

y

(x3, y3) u 3, v 3 3

u=?

2 (x2, y2) u 2, v 2

4 (x4, y4) u 4, v 4

x1 x2 x3 x4 x5 x6

1 1 1 1 1 1

x12 x 22 x 32 x 42 x 52 x 62

y1 y2 y3 y4 y5 y6

y12 y 22 y 32 y 42 y 52 y 62

x1 y 1 x2 y2 x3 y 3 x4 y4 x5 y 5 x6 y6

a1 a2 a3 a4 a5 au6

1

v1 u2 v2

U

u v

N1 0

0 N1

N 2 0

0 N 2

N 3 0

0 N N 3 0

4

0 N

4

N 0

5

0 N

5

N 0

6

0 N 6

u3 v3 u4 v4 u5 v5 u6 v6

28

3

Strain within an element: Displacement: u(x,y) = a1 + a2x + a3y + a4x2 + a5xy + a6y2 v(x,y) = b1 + b2x + b3y + b4x2 + b5xy + b6y2 Strain: xx

yy

xy

6

y x

(b2

a 3 ) (a5

5 u

1

2 4

u a 2 2a 4 x a5 y x v b3 b5 x 2b6 y y u y

v

2b4 ) x (2a 6

e

b5 ) y

B U

e 29

Constitutive Relation Stress and strain can be written in vector form and then expressed as

D Linear isotropic elasticity

30

3

Body forces and surface tractions applied to the element may be generalized into a set of forces acting at the nodes 6 Based on an appropriate variational principle (e.g. minimum potential energy) to derive element equations:

e

K U

e

where

K

e

T

B DBdv

P

e

P1x

5

1

2 4 P1Y

In order to get [Ke], integration (gaussian integration) must be performed for each element. Basically, the integral of the function is replaced by weighted sum of the function at a number of integration points 31

The stiffness for the complete mesh is evaluated by combining the individual element stiffness matrixes assembly) This produces a square matrix K of dimension equal to the number of degree-of-freedom in the mesh The global vector of nodal forces P is obtained in a similar way by assembling the element nodal force vectors The assembled stiffness matrix and force vector are related by:

K U

P

32

1 K11 1 K11

1 K12

1 K13

1 K14

1 K 22

1 K 23 1 K 44

1 K 24 1 K 43

1 K12 1 K 22

1 K13 1 K 23

K141 1 K 24

1 K 33

1 K 34 1 K 44

1 K 33

1 K11

K 332

K 342

K 362

K 352

2 K 44

2 K 46 K 662

K 452 K 652 K 552

K121 1 K 22

1 K13 1 K 23 1 K 33

K 332

1 K14 1 K 24 1 K 34 1 K 44

K 342 K 442

K 352 2 K 45

K 362 K 462

K 552

K 562 K 662 33

CL Footing (B/2)

1

4

10x(B/2)

10x(B/2)

2

3

Find symmetrical features, central line can be a roller boundary. (CL) (1) Soil domain needs to be large enough to avoid boundary effect. (10x(B/2), 10x(B/2)) The bottom boundary can be fixed boundary. (2) The side boundary can be roller boundary. (3) Top boundary is normally a free boundary. (4) 34

Element size: the smaller, the more accurate Element type: the higher order, the more accurate Boundary conditions: domain size, realistic Constitutive model: complexity

economy

Soil parameters: realistic, measurable Understanding of the real problem

numerical

model 35

Less elements to reduce computation time Smaller elements to increase accuracy

Optimum Mesh Combination of coarse and fine mesh

How ?

36

Footing (B/2)

37

Displacement control (prescribed displacement) or load control (prescribed load) ? 2-dimensional or 3-dimensional analysis ? Plain strain or axisymmetric ? Drained, undrained or consolidation analysis? Construction Stages

38

Pre-processing Define problem(2D or 3D? Plain strain or Axisymmetric? Soil model? Drained or undrained?); define domain (size?); define boundary condition; generate mesh (element type? mesh density?); input soil/foundation parameters (worked out soil parameter from site investigation).

2) Calculation FEM Calculation Steps

3) Post-processing Process calculation results, such as soil stress/strain distribution; soil deformations, et al.

39

View more...
Dr. Zhenhe (Song) Song [email protected] GHD Pty Ltd

1

Part 1 Geomechanics (Plaxis) Dr. Zhenhe Song [email protected] Part 2 Hydraulics (Fluent) A/Prof. Tongming Zhou (Unit coordinator) [email protected] Part 3 Structures (Multiframe) Mr. Philip Christensen [email protected] 2

Yusuke Suzuki

[email protected]

Wensu Chen

[email protected]

Wen Gao

[email protected]

3

All the students to set up PLAXIS Version 9 software before tutorial. If you get your laptop this year, you may have PLAXIS 2010, you need to reinstall Plaxis V9 Please try to run PLAXIS in your laptop and make sure it works well. Please ask help from the IT support if you have any problems to open PLAXIS. IT Support: Keith Russell [email protected]

4

2x2hrs sessions per week First 2hrs: Lecture (Theory) Second 2hrs: Tutorials (Practice) 4 weeks in total 6% Weekly Practice; 14% Assignment 40% Exam (combined) 5

This note has incorporated the note from previous teaching by Prof. Yuxia Hu

The development of tutorial questions by Dr. Long Yu

6

Finite element analysis in geotechnical engineering: theory, David M. Potts, Lidija Zdravkovi Finite element analysis in geotechnical engineering: application, David M. Potts, Lidija Zdravkovi Guidelines for the use of advanced numerical analysis, David Potts, Kennet Axelsson, Lars Grande, Helmut Schweiger and Michael Long

7

Modelling and FEM in Geotechnical Engineering

8

Stability Loading on Structure Movement

Footing; Retaining Wall and Deep Excavation; Piles and Bridge Abutment; Embankment, Dams and Seawalls; Tunnel; Stockpile; Dynamic (Seismic Analysis)

10

Soils are neither elastic, nor homogeneous. Soils around the world vary. Same soil with different saturations and consolidations behaves differently. Soil properties are difficult to measure. In situ vs laboratory testing …

11

New civil engineer: 14/04/2005

13

Geotechnical engineering is complex. It is not because you’re using the FEM that it becomes simpler; The quality of a tool is important, yet the quality of a result (mainly) depends on the user’s understanding of both the problem and the tool; The design process involves considerably more than analysis.

14

Traditional methods of analysis often use techniques that based on assumptions that over simplify the problem at hand. These methods lack the ability to account for all of the factors and variables the design engineer faces and may severely limit the accuracy of the solution.

15

Equilibrium (stress) Compatibility (strain) Constitutive Relationship (stressstrain) Boundary Condition

Solution of Geotechnical Problems

“Exact” or Closed Form

Finite/ Boundary Element

Limit Equilibrium

Boundary Element

Numerical

Finite Difference

Limit Analysis

Finite Element

Empirical, Based on Experience

Discrete Element

17

Method of Analysis

Solution Requirements Stress Equilibrium

Compatibility

Limit equilibrium

(P)

X

Slip-line method

(P)

X

Limit Analysis -Lower Bound -Upper Bound

Constitutive behaviour Rigid plastic Rigid plastic

Design Information Stability

Displacements X X

Perfectly plastic

X

X X

X

Displacement finite element

Any P– partially satisfied 18

Receive Design Prescriptions (from a client) Obtain Soil Properties (Site investigations and lab testing) Model Geotechnical Problem Verification Detailed Design Report 19

Numerical modelling Geotechnical model

Silo

http://www.cofs.uwa.edu.au/Researh/centrifugeprojects.html

Physical modelling

http://www.pbase.com/image/41209293

20

Plain strain or axisymmetric

21

CL Footing (B/2)

22

Discretisation (mesh):Divide the model field (soil and/or structure) into parts (nodes and elements) Displacement Approximation: Over each part (element), displacement is expressed as function of nodal values Element Equation: Use an approximate variational principle (e.g. minimum potential energy) to derive an element equation KUE=PE Global Equation: Then assemble the parts of element equation to form a global equation KU=P Boundary Condition: Formulate boundary conditions and modify global equations. Loads affect P, displacement affect U Solutions: Solve displacement values at nodes and then stress and strain can be evaluated 23

CL Footing (B/2)

Element x x

x

Node x

Gauss point (integration point)

24

Element Type

Degree of Freedom per Element

Plane Strain

Axisymmetric

Integration rule Gauss point

Constraints per Element

Ratio Degrees of Freedom Constraints

Suitable

Integration rule Gauss point

Constraints per Element

Ratio of Degrees of Freedom Constraints

Suitable

3-noded constant Strain triangle

1

1-point

1

1

Y

3-point

3

1/3

N

6-noded linear Strain triangle

4

3-point

3

4/3

Y

6-point

6

2/3

N

10-noded quadratic Strain triangle

9

6-point

6

3/2

Y

12-point

10

9/10

N

15-noded cubic Strain triangle

16

12-point

10

8/5

Y

16-point

15

16/15

Y

4-noded quadrilateral

2

2x2

3

2/3

N

3x3

5

2/5

N

8-noded quadrilateral

6

3x3

6

1

Y

3x3

9

2/3

N

12-noded quadrilateral

10

4x4

10

1

Y

4x4

13

10/13

N

17-noded quadrilateral

16

5x5

14

8/7

Y

5x5

19

16/19

N 25

Sloan, S. W. and Randolph, M. F. (1982) “Numerical prediction of collapse loads using finite element analysis”, Int. J. Num. Ana. Meth. Geo.

(x3, y3) u 3, v 3

3

y x

Function: u(x,y) = a1 + a2x + a3y v(x,y) = b1 + b2x + b3y

v u 1 (x1, y1) u 1, v 1

u=?

2 (x2, y2) u 2, v 2

u1 = u(x1, y1) = a1 + a2x1 + a3y1 u2 = u(x2, y2) = a1 + a2x2 + a3y2 u3 = u(x3, y3) = a1 + a2x3 + a3y3

u1 u2 u3

1 x1 1 x2 1 x3

y1 y2 y3

a1 a2 a3

Solve for a 1, a 2, a 3 26

(x 2 y 3 Function of (x,y) N

N1 N2 N3

(x 3 y1 (x1y 2

x 3 y 2 ) x(y 2 y 3 ) y(x 3 x 2 ) 2A x1y 3 ) x(y 3 y1 ) y(x1 x 3 ) 2A x 2 y1 ) x(y1 y 2 ) y(x 2 x1 ) 2A

u1

Function of (x,y)

U

v1 u2

u

N1

0

N2

0

N3

0

v

0

N1

0

N2

0

N3 v2 u3 v3

27

(x6, y6) u 6, v 6 6

x

(x1, y1) u 1, v 1

u1 u2 u3 u4 u5 u6

(x5, y5) 5 u ,v 5 5

v u

1

Function: u(x,y) = a1 + a2x + a3y + a4x2 + a5xy + a6y2 v(x,y) = b1 + b2x + b3y + b4x2 + b5xy + b6y2

y

(x3, y3) u 3, v 3 3

u=?

2 (x2, y2) u 2, v 2

4 (x4, y4) u 4, v 4

x1 x2 x3 x4 x5 x6

1 1 1 1 1 1

x12 x 22 x 32 x 42 x 52 x 62

y1 y2 y3 y4 y5 y6

y12 y 22 y 32 y 42 y 52 y 62

x1 y 1 x2 y2 x3 y 3 x4 y4 x5 y 5 x6 y6

a1 a2 a3 a4 a5 au6

1

v1 u2 v2

U

u v

N1 0

0 N1

N 2 0

0 N 2

N 3 0

0 N N 3 0

4

0 N

4

N 0

5

0 N

5

N 0

6

0 N 6

u3 v3 u4 v4 u5 v5 u6 v6

28

3

Strain within an element: Displacement: u(x,y) = a1 + a2x + a3y + a4x2 + a5xy + a6y2 v(x,y) = b1 + b2x + b3y + b4x2 + b5xy + b6y2 Strain: xx

yy

xy

6

y x

(b2

a 3 ) (a5

5 u

1

2 4

u a 2 2a 4 x a5 y x v b3 b5 x 2b6 y y u y

v

2b4 ) x (2a 6

e

b5 ) y

B U

e 29

Constitutive Relation Stress and strain can be written in vector form and then expressed as

D Linear isotropic elasticity

30

3

Body forces and surface tractions applied to the element may be generalized into a set of forces acting at the nodes 6 Based on an appropriate variational principle (e.g. minimum potential energy) to derive element equations:

e

K U

e

where

K

e

T

B DBdv

P

e

P1x

5

1

2 4 P1Y

In order to get [Ke], integration (gaussian integration) must be performed for each element. Basically, the integral of the function is replaced by weighted sum of the function at a number of integration points 31

The stiffness for the complete mesh is evaluated by combining the individual element stiffness matrixes assembly) This produces a square matrix K of dimension equal to the number of degree-of-freedom in the mesh The global vector of nodal forces P is obtained in a similar way by assembling the element nodal force vectors The assembled stiffness matrix and force vector are related by:

K U

P

32

1 K11 1 K11

1 K12

1 K13

1 K14

1 K 22

1 K 23 1 K 44

1 K 24 1 K 43

1 K12 1 K 22

1 K13 1 K 23

K141 1 K 24

1 K 33

1 K 34 1 K 44

1 K 33

1 K11

K 332

K 342

K 362

K 352

2 K 44

2 K 46 K 662

K 452 K 652 K 552

K121 1 K 22

1 K13 1 K 23 1 K 33

K 332

1 K14 1 K 24 1 K 34 1 K 44

K 342 K 442

K 352 2 K 45

K 362 K 462

K 552

K 562 K 662 33

CL Footing (B/2)

1

4

10x(B/2)

10x(B/2)

2

3

Find symmetrical features, central line can be a roller boundary. (CL) (1) Soil domain needs to be large enough to avoid boundary effect. (10x(B/2), 10x(B/2)) The bottom boundary can be fixed boundary. (2) The side boundary can be roller boundary. (3) Top boundary is normally a free boundary. (4) 34

Element size: the smaller, the more accurate Element type: the higher order, the more accurate Boundary conditions: domain size, realistic Constitutive model: complexity

economy

Soil parameters: realistic, measurable Understanding of the real problem

numerical

model 35

Less elements to reduce computation time Smaller elements to increase accuracy

Optimum Mesh Combination of coarse and fine mesh

How ?

36

Footing (B/2)

37

Displacement control (prescribed displacement) or load control (prescribed load) ? 2-dimensional or 3-dimensional analysis ? Plain strain or axisymmetric ? Drained, undrained or consolidation analysis? Construction Stages

38

Pre-processing Define problem(2D or 3D? Plain strain or Axisymmetric? Soil model? Drained or undrained?); define domain (size?); define boundary condition; generate mesh (element type? mesh density?); input soil/foundation parameters (worked out soil parameter from site investigation).

2) Calculation FEM Calculation Steps

3) Post-processing Process calculation results, such as soil stress/strain distribution; soil deformations, et al.

39

Thank you for interesting in our services. We are a non-profit group that run this website to share documents. We need your help to maintenance this website.