Lecture 01 Introduction Lecture Handout

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Dr. Zhenhe (Song) Song

GHD Pty Ltd

Civil Engineering Analysis and Modelling (CIVL3140)

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Part 1 Geomechanics (Plaxis)

Dr. Zhenhe Song

Part 2 Hydraulics (Fluent)

A/Prof. Tongming Zhou (Unit coordinator)

Part 3 Structures (Multiframe)

Mr. Philip Christensen

Yusuke Suzuki

Wensu Chen

Wen Gao

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]

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2x2hrs sessions per week

First 2hrs: Lecture (Theory)

Second 2hrs: Tutorials (Practice)

4 weeks in total

6% Weekly Practice; 14% Assignment

40% Exam (combined)

(6)

This note has incorporated the note from previous

teaching by

Prof. Yuxia Hu

The development of tutorial questions by

Dr. Long

Yu

(7)

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

(8)

Modelling and FEM in Geotechnical Engineering

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Stability

Loading on Structure

(10)

Footing;

Retaining Wall and Deep Excavation;

Piles and Bridge Abutment;

Embankment, Dams and Seawalls;

Tunnel;

Stockpile;

Dynamic (Seismic Analysis)

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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 …

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13

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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.

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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.

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Equilibrium (stress)

Compatibility (strain)

Constitutive

Relationship

(stress-strain)

Boundary Condition

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Solution of Geotechnical Problems Numerical “Exact” or Closed Form Empirical, Based on Experience Limit Analysis Discrete Element Finite Element Finite Difference Boundary Element Finite/ Boundary Element 17 Limit Equilibrium

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Method of Analysis Solution Requirements Design Information Stress Equilibrium Compatibility Constitutive behaviour Stability Displacements Limit equilibrium (P) X Rigid plastic X Slip-line method (P) X Rigid plastic X Limit Analysis -Lower Bound -Upper Bound X X Perfectly plastic X X Displacement finite element Any P– partially satisfied 18

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Receive Design Prescriptions

(from a client)

Obtain Soil Properties

(Site investigations and lab testing)

Model Geotechnical Problem

Detailed Design Report

19

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

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

Geotechnical model

Numerical modelling

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Plain strain or axisymmetric

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Footing (B/2)

CL

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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

KU

E

_=P

E

23

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

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Footing (B/2) Element x x x Node

Gauss point (integration point)

x

CL

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25

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

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26 x y u v 1 2 3 Function: u(x,y) = a₁ + a₂x + a₃y v(x,y) = b₁ + b₂x + b₃y (x₁, y₁) u₁, v₁ (x₃, y₃) u₃, v₃ (x₂, y₂) u₂, v₂ u₁ = u(x₁, y₁) = a₁ + a₂x₁ + a₃y₁ u₂ = u(x₂, y₂) = a₁ + a₂x₂ + a₃y₂ u₃ = u(x₃, y₃) = a₁ + a₂x₃ + a₃y₃ 3 2 1 3 3 2 2 1 1 3 2 1

1

1 a

a

y

x

y

x

y

x

u

u = ?

Solve for a₁, a₂, a₃

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27

2A

)

x

y(x

)

y

x(y

)

y

x

y

(x

2A

)

x

y(x

)

y

x(y

)

y

x

y

(x

2A

)

x

y(x

)

y

x(y

)

y

x

y

(x

N

1 2 2 1 1 2 2 1 3 1 1 3 3 1 1 3 2 3 3 2 2 3 3 2 3 2 1 3 3 2 2 1 1 3 2 1 3 2 1

N

0 N

0

0 N

v

u

U

v

u

v

u

v

u

Function of (x,y) Function of (x,y)

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6 6 5 5 4 4 3 3 2 2 1 1 6 5 4 3 2 1 6 5 4 3 2 1 N 0 N 0 N 0 N 0 N 0 N 0 0 N 0 N 0 N 0 N 0 N 0 N v u U v u v u v u v u v u v u 28 x y u v 1 2 3 (x₁, y₁) u₁, v₁ (x₃, y₃) u₃, v₃ (x₂, y₂) u₂, v₂

u = ?

6 ₅ 4 (x₆, y₆) u₆, v₆ (x5, y5) u₅, v₅ (x₄, y₄) u₄, v₄ Function: u(x,y) = a₁ + a₂x + a₃y + a₄x2 _{+ a} 5xy + a6y2 v(x,y) = b₁ + b₂x + b₃y + b₄x2 _{+ b} 5xy + b6y2 6 5 4 3 2 1 2 6 6 6 2 6 6 6 2 5 5 5 2 5 5 5 2 4 4 4 2 4 4 4 2 3 3 3 2 3 3 3 2 2 2 2 2 2 2 2 2 1 1 1 2 1 1 1 6 5 4 3 2 1 1 1 1 1 1 1 a a a a a a y y x x y x y y x x y x y y x x y x y y x x y x y y x x y x y y x x y x u u u u u u

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Strain within an element: Displacement: u(x,y) = a₁ + a₂x + a₃y + a₄x2 _{+ a} 5xy + a6y2 v(x,y) = b₁ + b₂x + b₃y + b₄x2 _{+ b} 5xy + b6y2 Strain: 29 u v 1 2 3 6 ₅ 4

y

a

x

a

x

u

5 4 2 xx

2 y

b

x

b

y

v

6 5 3 yy

b

2

y b a x b a a b x y y u ) 2 ( ) 2 ( ) ( ₂ ₃ ₅ ₄ ₆ 5 xy

e

U

B

e

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30

Constitutive Relation

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

D

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31 1 2 3 6 ₅ 4 P_1x P_1Y

Body forces and surface tractions applied to the element may be generalized into a set of forces acting at the nodes

Based on an appropriate variational

principle (e.g. minimum potential energy) to derive element equations:

e

P

U

e

K

where

v

DBd

B

K

e T

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

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32

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:

P

U

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33 1 44 1 34 1 33 1 24 1 23 1 22 1 14 1 13 1 12 1 11 1 33 1 43 1 44 1 24 1 23 1 22 1 14 1 13 1 12 1 11 K K K K K K K K K K K K K K K K K K K K 2 66 2 56 2 55 2 46 2 45 2 44 1 44 2 36 2 35 2 34 1 34 2 33 1 33 1 24 1 23 1 22 1 14 1 13 1 12 1 11 2 55 2 65 2 66 2 45 2 46 2 44 2 35 2 36 2 34 2 33 K K K K K K K K K K K K K K K K K K K K K K K K K K K K K K

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Find symmetrical features, central

line can be a roller boundary.

(CL)

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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 CL Footing (B/2) 10x(B/2) 10x(B/2) 1 2 3 4

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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

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Less elements to reduce computation time

Smaller elements to increase accuracy

36

Optimum Mesh

Combination of coarse and fine mesh

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37

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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

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