Plaxis_advanced_course_hong_kong_2012.pdf

Advanced Computational Geotechnics

AD A D A ANCED NCED C

MPU A ATI TI

NAL

EOTE EOT ECH ICS

2012 HO G K

NG

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DV NCED

OURS

N COMPU A ATIO TIO A AL L

EOT CHNICS

HONG K ON

Venue Dates

HKMA, S op 2, G/F, Pico Tower, 6 6 Gloucester Road, Wan Chai, Hong Kong

6‐8 November 2012: Advanced Course 6 9 November 2012 : 3D Applica ion Course

Le turers

Pr fessor Helmut Schweiger (Cours leader)

Graz Univ rsity of rsity of Tec Tec nology, Gra z, Austria z, Austria

Pr fessor Antonio Gens

Technical niversity of niversity of Catalonia ( PC), Barcel Barcel na, Spain

Dr Lee Siew

Golder As Golder As ociates (HK Ltd.

ei

Dr Johnny Ch uk

Aecom Asi Aecom Asi a, Hong Kon

Dr William Cheang

Plaxis Asi Plaxis Asi Pac, Singa ore

Organised by So lutions Re search Centre Ltd

1709-11, Leighton Centr 77 Leighton oad Causeway B y, Hong K ng Pl xis AsiaP ac Pte Ltd

16 Jalan Kila g Timor 05-07 Redhill Forum Si gapore

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CONTENTS CG

MODULES

PAGE

1

Geotechnical Finite Element Analysis

7

2

Elasto‐plasticity and Mohr‐Coulomb

23

3

Exercise 1:Elasto‐plastic Analysis of Shallow of Shallow Foundations

43

4

Critical States Soil Mechanics and Soft Soil Model

81

5

Hardening Soil and HS‐small Model

101

6

Exercise 2:Simulation of Triaxial of Triaxial and Oedometer Tests

131

7

Modelling of Deep of Deep Excavations

179

8

Structural Elements in Plaxis

203

9

Exercise 3: Excavation of a of a Building Pit

217

10

Drained and Undrained Analysis

253

11

Consolidation Analysis

273

12

Modelling of Groundwater of Groundwater in Plaxis

289

13

Exercise 4: Excavation and Dewatering

317

14

Unsaturated Soil Mechanics and Barcelona Basic Model

329

15

Initial Stresses, Phi‐C’ Reduction and Slope Stability Analysis

361

16

Exercise 5: Stability of a of a Slope Stabilised by Soil Nails

389

17

Hoek‐Brown and Rock Jointed Model

403

18

Tunnelling in 2D

427

19

Exercise 6: Excavation of a of a Tunnel in Rock

439

Special 3D Modelling Modules 20

Plaxis 3D

450

21

3D Modelling of Tunnels of Tunnels

468

22

Exercise 7: Stability of a of a NATM Tunnel

496

23

3D Modelling of Deep of Deep Foundations

518

24

3D Modelling of Deep of Deep Excavations

552

25

Exercise 8: Modelling of Excavation of Excavation

578


COURSE STRUCT RE

ND L EC UR RS

The course is divided in o nineteen lectures (CG) which includes six 2-D modelling e ercises for the 3-day advanced course. For t e add-on s ecial course on 3-D mo elling, there is an additi n of six lect res and it includes tw 3-D modelling exercises. The modu les and exercises are gr ouped into themes to orm the Advanced omputation l Geotechnics course in Hong Kong using Plaxi finite elem nt program Them e 1 Advanced Computational Geotechnics and Soil Behavi ur Them e 2 Applic tions I: Exc vations and Modelling f Groundwa er Them e 3 Applic tions II: Initial Stresses, Unsaturate Soils and odelling of roblems in Rock Them e 4 Applic tions III: 3D Analysis of Deep Foun ations, Exc vations & T unnels

Professor Helmut Schweiger

OURSE L ADER

Pr fessor Antonio Gens T chnical Uni ersity of Catalonio ( PC)

r William Cheang Plaxis Asi Pac

Helmut obtained his Ph.D f om the Unive rsity College f Swansea, U.Kingdom a d teaches courses on Adv nced Soil M chanics and Computation l Geomechanics at the Gra z University of Technology, ustria. He h s over 15 ye rs of experie ce in develo ment and application of the finite ele ent method. As a member of several committees Helmut is invol ed in formula ting guidelines and recommendations for the u e of finite elements in practical eotechnical ngineering.

ntonio is a f culty member at the Techn ical Universit of Catalonia since 1983 af er a Ph.D. at Imperial Coll ge, London. e has been iinvolved in geotechnical re earch, education and prac ice for more han 25 years with special r eference to t e application of numerical analysis to engineering pro lems. He ha consulted in a variety of projects invol ing deep excavations, tun els, ground i provement echniques, dams, power s ations, foundations and slopes. He elivered the the 47th Ran ine lecture in 2007

illiam obtai ed his Ph.D f om the Natio nal University of Singapore. His interest is in Computati nal Geotech ics. He has orked as a Geotechnical Engineer in alaysia, Singapore and Thailand. He is iinvolved with many semina s and worksh ops around sia for the promotion of ood and effe tive usage of Plaxis Finite Element Codes.

COORDIN TOR

Dr Lee Sie Wei Technical Director G lder Associ tes (HK)

Dr Johnny heuk Senior Engiineer ECOM Asia o. Ltd.

SW obtained his PhD in Geotechnical Engineering fro Cambridge University in 001 and has been working for Geotechnical Consulting G oup (Asia) Lt (now Golder Associates ( K) Ltd) since. He is a qualified civil geotechnical engineer, an has applied numerical modelling to the esign, assessment and in estigation of eotechnical roblems in Hong Kong an other Asian countries ohnny obtai ed his PhD degree from Cambridge niversity. H had lectured at City Unive sity of Hong Kong and th University of Hong Kong before joining AECOM in 2 009. Johnny has extensiv experience in research and practic in offshor and slop engineering. He is a member of the In ternational S ciety for Soil Mechanics nd Geotechnical Engineering (ISSMGE)– TC10 (Numerical ethods in eomechanics) and TC104 (Physical Modelling in Geotechnics . Johnny is a member nd Assistant Secretary of Geotechnical Division Co mittee of th Hong Kong. ohnny has been an Hono ary Assistant Professor at the Universit f Hong Kong since 2009.

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Time 9:00

10:00

Day 1 Advanced Computational Geotechnics CG1 Geotechnical Finite Element Analysis

Gens

10:00

11:00

CG2

Gens

11:00

11:15

11:15

1:00

1:00

2:00

2:00

3:00

CG4

Critical State Soil Mechanics and Soft Soil Model

Schweiger

3:00

4:00

CG5

Hardening Soil and HS‐small Model

Schweiger

4:00

4:15

4:15

5:30

CG6

9:45 10:30 10:45 12:00 1:00 1:45 2:30 3:15 3:30 5:00

Day 2: CG Applications 1: Excavations CG7 Modelling of Deep of Deep Excavations CG8 Structural Elements in Plaxis Break CG9 Exercise 3: Tied‐Back Excavation Lunch CG10 Drained and Undrained Analysis CG11 Consolidation Analysis Break CG12 Modelling of Groundwater of Groundwater in Plaxis CG13 Exercise 4: Dewatering in Excavation

10:30 11:30 11:45 1:30 2:30 3:30 4:00 4:15 5:30

Day 3: CG Applications 2: Unsaturated Soils and Rock CG14 Unsaturated Soils and Barcelona Basic Model CG15 Initial Stresses and Slope Stability Analysis Break CG16 Exercise 5: Slope Stability Exercise Lunch CG17 Hoek‐Brown and Rock Jointed Models CG18 Modelling of Tunnels of Tunnels in 2D Break CG19 Exercise 6: Tunnelling in Rock

10:00 11:00 11:15 1:00 2:00 3:00 4:00 4:15 5:30

Day 4: CG Applications 3: 3D Analysis (Optional) CG20 Introduction to Plaxis 3D CG21 3D Modelling of Tunnels of Tunnels Break CG22 Exercise 7: Tunnel Stability Lunch CG23 3D Modelling of Deep of Deep Foundations CG24 3D Modelling of Deep of Deep Excavations Break CG25 Exercise 8: Modelling of Excavations of Excavations

Time 9:00 9:45 10:30 10:45 12:00 1:00 1:45 2:30 3:15 3:30

Time 9:00 10:30 11:30 11:45 1:30 2:30 3:30 4:00 4:15

Time 9:00 10:00 11:00 11:15 1:00 2:00 3:00 4:00 4:15

Elasto‐Plasticity and Mohr‐Coulomb Break

CG3

Exercise 1: Foundation on Elasto‐Plastic Soils

Cheang

Lunch

Break Exercise 2: Triaxial and Oedometer

Cheang

Schweiger Cheuk Lee Gens Gens Lee Cheuk

Gens Schweiger Cheuk Schweiger Schweiger Cheang

Cheang Schweiger Cheang Schweiger Schweiger Cheang

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KCG 1 GEO GEOTE HNIC L FI ITE

L EM NT A NALYSIS

rofess r Anto io Gen

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CG1: GEOTEC GEOTECHNICAL HNICAL FINI FINITE TE EL EMENT ANALYSIS

Antonio Gens Technical University of Catalunya, Barcelona

some of the slides were originally created by: Andrew Abbo (University of Newcastle) Cino Viggiani (Laboratoire 3S, Grenoble, France) Dennis Waterman (Plaxis)

Outline • In Intr trod oduc ucti tion on • Fi Fini nite te El Elem emen ents ts di disp spla lace ceme ment nt an anal alys ysis is Elements Elements for two-dimen two-dimensiona sionall analysis analysis Displacement interpolation Strains Constitut Constitutive ive equation equation Element stiff stiffness ness matr matrix ix Global Global stiffn stiffness ess matrix matrix Soluti Solution on of the global global stiffn stiffness ess equati equations ons • Ela last stic icit ity y as app ppli lie ed to so soil ilss Fundamen Fundamentals, tals, and elastic elastic paramete parameters rs Two-dimens wo-dimensiona ionall elastic elastic ana analysis lysis

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design requirements in geotechnical engineering •

•

Stab Stabil ilit ity y (loc (local al and and gen gener eral al))

Admiss Admissibl ible e defor deformat mation ion and displa displacem cement entss

design requirements in geotechnical engineering •

•

Flow problems

Sometime Sometimess flow and and stabilit stability/de y/deform formation ation problems problems are are solved solved toget together her 

See tomorrow’s lecture on consolidation (CG11)

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geotechnical analysis: basic solution requirements • Unknowns: 15 (6 stresses, 6 strains, 3 displacements)

• Equilibrium (3 equations) • Compatibility (6 equations) • Constitutive equation (6 equations)

Potts & Zdravkovic

(1999)

geotechnical numerical analysis •

methods for numerical analysis 

Finite difference method



Boundary element method (BEM)



Discrete element method (DEM)



Finite element method (FEM)



Others (meshless methods, particle methods…)

•

While the FEM has been used in many fields of engineering practice for over 40 years, it is only recently that it has begun to be widely used for analyzing geotechnical problems. This is probably because there are many complex issues which are specific to geotechnical engineering and which have been resolved relatively recently.

•

when properly used, this method can produce realistic results which are of value to practical soil engineering problems

•

A good analysis, which simulates real behaviour, allows the engineer to understand problems better. While an important part of the design process, analysis only provides the engineer with a tool to quantify effects once material properties and loading conditions have been set

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geotechnical finite element analysis •

Objectives of the numerical (finite element) analysis 

Selection of design alternatives



Quantitative predictions



Backcalculations



Understanding! 

Identification of critical mechanisms



Identification of key parameters


•

Advantages of numerical (finite element) analysis 

Simulation of complete construction history



Interaction with water can be considered rigorously



Complex geometries (2D-3D) can be modeled



Structural elements can be introduced



No failure mechanism needs to be postulated (it is an outcome of the analysis)

(Nearly) unavoidable uncertainties 

Ground profile



Initial conditions (initial stresses, pore water pressure…)



Boundary conditions (mechanical, hydraulic)

 

Appropriate model for soil behaviour Model parameters

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Some requirements for successful numerical modelling 

Construction of an adequate conceptual model that includes the basic features of the model. The model should be as simple as possible but not simpler



Selection of an appropriate constitutive model. It depends on: 

type of soil or rock



goal of the analysis



quality and quantity of available information



Pay attention to patterns of behaviour and mechanisms rather than just to quantitative predictions



Perform sensitivity analyses. Check robustness of solution



Model calibration (using field results) should be a priority, especially of quantitative predictions are sought



Check against alternative computations if available (even if simplified)

three final remarks

1. geotechnical engineering is complex. It is not because you’re using the FEM that it becomes simpler 2. the quality of a tool is important, yet the quality of a result also (mainly) depends on the user’s understanding of both the problem and the tool 3.

the design process involves considerably more than analysis

Borrowed from C. Viggiani, with thanks

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introduction: the Finite Element Method

the FEM is a computational procedure that may be used to obtain an approximate solution to a boundary value problem the governing mathematical equations are approximated by a series of algebraic equations involving quantities that are evaluated at discrete points within the region of interest. The FE equations are formulated and solved in such a way as to minimize the error in the approximate solution this lecture presents only a basic outline of the method attention is focused on the "displacement based" FE approach

introduction: the Finite Element Method

The FEM is a computational procedure that may be used to obtain an approximate solution to a boundary value problem What kind of problem? Apply load

obtain displacements stiffness matrix

Apply head

obtain flow permeability matrix

Though we would like to know our solution at any coordinates in our project, we will only calculate them in a certain amount of discrete points (nodes) and estimate our solution anywhere else

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introduction: the Finite Element Method the first stage in any FE analysis is to generate a FE mesh

The first stage in any FE analysis is to generate a FE mesh

Footing width = B

A mesh consists of elements connected together at nodes Node

We will calculate our solution in the nodes, and use some sort of mathematical equation to estimate the solution inside the elements.

Gauss point

examples: embankment

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examples: multi-anchored diaphragm wall

introduction: the Finite Elements Method

the nodes are the points where values of the primary variables (displacements) are calculated

Footing width = B

Node

the values of nodal displacements are interpolated within the elements to give algebraic expressions for displacement and strain throughout the complete mesh

Gauss point

a constitutive law is then used to relate strains to stresses and this leads to the calculation of forces acting at the element nodes

the nodal forces are related to the nodal displacements by equations which are set up and solved to find values of the nodal displacements

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Apply load

obtain displacements stiffness matrix

K u  F For soil we don’t have a direct relation between load and displacement, we have a relation between stress and strain. Displacements

Strains

Material model

Differentiate

  Bu Combine these steps:

Stresses

  D 

Loads

Integrate



F   d V



T K  B DB dV


The FEM involves the following steps (1/2) Elements discretization This is the process of modeling the geometry of the problem under investigation by an assemblage of small regions, termed finite elements. These el em ents have nodes defined on the element bound aries, or withi n the el ements

Primary variable approximation A primary variable must be selected (e.g., displacements) and rules as how it s hould vary over a finite element es tabl ished. This variation is exp ressed i n terms of nodal values  

A polynomial form is assumed, where the order of the polynomial depends on the number of nodes in the element The higher the number of nodes (the order of the polynomial), the more accurate are the results (the longer takes the computation!)

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The FEM involves the following steps (2/2) Element equations Derive element equations:

where is the element stiffness matrix, d isp lacements and is the vector of nodal forc es

is the vector of nodal

Global equations Combine element equations to form global equations

Boundary conditions Formulate boundary conditions and modify global equations. Loads while displacements affect U

affect

P ,

Solve the global equations to obtain the displacements at the nodes

Compute additional (secondary) variables From nodal displacements secondary quantities (stresses, strain) are evaluated

displacement interpolation two-dimensional analysis of continua is generally based on the use of either triangular or quadrilateral elements the most used elements are based on an iso-parametric approach

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Displacement interpolation primary unknowns: values of the nodal displacements displacement within the element: expressed in terms of the nodal values using polynomial interpolation

u ( ) 

n

 N ( ) u i

i

, N i  shape function of node i

i 1

Shape function of node i Is a function that has value “1” in node i and value “0” in all other n-1 nodes of the element

Shape functions for 3-node line element N 1  

1 2

(1   ) 

,

N 2  (1   )(1   )

, N 3 

1 2

(1   ) 


Illustration for the six-noded triangular element

6

v

x

5

u ( x, y )  a0  a1 x  a2 y  a3 x  a4 xy  a5 y 2

2

v ( x, y )  b0  b1 x  b2 y  b3 x 2  b4 xy  b5 y 2

u

1

quadratic interpolation

y

3

2 4

12 coefficients, depending on the values of the 12 nodal displacements

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Illustration for the six-noded triangular element Strains may be derived within the element using the standard definitions

u  a1  2a3 x  a4 y  x ε  Lu v  b2  b4 x  2 b5 y  yy   y u v   (b1  a2 )  ( a4  2 b3 ) x  (2 a5  b4 ) y  xy   y x  xx 

ε



Lu



LNU

e



BU

e

ε

 BU e


Constitutive relation (elasticity) Elasticity: one-to-one relationship between stress and strain in a FE context, stresses

and strains

are written in vector form

the stress-strain relationship is then expressed as:

linear isotropic elasticity in plane strain

= D material stiffness matrix

 1   0  v  v E  v 1 v 0  D  (1  2v )(1  v ) 1  2v   0  0 2   in this case the coefficients of the matrix are constants, which means that (for linear kinematics) the resulting F.E. equations are linear

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What happens with inelastic constitutive relations? Advantage with elasticity: the coefficients of the matrix are constants, the resulting F.E. equations are linear, hence the problem may be solved by applying all of the external loads in a single calculation step soils usually do not behave elastically

 

D

 

with D depending on the current and past stress history

It is necessary to apply the external load in separate increments and to adopt a suitable non-linear solution scheme

Element stiffness matrix body forces and surface tractions applied to the element may be generalized into a set of forces acting at the nodes (vector of nodal forces)

nodal forces may be related to the nodal displacements by: e

K U

e

P

e

6

K e element stiffness matrix e

K





3

P1 x

5

1 4

T

B DBdv

recall

P1 y

2

 P1 x     P1 y  P   2 x   P2 y  e P            P6 x  P   6 y

D material stiffness matrix B matrix relating nodal displacements to strains

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Gauss points e

K





T

B DBdv

To evaluate K e, integration must be performed for each element A numerical integration scheme must be employed (Gaussian integration) Essentially, the integral of a function is replaced by a weighted sum of the function evaluated at a number of integration points


Global stiffness matrix (1) The stiffness matrix 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 degrees-offreedom 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: KU

 P

where vector U contains the displacements at all the nodes in the mesh

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Global stiffness matrix (2) if D is symmetric (elasticity), then K e and hence K will K will be symmetric The global stiffness matrix generally contains many terms that are zero if the node numbering scheme is efficient then all of the non-zero terms are clustered in a band along the leading diagonal

assembly schemes for

storage solution

take into account its sym and banded structure number of dofs


Solution of the global stiffness equations Once the global stiffness equations have been established (and the boundary conditions added), added), they mathematically form a large system of symultaneous (algebraic) equations

KU

 P

Thes These e have have to be solv solved ed to give give value aluess for for the the noda nodall disp displa lace ceme ment ntss It is adv advanta antage geou ouss to adop adoptt specia speciall techni techniques ques to redu reduce ce comp computa utatio tion n time time (e.g . band bandwid width th and and fron fronta tall tech techni niqu ques es)) Deta Detail iled ed disc discus ussi sion on of such such tech techni niqu ques es is beyo beyond nd the the scop scope e of this lectur lecture e

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Compute additional (secondary) values once the nodal displacements have been obtained from the inversion of the matrix K

KU

P

e

The The comp comple lete te disp displa lace ceme ment nt fiel field d can can be obta obtain ined ed::

u ( x, y ) 

n

 N ( x, y ) u i

i

, N i  shape function of node i

i 1

Stra Strain inss and and stre stresse ssess are are comp comput uted ed at the the Gaus Gausss poin points ts::

ε

 BU e

Δσ = D Δε


some practical issues

1. A good finite element mesh is important. A poor mesh will give a poor (inaccurate) solution. 2. Post Post proces processing sing – Stress Stress are are compute computed d at Guass Guass points points only only.. Contour plots of stresses involve further processing of the results. 3. Do the the resu results lts ma make ke sens sense? e? 4. FEA FEA can can be ver very y time time con consum sumin ing! g!

Page 22


KCG 2 EL A AST ST

-PLA -PL A TICI Y AN A ND MO R-C F ILUR E CRITERI

UL O

B

N

rofess r Anto io Gen

Page 23


CG2: ELASTO-PLASTICTY EL ASTO-PLASTICTY AND MOHR COULOMB Antonio Gens Technical University of Catalunya, Barcelona

some of the slides were originally created by: Cino Viggiani (Laboratoire 3S, Grenoble, France) S.W. Lee (GCG Asia – Golder Associates) Associates) Helmut Schweiger (Technical Schweiger (Technical University of Graz, Austria)

Contents

• A quic quick k reminder of (linear isotropic) Elasticity • Motivations for plasticity (elasticity vs. plasticity) • Basic ingredients of any elastoplastic model 

elastic properties (how much recoverable deformation?) deformation?)



yield surface (is plastic deformation occurring?)



plastic potential (direction of plastic strain increment?)



consistency condition (magnitude of plastic strain increment?)



hardening rule (changes of yield surface?)

• Element tests: tests: (drained) simple shear & triaxial tests • Tips and tricks • Advantages and limitations limitations

Page 24


Constitutive models

Constitutive models models provide us with a relationship with stresses and strains expressed as: Δσ = D Δε

Elasticity



Linear-elastic



Non-linear elastic



 σ = Dε

Hooke’s law ε

  xx   1     yy      zz  1      E  0  xy    yz   0     zx   0



 Cσ 0

0

1

   

0

0

 

1

0

0

0

0

2  2

0

0

0

0

2  2

0

0

0

0

 0   0   0  0   2  2  0

 xx     yy   zz     xy   yz     zx 

Page 25


Model parameters in Hooke’s law:

d 1

Two parameters:

- d 1

- Y Young’s oung’s modulus E - Poisson’s ratio 

d 3

- 1

 

Meaning (axial compr.): E 



1

d  1

  

E

E

d  1

d  3

- 1

 1

d  1

0 ; - 1  



3

0 .5

Alternative parameters in Hooke’s law: In spherical and deviatoric stress / strain components:

 v  1 / K     0  s 

p

  p   1 / 3G   q  0

1



 3  1   2   3

q

1 2



(1   2 )2  ( 2   3 )2  ( 3   1 )2

d xy

Shear modulus: G 

d  xy



d  xy



E

21   

Bulk modulus: K 

E



dp d  v



G

dp

E

d v

31  2 

9KG

 3 K

d xy

v



3K 6K

 2G  2G

Page 26


Hooke’s law = Dε

σ

Inverse: 1    xx        yy     zz  E    (1   )(1  2 )  0   xy   0  yz      zx   0

K    xx      K  yy     zz      K   xy     yz        zx    

4G



3 2G



3 2G



3

K





0

0

1



0

0



1  

0

0

0

0

K



K



0

0

0

0

0

3 4G 3 2G 3

K



K



K



2G

1 2

3 2G 3 4G 3







0

 

0   yz 

1 2

0 0



0    zz  0   xy 

 

0

2G



1 2

0    xx  0   yy 

   0   0  0  0  G

0

0

0

0

0

     zx 

0

0

0

0

G

0

0

0

0

0

G

0

0

0

0

0

  xx      yy    zz      xy    yz      zx 

Hooke’s law σ = Dε

Plane strain

D



E

(1   ) (1  2 )

 1        0 



1   0

4G   K  3 0     2G 0   K   3 1  2   0   2  

K



K



2G 3 4G 3

0

   0   G  0

Axisymmetry

D

1      E    (1   ) (1  2 )   0 





1 





1  

0

0

K  0    0  K 0      1  2  K   2  

   0

4G 3 2G 3 2G 3

K



K



K

 0

2G 3 4G 3 2G 3

K



K



K

 0

2G 3 2G 3 4G 3

   0   0  G  0

Page 27


Elasticity vs. Plasticity (1)

In elasticity elasticity,, there is a one-to-one relationship between stress and strain.. Such a relationship may be linear or non-linear. strain non-linear. An essential feature is that the application and removal of a stress leaves the material in its original condition

Elasticity vs. Plasticity (2)

for elastic materials, materials, the mechanism of deformation depends on the stress increment for plastic materials which are yielding, the mechanism of (plastic) deformation depends on the stress reversible = elastic

irreversible = plastic

Page 28


Plasticity: some definitions (1)

One-dimensional

LINEAR ELASTIC - PERFECTLY PLASTIC Y0 = yield stress

IMPORTANT: y ield str ess = failure str ess for perfect plasti city



  e   p

General three-dimensional str ess state

    e   p


One-dimensional

LINEAR ELASTIC – PLASTIC HARDENING Y0 = yield stress YF = failure stress

IMPORTANT: yield str ess



fail ure str ess

  e   p

General three-dimensional str ess state

    e   p

Page 29



LINEAR ELASTIC - PLASTIC WITH SOFTENING One-dimensional

Y0 = yield stress YF = failure stress

yield function (1)

when building up an elastic-plastic model, the first ingredient that we need is a yield surface (is plastic deformation occurring?)

Page 30


yield function (2)

F = 0 represents s urface in str ess space

f  

 f 

 1 ,  2 ,  3



f  

0

stress state is elastic

f  

0

stress state is plastic

f  

0

stress state not admissible

The yield surface bounds all elastically attainable states (a generalized preconsolidation pressure)

yield function (5)

Basically: changes of stress which remain inside the yield surface are associated with stiff response and recoverable deformations, whereas on the yield surface a less stiff response is obtained and irrecoverable deformations are developed Where do we get this function f ? The dominant effect leading to irrecoverable changes in particle arrangement is the stress ratio, or mobilized friction The mean normal effective stress p ’ is of primary importance. The range of values of q for stiff elastic response is markedly dependent on p ’ Tresca

& Von Mises yield functions are not appropriate

Page 31


Mohr-Coulomb Model, yield function

To most engineers the phrase “strength of soils” conjures up images of Mohr-Coulomb failure criteria

frictional resistance independent of normal stress

Classical notions of Mohr-Coulomb failure can be reconciled with the patterns of response that we are modeling here as elasto-plastic behavior

Mohr-Coulomb Model, yield function

1 and 3 : major and minor principal stresses

Page 32


The Mohr-Coulomb failure criterion

t*c’ cos’ - s* sin’

MC criterion:

t* = ½(’ 3 - ’ 1) s* = ½(’ 3+’ 1) 1 2

 '3  '1   c' cos   '1 

'

2c' cos  ' 1  sin  '



 '3  '1 sin

1 2

1  sin  ' 1  sin  '

'

 ' 3

Note: Compression is negative, and ’ 1: major, ’ 2: intermediate, ’ 3: minor principal stress 19

Mohr-Coulomb Model, yield function MOHR COULOMB IN 3D STRESS SPACE

f



1 2

 '  '    '  ' sin ' c ' cos  ' 1

1

3

2

1

3

-1

f > 0 Not acceptable f = 0 Plasticity f < 0 Elasticity

-2

-3

Page 33


plastic potential (1)

Summing up:

Plastic strain increment arises if: 1) the stress state is located on the yield surface (f = 0) AND 2) the stress state remains on the yield surface after a stress increment

knowledge of function f tells us whether plastic strain is occurring or not But, this is only one part of the story: We would also like to know direction and magnitude of plastic strain • will we get plastic volume changes? • and plastic distortion? 

for that, we need another concept (another function: g )

plastic potential (2)

flow rule Recall: plastic deformations depend on the stress state at which yielding is occurring, rather than on the route by which that stress is reached

we have now two functions, f and g 

the question is: where do we get g ?

Page 34


associated and non associated flow rules

it would be clearly a great advantage if, for a given material, yield locus and plastic potential could be assumed to be the same f = g  only 1 function has to be generated to describe plastic response also advantageous for FE computations: • the solution of the equations that emerge in the analyses is faster • the validity of the numerical predictions can be more easily guaranteed

is f = g a reasonable assumption? for metals, it turns out that YES, it is  for geomaterials, NOT Where is the problem? The assumption of normality of plastic strain vectors to the yield locus would result in much greater plastic volumetric dilation than actually observed

Mohr-Coulomb Model, plastic potential

dilatancy angle

Page 35


plastic dilatancy

how to understand dilatancy i.e., why do we get volume changes when applying shear stresses?

 =  + i the apparent externally mobilized angle of friction on horizontal planes () is larger than the angle of friction resisting sliding on the inclined planes (i)

strength = friction + dilatancy

consistency condition

Page 36


Parameters of MC model

E

Young’s modulus Poisson’s ratio (effective) cohesion (effective) friction angle Dilatancy angle

 c ’ ’



[kN/m2] [-] [kN/m2] [º] [º]

MC model for element tests

tan  

 yy  xy

Page 37




Page 38


limitations of MC model (1)

limitations of MC model (2)

Page 39


warning for dense sands

Possibilities and limitations of the Linear Elastic- Perfectly Plastic (LEPP) Mohr-Coulomb model Possibilities and advantages – Simple and clear model – First order approach of soil behaviour in general – Suitable for a good number of practical applications (not for deep excavations and

1

tunnels)

– Limited number and clear parameters – Good representation of failure behaviour (drained) – Dilatancy can be included

2

3

34

Page 40


Possibilities and limitations of the Linear Elastic- Perfectly Plastic (LEPP) MohrCoulomb model

Limitations and disadvantages – Isotropic and homogeneous behaviour – Until failure linear elastic behaviour 1 – No stress/stress-path/strain-dependent stiffness – No distinction between primary loading and unloading or reloading – Dilatancy continues for ever (no critical state) – Be careful with undrained behaviour 2 – No time-dependency (creep)

3

35

Page 41


KCG 3 EXERCI E 1 OUN ATION ON ELA TO-P ASTIC SOIL

Dr Wiilliam C eang

Page 42


Elastoplastic analysis of a footing

ELASTOPLASTIC ANALYSIS OF A FOOTING

Computational Geotechnics

1 Page 43



INTRODUCTION One of the simplest forms of a foundation is the shallow foundation. In this exercise we will model such a shallow foundation with a width of 2 meters and a length that is sufficiently long in order to assume the model to be a plane strain model. The foundation is put on top of a 4m thick clay layer. The clay layer has a saturated weight of 18 kN/m 3 and an angle of internal friction of 20°.

Figure 1: Geometry of the shallow foundation. The foundation carries a small building that is being modelled with a vertical point force. Additionally a horizontal point force is introduced in order to simulate any horizontal loads acting on the building, for instance wind loads. Taking into account that in future additional floors may be added to the building the maximum vertical load (failure load) is assessed. For the determination of the failure load of a strip footing analytical solutions are available from for instance Vesic, Brinch Hansen and Meyerhof: Qf B

= c ∗ N c + 21 γ B ∗ N γ N q = e π tan ϕ tan2 (45 + 21 ϕ ) N c = (N q − 1) cot ϕ 2(N q + 1) tan ϕ N γ = 1.5(N q − 1) tan ϕ (N q − 1) tan(1.4 ϕ ) 







  







(V esic) (Brinch Hansen) (Meyerhof )

This leads to a failure load of 117 kN/ (Meyerhof) respectively.

2

2

(Vesic), 98 kN/

2

(Brinch Hansen) or 97 kN/

2

Computational Geotechnics Page 44



SCHEME OF OPERATIONS This exercise illustrates the basic idea of a finite element deformation analysis. In order to keep the problem as simple as possible, only elastic perfectly-plastic behaviour is considered. Besides the procedure to generate the finite element mesh, attention is paid to the input of boundary conditions, material properties, the actual calculation and inspection of some output results.

Aims • Geometry input • Initial stresses and parameters • Calculation of vertical load representing the building weight • Calculation of vertical and horizontal load representing building weight and wind force • Calculation of vertical failure load. A) Geometry input • General settings • Input of geometry lines • Input of boundary conditions • Input of material properties • Mesh generation B) Calculations • Initial pore pressures and stresses • Construct footing • Apply vertical force • Apply horizontal force • Increase vertical force until failure occurs C) Inspect output

Computational Geotechnics

3 Page 45

Plaxis_advanced_course_hong_kong_2012.pdf

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