Plaxis Advanced Course, New Delhi, India, (2014).pdf

PLAXIS Advanced Course
New Delhi, India.
Venue Department of Civil Engineering
IIT Delhi, New Delhi, India
Date 29‐31 October 2014
Lecturers
Professor Antonio Gens Technical Univ. of Catalonia
Professor K Rajagopal IIT‐Madras
Professor Helmut Schweiger Graz Univ. of Technology
Prof. G.V. Ramana IIT‐New Delhi
Dr William Cheang Plaxis AsiaPac, Singapore
Plaxis Advanced Course, New Delhi, India (29-31 October 2014) 1/448

CONTENTS
ADVANCED COMPUTATION GEOTECHNICS, NEW DELHI 2014 PAGE
Session 1: Geotechnical Finite Element Analysis & Soil Behaviour 1
CG1 Geotechnical Finite Element Analysis 4
CG2 Elasto‐plasticity and Mohr‐Coulomb 22
CG3 Exercise 1: Foundation on Elasto‐plastic Soils 44
Session 2: Soil Behaviour 2
CG4 Critical State Soil Mechanics and Soft Soil Model 69
CG5 Hardening Soil and HS‐small Models 88
CG6 Exercise 2:  Triaxial & Oedometer Simulation 119
Session 3: Modelling of Deep Excavations
CG7 Modelling of Deep Excavations   143
CG8 Structural Elements in PLAXIS 169
CG9 Exercise 3: Modelling of an Anchored Retaining Wall in 2D 194
Session 4: Modelling of Groundwater, Undrained Conditions & Consolidation
CG10 Modelling of Groundwater in PLAXIS 216
CG11 Drained and Undrained Analysis 251
CG12 Consolidation Analysis 272
CG13 Exercise 4: Modelling of an Strutted and Embedded Excavation in 3D 289
Session 5: Initial Stresses, Slope Stability & Unsaturated Soils
CG14 Unsaturated Soils and Barcelona Basic Model 306
CG15 Initial Stresses and Slope Stability Analysis 337
CG16 Exercise 5: Slope Stability Analysis 375
Session 6: Modelling of Tunnels in Rock
CG17 Hoek‐Brown and Rock Jointed Models 392
CG18 Modelling of Tunnels in 2D 416
CG19 Exercise 6: Tunnelling in Rock   437

PLAXIS ADVANCED COURSE
29‐31 October 2014, New Delhi, India.
Time Wednesday 29 October 2014
Session 1: Geotechnical Finite Element Analysis & Soil Behaviour 1
9:00 10:00 CG1 Geotechnical Finite Element Analysis Gens
10:00 11:00 CG2 Elasto‐Plasticity and Mohr‐Coulomb Rajagopal
11:00 11:15 Break
11:15 1:00 CG3 Exercise 1: Foundation on Elasto‐Plastic Soils Cheang
1:00 2:00 Lunch
Session 2: Soil Behaviour 2
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 Break
4:15 5:30 CG6 Exercise 2: Triaxial and Oedometer Cheang

Time Thursday 30 October 2014
Session 3: Modelling of Deep Excavations
9:00 9:45 CG7 Modelling of Deep Excavations Gens
9:45 10:30 CG8 Structural Elements in Plaxis Rajagopal
10:30 10:45 Break
10:45 12:00 CG9 Exercise 3: Simulation of an Anchored Retaining Wall Cheang
12:00 1:00 Lunch
Session 4: Modelling of Groundwater, Undrained Conditions and Consolidation
1:00 1:45 CG10 Modelling of Groundwater in Plaxis Cheang
1:45 2:30 CG11 Drained and Undrained Analysis Gens
2:30 3:15 Break
3:15 3:30 CG12 Consolidation Analysis Gens
3:30 5:00 CG13 Exercise 4: Dewatering in Excavation Siva

Time Friday 31 October 2014
Session 5: Initial Stresses, Slope Stability and Unsaturated Soils
9:00 10:30 CG14 Unsaturated Soils and Barcelona Basic Model Gens
10:30 11:30 CG15 Initial Stresses and Slope Stability Analysis Schweiger
11:30 11:45 Break
11:45 1:30 CG16 Exercise 5: Slope Stabilised by Soil Nails Siva
1:30 2:30 Lunch
Session 6: Modelling of Tunnels in Rock
2:30 3:30 CG17 Hoek‐Brown and Rock Jointed Models Schweiger
3:30 4:00 CG18 Modelling of Tunnels in 2D Schweiger
4:00 4:15 Break
4:15 5:30 CG19 Exercise 6: Tunnelling in Rock Cheang


New Delhi Advanced 2014
Computation Geotechnics 1
Geotechnical Finite Element Analysis
Professor Antonio Gens


CG1: GEOTECHNICAL FINITE ELEMENT 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
Introduction
• Design requirements in geotechnical engineering
• Geotechnical methods of analysis
• Geotechnical finite element analysis: some remarks
The Finite Element Method
• Introduction and general overview
• Domain discretization
• Element formulation
• Constitutive law
• Element stiffness matrix
• Global equations: assembly and solution
• Compute secondary variables
Final remarks

design requirements in geotechnical engineering
• Stability (local and general)
• Admissible deformation and displacements
design requirements in geotechnical engineering
• Flow problems
• Sometimes flow and stability/deformation problems are solved together
 See tomorrow’s lecture on consolidation (CG12)

geotechnical analysis: basic solution requirements
• Equilibrium (3 equations)
• Compatibility (6 equations)
• Constitutive equation (6 equations)
• Unknowns: 15
(6 stresses, 6 strains, 3 displacements)
Potts & Zdravkovic
(1999)
• 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
geotechnical methods of numerical analysis
• methods for numerical analysis
 Finite difference method
 Boundary element method (BEM)
 Discrete element method (DEM)
 Finite element method (FEM)
 Others (meshless methods, material point method, particle methods…)

• Objectives of the numerical (finite element) analysis
 Selection of design alternatives
 Quantitative predictions
 Backcalculations
 Understanding!
 Identification of critical mechanisms
 Identification of key parameters
geotechnical finite element analysis
• 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

• 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

The Finite Element Method: introduction and overview
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
Governing mathematical equation: (equilibrium)
Algebraic equation:
x
xz
xy
x
b
z
y
x








 


1
1
2
12
1
11 c
x
a
x
a
x
a n
n 


 
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
this lecture presents only a basic outline of the method
attention is focused on the first problem using the "displacement based" FE
approach Plaxis Advanced Course, New Delhi, India (29-31 October 2014) 10/448

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 elements have nodes defined on the element
boundaries, or within the elements
Primary variable approximation
A primary variable must be selected (e.g., displacements) and rules
as how it should vary over a finite element established. This
variation is expressed in 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!)
The FEM involves the following steps (2/2)
Element equations
Derive element equations:
where is the element stiffness matrix, is the vector of nodal
displacements and is the vector of nodal forces
Global equations
Combine (assemble) element equations to form global equations
Boundary conditions
Formulate boundary conditions and modify global equations. Loads
affect P, while displacements affect U
Solve the global equations
to obtain the displacements at the nodes
Compute additional (secondary) variables
From nodal displacements secondary quantities (stresses, strain) are
evaluated

stiffness matrix
For soil we don’t have a direct relation between load and displacement,
we have a relation between stress and strain.
Displacements Strains Stresses Loads
Differentiate Integrate
Material
model
Bu
  D
 
 d
F V

 
T
B DBd
K V
 
K F
u 
Combine these steps:
Domain discretization
Footing
width = B
Node
Gauss point
The first stage in any FE analysis
is to generate a FE mesh
A mesh consists of elements
connected together at nodes
We will calculate our solution in
the nodes, and use some sort of
mathematical equation to
estimate the solution inside the
elements.

examples: embankment
examples: multi-anchored diaphragm wall
There is a whole zoo of different finite elements available!

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
Element formulation
Displacement interpolation
primary unknowns: values of the nodal displacements
displacement within the element: expressed in terms of the nodal values
using polynomial interpolation
1
( ) ( ) , shapefunctionof node
n
i i i
i
u N u N i
 

 

Element formulation

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
1 2 3
1 1
(1 ) , (1 )(1 ) , (1 )
2 2
N N N
     
       
Element formulation
Shape functions for 5-node line element
Illustration for the six-noded triangular element
12 coefficients, depending on the values of the
12 nodal displacements
u
v x
y
1 2
3
4
5
6
quadratic interpolation
2 2
0 1 2 3 4 5
2 2
0 1 2 3 4 5
( , )
( , )
u x y a a x a y a x a xy a y
v x y b b x b y b x b xy b y
     
     
e
NU
u 
Element formulation

Illustration for the six-noded triangular element
Strains may be derived within the element using the standard definitions
1 3 4
2 4 5
1 2 4 3 5 4
2
2
( ) ( 2 ) (2 )
xx
yy
xy
u
a a x a y
x
v
b b x b y
y
u v
b a a b x a b y
y x




   


   

 
       
 
e e
  
ε Lu LNU BU
e

ε BU

ε Lu
Element formulation
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:  = D 
material stiffness matrix
linear isotropic elasticity in plane
strain


















2
2
1
0
0
0
1
0
1
)
1
)(
2
1
( v
v
v
v
v
v
v
E
D
in this case the coefficients of the matrix are constants, which means
that (for linear kinematics) the resulting F.E. equations are linear
Constitutive law

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
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
What happens with inelastic constitutive relations?
 
  
D
Constitutive law
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)



























y
x
y
x
y
x
P
P
P
P
P
P
6
6
2
2
1
1


e
P
1
2
3
4
5
6
P1x
P1y
D material stiffness matrix
B matrix relating nodal displacements to strains
dv
B
D
B
K T
e


Ke element stiffness matrix
recall
nodal forces may be related
to the nodal displacements by:
e
e
e
P
U
K 

stiffness matrix
For soil we don’t have a direct relation between load and displacement,
we have a relation between stress and strain.
Displacements Strains Stresses Loads
Differentiate Integrate
Material
model
Bu
  D
 
 d
F V

 
T
B DBd
K V
 
K F
u 
Combine these steps:
Gauss points
dv
B
D
B
K T
e


To evaluate Ke, 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-of-freedom in the mesh
• in 2D number of d.o.f = 2 x number of nodes
• in 3D number of d.o.f = 3 x number of nodes
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:
where vector U contains the displacements at all the nodes in the mesh
P
KU 
Global equations: assembly and solution
Global stiffness matrix (2)
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
storage
solution
schemes for
take into account its sym and
banded structure
if D is symmetric (elasticity), then Ke and hence K will be symmetric
number of dofs

Solution of the global stiffness equations
Once the global stiffness equations have been established
(and the boundary conditions added), they mathematically
form a large system of symultaneous (algebraic) equations
These have to be solved to give values for the nodal displacements
It is advantageous to adopt special techniques to reduce
computation time (e.g. bandwidth and frontal techniques)
Detailed discussion of such techniques is beyond the scope of
this lecture
P
KU 
Compute additional (secondary) values
Computation of secondary variables
once the nodal displacements have been obtained from the inversion
of the matrix K
The complete displacement field can be obtained:
e

KU P
Strains and stresses are computed at the Gauss points:
1
( , ) ( , ) , shapefunctionof node
n
i i i
i
u x y N x y u N i

 

Δσ = DΔε
e

ε BU

some practical issues
1. A good finite element mesh is important. A poor mesh will give
a poor (inaccurate) solution.
2. Post processing – Stress are computed at Guass points only.
Contour plots of stresses involve further processing of the results.
3. Do the results make sense?
4. FEA can be very time consuming!
Final remarks

Elasto-plasticity and Mohr-Coulomb
Professor K.Rajagopal


CG2: ELASTO-PLASTICTY AND MOHR COULOMB
some of the slides were originally created by:
Cino Viggiani (Laboratoire 3S, Grenoble, France)
Antonio Gens (UPC, Spain)
S.W. Lee (GCG Asia – Golder Associates)
Helmut Schweiger (Technical University of Graz, Austria)
K. Rajagopal (IIT Madras) – additional slides September 2013
Professor K Rajagopal
IIT Madras
• A quick reminder of (linear isotropic) Elasticity
• Motivations for plasticity (elasticity vs. plasticity)
• Basic ingredients of any elastoplastic model
 elastic properties (how much recoverable 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: (drained) simple shear & triaxial tests
• Tips and tricks
• Advantages and limitations
Contents

Constitutive models
Linear-elastic


Non-linear elastic


Δσ = DΔε
Constitutive models provide us with a relationship with stresses and
strains expressed as:
Elasticity
σ = Dε






 
 
 









xx
yy
zz
xy
yz
zx
xx
yy
zz
xy
yz
zx
E



















 
 
 







































1
1 0 0 0
1 0 0 0
1 0 0 0
0 0 0 2 2 0 0
0 0 0 0 2 2 0
0 0 0 0 0 2 2
Hooke’s law

ε Cσ

Two parameters:
- Young’s modulus E
- Poisson’s ratio 
Meaning (axial compr.):



 
d
d
3
1


- d1
d3
- 1
- 1
3
E
1
1

Model parameters in Hooke’s law:
d1
E
d
d



1
1
0 ; -1 0.5
E 
  
Shear modulus:

dxy
dxy
 
K
dp
d
E
v
 

 
3 1 2
Bulk modulus: dp
dv
Alternative parameters in Hooke’s law:
 
G
d
d
E
xy
xy
 


 
2 1
9
3
K G
E
G K


3 2
6 2
K G
v
K G



1/ 0
0 1/ 3
v
s
K p
G q


     

     
   
 
In spherical and deviatoric stress / strain components:
 
1
1 2 3
3
p   
  
2 2 2
1 2 2 3 3 1
1
( ) ( ) ( )
2
q      
     







 
  
  
  









xx
yy
zz
xy
yz
zx
xx
yy
zz
xy
yz
zx
E



















 










































( )( )
1 1 2
1 0 0 0
1 0 0 0
1 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1
2
1
2
1
2
Inverse:
Hooke’s law
σ = Dε
4 2 2
0 0 0
3 3 3
2 4 2
0 0 0
3 3 3
2 2 4
0 0 0
3 3 3
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
xx xx
yy yy
zz zz
xy xy
yz yz
zx zx
G G G
K K K
G G G
K K K
G G G
K K K
G
G
G
 
 
 
 
 
 
 
  
 
   
 
   
 
  
   
 
   
 

   
 
  
   
 
   
 
   
 
   
   
 
 
 
Hooke’s law
4 2
0
3 3
1 0
2 4
1 0 0
(1 )(1 2 ) 3 3
1 2
0 0
0 0
2
G G
K K
E G G
K K
G
 
 
 

 
 
   
 
  
   
    
   
   
  
   
 
 
 
D
σ = Dε
4 2 2
0
1 0 3 3 3
2 4 2
1 0
0
3 3 3
1 0
(1 )(1 2 )
2 2 4
1 2 0
0 0 0 3 3 3
2
0 0 0
G G G
K K K
G G G
K K K
E
G G G
K K K
G
  
  
  
 

 
  
 

 
 
 
  
    
 
 
 

   
 
  
  
 
 
 
 
 
 
D
Plane strain
Axisymmetry

Elasticity vs. Plasticity (1)
In elasticity, there is a one-to-one relationship between stress and
strain. Such a relationship may be linear or 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, 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

Plasticity: some definitions (1)
One-dimensional
IMPORTANT: yield stress = failure stress for perfect plasticity
p
e


 

     p
e


 

General three-dimensional stress state
Y0 = yield stress
LINEAR ELASTIC - PERFECTLY PLASTIC
One-dimensional
IMPORTANT: yield stress  failure stress
p
e


 

     p
e


 

General three-dimensional stress state
Y0 = yield stress
YF = failure stress
LINEAR ELASTIC – PLASTIC HARDENING

One-dimensional
Y0 = yield stress
YF = failure stress
LINEAR ELASTIC - PLASTIC WITH SOFTENING
yield function (1)
when building up an elastic-plastic model,
the first ingredient that we need is a yield surface
(is plastic deformation occurring?)

yield function (2)
The yield surface bounds all elastically attainable states
(a generalized preconsolidation pressure)
F = 0 represents surface in stress space
stress state is elastic
 
  0


f
 
  0


f
stress state is plastic
stress state not admissible
 
   
3
2
1 


 ,
,
f
f 
 
  0


f
yield function (5)
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
Basically:
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

Mohr-Coulomb Model, yield function
To most engineers the phrase “strength of soils” conjures
up images of Mohr-Coulomb failure criteria
Classical notions of Mohr-Coulomb failure can be
reconciled with the patterns of response that we are
modeling here as elasto-plastic behavior
frictional
resistance
independent of
normal stress
1 and 3 : major and minor principal stresses

MC criterion: t*c’ cos’ - s* sin’
t* = ½(’3 - ’1)
s* = ½(’3+’1)
    '
sin
'
'
'
cos
'
'
' 1
3
2
1
1
3
2
1





 


 c
3
1 '
'
sin
1
'
sin
1
'
sin
1
'
cos
'
2
' 











c
Note: Compression is negative, and ’1: major,
’2: intermediate, ’3: minor principal stress
The Mohr-Coulomb failure criterion
19
f < 0 Elasticity
f = 0 Plasticity
f > 0 Not acceptable
-1
-3
-2
MOHR COULOMB IN 3D STRESS SPACE
    '
cos
'
'
sin
'
'
'
' 




 c
f 



 3
1
2
1
3
1
2
1

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
we have now two functions, f and g
 the question is: where do we get g ?
Recall: plastic deformations depend
on the stress state at which yielding
is occurring, rather than on the route
by which that stress is reached

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

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

Parameters of MC model
E Young’s modulus [kN/m2]
 Poisson’s ratio [-]
c’ (effective) cohesion
[kN/m2]
’ (effective) friction angle [º]
 Dilatancy angle [º]
MC model for element tests
xy
yy






tan






sin
sin
1
cos
sin
max


n

limitations of MC model (1)
limitations of MC model (2)

warning for dense sands
Tension cut-off
Compression
zone
Tension
zone
Tension cut-off
Tension cut-off: if c>o, MC model allows tensile stresses to be
developed

Simple Shear Test
• Given:
• Initialconditions:
• Boundaryconditions:
• Implicitstress increment :
• With Dep evaluated at the beginning of each step (using σold)
o
o
0 yy
o
xx
xy σ'
K
σ'
0,
τ 

o
yy
xy σ
,
γ
ψ,
,
'
,
c'
,
ν'
,
E' 
0
Δε
0,
Δσ xx
yy 

ε
D
σ
σ
σ
σ






ep
old
new '
'
1
'

3
'

3
'

1
'

2
45


At failure for
simple shear test
Dilatancy
For simple shear
 is positive – volume increase in shear - Dilatancy
ψ is negative – volume decrease in shear - Contractancy
   
p
xy
p
yy
2
p
xy
2
p
yy
p
yy
γ
ε
tan ψ
γ
ε
ε
sin ψ











0
ε
and
0
ε
ε
ε
Since,
e
xx
p
xx
e
xx
xx








0
ε
p
xx 

ψ
p
yy
ε

p
xy

τ
σyy
   
sinψ
γ
ε
ε
ε
ε
γ
ε
2
p
xy
2
p
yy
p
xx
p
yy
p
xx
p
p
v













Plastic volumetric
strain
Plastic sheardistortion

Results of undrained simple shear test
Non-associated plasticity, f ≠ g
Parameter Value
Bulk unit weight of soil,γ 0 kN/m3
Effectiveangle of internal friction, ′ 43o
Effectivecohesion, c′ 0 kPa
Poisson’s ratio, υ′ 0.20
Young’s modulus, E′ 45000 kN/m2
Angle of dilation , ψ -3o ,0o ,15o
Bulk modulus of soil, Ke = Kw/n 1.86 * 106 kN/m2
Bulk unit weight of gravityelements,γ 20 kN/m3
Normal stress in y-direction, σyy 100 kN/m2
Shear strain, γxy 0.025
Ko 0.25
Comparison between drained and undrained results
Property Value
Bulk unit weight of soil, γ 0 kN/m3
Effective angle of internal
friction, ′
35o
Effective cohesion, c′ 0 kPa
Poisson’s ratio, υ′ 0.30
Young’s modulus, E′ 26000 kPa
Angle of dilation , ψ -3o, 0o, 5o
Bulk modulus of soil, Ke =
Kw/n
1.85* 106
kN/m2
Bulk unit weight of gravity
elements, γ
20 kN/m3
Normal stress in y-
direction, σyy
100 kN/m2
Shear strain, γxy 0.025

Influence of dilation angle on behaviour of circular footing
-0.45
-0.4
-0.35
-0.3
-0.25
-0.2
-0.15
-0.1
-0.05
0
0 500 1000 1500 2000 2500 3000 3500
settlement
(m)
Pressure (kPa)
psi =0
psi=3
psi=6
psi=10
E = 35000 kPa
ν  = 0.3
c = 1 kPa
φ  =35
Diameter=4m
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
tunnels)
– Limited number and clear parameters
– Good representation of failure behaviour
(drained)
– Dilatancy can be included
1
3
2
40

Limitations and disadvantages
– Isotropic and homogeneous behaviour
– Until failure linear elastic behaviour
– 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
– No time-dependency (creep)
1
3
2
41
Possibilities and limitations of the Linear Elastic- Perfectly Plastic (LEPP) Mohr-
Coulomb model

Exercise 1: Shallow Foundation on Elasto-plastic Soil
Dr William Cheang


Elastoplastic analysis of a footing
ELASTOPLASTIC ANALYSIS OF A
FOOTING
Computational Geotechnics 1

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/m3
and an angle of internal
friction of 200
.
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 ∗ Nc + 1
2
γ0
B ∗ Nγ
Nq = eπ tan ϕ0
tan2
(45 + 1
2
ϕ0
)
Nc = (Nq − 1) cot ϕ0
Nγ =





2(Nq + 1) tan ϕ0
(V esic)
1.5(Nq − 1) tan ϕ0
(Brinch Hansen)
(Nq − 1) tan(1.4 ϕ0
) (Meyerhof)
This leads to a failure load of 117 kN/m2
(Vesic), 98 kN/m2
(Brinch Hansen) or 97 kN/m2
(Meyerhof) respectively.
2 Computational Geotechnics

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
• Input
– Start new project
– Soil mode
* Create soil layers
* Create and assign soil material sets
– Structures mode
* Create footing
* Create load
– Mesh mode
* Generate mesh
– Staged construction mode
* Determine initial situation
* 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.
• Output
– Inspect deformations
– Inspect failure mechanism
– Inspect load-displacement curve

INPUT
Introduction
Start PLAXIS 2D by double-clicking the icon of the PLAXIS 2D Input program. The Quick
select dialog box will appear in which you can select to start an new project or open an
existing one. Choose Start a new project (see Figure 2). Now the Project properties window
appears, consisting of the two tabsheets Project and Model (see Figure 3 and Figure 4).
Figure 2: Quick select dialog
Project properties
The first step in every analysis is to set the basic parameters of the finite element model.
This is done in the Project properties window. These settings include the description of the
problem, the type of analysis, the basic type of elements, the basic units and the size of the
drawing area.
Project tabsheet
In the Project tabsheet, enter “Exercise 1” in the Title box and type “Elasto-plastic analysis of
a drained footing” or any other text in the Comments box.
Model tabsheet
In the Model tabsheet several model specific parameters can be specified
• In the Type box the type of the analysis (Model) and the basic element type (Elements)
are specified. As this exercise concerns a strip footing, choose Plane strain from the
Model combo box. Select 15-node from the Elements combo box.

Figure 3: Project tabsheet of the Project Properties window
Figure 4: Model tabsheet of the Project properties window
• The Units box defines the units for length, force and time that have to be used in this
project. There is a choice for several units, both metric and emperial.
For this project use the default units (Length = m; Force = kN; Time = day).
• In the Contour box the size of the considered geometry must be entered. The values
entered here determine the size of subsoil input window. PLAXIS will automatically add
a small margin so that the geometry will fit well within the draw area. Enter xmin=0.00,
xmax=14.00, ymin=0.00 and ymax=4.25, see figure 4
• Click on the OK button below the tabsheets to close the Project properties window.
Hint: In the case of a mistake or for any other reason that the project properties
should be changed, you can access the Project properties window by
selecting the Project properties option from the File menu.

Soil Mode
The program is now in Soil mode in which the subsoil should be created. As shown in figure
1 the subsoil consist of a single 4m thick clay layer and creating this layer is done in 2 steps:
first the soil layer is defined through the definition of a borehole, after which the material set
representing the clay is defined and assigned to the appropriate layer.
Create soil layer
• Select the button Create borehole ( ) and click in the drawing area on the origin to
indicate a borehole should be created there. The Modify soil layers window opens, see
figure 5. Intially this window is empty as no boreholes have been defined yet for this
project.
Figure 5: The initial Modify soil layers window
• Now click the Add button in order to add a layer to the borehole.
• On the Soil layers tabsheet the different soil layers present in the borehole must be
defined. In this exercise there is only 1 soil layer with the Top at 4.0m and the Bottom at
0.0m, see figure 6.
• On the left side of the Modify soil layers window there is a graphical representation of
the borehole. Note that the soil layer does not have a soil material assigned yet.
• Above the borehole the Head option specifies the position of the global water level in
this borehole. In this exercise it is assumed that the phreatic level is at groundlevel, so
the Head must be set equal to 4.0 to indicate that the phreatic level is at ground level.
• Now press the <OK> button to close the Modify soil layers window. The drawing area
now shows a grey rectangular subsoil.

Figure 6: The Modify soil layers window with 1 borehole containing 1 soil layer
Create and assign material sets
In this exercise 2 material sets will be used: one material set for the clay layer, and the second
material set will be used to model the concrete footing. To create the material sets, follow
these steps:
• Select the Materials button ( ) - the Material sets window will open. The list of
material sets available for this project is still empty.
• Click on the New button at the lower side of the Material Sets window. A new dialog box
will appear with five tabsheets: General, Parameters, Flow parameters, Interfaces and
Initial (see figure 7).
• In the Material Set box of the General tabsheet, write “Clay” in the Identification box.
• Select Mohr-Coulomb from the Material model combo box and Drained from the Material
type combo box.
• Enter the proper values for the weights in the General properties box according to the
material properties listed in table 1
• Click on either the Next button or click on the Parameters tabsheet to proceed with
the input of model parameters. The parameters appearing on the Parameters tabsheet
depend on the selected material model (in this case the Mohr-Coulomb model).
• Enter the model parameters of table 1 in the corresponding edit boxes of the Parameters
tabsheet. The parameters in the Alternatives and Velocities group are automatically
calculated from the parameters entered earlier.
• See also figure 8. In this figure the Advanced parameters part has been collapsed.

Figure 7: General tabsheet of the soil and interface data set window for Clay
• Since the geometry model does not include groundwater flow or interfaces, the third and
fourth tabsheet can be skipped. Click on the OK button to confirm the input of the current
material data set.
• Now the created data set will appear in the tree view of the Material Sets window.
Table 1: Material properties of the clay layer and the concrete footing.
Parameter Symbol Clay Concrete Unit
Material model Model Mohr-Coulomb Linear elastic —
Type of behaviour Type Drained Non-porous —
Weight above phreatic level γunsat 16.0 24.0 kN/m3
Weight below phreatic level γsat 18.0 — kN/m3
Young’s modulus E0
5.0·103
2.0·107
kN/m2
Poisson’s ratio ν0
0.35 0.15 —
Cohesion c0
ref 5.0 — kN/m2
Friction angle ϕ0
20 — °
Dilatancy angle ψ 0 — °
For the concrete of the footing repeat the procedure, but choose a Linear Elastic material
behaviour and enter the properties for concrete as shown in table 1 (see also figures 9 and
10).

Figure 8: Parameters tabsheet of the soil and interface data set window for Clay
Figure 9: General tabsheet of the soil and interface data set window for Concrete

Figure 10: Parameters tabsheet of the soil and interface data set window for Concrete
• Now from the Material sets window drag the Clay material set with the mouse over the
grey subsoil and drop it. The subsoil should now get the colour of the material set, see
figure 11.
Figure 11: Subsoil before (left) and after (right) assigning the Clay material set
This ends the creation of the subsoil in Soil mode. By clicking on the Structures tabsheet now
move to Structures mode.

Structures mode
Introduction
In Structures mode the footing as well as the point load acting on the footing will be created.
However, first an adjustment to the snapping interval must be made in order to be able to draw
the 0.25m thick footing. By default, the snapping interval is set to 1m.
• From the vertical toolbar select the Snapping options button ( ). The Snapping window
now opens.
• Make sure the options Enable snapping and Show grid are selected
• Leave the Spacing to 1 m
• Set the Number of snap intervals to 4. This means that every spacing of 1 meter is
divided in 4, hence the snapping distance will be 0.25m.
• Click the <OK> button to confirm the new settings and close the window.
Create footing
1. Select the Create soil button ( ) and from the drop-down list that opens now select
the Create soil rectangle button ( ).
2. Move the mouse cursor to the coordinates (x y) = (6 4) and single-click the left mouse
button
3. Now move the mouse cursor to the coordinates (x y) = (8 4.25) and single-click the left
mouse button again. We have now created the footing.
4. Select the Show materials button ( ), the Material sets window will open.
5. Drag-and-drop the Concrete material set onto the footing.
Create load
1. Select the Create load button ( ) and from the drop-down list that opens select the
Create point load option ( ).
2. Move the mouse cursor to the coordinates (x y) = (7 4.25) and single-click the left mouse
button to insert the point load.
This concludes the creation of the footing and loads. By clicking on the Mesh tabsheet now
move to Mesh mode.

Mesh mode
In Mesh mode the user can specify necessary mesh refinements and generate the mesh. In
this exercises no additional mesh refinement will be used.
• Select the Generate mesh button ( ). The Mesh options window will open.
• Leave the Element distribution to Medium and press <OK> to start mesh generation
• If mesh generation finished succesfully this will be confirmed in the Command explorer
with the message "Generated XX elements, YY nodes" where XX and YY stand for the
amount of elements and nodes respectively.
• Select the View mesh button in order to view the generated mesh, see figure 12.
Figure 12: Generated mesh
Close the mesh window by selecting the green <Close> button. This ends the Mesh mode.
As no water levels will be used in this exercise, the Water levels mode can be skipped and we
can move directly to Staged construction mode to define the calculation phases.
Staged construction mode
In Staged construction mode all calculation phases will be defined. In this exercise we will use
5 calculation phases, which includes the initial phase.
Initial phase
The initial phase represents the field conditions that exist at the moment our project starts.
This means that only the subsoil exists in the initial conditions whereas the footing should be
deactivated.

• Right-click on the footing. The footing will become red (indicating it is selected) and a
drop-down menu appears.
• From the drop-down menu select the option Deactivate in order to deactivate the footing.
Phase 1: Construction of the footing
• In the Phase explorer select the Add phase button ( ) so that a new phase will be
added.
• Right-click on the footing and from the drop-down menu that appears select the option
Activate to activate the footing, see figure 13.
Figure 13: Geometry configuration for the initial phase (left) and phase 1 (right)
Phase 2: Apply vertical load
added.
• Click on the point on which the load acts so that it becomes red. On the left side the data
of the load now appears in the Selection explorer.
• Activate the point load and set the value of the vertical component, Fy,ref = -50 kN (=
downwards), see figure 14.
Phase 3: Add horizontal load
added.
of the load again appears in the Selection explorer.
• Set the value of the vertical component of the point load, Fx,ref = 20 kN, see figure 14.

Figure 14: Activating and changing the point load through the Selection explorer in phase 2
(left) and phase 3 (right)
Phase 4: Vertical failure load
In this phase we will calculate the vertical failure load as if no horizontal load has been applied.
This means that phase 4 must be a continuation of applying the vertical load in phase 2.
• In the Phase explorer select phase 2 so that it will show in bold letter type
• Now select the Add phase button ( ) so that a new phase will be added that follows on
phase 2 rather than on phase 3.
of the load again appears in the Selection explorer.
• Set the value of the vertical component of the point load, Fy,ref = -500 kN. Note that
Fx,ref should remain 0 (zero).
This finishes the definition of the calculation phases for this project.
Calculation
Load-displacement curves
As a calculation result we would like to draw a load-settlement curve for the footing. In order
to do so, the user must select one or more points for which Plaxis has to gather data during
the calculation:
• Select the Select points for curves button ( ). The output program now opens, showing
the mesh with all nodes.

• Select the node in the middle underneath the footing, hence at or very close to (x y) = (7
4). The node will appear in the Select points list, see figure 15.
• Close Plaxis Output by clicking the green <Update> button.
Figure 15: Selecting points for node displacement curves
Calculate
Press the Calculate button ( ) to start the calculation.
Note that the last calculation phase fails: the intended vertical load of 500 kN cannot be fully
applied due to failure of the subsoil underneath the footing.

RESULTS
Output
After the calculation finishes, click the View calculation results button ( ). Plaxis Output
will open, showing the calculation results of the last calculation phase.
By default Plaxis Output will show the Defomed mesh, see figure 16. If this is not the case the
Deformed mesh can be shown by choosing the menu Deformations → Deformed mesh |u|.
Figure 16: Deformed mesh after phase 4
Now choose the menu option Deformations→ Incremental displacements→|∆u|, see figure
17.
The incremental displacements is the change in displacements in the current calculation
step (here that is the last calculation step of the phase 4). Under working conditions the
change of displacement per calculation step is quite small, but in case of failure, the change of
displacements can be large inside the failure zone. Therefore the Incremental displacements
graph can be very suitable for detecting whether failure occurs and what the failure zone may
look like. Figure 17 shows the typical Prandtl-like failure zone.
Figure 17: Incremental displacements for the final calculation step of phase 4
Finally, we will inspect the load-settlement curve and determine the failure load. To do so,
follow these steps:

• From the button bar select the Curves manager button ( ). The Curves manager will
open.
• In the Curves manager select the <New> button in order to generate a new curve. Now
the Curve generation window opens.
• In the Curve generation window, select for the x-axis data from point A (instead of Project
data) from the drop down list.
• Now in the tree below, select Deformations → Total displacements→ |u|
• For the y-axis we will plot a Project value, and that is the Multiplier ΣMstage.
• Press <OK>. A curve as can be seen in figure 18 will show.
Figure 18: Load-settlement curve
In a Plaxis calculation any change made in a construction phase leads to a so-called unbalance,
that is a disturbance between the total of the internal stresses and the external load. This
unbalance is gradually solved using the ΣMstage multiplier. The ΣMstage multiplier indicates
how much of the unbalance has been solved, where ΣMstage = 0 indicates that no unbalance
was solved and ΣMstage = 1 that the full unbalance has been solved.
In the curve shown in figure 18 the lines at the left indicate the variation of ΣMstage for the first
3 calculation phases, where as the long curved line shows the variation of ΣMstage during
the final phase.
It shows that at failure occurs when ΣMstage = 0.38, hence 38% of the unbalance was solved.
In this case the unbalance applied was the increase of the vertical load from 50 kN/m to 500
kN/m. Hence, at failure the total load applied is the load at the beginning of the phase (50
kN/m) plus 38% of the change of load that could be applied: Fmax = 50+0.38·(500−50) = 221
kPa
The exact value of the ΣMstage multiplier can be inspected by moving the mouse cursor over
the plotted line. A tooltip box will show up with the data of the current location.

Comparison
In addition to the mesh used in this exercise calculations were performed using a very coarse
mesh with a local refinement at the bottom of the footing and a very fine mesh. Fine meshes
will normally give more accurate results than coarse meshes. Instead of refining the whole
mesh, it is generally better to refine the most important parts of the mesh, in order to reduce
computing time. Here we see that the differences are small (when considering 15-noded
elements), which means that we are close to the exact solution. The accuracy of the 15-
noded element is superior to the 6-noded element, especially for the calculation of failure
loads.
Hint: In plane strain calculations, but even more significant in axi-symmetric
calculations, for failure loads, the use of 15-noded elements is recommended.
The 6-noded elements are known to overestimate the failure load, but are ok
for deformations at serviceability states.
Table 2: Results for the maximum load reached on a strip footing on the drained sub-soil for
different 2D meshes
Mesh size Element
type
Nr. of
elements
Max.
load
Failure
load
[kN/m] [kN/m2
]
Medium mesh 15-noded 212 221 117
Very coarse mesh 6-noded 84 281 147
Medium mesh 6-noded 212 246 129
Very fine mesh 6-noded 626 245 129
Very coarse mesh 15-noded 84 224 118
Very fine mesh 15-noded 626 221 117
Analytical solutions of:
- Vesic
- Brinch Hansen
- Meyerhof
117
98
97
In this table the failure load has been calculated as:
Qu
B
= Maximum force
B
+ γconcrete ∗ d = Maximum force
2
+ 6
From the above results it is clear that fine FE meshes give more accurate results. On the other
hand the performance of the 15-noded elements is superior over the performance of the lower
order 6-noded elements. Needless to say that computation times are also influenced by the
number and type of elements.

ADDITIONAL EXERCISE:
UNDRAINED FOOTING

INTRODUCTION
When saturated soils are loaded rapidly, the soil body will behave in an undrained manner, i.e.
excess pore pressures are being generated. In this exercise the special PLAXIS feature for
the treatment of undrained soils is demonstrated.
SCHEME OF OPERATIONS
In PLAXIS, one generally enters effective soil properties and this is retained in an undrained
analysis. In order to make the behaviour undrained one has to select ‘undrained A’ as the type
of drainage. Please note that this is a special PLAXIS option as most other FE-codes require
the input of undrained parameters e.g. Eu and νu.
Aims
• The understanding and application of undrained soil behaviour
• How to deal with excess pore pressures.
• Use previous input file and ave as new data file
• Soil mode
– Change material properties, undrained behaviour for clay
• Mesh mode
– Mesh generation, global mesh refinement B)
• Staged construction mode
– Re-run existing calculation phases
• Output
– Inspect excess pore pressures
Soil mode

INPUT
Use previous input file
If PLAXIS Input is no longer open, start PLAXIS by clicking on the icon of the Input program
and select the existing project file from the last exercise (drained footing). From the File menu
select Save As and save the existing project under a new file name (e.g. ‘exercise 1b’).
Change material properties
• Change material properties by selecting the Show materials button ( ). Please note
that this button is only available in Soil mode, Structures mode and Staged construction
mode.
• From the Material sets window, select the ’Clay’ and click on the <Edit> button.
• In the Soil window that opened on the first tab sheet (General) change the Drainage
type to "Undrained A" and close the data set.
Mesh generation
The mesh generator in PLAXIS allows for several degrees of refinement. In this example
we will globally refine the mesh, resulting in an increased number of finite elements to be
distributed along the geometry lines:
• Go to the Mesh mode
• Select the Generate mesh button ( ) and in the Mesh settings window choose Fine
for the Elements distribution.
Calculation
• Go to the Staged construction mode. All phases are indicated by (blue arrows)
After mesh (re)generation, staged construction settings remain and phase information is rewritten
automatically for the newly generated mesh. However, this is not the case for points for load
displacement curves due to the new numbering of the mesh nodes.
• Click on the Select points for curves button ( ) in the toolbar. Reselect the node
located in the centre directly underneath the footing

• Click on the Calculate button ( ) to recalculate the analysis. Due to undrained
behaviour of the soil there will be failure in the 3rd and 4th calculation phase.
OUTPUT
As mentioned in the introduction of this example, the compressibility of water is taken into
account by assigning ’undrained’ behaviour to the clay layer. This normally results, after
loading, in excess pore pressures. The excess pore pressures may be viewed in the output
window by selecting:
• Select in the Phases explorer the phase for which you would like to see output results.
• Start the output program by clicking the View calculation results button ( ).
• In PLAXIS Output, select from the Stresses menu the option Pore pressures and then
pexcess, this results in figure 19.
The excess pore pressures may be viewed as contour lines ( ), shadings ( ), stress
crosses ( ) or as tabulated output ( ). If, in general, stresses are tensile stresses the
principal directions are drawn with arrow points. It can be seen that after phase 3 on the
left side of the footing there are excess pore tensions due to the horizontal movement of the
footing. The total pore pressures are visualised using the option of active pore pressures.
These are the sum of the steady state pore pressures as generated from the phreatic level
and the excess pore pressures as generated from undrained loading.
Figure 19: Excess pore pressures at the end of the 3rd phase
• Select from the Stresses menu the option Pore pressures and then pactive. The results
are given in figure .

From the load displacement curve it can be seen that the failure load in the last phase is
considerably lower for this undrained case compared to the drained situation, as expected.
For the undrained case the failure load is just under 70 kPa.
Figure 20: Active pore pressures at the end of the 3rd phase

APPENDIX A: BEARING CAPACITY
CALCULATION
Given the formula for bearing capacity of a strip footing:
Qf
B
= c · Nc + 1
2
γ0
B · Nγ
Nq = eπ tan ϕ0
tan2
(45 + 1
2
ϕ0
)
Nc = (Nq − 1) cot ϕ0
Nγ =





2(Nq + 1) tan ϕ0
(V esic)
1.5(Nq − 1) tan ϕ0
(Brinch Hansen)
(Nq − 1) tan(1.4 ϕ0
) (Meyerhof)
Filling in given soil data:
Nq = eπ tan(20)
tan2
(55) = 6.4
Nc = (6.4 − 1) cot(20) = 14.84
Nγ =





2(6.4 + 1) tan(20) = 5.39 (V esic)
1.5(6.4 − 1) tan(20) = 2.95 (Brinch Hansen)
(6.4 − 1) tan(28) = 2.97 (Meyerhof)
The effective weight of the soil:
γ0
= γw − 10 kN/m3
= 18 − 10 = 8 kN/m3
For a strip foundation this gives:
Qf
B
= c · Nc + 1
2
γ0
B · Nγ =





5 ∗ 14.83 + 1
2
∗ 8 ∗ 2 ∗ 5.39 ≈ 117 kN/m2
(V esic)
5 ∗ 14.83 + 1
2
∗ 8 ∗ 2 ∗ 2.95 ≈ 98 kN/m2
(Brinch Hansen)
5 ∗ 14.83 + 1
2
∗ 8 ∗ 2 ∗ 2.87 ≈ 97 kN/m2
(Meyerhof)

Critical State Soil Mechanics and Soft Soil Model
Professor Helmut Schweiger


1
CG 04
CRITICAL STATE SOIL MECHANICS
SOFT SOIL MODEL
Helmut F. Schweiger
Computational Geotechnics Group
Institute for Soil Mechanics and Foundation Engineering
Graz University of Technology
Advanced Course on Computational Geotechnics, New Delhi, India, 29 - 31 October 2014
S C I E N C E P A S S I O N T E C H N O L O G Y
Critical State / Plaxis Soft Soil model
2
 Direct shear test
 Triaxial tests
 Critical state line
 Modified Cam Clay model (MCC)
 Drained and undrained triaxial stress paths (NC / OC)
 Plaxis Soft Soil model
 Possible enhancements of Critical State Models
CONTENTS
Direct Shear Test | Triaxial Test | Critical State Line | Modified Cam Clay | Stress Paths | Plaxis Soft Soil | Possible Enhancements

2
3
DIRECT SHEAR TEST
v’

Direct Shear Box (DSB)
v’

Direct Simple Shear (DSS)
 

s
Slow Direct Shear Tests on Triassic Clay,NC
0
20
40
60
80
100
120
140
0 1 2 3 4 5 6 7 8 9 10
Displacement,  (mm)
Shear
Stress,

(kPa)
n'
(kPa)=
214.5
135.0
45.1
Slow Direct Shear Tests on Triassic Clay, Raleigh, NC
0
20
40
60
80
100
120
140
0 50 100 150 200 250
Effective Normal Stress, n' (kPa)
Shear
Stress,

(kPa)
0.491 = tan  '
Strength Parameters:
c' = 0;  ' = 26.1
o
Peak
Peak
Peak
Mayne, 2006
4
Log v'
Effective stress v'
Shear
stress

Void
Ratio,
e
NC
CC
tan'
CSL
NC
CSL
CSSM Premise:
“All stress paths fail on the
critical state line (CSL)”
CSL

c=0
Mayne, 2006

3
5
Log v'
Shear
stress

Void
Ratio,
e
NC
NC
CC
tan'
CSL
CSL
CSL
STRESS PATH No.1
NC Drained Soil
Given: e0, vo’, NC (OCR=1)
e0
vo
vo
Drained Path: u = 0
max = c +  tan
ef
e
Volume Change is Contractive:
vol = e/(1+e0) < 0
c’=0
Mayne, 2006
6
NC
NC
CC
tan'
CSL
CSL
CSL
STRESS PATH No.2
NC Undrained Soil
Given: e0, vo’, NC (OCR=1)
e0
vo
vo
Undrained Path: V/V0 = 0
+u = Positive Excess Porewater
Pressures
vf
vf
u
max = cu = su
Shear
stress

Void
Ratio,
e
Log v'
Mayne, 2006

4
7
NC
NC
CC
tan'
CSL
CSL
CSL
CS
p'
p'
OC
Overconsolidated States:
e0, vo’, and OCR = p’/vo’
where p’ = vmax’ = Pc’ =
preconsolidation stress;
OCR = overconsolidation ratio
Shear
stress

Void
Ratio,
e
Log v'
Mayne, 2006
8
NC NC
CC
tan'
CSL
CSL
CSL
CS
OC
Stress Path No. 3
Undrained OC Soil:
e0, vo’, and OCR
vo'
e0
vo'
Stress Path: V/V0 = 0
Negative Excess u
vf'
u
Shear
stress

Void
Ratio,
e
Log v'
Mayne, 2006

5
9
NC
NC
CC
tan'
CSL
CSL
CSL
CS
OC
Stress Path No. 4
Drained OC Soil:
e0, vo’, and OCR
Stress Path: u = 0
Dilatancy: V/V0 > 0
vo'
e0
vo'
Mayne, 2006
Void
Ratio,
e
Log v'
10
Typical results from drained (a) and undrained (b) triaxial tests on normally consolidated soils
(from Atkinson & Bransby, 1978)
a) b)

6
11
Typical results from drained (a) and undrained (b) triaxial tests on overconsolidated soils
(from Atkinson & Bransby, 1978)
a)
b)
12
1+e
DRAINED TRIAXIAL TEST (NC)

7
13
UNDRAINED TRIAXIAL TEST (NC)
14

8
15
NCL AND CSL IN p-q-v - SPACE
16
UNDRAINED PLANES

9
17
DRAINED PLANES
18
STATE BOUNDARY SURFACE

10
19
OCR is very important for
soil behaviour
OVERCONSOLIDATION
20
Right from the M-line (“wet side”): q < M p’ 0, 0
p p
v s
d d
 
  (contraction, hardening)
Left from the M-line (“dry side”): q > M p’ 0, 0
p p
v s
d d
 
  (dilatancy, softening)
On the ellipse top: q = M p’
0,
p
v
p
s
d
d 
    Failure!
f=0
M
1
CSL

11
21
B-C-D-E-F: slope of yield locus becomes flatter
ratio distortional/volumetric strain becomes larger
D. Muir Wood, 1990
normally consolidated
drained compression
Stiffness: primary loading
22
lightly overconsolidated
drained compression D. Muir Wood, 1990
Stiffness:
- unloading / reloading
- primary loading

12
23
heavily overconsolidated
drained compression
D. Muir Wood, 1990
24
v
normally consolidated
undrained compression
D. Muir Wood, 1990
due to change of stress
due to plastic soil behaviour

13
25
lightly overconsolidated
D. Muir Wood, 1990
26
heavily overconsolidated
D. Muir Wood, 1990

14
27
Elastic deformation is generated according to:
unloading/reloading
Total deformation is generated according to:
primary compression
e = void ratio
 = swelling index
 = compression index
0
0
'
ln
e e p
e e
p

 
    
 
0
0
'
ln
p
e e
p

 
    
 
28
Generally we prefer notation in strains:
0 * *
0
0 * * *
0
'
ln ,
1
'
( ) ln ,
1

   

    
 
   
  
 
 
   
  
 
e e
v v
p p
v v
p
p e
p
p e
εv = volumetric strain
* = modified swelling index
* = modified compression index

15
29
pp
q
p’
M
1
α
K0
NC
Soft Soil model:
• Mohr-Coulomb failure surface for strength
• M-line for determining K0
NC
(no longer acts as CSL, only determines shape of cap)
MC-line
PLAXIS SOFT SOIL MODEL
30
“MODIFIED CAM CLAY” WITH MOHR COULOMB

16
31
Input Parameters:
* =  / 1+e …….. Modified compression index
* =  / 1+e …….. Modified swelling index
c ………………… Cohesion
 ………………… Friction angle
 ………………… Dilatancy angle
ur ……………….. Poisson's ratio for unloading
K0
nc ………………. Coefficient of lateral earth pressure in normal consolidation
M …………………. K0
nc parameter
p
*
*
32
-600
-500
-400
-300
-200
-100
0
-0.5
-0.4
-0.3
-0.2
-0.1
0
vertical stress [kN/m2]
vertical strain
Chart 1
SS
MC
vertical stress vs vertical strain
SS vs MC MODEL - OEDOMETER TEST

17
33
-350
-300
-250
-200
-150
-100
-50
0
-600
-500
-400
-300
-200
-100
0
horizontal stress [kN/m2]
vertical stress [kN/m2]
Chart 1
SS
MC
horizontal stress vs vertical stress
SS vs MC MODEL - OEDOMETER TEST
34
Stiffness: primary loading
Stiffness: unloading / reloading
Stiffness: unloading / reloading
PRIMARY LOADING - UNLOADING / RELOADING
elastic
region
current yield surface

18
35
MCC-MODEL - FURTHER DEVELOPMENTS
"Bubble models" with
kinematic hardening
e.g. 3-SKH Model
(Baudet & Stallebrass, 2004)
Anisotropic models based on
Modified Cam Clay (rotated yield
surfaces)
e.g. Wheeler, Näätänen, Karstunen
& Lojander (2003)
36
e.g. Leroueil & Vaughan (1990)
Atkinson & Sallfors (1991)
MCC-MODEL - FURTHER DEVELOPMENTS

Hardening Soil and HS-small Models
Professor Helmut Schweiger


1
S C I E N C E ■ P A S S I O N ■ T E C H N O L O G Y
Computational Geotechnics Group
Institute for Soil Mechanics and Foundation Engineering
Graz University of Technology
Helmut F. Schweiger
CG5
HARDENING SOIL SMALL MODEL
2
CG5 - Hardening Soil Small Model
Advanced Course on Computational Geotechnics, New Delhi, India, 29 -31 October 2014
CONTENTS
 Introduction (why advanced model?)
 Short description of Hardening Soil Model
 Parameters of Hardening Soil Model
 Comparison with experimental data
 Influence of important parameters
 Extension to account for small strain stiffness (HS-Small)
 Summary
Introduction | Description of HS-Model | Parameters | Comparison with Experiments | Influence of Parameters | HS-small | Summary

2
3
Soil behaviour includes:
 difference in behaviour for primary loading – reloading/unloading
 nonlinear behaviour well below failure conditions
 stress dependent stiffness
 plastic deformation for isotropic or K0-stress paths
 dilatancy is not constant
 small strain stiffness (at very low strains and upon stress reversal)
 influence of density on strength and stiffness
cannot be accounted for with simple
elastic-perfectly plastic constitutive models
4
HS
oedometer test

1-
MC

3
5
0 0.01 0.02 0.03 0.04 0.05
0
50
100
150
200
250
eps_axial
q [kN/m2]
Mohr Coulomb Model
HS-Model
6
-0,2
0
0,2
0,4
0,6
0,8
1
0 3 6 9 12 15
distance [m]
s
/
s
max
[-]
Linear Elastic
Mohr Coulomb
Hardening Soil
Model
smax
[mm]
LE 33
MC 36
HS 60
• All models calculate settlements
• Differences in shape of trough
and maximum values

4
7
Example for vertical displacements behind a retaining wall
Typical vertical displacements behind a retaining wall
(sheet pile wall in clay)
-40
-20
0
20
40
60
80
100
120
0 5 10 15 20
distance from wall [m]
vertical
displacements
[mm]
Mohr Coulomb
Hard. Soil
> Hardening Soil Model calculates Settlements
> Mohr-Coulomb Model calculates Heave
8
TRIAXIAL TEST
q = failure value
f
q = residual value
r
1
3
1
qf
qr
dense soil
loose
3
1
q 



isotropic loading
1
1
= constant
1

3

3

Applied stress path and results for standard drained triaxial test
1
vol
dense
loose
3
2
1
volumetric
vol 









5
9
HYPERBOLIC APPROXIMATION OF STANDARD DRAINED TEST
= reference modulus for primary loading at 50% of strength
ref
50
E
3
1 σ
σ
q 



1

1
E50
50%
50%
Hyperbola
q
q
q
E
2
q
ε
a
50
a
1




m
ref
3
ref
50
50
a
p
a
σ
E
E 












msand  0.5 ; mclay  1
10
hyperbolic for q < qf
  










 cot
c
a
sin
1
sin
2
a
q 3
f
1

Asymptote
Hyperbola
3
1 σ
σ
q 



qf
otherwise q = qf
f
f
a
R
q
q  0.9
Rf 
MC failure criterion

6
11
1

Asymptote
Hyperbola
3
1 σ
σ
q 



a
q
q
q
q
E
2
q
a
50
a




1
3
1 ε
2
3
ε
ε
strain
shear 



γ
q
q
q
E
q
4
3
a
50
a



γ
12
SHEAR STRAIN CONTOURS IN P-Q-PLANE
q
p´



05
.
0


01
.
0


a
50
a
q
q
q
E
4
q
3



m
ref
3
ref
50
50
a
p
a
σ
E
E 











a
a
3
a
φ
sin
1
sinφ
2
a)
σ
(
q




lines
curved
0.5
m
:
sands 
p´
q
0
c 

lines
straight
1
m
:
clays 

7
13
LINES OF EQUAL SHEAR STRAINS IN TRIAXIAL TEST
Ref. : Ishihara, Tatsuoka and Yasuda (1975). “Undrained deformation and liquefaction of sand under
cyclic stresses“. Soils and Foundations, Vol. 15, No. 1.
14
3
50 50
'cos ' ' sin '
'cos ' sin '
m
ref
ref
c
E E
c p
  
 
 

  

 
3
'cos ' ' sin '
'cos ' sin '
m
ref
ur ur ref
c
E E
c p
  
 
 

  

 
Note: Stress-dependent stiffness based on 3’
3
2 'cos ' 2 ' sin '
1 sin '
f a
c
R q
  




Failure according to MC criterion

8
15
DEFINITION OF E50
 
kPa
3
1 


300
kPa
600
3́ 

0 0.1 0.2 0.3
500
1500
100
E50, reference pressure = 200 kPa
E50, reference pressure = 100 kPa
16
STIFFNESS IN UNLOADING-RELOADING
Triaxial tests:
Unloading is purely
elastic in HS model

9
17
LINES OF EQUAL VOLUMETRIC STRAINS IN TRIAXIAL TEST
Biarez, J. & Hicher, P.-Y. (1994), Elementary Mechanics of Soil Behaviour, Balkema - Publishers.
18
LINES OF EQUAL VOLUMETRIC STRAINS IN TRIAXIAL TEST
q [MN/m²]
0.25
vol 

0.20
vol 

0.14
0.07
p’ [MN/m²]

10
19
DENSITY HARDENING IN THE HS MODEL
1
'cot '
1 'cot '
m
p
p
v ref
c p
m c p






 
  
 
 
 is determined by K0
nc
 is determined by Eoed
ref
p’
q MC failure line
pp
 pp
Cap
20
DEFINITION OF Eoed
m
ref
ref
oed
oed
p
c
c
E
E


















cot
cot 1
holds strictly for K0-stress paths only

11
21
ELASTIC REGION
f < 0 + fc < 0
REGION 1
no yield surface active > elastic
f
fc
pc
1
p
q
p‘ = (‘1 + ‘2 + ‘3) / 3
q = 1 - 3
22
SHEAR HARDENING
REGION 2
shear hardening surface active
f > 0 + fc < 0
2
p
q
f
fc
pc

12
23
SHEAR AND VOLUMETRIC HARDENING
REGION 3
shear hardening and volumetric hardening surfaces active
f > 0 + fc > 0
3
p
q
f
fc
pc
24
VOLUMETRIC HARDENING
REGION 4
volumetric hardening surface active
f < 0 + fc > 0
4
f
pc p
q

13
25
HARDENING SOIL MODEL IN PRINCIPAL STRESS SPACE
26
FLOW RULE
Volumetric behaviour
"stress dilatancy theory" (Rowe, 1962)
p
m
p
v 

 
 sin

cv
m
cv
m
m





sin
sin
1
sin
sin
sin









cot
2
sin '
3
'
1
'
3
'
1





c
m





sin
sin
1
sin
sin
sin



cv
dilatancy angle > non-associated flow rule

14
27
FLOW RULE
q
p'
cv
mob
mobilized dilatancy angle for  = 35°
mobilized friction angle [°]
0 5 10 15 20 25 30 35
mobilized
dilatancy
angle
[°]
-35
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
30
35
 = 0
 = 5
 = 20
 = 35
cv cv
dilation
contraction
negative values of  are cut-off in Plaxis
Note: flow rule in HS-small
model is slightly different >
undrained shear strength
predicted is different (lower)
28
FLOW RULE
q
p'
cv
cv
m
cv
m
m





sin
sin
1
sin
sin
sin



m < cv   = negative > contraction
m = cv   = 0
m > cv   = positive > dilation
plastic potential Q
x
x
x
x
Volumetric behaviour
"stress dilatancy theory" (Rowe, 1962)

15
29
horizontal stress [kN/m2
]
-350
-300
-250
-200
-150
-100
-50
0
vertical
stress
[kN/m
2
]
-600
-550
-500
-450
-400
-350
-300
-250
-200
-150
-100
-50
0
Hardening Soil Model
Soft Soil Model
Mohr Coulomb Model
Mohr-Coulomb model:
ratio 3/1 determined by 
Hardening (Soft) Soil model:
ratio 3/1 determined
by K0
nc
Unloading: ur
OEDOMETER TEST - COMPARISON MC / SS / HS
30
vertikal stress [kN/m2
]
-600
-500
-400
-300
-200
-100
0
vertical
strain
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
Hardening Soil Model
Soft Soil Model
Mohr Coulomb Model
OEDOMETER TEST - COMPARISON MC / SS / HS

16
31
OVERCONSOLIDATION IN HS-MODEL
p’
q
pp
 pp
Cap position based on
previous stress history
(p*, q*)
Stress point due to current initial
stress state
(p*,q*)
Initial pre-consolidation pressure
pp0 relates to initial p0
Calculation of p0 based on:
• OCR (Over-Consolidation Ratio)
• POP (Pre-Overburden Pressure)
pP = Isotropic pre-cons. pressure
p = Vertical pre-cons. pressure
0
,
' p
yy p
 

0 0
, , 0
' ' nc
p
xx p zz p K
  
 
 
0 0 0
1
, , ,
3
0 0
, ,
2
2
*= ' ' '
* | ' ' |
*
( *)
xx p yy p zz p
xx p yy p
p
p
q
q
p p
  
 

  
 
 
   
 
32
OVERCONSOLIDATION
Calculation of p0 based on OCR: Calculation of p0 based on POP:
POP
'
yy0
p0
σ
σ
OCR 
'
yy0
σ '
yy0
σ
p0
σ p0
σ
0
0 'yy
p OCR 
  POP
yy
p 
 0
0 '



17
33
PLASTIC POINTS
-σ3
-σ1
Mohr-Coulomb point
f<0
Cap point
Hardening point
Cap & Hardening point
-σ3
-σ1
Tension point
Tension cut-off: Principal tensile stress is set to zero
34
PLASTIC POINTS
elastic
elastic-plastic
double hardening

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