NUMERICAL SIMULATION OF EXCAVATIONS
~rrTHIN JOINTED ROCK OF INFINITE EXTENT
by
ALEXANDROS I. SOFIANOS
(M.Sc. ,D.LC.)
May 1984
A thesis submitted for the degree of Doctor
of Philosophy of the University of London
Rock Mechanics Section,
R.S.M.,Imperial College,
London, SW7 2BP.
I
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Abstract
The subject of the thesis is the development of a program to study
the behaviour of stratified and jointed rock masses around
excavations.
The rock mass is divided into two regions,one which is, supposed to
exhibit linear elastic behaviour,and the other which will include
discontinuities that behave inelastically. The former has been
simulated by a boundary integral plane strain orthotropic module,and
the latter by quadratic joint,plane strain and membrane elements.The
two modules are coupled in one program.Sequences of loading include
static point,pressure,bodY,and residual loads,construction and
excavation, and quasistatic earthquake load.The program is
interactive with graphics. Problems of infinite or finite extent may
be solved.
Errors due to the coupling of the two numerical methods have been
analysed. Through a survey of constitutive laws,idealizations of
behaviour and test results for intact rock and
discontinuities,appropriate models have been selected and parameter
ranges identified.The representation of the rock mass as an
equivalent orthotropic elastic continuum has been investigated and
programmed.
Simplified theoretical solutions developed for the problem of a
wedge on the roof of an opening have been compared with the computed
results.A problem of open stoping is analysed.
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ACKNOWLEDGEMENTS
The author wishes to acknowledge the contribution of all members
of the Rock Mechanics group at Imperial College to this work, and
its full financial support by the State Scholarship Foundation of
Greece.
furstudy,and
Furthermore thanks are due to:
Dr. J.O.Watson,for his supervision of this
introducing me to the boundary element method.
Dr. J.W.Bray,for discussions.
Professor E.T.Brown for useful commments.
Professor R.E.Goodman and Dr. Nicholas Barton for providing me
with useful information.
Dr. A.M. Crawford for his supervision during the first year of
this work, and his friendship.
Messrs S.Budd and T.Sippel,for suggestions on programming.
Mr. J.A.Samaniego,for his friendship and exchange of ideas.
Last but not least I whish to express my thanks to my wife for
her patience during the hard period of the study, and to my mother
for dealing with all my interests during my absence from home.
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TABLE OF CONTENTS
ABSTRACT
ACKNOWLEDGEMENTS
TABLE OF CONTENTS
LIST OF FIGURES
LIST OF TABLES
NOTATION AND CONVENTIONS
CHAPTER 1 - INTRODUCTION
CHAPTER 2 - NUMERICAL MODELLING OF JOINTED ROCK
2,0 Distribution of stresses and displacements
2.1 The continuum
2.1.1 Mechanical properties
2,1.2 Simulation
2,2 Discontinuities - A literature survey
2,2,1 Mechanical properties
2.2,2 Simulation
2,3 The joint element
2,3,1 The element
2,3,2 The constitutive law
2,3,3 Iterative solution
2.3.4 Examples
2.4 Change of the geometry
2.4.1 Excavation
2.4.2 Construction
2,5 Types of activities
Page
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3
4
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14
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23
24
24
26
34
34
39
47
47
50
63
72
77
77
79
80
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CHAPTER 3 - THE ELASTIC REGION
3.0 General
3.1 Equivalent elastic properties of a jointed rock ma~s
3.1.1 Three orthogonal sets of joints
3.1.2 Two oblique sets of joints
3.2 Implementation of the direct boundary integral method
3.2.1 The integral equation for the complementary
function
3.2.2 Kernels U and T
3.2.3 Isoparametric element
3.2.4 Nodal collocation
3.2.5 Numerical integration
3.2.6 Rotation of axes
3.2.7 Particular integral
3.2.8 Infinite domain
3.3 Example
CHAPTER 4 - COUPLING REGIONS WITH CONTINUOUS AND
DISCONTINUOUS DISPLACEMENT FIELDS
4.0 General
4.1 Symmetric coupling
4.2 Validation
4.3 Inherent errors
4.3.1 Causes of errors
4.3.2 Examples
Page
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84
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87
89
91
93
95
96
97
98
103
107
107
108
110
125
125
130
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CHAPTER 5 - STABILITY OF AN OVERHANGING ROCK WEDGE
IN AN EXCAVATION
5.0 General
5.1 Idealised behaviour
5.1.1 Symmetric wedge
5.1.2 Asymmetric wedge
5.2 Numerical solution
5.2.1 Symmetric wedge
5.2.2 Asymmetric wedge
CHAPTER 6 - APPLICATION OF THE PROGRAM TO ORE STOPING
CHAPTER 7 - SUMMARY AND CONCLUSIONS
APPENDIX 1 - Description of input for program AJROCK
APPENDIX 2 - Overall structure of the program
APPENDIX 3 - Additional information relevant to Chapter 3
A3.1 Orthotropic kernels
A3.2 Integration of kernel - shape function products
over an element containing the first argument
A3.3 Particular integral
APPENDIX 4 - Estimate of error due to the assumption of
continuous tractions at nodes
APPENDIX 5 - Graphs for estimating the stability of a wedge
in a tunnel roof.
REFERENCES
Page
142
142
142
144
161
169
170
182
188
200
206
231
233
233
243
248
252
254
279
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LIST OF FIGURES
Page
CHAPTER 2
Fig. 2.1 Eight node serendipity element
Fig. 2.2 Axes for transverse isotropy and global cartesian
system
Fig. 2.3 Sign convention for internal forces
Fig. 2.4 Isoparametric three node membrane element
Fig. 2.5 Peak shear strength
Fig. 2.6 Peak shear strength
Fig. 2.7 First joint element
Fig. 2.8 Three dimensional joint elements
Fig. 2.9 Isoparametric quadratic joint element
27
29
29
31
36
37
43
43
48
Fig. 2.10 Failure criteria and parameters 51
Fig. 2.11 Load history effect on current peak shear strength 55
Fig. 2.12 Normal stress vs normal strain law 57
Fig. 2.13 Shear strain vs shear stress 58
Fig. 2.14 Three dimensional sketch for E ,a, T •s
Fig. 2.15 Dilation vs shear strain law for the two models
Fig. 2.16 Iterative process (for compression) - Joint 1,
no dilation
Fig. 2.17 Iterative process - Joint 2 without dilation
Fig. 2.18 Iterative process - Joint 1 with dilation
Fig. 2.19 Iterative process - Joint 2 with dilation
Fig. 2.20 Iterations for simple examples
Fig. 2.21 Strain softening joints (examples)
60
62
65
66
67
68
71
73
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Page
CHAPTER 3
Fig. 3.1 Three orthogonal sets of joints 83
Fig. 3.2 Two oblique sets of joints 83
Fig. 3.3 Conventions for kernel arguments 90
Fig. 3.4 Isoparametric boundary element 92
Fig. 3.5 Coordinate systems H,V and 1,2 92
Fig. 3.6 Integration over remote boundary 99
Fig. 3.7 Initial meshes for the examples of Section 3.3 104
Fig. 3.8 Boundary element region subjected to gravitational
field
Fig. 3.9 Plane strain and joint elements subjected to
gravitational field
CHAPTER 4
Fig. 4.1 Square block in tension
Fig. 4.2 A circular hole under pressure
Fig. 4.3 Hole within infinite rock mass modelled
by boundary elements only
Fig. 4.4 Hole within infinite rock mass modelled
by boundary and finite elements
Fig. 4.5 Tension of a long plate
Fig. 4.6 Lined opening
Fig. 4.7 Excavation of a circular tunnel
Fig. 4.8 Excavation of a circular tunnel
Fig. 4.9 Various methods to determine the limiting values
of tractions at the two sides of a corner
Fig. 4.10 Two boundary element regions
Fig. 4.11 Circular disc
105
106
111
111
113
113
116
119
122
122
126
131
131
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Fig. 4.12 Square block modelled by 32 boundary elements
Fig. 4.13 Square block modelled by boundary and
plane strain elements
Fig. 4.14 Large problem with boundary and finite elements
CHAPTER 5
Fig. 5.1 Wedge idealization
Fig. 5.2 Symmetric wedge - Friction angle greater than a
Fig. 5.3 Symmetric wedge - Friction angle less than a
Fig. 5.4 Examples for very low stiffness ratio joints
Fig. 5.5 Behaviour of a symmetric rigid wedge
Fig. 5.6 Effect of intact rock flexibility
Fig. 5.7 Models for elastic wedge
Fig. 5.8 Stress redistribution
Fig. 5.9 Asymmetric rigid wedge
Fig. 5.10 Oblique wedge
Fig. 5.11 Wedge with rotation
Fig. 5.12 Symmetric flexible wedge within rigid rock
Fig. 5.13 Symmetric elastic wedges within elastic
rock;0
a=20
Fig. 5.14 Example of stress redistribution in a
symmetric wedge
Fig. 5.15 Asymmetric wedge
Page
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183
/
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Page
CHAPTER 6
Fig. 6.1 Stope and drive geometry 189
Fig. 6.2 Stope,drive,and surrounding rock discretization 189
Fig. 6.3 Initial mesh 193
Fig. 6.4 Gravitational loading 194
Fig. 6.5 Excavation of the drive 195
Fig. 6.6 First level ore excavation 196
Fig. 6.7 Second level ore excavation 197
Fig. 6.8 Third level ore excavation 198
APPENDICES
Fig. Al.1 Boundary element convention 213
Fig. Al.2 Plane strain element convention 219
Fig. Al.3 Joint element convention 219
Fig. A2.1 Flow chart of program AJROCK 232
Fig. A3.1 Lines on which the orthotropic kernel U is undefined 240
Fig. A3.2 Analytical integration of a logarithm - polynomial
product over a straight line element
Fig. A3.3 Spiral method used for the determination of
the diagonal terms of matrix T
244
244
/
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LIST OF TABLES
Page
CHAPTER 2
Table 2.1 Shear strain regions
Table 2.2 Two plane strain and two joint elements
Table 2.3 One plane strain and three joint elements
CHAPTER 3
Table 3.1 Integration of kernel shape function products
over an element containing the first argument
59
74
76
95
Table 3.2 Displacements at the nodes of a brick (example) 103
CHAPTER 4
Table 4.1 Square block in tension
Table 4.2 Circular hole modelled by boundary
elements only
Table 4.3 Circular hole modelled by boundary and
finite elements - Displacements at nodes
Table 4.4 Circular hole modelled by boundary and finite
112
114
114
elements - Stresses within plane strain elements 114
Table 4.5 Tension of a long plate modelled by
symmetric mesh
Table 4.6 Tension of a long plate modelled by
asymmetric mesh
Table 4.7 Lined circular tunnel with full adhesion
on interface
115
117
123
Table 4.8 Lined circular tunnel with free slip on interface 123
Table 4.9 Excavation of a circular tunnel 124
Table 4.10 Prescribed values in finite and boundary elements 129
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Table 4.11 Equivalent nodal forces and displacements
for various particular solutions
Table 4.12 Equivalent nodal forces given by use of
stiffness matrices K' .K1,(K1)T
Table 4.13 Square block modelled by 32 boundary elements
Table 4.14 Square block modelled by finite and
boundary elements
Table 4.15 Stresses at centres of plane strain elements
of large problem for KA=O
Table 4.16 Stresses at centres of plane strain elements
of large problem for KA=l
CHAPTER 5
Table 5.1 Symmetric almost rigid wedge within
infinitely stiff rock
Table 5.2 Symmetric elastic wedge within rigid
surrounding rock
Table 5.3 Symmetric elastic wedge within elastic rock,
without excavation sequence
Table 5.4 Symmetric elastic wedge within elastic rock,
with excavation sequence
Table 5.5 Asymmetric wedge surrounded by rigid
rock; 0 50a1=35 ,a2=
Table 5.6 Asymmetric wedge surrounded by rigid
rock; 0 50a1=20
,a2=
Table 5.7 Asymmetric wedge surrounded by rigid
rock;o 0a
1=35,a2=20
Page
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135
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139
140
172
176
177
179
184
185
186
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Page
CHAPTER 6
Table 6.1 Material properties 190
Table 6.2 Discretization 191
Table 6.3 Activities 191
APPENDICES
Table A3.1 Determination of function arctan 241
Table A3.2 Integrals I for isotropy 243n
Table A3.3 Integrals I for orthotropy 245n
Table A3.4 Analytically evaluated integrals of kernels U 246g
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NOTATION AND CONVENTIONS
The following notation is used unless defined otherwise
A Support force in Chapter 5
A Cross sectional areas
B Strain shape function
BO Ratio of residual to peak strength at very low
normal pressure
BE Boundary element region
C Transformation matrix from tractions to equivalent
nodal forces
In Chapter 5 equation coefficient matrix
D Elasticity matrix
In Chapter 5 strength parameter
E Young's modulus
F Compliance
Fr Frequency of joints in a set
FS Factor of safety
G Shear modulus
H Horizontal coordinate
In Chapter 5 horizontal force
I Identity matrix of order nn
J Jacobian matrix
In Chapter 5 force
JCS Joint compressive strength (same as q )u
JRC Joint roughness coefficient (similar quantity to i)
K Stiffness
KA Ratio of horizontal to vertical stress
L Length ; In Chapter 5 base length of wedge
M In Chapter 5 strength parameter
N
P
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Shape function
In Chapter 5 normal force
Equivalent nodal force
In Chapter 5 resultant force
Pr Persistence of joints
Q External nodal force
R Radius
S Traction - displacement stiffness matrix
In Chapter 5 shear force
Scf Stress concentration factor
T Kernel function
TO Tensile strength of wall rock
TR Coordinate transformation matrix
U Kernel function
V Vertical coordinate
V Maximum closure for a jointmc
W Work
In Chapter 5 Weight
x
y
a
Solution vector
Load vector
Nodal displacement
In Chapter 5 wedge angle
b body force
b. Boundary element il
c In Chapter 5 non-dimensional support force
c. . Coefficient of the free termlJ
d Total displacement
d(b,e): External node number of node e of boundary element b
e Internal node number for boundary elements
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In Chapter 5 angle
f
g
h
i
kn
ks
m.1
n
nG
r
Internal finite element nodal force
Acceleration of gravity
height
In Chapter 5 the height of the wedge
height to free surface
Dilation angle
Joint element i
Joint normal stiffness
Joint shear stiffness
Membrane element i
Normal direction to a surface
Number of Gauss points
Pressure at the free surface
Plane strain element i
Unconfined compressive wall strength
Distance between the first and the second argument of
the kernel
s Tangential direction to a surface
So Cohesion
t Traction on a surface
t Given tractions on a boundary
u Displacement component(in shear direction if direction
not specified)
v
v
w
x
y
Displacement component in normal direction
Dilation rate
Weight function
First argument of the kernel (or coordinate)
Second argument of the kernel (or coordinate)
z
r
B
y
6..lJ
6(x)
e
1
u
v
'IT
p
a
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Depth of excavation
Boundary
Difference operator
Stress component ij due to a unit force in direction k
Total potential energy
Addition operator
Domain
Acceleration
Angle
Engineering shear strain
Kronecker delta
Virtual displacement operator
Dirac's delta
Strain
Scaled coordinate
Second curvilinear coordinate
Angle between axes Hand 1
Angle between two joint sets
Square root of (-1)
Lame constant
Lame constant
Poisson's ratio
First curvilinear coordinate
Very low stress
Ratio of length of circle circumference to diameter
Density
Unit weight (peg)
Stress ; in Chapters 2 and 5 , normal stress
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cr' Effective stress
cr Unconfined compressive strength of unweathered rockc
L Shear stress
Peak shear strength, (same as L )P
Residual shear strength , (same as
¢ Friction angle
L )r
¢b Basic friction angle of unweathered dry smooth surface
¢ Residual friction angler
¢~ Angle of frictional sliding resistance along the
contact surfaces of the teeth
w angle
Superscripts
c Complementary function
p Particular integral
o Initial
T Transpose
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CONVENTIONS
If A is any alphabetical symbol,then
A Matrix A
A Vector A
(A)HV Components of A in HV coordinate system
(A)12 Components of A in 12 coordinate system
(A)sn Components of A in sn coordinate system
(A\y Components of A in xy coordinate system
The coordinate systems used are :
H,V Horizontal and vertical axes
1,2 Axes 1 and 2 parallel with the principal axes for orthotropy
s,n n is axis of sYmmetry for transverse isotropy and
s is axis parallel with the strata
x,y General Cartesian coordinate system
The repeated index summation convention is used.
Units
Any consistent set of units may be used by the program.The following
set of units is used unless otherwise specified:
Quantity
Length
Force
Stress
Unit
mm
N
MPa
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CHAPTER 1 - INTRODUCTION
The aim of this study is to simulate the behaviour of fractured
rock masses near underground excavations in hard rock. This requires
modelling of intrinsic structural features such us joints, bedding
planes, faults, etc. in the near field, and the rock mass equivalent
continuum behaviour in the far field, where appropriate boundary
conditions should also be satisfied.
The constitutive law parameters for the various materials
involved, used as input by the models, are usually given by
laboratory tests.These data may not be representative of the
deformability of the rock in place, since scale effects are
important, and in-situ measured parameters would be more
appropriate. In view of the difficulties involved in the
determination of a large number of parameters, additional
assumptions are often made, in order to reduce their number.
Apart from representing the fractured rock mass by an equivalent
continuum, special numerical techniques have recently been developed
in order to model properly discontinuous behaviour. Among these
techniques,the one presented by Goodman et al., which is a finite
element formulation, appears capable of properly modelling the true
mechanical characteristics of a fractured rock mass. in which no
topological change of contacts between rock blocks occurs.
Modelling of the exterior problem, that is the imposition of
appropriate boundary conditions to the near field can be achieved by
a boundary integral formulation over the far field.
The present work is an extension of Goodman's original
development, in order to allow for discontinuous behaviour between
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quadratic boundary and/or finite elements in plane strain problems,
and to take into account the effect of the far field. The program
developed is interactive and is supported by graphical facilities.
In Chapter 2 is presented a literature survey on the behavioural
models for the materials, and appropriate numerical methods,
available for the representation of discontinuous rock in the near
field. The quadratic joint element, and its behavioural models
chosen to simulate the discontinuities are described in the latter
part of the chapter.
Chapter 3 deals with the far field. This is taken to be
homogeneous linear orthotropic elastic. The equivalent orthotropic
properties of the rock mass are derived from the individual
properties of the intact rock and its discontinuities. Then the
boundary integral method used for the formulation of the system of
equations is described. The integration of the kernels of the
integral equation over a boundary tending to infinity is examined.
In Appendix 3 is provided additional information for this Chapter.
There, new numerical techniques are shown, for the evaluation of
integrals of the kernels over elements of which the first argument
is a node.
Chapter 4 deals with the complete problem. In the first part the
symmetric coupling method is used to couple the two numerical
methods. Then validation of the combined code is presented. by
analysing various problems, for which numerical or analytical
solutions are known.In the last part, inherent errors due to the
implementation of the coupling are identified and illustrated by
simple examples. In Appendix 4, the error due to known
discontinuous tractions is evaluated.
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In Chapter 5. the specific problem of the stability of a wedge in
a tunnel roof. subjected to a horizontal stress field is presented.
Initially closed form solutions are derived for symmetric and
inclined wedges on the basis of principles developed by Bray. The
importance of parameters not included is investigated. Then the
various wedge configurations are analysed numerically and the
results correlated with the closed form solution. In Appendix 5,
graphs showing the analytical solution of the symmetric wedge are
presented.
In Chapter 6 the application of the algorithm is illustrated to
an underground excavation problem, in which the stability of an
overhanging wedge is reduced due to shadowing.
Finally in Appendices 1 and 2 are presented a description of
input data and a description of the overall structure of the
program.
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CHAPTER 2 - NUMERICAL MODELLING OF JOINTED ROCK
2.0 Distribution of stresses and displacements
Discontinuities are a fundamental characteristic in rock. The para-
meters characterizing its discontinuous behaviour are many and not well
known,and an appropriate behavioural model is complex.Quantitative resu
lts are usually obtained from continuum. elastic models.Maury(I970) sho
wed that stress concentrations develop in the heart of the rock mass,
which can cause large deformations and failure,as well as making the rock
impervious through closure of the fissures and change seepage forces co
mpletely.These stress concentrations cannot be evaluated using elastic
models.His suggested experimental stress analysis method based on photo
elastic and interferometric models with friction between strata was sui-
table for layered ground.
Working on a biaxial compression rig Ergtin(I970),demonstrated pho
toelastically that the stress distribution in a rock mass with joints
that slip or separate or have voids may be complicated.Stresses might
be tensile or compressive and concentrations up to many times that of
the applied compressive stress. The displacements for various angles of
orthogonal sets of joints (continuous or staggered) were, measured by Ga
ziev and Erlikhman(I97I) on a model test for foundations.They calculated
the stresses corresponding to the above displacements,and equivalent mo
duli of elasticity,and compared the stress distribution to the measured
one (strain rosettes) and found great discrepancies increasing with the
degree of discreteness.Rotation and jamming caused tensile stresses in
separate blocks of the foundation. Interaction between blocks were exami
ned by Chappell(I979).Slip,rotation,and constraints controlled the load
transmission pattern of the discrete model. The material mass was concei
ved as a structure with a finite number of redundancies,progressively~
reduced as hinges form within the mass, till the material becomes a me-
chanism and collapses.
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The main parameters determining the stress and displacement distribution
pattern are:
a. Directions of joint systems defining the anisotropy.
b. Type of discontinuity, shape of blocks and their arrangement in the
rock system.
c. Characteristics of joint contact surfaces.
d. Shear strength of joints.
e. Deformability and strength of intact rock.
f. Type of loading, or interaction between rock and structure.
g. Number of rock blocks in direct contact with structure.
2.1 The continuum.
By continuum we mean the parts of the structure on which the displa
cements are continuous, and hence the strain finite.Parts of the stru
cture that have been modelled as continuous,are the intact rock and the
structural elements as timber, steel ribs with wall or liner plates,rock
bolts and concrete lining placed within forms or pneumatically.
2.1.1. Mechanical prpperties.
Attention is restricted to deformability only.
Intact rock.
Extensive literature exists on the elastic properties of intact rock.
These data have been summarised by Kulhawy (1975) and Hoek and Brown (1980).
Two types of Young's moduli appear;the modulus of the deformation which is
a secant one and a modulus of elasticity which is an initial tangent one
and hence greater than the former. Values range between I and 100 GPa.Hi
ghest values are for plutonic igneous rocks, intermediate values are for
clastic sedimentary rocks and non foliated metamorphic rocks,and lowest
for volcanic igneous rocks as tuff.Poisson's ratios vary between 0.02 to
-25-
0.73 with an average of 0.20.Variation with stress of elastic modulus
under triaxial conditions is small for hard crystalline rocks, but signi-
ficant for porous clastic or closely jointed rocks. The Poissonsratio va-
ries with stress level.
Another important factor is anisotropy, i.e. variation of the rock
modulus with direction. The deformational behaviour of schistous rocks (Lou-
reiro,(1970)) in laboratory experiments was found well described by a
transversely isotropic idealization. In situ tests on greywake indicated
the likelihood of similar behaviour.
The variation of the elastic modulus with direction for granites
Peres, (1966)) was found to be adequately represented by the formula
x2/A2ty2/B2tZ2/(i=I, E=x2ty2tz2, which is an ellipsoid.The maximum ani-
sotropy varied from 1.25 to 2.54.The same investigation for sedimentary
and metamorphic rocks ( Peres) (1970)) showed the need for a higher
. 222222224224.order equatlon of the form;x /1 ty /m tz /n tx·z /p ty.z /q =1 ,whlch
is a quartic.
Structural material
The mechanical characteristics of the various materials are well de-
fined and may be found in appropriate codes as GPIIOor~DIN~I045:::forr;cQR-
crete,BS 449 for steel and BSCP 112 for timber. They may be found also in
hand-books as Kempe's Engineers year book, Morgan-Grampian book publishing
Co. Ltd and Beton Kalender, W.Ernst und Sohn,issued annually.The elastic
modulus of shotcrete for various projects (Hoek and Brown, (1980)) ranged
between 17.8-35.9 GPa i.e. in the same range as the intact rock.Its value
is usually between 2I~7 GPa (Hoek and Brown,(I980)) with Poisson's ratio
0.25 • The Young's modulus for steel sets or rock bolts is 207 GPa •
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2.I.2.Simulation.
Intact rock will be simulated with plane strain elements,and concre-
te lining and rock bolts with membrane elements.
2.I.2.I.Plane strain element.
An eight node serendipity element is used. The element and the shape
functions are shown in fig.2.I.The strain displacement matrix for node i
is given by:
a/ax 0
B.= 0 a/ay N.-J. J.
a/aya/ax
N= (I 2Nl' I 2N2' •••• , I 2N8 )
B= (~'~2".""~8)
The Jacobian matrix is
J=~a.x/a~ ay/aJ= GNi/a~'Xiax/an ay/an aN./an·x.
J. J.
=J- ~ra/a~lN.la/a~ J.
(2.1)
(2.2)
(2.3)
aN/a~'.YiJ ,i=I,2, •• ,8 (2.4)aN. /an- y.
J. J.
The inverse Jacobian becomes
- L ~~/ax an/dx]. bay/anJ- = (l/detJ) •a~/ay an/ay -ax/an
-aY/a~Jax/a~
The constitutive law is transverse isotropy.In three dimensions the
law is given by:
E: s 1 lEI -V2/E2 -V/EI 0 0 0 Os
E:n -V2/E2 1/E2 -V/E2 0 0 0 on
E: -V/EI -V/E2 l/E l 0 0 0 oz z (2.6)=
Ynz 0 0 0 1/G2 0 0 Lnz
Ysz 0 0 0 0 l./GI 0 LSZ
Ysn 0 0 0 0 0 1/G2 Lsn
where the axis of symmetry is n,subscripts 1 describe behaviour within
the isotropy plane sz,subscripts 2 describe behaviour within the aniso-
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-7'5
Node coordinates
in ~, 11 plane.
Node~i 11i
I -I -I
2 0 -I
3 I -I
4 I 0
5 I I
6 0 I
7 -I I
8 -I 0
7
8
'8
12
6
4
5
Element
4
Mapping
Shape functions
1 2 3
Corner nodes N~=0.25(1+E,;·E,;.)·(H11·11.)·(E,;·';:.+n·11.-1),i=1,3,5,7~ ~ ~ ~ ~
Midside nodes N~=0. 50( (E,;~ ~(l+E,;. E,;. ). (1-11 2) +n~. (1 -n-n .)·(1_/:,"2», 5..=2 ,4.6,8.~ 1 - - ~ ~ -- 1 ~
Figure 2.1 Eight node serendipity element (Owen & Hinton 1980)
(E:) =(D) • (0) ,ndE1!E2,m=G2!E2.... sn - sn N sn
-28-
tropy planes sn,zn and G1 is dependent i.e. G1=E1!(2 '(1+vi)).If axis z
coincides with the axis of the excavation, the relation for plane strain
becomes (Zienkiewicz,I977)
If axis slis inclined to x at an angle a (Fig.2.2),then a rotation of
axes is needed for D,
(D) =TT.(D) .T- xy - - sn-
~os2a sin2a sina-cosa J
T= sin2a cos2a -sina'cosa- " " 2, 2,
-2sin&cosa ~sina~osa cos a-sin a
(2.8)
where T the engineering strain coordinate transformation matrix
i. e .(t:) =T· (E:).... sn - ,... xy
Imposing static equilibrium
K'a+f=Q-/\,,..,.. nGnG
K, .> [1 J1B:·D'B.'detJ'd~n =L LT'(~ .n ). ,-w ·w1J - 1 - 1 ---1 - ,., J 1 1 P q 1J P q
T~ ,=B:"D~B.~detJ p q1J "'1 - "'J
The integration above has been performed numerically using a Gaussian
quadrature formula. The internal forces f are given by (Fig. 2.3) :
f=fb+fs+fO (2.12)
where b denotes body forces,s surface tractions, and 0 initial stresses.T 1 1 nGnG,
fb,=(P "P ')b= J J N,'b'detJ'd~dn =LLQ,(~ .n )'w·w (2.13)1 X1 Y1 - 1 - 1 1 11.. 1 P q P q
Q'. =N. -b-de t-I • p q1 1
For gravitational loading b is given by
b=- r:g , (0,1) T
For quasistatic earthquake loading with -a ,-a accelerations,b isx yTb=pg-(a,a)x y (2.15)
-29-
x
s axis on isotropy plane
n axis of symmetry for transverse isotropy
Figure 2.2 Axes for transverse isotropy and global cartesian system
t: trac"tions
initial stresses
b: acceleration of gravity
y
one element
x
Figure 2.3 Sign convention for internal forces.
t l1rax/a~Jt~La y/a~
-30-
The nodal force at node i due to distributed
T ~t.f .=(p .,P .) = !.N.. tSl Xl Yl s r: 1
-tn
The nodal force due to residual stresses is
tractions on surface r is
(2.16)
T :J,.J,. T TfO'=(P .,P ')0= JJ B.-(o 0,0 O,T 0) -J'd~ dn
1 Xl Yl ~~ ~1 X Y xy
The external loads at node i are
(2.17)
TQ.=(p .,P .) (2.18)1 Xl Yl
All integrations are implemented numerically.A two point Gauss fo-
rmula is used,Le •. 2 points for the edges,4 points for the area.It is
interesting that when only one element is used with 3 constraints, ,
the stiffness matrix becomes singular.Zienkiewicz(I977) gives the fol-
lowing explanation:
There are 13 degrees of.freedom and there are 4(Gauss points)X3
(stress components per Gauss point)=12 independent relationships.
Because the former are more than the latter)the structure behaves as a
mechanism.For more than one element the number of Gauss points increases
more rapidly than the number of nodes and the problem ceases to exist.
2.I.2.2.Membrane element.
The characteristic of a membrane iSJit is thin and stresses do not
vary accross the thickness.A thin shell or disc transmitting stresses
normal to their cross section only)are examples of bodies exhibiting
membrane action.In the plane strain problem under consideration.)the
concrete lining or any rockbolts evenly distributed along the axis of
the excavation,may be considered as line membrane bodies transmitting
stresses within their line.For the lining,expansion joints are assumed
so that plane stress conditions exist (the effect of ~oisson's ratio is
neglected). The element is three node isoparametric(Fig.2.4).
Geometry.
Three nodes define the position of the element as
-31-
£ =a u /ass
a =E·'t
Mapping on line ~
3
x
Element
.~ -1;=-1 ~=O g=+10 e e1 2 3
0-/2, -1/8)
~1.0
Figure 2.4 Isoparametric three node membrane element.
-32-
T Tx=(x,y) =N.·x.=N.·(x.,y.)11111
The shape functions ~are defined on line ~ in Fig.2.4.
The first derivatives of the shape functions are
(2.20)
The displacements are defined from
T Tu=(u ,u ) =N.·a.=N.·(a .,a .), i=1,2,3. (2,21)x y 1 1 1 Xl yl
The mapping of line s of the element to line ~ is through the Jacobian
J(~)=ds/d~=«dx/d~)2+(dy/d~)2)1/2,
(d/d~) x=(d/d~) N.·x.=N.·x.1 1 1 1
The direction of line s is defined from
(2.22)
(2.23)
The strain is defined as the ~hange in length per unit length of an
infinitesimal element.
E=(d/ds) xT.(d/ds) u= J-2.(x:.N:).N~.a.J J 1 1
The strain. displacement matrix ~ is defined from
E=B·a ~=(Bl,B2,B3)
B.=J-2.(X:.N~).N~1 J J 1
T~=(al,a2,a3)
The constitutive law is given by
Imposing static equilibrium
T T Toa ·Q+fOu ·b·dV = fOE ·o·dV..., ....
Substituting for U,E,O from 2.21,2.26,2.29 we get
oaT.Q+foaT.NT.b.dV = foaT.BT.(E.B.a+o ).dV......... r- -., ~ /'ftJ,w N""J 0
K·a+f=Q_ IV I"'J _
(2.25)
(2.26)
(2.27)
(2.28)
The stiffness matrix eomponent K.. relating nodes i and j of onelJ
element is given by,
-33-
T T T T nG T TK.. =E·A· JB.·B.·ds=E·A·JB.·B.·J·d t,;=E'Ao E(B.·B.·Jow) - (2.33)
J.J 5 J. J S J. J p= 1 J. J p
The internal forces due to body forces b and initial stresses are
f=fb+fO (2.34)
f =A'JNToboJ'dt,; = b'A'L:(NT'J'w) (2.35)Nb S N S ~ P
b=_pg(O,I)T or b=pg(a ,a )Tx y
T nG Tf =A· JB 0 a .J' dt,; = a .A. L: (B . J ·w) (2.36 ),..0 5 ,." 0 0 p~l '" P
ASis the unff'orm'. thickness of the membrane.For rock bolts Asis the sum
of the cross sectional areas of the bolts per unit length of the tun-
nel. 2 to 5 point Gaussian quadrature formulae are used in the program.
2.1.2.3 Other compatible elements not included in the program.
Beam elements to model steel ribs are compatible as are also fluid
elements to simulate water drag forces and flow.AII solid elements may
be made plastic by changing the appropriate modUk to accept post yield
behaviour.
-34-
2.2 Discontinuities - A literature survey.
The variation of strength with direction may be termed strength ani
sotropy.If this variation is continuous, the anisotropy may be termed co
ntinuous,if not it is termed discontinuous.Continuous strength anisotropy
for rock has been investigated by Jaeger (1960), McLamore and Gray(I967)
and others.A special case of discontinuous anisotropy,is a rock isotropic
in strengthJcut by a continuous joint set (Jaeger (I960)).His theory was
named the plane of weakness theory.Bray (1967) investigated discontinuous
strength anisotropy with one, two, six and multiple joint sets.In the last
cape if the strength of the joints is the same or continuously varied
with angle,the strength may be modelled as continuous. This work deals with
discontinuous strength anisotropy exhibited in a rock mass cut by sets of
joints, cleavage, or crushed fault zones,to form the planes of weakness.The
strength of the intact rock is assumed not to be critical. The discontinui
ties are assumed to play the dominant role in the collapse and deformatio
nal behaviour of the excavation. Many sets of them will give the rock a co
mplex fabric.
2.2.1. Me chanical pr:<>?p.erti.es.
It is argued that experimental results obtained on isolated rock joints
can be used effectively in the models.Joints may be filled or unfilled.
Filled joints with thickness of infIlling material of more than twice the
height of the asperities, have the properties of the infilling material.
Their behaviour falls in the context of soil mechanics and they will not
be further considered.Unfilled joints have been found to have the follo
wing characteristics:
a. Tension cannot be carried in the normal direction.
b. Shear strength is a function of normal stress and material properties
parameters.
c. Elastic behaviour is exhibited within the yield envelope.
-35-
2.2.1.1. Shear strength
The first successful model for the shear strength of a joint was
conceived by Patton (1966).He concluded that,
a. Failure envelopes for rough joints are curved.
b. Changes in the slope of the failure envelope reflect changes in
the mode of failure.
c. Changes in the mode of failure are related to physical properties
of the irregularities along the failure surface.
His first model (Fig. 2.5) is a bilinear envelope,fitted to the curved
ones.The joint surface is idealised as a saw tooth moae~,the teeth being
inclined at an angle i.At high stresses the teeth are assumed to break.
He stressed the need for a curved envelope to reflect the multiple modes
of shear failure.Ladanyi and Archambault (1970) proposed a curved failure
envelope for the peak strength.ln Fig 2.6 the law is shown. The law is used
in the program and will be dealt in detail in section 2.3.2••Ladanyi and
Archambault (1980)having more results from laboratory tests adjusted some
constants into their failure equation (power constants for a and v).s
They derived also avaryingi Patton law (Fig. 2.5).The prediction of stre-
ngth in biaxial tests was also investigated.Jaeger (1971) suggested~ava-
rying cohesion law (Fig. 2.5).A general criterion for rough joints was
developed by Barton(1971,1973,1974).lt is in the form(Fig. 2.6)
1 1 1T/an=tan«JRC) .logI0(JCS/an)+$b) , O.01<a/JCS<I. 0 (2.37)
1For a/JCS>I.0 a Mohr Coulomb failure criterion was suggested.
The joint roughness coefficient(JRC) can be taken as
20 for rough joints Class A.
10 for smooth undulating joints Class B.
5 for smooth nearly planar joints Class C (foliation and bed-
ding joints).
o for smooth planar joints.
o 0The basic friction angle $b normally falls between 25 and 35 and can
-36-
Patton's Law
low stress T p=otan(<I>,..+i)
high stress Tp=sO+otan<l>O
T
T1
a
T
Jaeger's Law (varying cohesion)
( -b<J)T =S' i-e + tan<l>Op 0
TPatton's Law with varying i
101<loiT =a-tan (<I> +i)
p 1.1
tani=( 1- (a lOT) O.25).tanioi
O~J30
",/,
//
/,/
"
/,,/
aT
a
a
Figure 2.5 Peak shear strength.
-37-
~adanyi and Archambault
Tp =(o(l~ )·(v+tan¢ )+0. 'TO
) / (I - ( I - a. ).v.tan¢ ). s k .r s s L rv=(1-0/(TP
T)).tamo' a. s =1-(1-0/(naT))
v=O , a. s =1 for a/a? 1.0·n
K=4,L=I.5
Barton's Law.
in Region II
TJOn=t an (J RC.l og1 0(J CS,u n)-I<1> b)
in Region III
T-pO n=tan(JRC'logI 0(OI-03)/On~ b)
in Region I
straight line
qu and JCS are the same
a I axial stress at failure
a 3 confinement
Figure 2.6 Peak shear strength.
-38-
be taken 300 if not known.The joint wall compressive strength(JCS=q )u
varies from ° for unweathered joints to 0.25-0 for weathered ones.c c
If the dilation angle at peak i and the maximum dilation angle i atp 0
extremely low normal stress are known,then the formula might be written
tT/o =tan«9d'-i )-(i Ii )+i )n 0 p 0 0
No significant scale effect exists for peak shear strength of tension
joints. Scale effect is significant for displacements.t
For o/JCS<O.OI Barton(I976) suggested a straight line to zero.A curve
with normal tangent at ° =0 seemed also appropriate.For high stressesn
ti.e. o/JCS>I.O he assumed that confinement was of importance and hence
JCS was no longer appropriate.He introduced in his formula instead of
Jes, 01-03 where °1 is" the axial stress at failure and °3 the confinement.
This formula reduces to the previous one for °3
=0 in which case 0I=JCS.
Barton and Bandis(I982) concluded that the shear strength and shear
stiffness reduce with increase of the block size. This may be used for the
scaling of laboratory tests,when no rotational or kink band deformations
occur.Krsmanovic(I967) conducted a series of direct shear tests in sand-
stone, conglomerate and limestone and determined the initial and residual
shear strength of the discontinuities in hard rock. The parametert
n=Tp/Tult was plotted for various displacements and found to be as high as
10 for small normal stresses(0.3MPa).For higher normal stresses the ra-
tio tended to I.O.The effect of normal stiffness on the shear strength
of the rock was examined by Obert,Brady,and Schmechel(I976).
2.2.1.2 Deformability.
The deformability of joints has been discussed by Goodman(1970,1974,
I977).He proposed a hyperbolic compression curve,a constant stiffness
or constant peak displacement model for shear and a model for dilation.
These models will be dealt in section 2.3.2.Celestino(I979) performed
cycled tests on artificial specimen joints with very regular geometric
form.Hungr and Coates(I978) studied the relation of deformability of
-39-
joints to rock foundation sett1ements.Wa1sh and Grosenbauch (1979) model
led the compressibility of fractures.Swan (1983) showed the functional re
lationship between .norma1 stress,norma1 stiffness and true contact area.
Estimates for in situ joint deformation parameters are given by Barton (1972,
1980) ,BandiS'et al(1983) ,and Barton et a1 (1983).
2.2.2. Simulation
Four approaches have been investigated to select the simulation method
used.
a. No tension and laminar elements
b. Discrete elements
c. Displacement discontinuity elements
d. Joint elements
2.2.2 •.1. No tension and laminar elements.
Zienkiewicz,Va11iapan and King (1968) used no tension elements to
simulate rock behaviour. This technique was used first for concrete. The
laminar element is a thin no tension element used in composite materials.
An eight noded plastic plane strain e1emeht(Pande,(1979)) with length
several thousand times its thickness might also be used to simulate
joint behaviour.
2.2.2.2. Discrete elements
These are suitable for closely spaced joints in hard rock.The joint defor
mations overshadow the intact rock deformations and the intact rock may be
considered rigid. The block centroids having only three degrees of freedom
-40-
determine the geometric position of the block, thus reducing the size of
the problem.Further,no stiffness matrix factorization is performed as the so
lution is sought through successive relaxations. The method is suitable for
large movements and changes of contacts. Two methods are used to find a sta
tic equilibrium position.
Dynamic relaxation (Cundall (1971), Vargas (1982)) inputs incremental forces
at the joints,which are transformed to incremental forces and moments at
the centroids of each block. The displacements then are followed in the
time domain,by integrating the accelerations (- acting force/block inertia).
New contact forces will correspond to the displacements,and a new cycle be
gins.New contacts may arise and others will cease to exist. The cycles will
continue till a stable position is attained.,
Static relaxation (Stewart, (I98n) is similar to the well known Hardy Cross
method for the solution of frames in statics with relaxation. Small incre
ments of force must be used in order to follow large displacements. This
method is better than the previous one as far as computer time is con
cerned.
It is argued that any type of constitutive law may be used by the methods.
2.2.2.3. Displacement discontinuity (D.D.)
The method is especially suitable for dealing with cracks (Roberts and
Einstein, (1979)) and slit like openings.It is based on the solution
to the problem of a constant discontinuity in displacement over a finite
line segment in the x,y plane of an infinite elastic body,derived by Crouch
in I976.Any distribution of relative displacements between opposing sides
of a segment may be discretized by displacement discontinuity elements.
The displacements and stresses at any point are the sum of the displacements
or stresses due to all displacement discontinuities.Although usually constant
D.D elements are used (Crouch and Starfield (1983)), higher order elements
-41-
have also been used(Crawford and Curran, (1982)).
The system of algebraic equations is formed by considering the
boundary conditions for
each element.
- +D =u-udn n n
>--\
If tractions are prescribed,
ti=A~j.Di
If displacements are prescribed,
i_Bij Dju - d • d
If mixed conditions eXist,Jthe .above equations may be rearranged to
form a system of linear equations with bd the known quantities,
The displacement discontinuities are defined in the sketch above and
can be written in vector form
where s,n is the local coordinate system;-/+ indicates the side of
the element;i,j are nodal points;and A~j ,B~j are boundary influence
coefficients for the tractions and the displacements respectively.
Ddmust be computed first. Tractions and displacements will be computed
by substitution of Dd into 2.39 and 2.40.In this sense it is an indirect
boundary integral method, where instead of fictitious forces we have fi-
ctitious displacement discontinuities.
-42-
2.2.2.4.Joint element.
The joint element is a linkage element between faces of blocks.
It was developed by Goodman,Taylor ,and Brekke(I968).Their model shown
in Fig.2.7 is afour noded two dimensional element.Two independent com-
ponents for stress and strain exist i.e a 'l and e ,£ .Applying stan-n n s
dard finite element procedures,the stiffness matrix becomes,
o
o
K= L- .6
2ks
o
o
-ks
o ks
2kri
0
o 2ks
kn 0 2~
o -2k 0s
-ks
o
-2ks
o
o
-kn
o
-2kn
o
-2ks
o
-ks
o
o
-2kn
o
-kn
o
o -k 0 -2k 0n n
o -2~ 0 -~ 0 ~ 0
o
where k and ks are functions of£ ,£ , and the load history.n n s
This joint could model adequately features such as failing
in tension or shear,rotation of blocks,development of arches and to a
certain extent the collapse pattern of structures.The element was ex-
tended then to three dimensions(Mahtab and Goodman, (1970)) as shOvffi
in Fig.2.8.The no tension element technique has been compared to the
joint element one by Heuze et al.(I97I) for the case of borehole jack
deformability tests.Goodman and Dubois(I97I,I972) coupled shear and
normal stresses by introducing dilational properties to the joint ele-
ment.Thus roughness that increases the strength of the joint was intro-
duced as a factor affecting deformability.The constitutive matrix~
has become full. The constitutive law for dilatancy was formulated by
transforming Efor an assumed smooth plane parallel to the direction
of the asperities.For a smooth plane ~Ois diagonal and given by,
D= [ks 0]-0 0 k
n
-43-
ty.n
. Cp
1 bottom
f- L/2-+-
x,s2
L/2-*
1.0 [ _______ N1=N
4=0. 5-x/L
Figure 2.7, First joint element (Goodman et al.1968)
z
6_-----------:7'i
x :t
2
Figure 2.8 Three dimensional joint elements (Mahtab & Goodman 1970)
k =kns sn
-44-
By rotating the coordinate system by an
T ~01 [k ·cos
2i+ k 'sin
2i
D=TR' s . TR= s n- - 0 k - sini-cosi' (k -k )
n s n
[k kj'ss sn
=k fi.9' k,
angle i,~ becomes
sini-cosi'(k -k js n _
k 2· tk· 2. --cos J. . san J.n s
The solution may be approached either by a variable stiffness method
or a load transfer one.A constant energy perturbation would speed con-
vergence but would not always converge.It is argued that if the stif~
fness matrix is to be altered at each iteration, it would be only slig-
htly more expensive to modify the load vector as well. Thus the load
vector wo~ld be modified as for the load transfer method,whereas the
stiffness would be modified so that the energy spent would not change.
In the case in which k and k exist and the variable stiffnessns sn
technique is used,D will become asymmetric during solution , and if
a symmetric solver is used it will need symmetrization.
Stiffnesses may be determined experimentally according to their
definitions,
k :(aT/aU)ss k .=(aO/av) , k =(aT/dV) , k, =(ao/au) (2.46)nn sn ns
A machine in Prof.Muller's laboratory at Karlsruhe,Germany can deter-
mine directly these coefficients by doing controlled normal and shear
displacement,direct shear tests.Usually these coefficients are deter-
mined indirectly from controlled normal stress-direct shear tests.
Numerical simulation of crack initiation of a biaxially loaded
sand plaster plate with two perpendicular joint sets
(De Rouvray and Goodman, (I972»J showed the sUitability of the method
for parametric study.Ghabousi,Wilson and Isenberg(I973) developed a
joint element by defining the displacement degrees of freedom at the
nodes of the element to be the relative displacements between opposing
sides of the slip surface. This technique according to the authors,over-
comes numerical difficulties associated with high joint stiffness.
-45-
Let us consider the following one dimensional problem in the sketch
to illustrate their approach.
The stiffness matrix iste+ ~ -kj J (2.47)K=
- kj ke + kj;-~
For large kj/ke this matrix becomes ill conditioned.lf now the unknownsJ
b t b [] tare changed to be u and ~u=u -u ,and a= ~ ~ , hen
u=(ut,ub)T=a.ul=a.(~u,ub)Tand,-J -,.., -
, I TITK'u=K'a'u =(Ko'a)'u =P, (a 'Koa)'u =a'P_"'_"',.., _~ t"'J """,_,..J
(2.48)
The new stiffness matrix avoids the previous problem.It is given by
IT. rke+kj! =~ .~.~= Lke
ke}2ke
For the two dimensional case they considered they defined a as
where 14 and 04 are the 4X4 unit and zero
matrices respectively. The housekeeping of the global stiffness matrix
becomes more complicated.
Goodman(I975) included a two dimensional joint element program in his
book.Desai,{pesai,(1977) ) used 3-D curved joint(interface) elements
for the solution of foundation problems.Sharma,Nayak and Maheshwari
(1976),in analysing a rockfill dam took account of interfacial sliding
by using quadratic joint elements at the interfaces.Goodman and st.John
(1977) elaborated the use of F.E. for the analysis of discontinuous
rocks. They included a new type of joint element with three degrees of
freedom instead of the four of the previous linear element, the relation
between stress and strain being now,
T
T = a
k s
= 0
o
o
o
o
o
kw
-46-
!::J.u
!::J.v (2.50)
The element behaves as a linear element in the normal direction, and as
a constant one in the shear direction(i.e. it cannot accept change in
length).MO
is the moment about the centre caused by the linear variation
of a ,!::J.w is the rotation of the element caused by the linear variation
of !::J.v.The rotational stiffness k =k 'L 3/4,where L the length of theW n
element.Hittinger and Goodman(I978) presented a computer program for
stress analysis which included linear type joint elements with constant
shear stiffness. Their program has been the basis for the joint element
module developed here.
A comprehensive rock discontinuity model has been developed by Ro-
berts and Einstein(I978),that can treat the entire behavioural history
of a rock discontinuity without dilation.Ke Hsujun.(I979,I98I) used joint
elements,non linear material properties and load cycling.Heuze(I979)
illustrated. the significance of dilation in the analysis of jointed
structures. The increase in the shear stiffness and the normal stress
of a joint,subjected to transverse restraint during shearing were shown
to be of great importance. This would also increase its shear strength.
Heuze and Barbour(I98I,I982) presented a new model for axisymmetric in-
terraces, such as found in shaft and footing design.A new model for di-
latational effects was also included.VanDillen and Ewing(I98I) discussed
a new version of BMINES,a static three dimensional non-linear program
with joint elements,whose constitutive relations are posed in terms of
plasticity the.ory;i.e. dilation is considered to be plastic strain in
the normal direction and slip is taken to be plastic shear strain.A non
associated/flow rule allows slip and dilation to be specified separately.
Desai €It al (1983,1984) presented a thin layer element.
Carol and Alonso(I983) presented an isoparametric quadratic joint element
using a constant peak shear displacement law and dilation.
-47-
2.3.The joint element.
2.3.1.The element.
This is a quadratic isoparametric element, the shape functions and
the sign conventions for which are shown in Fig.2.9. The coordinates of
the middle line nodes are defined in terms of those of the interface e-
lement nodes by
x =0.5(x(1)+x(2»..1 - - (2.51)
The coordinates at any point in the middle line may be calculated as
x(/:")=N.(/:").x. , i=i,2,3.- ':> 1 ':> -1
The coordinates at any point of the boundary of the element may be
calculated similarly.The relative movement between the faces of the
element at node 1 in the xy plane is given by,
Li~1=(LiulfLivl)T=(U(2)-u( 1) ,v(Z) -v( l»T=
= r-10 1 0] (U(1),v(1),U(2),V(2»T= T'~alL0 -1 0 1 - <,
T' = (-12,1JFor all three nodes
TLia=(Lial,Lia2,Liaa)- ~ ."" -6x1
T, a=(al,a2,aa)...... - - -
12Xi
The relative displacements at any point are
@ =(Liu .t» )T=N.·Lia.=N.·T' ··a.=N.·a.xy x y 1 -1 1 - -1 1-1
[_N. 0 N. 0]
N.= 1 1
1 0 -N. 0 N.1 1
The jacobian determinant is given by
J=ds/d~= 1«dx/d~)2+(dy/d~)2)= 1«N~x.)2+(N~y.)2)1 1 1 1,
where N is the derivative of N with respect to ~
The direction of s is defined from
dx/ds=(dx/d~)· (d~/ds)= J_1'N~ (~)·x. ,i=1,2,3.- - 1 ""1
(2.57)
-48-
i= 1, 2,3.x ( s ) =x :N . (E,;)1 1
(2) .
(1)Element
Shape functions 1.0
N1 =0. 5·~· (~-l)
1 2 3
N =1 c- 22 . ......"
1 2 3
1 2 3
E,; =-1 E,; =0
~Sign conventions for stresses
and strains.
Figure 2.9 Isoparametric quadratic joint element.
-49-
The coordinate transformation matrix is
fdX/d S
E= L-dY/dSdy/dsJdx/ds
(2.60)
The strain is defined as
s=(s ,s )=(i.O/t)(lm,~v)T=(1.0/t)·R·(tm,~v)T=(1.0/t)-R·N.·a.=- s n Sn".,. xy - J. - J.
=B.-a. (2.61)-J. -1
where B the strain displacement matrix of the joint and t the thickness
of the joint.We accept constant thickness of the joint and t=i.o •
The constitutive law can be written in the form,..Tdo=d(T,O) =D'ds=D-B'da,. - - - - - ,
12= r;sGns
ksn
.}k..nn
(2.62a)
for constant D the relationship becomes,
(2.62b)
Imposing static equilibrium by equating the virtual work of the external
forces to the work of the internal forces,
Of}?' ,g=Jo~T- g -ds
foaT'BT'Oo-ds +- - ..fo=fBT.Oo-ds- - -
= fOaT·BT'(Oo+D-B.a)-ds=- - - ---T Tfoa -B -D-B-a'ds -+ Q=fo+K·a- - --- - - ".
K=fBT·D·B-ds
nG TK= ~ (B -D'B'J'w)- P-l - - - p
where f o the initial loads and K the stiffness matrix.- -Integration is performed numerically as,
nGf 0= ~ (B"'='Oo •.J ··w)- P-l - - P
The order of the Gaussian quadrature formula used is between 2 and 5.
In the first iteration D is constant within an element and hence the
polynomials are of degree 5.The exact answer can be obtained with 3
points of integration.After the first iteration ,the state determination
is performed at the Gaussian points.Since the ~ matrix varies continu-
ously along the joint in a form differing from a polynomial ,the stif-
fness matrix would not be computed exactly,although a better approxima-
tion would be obtained by a higher order formula.
-50-
Any unbalanced stresses at the Gauss points after the state determina-
tion has been computed,are transformed to unbalanced nodal forces as,
(2.65)
where superscripts e and r denote,due to elastic solution, and real va-
lues due to constitutive relation for the same strain, respectively.
2.3.2. The constitutive law.
The joint element is completely non-linear,with two independent
non-associated flow rules for normal stressvs normal strain, shear
stress vs shear strain.It also includes dilation that couples shear
to normal strain, accounting for roughness. The testing of compatibility
of displacements vs stresses through the constitutive law is called
state determination.Normal stress vs normal strain is no tension elastic
for compression. Shear stress vs shear strain is elastoplastic.Flow is
determined by the direction of the joint and the dilation law.
The peak shear strength is given by an envelope relating peak shear
strength to the normal stress.Two types of joint models are available.
Both models have the same shear stress-shear strain behaviour.
2.3.2.1. Shear strength (Fig.2.IO)
Peak shear strength of joint I.
Ladanyi and Archanbault's(I970) failure criterion is used.The
shear strength of the joint is assumed to be the sum of four separate
strength components,SI,S2,S3,S4.The three first components assume no
shearing occurs. From static and limit equilibrium we have,
N-cosi+S'sini=V
\ (2.66)
Nt ~~P.-
i \V""""'''''
-51-
Joint!
Ladanyi & Archambault
lI?~.l;~perc+T;r·U -per-c )
perc depends on time history
qu 0
- ... ---_.----- ---~_"":"__:":'_~--~-
1.0v/tani
o.ak-----------=:::===--+--
shear area ratio c s= 'f.I:iA/A as, s'1.0
dilation rate v=dy/dx
Joint2.
Mohr Coulomb-Patton
1.0
Residual to peak
relation
1.0 o/qu
Figure 2.10 _'l!'lf~11.:1re criteria and param eters.
-52-
From kinematic consideration we get
b.v/ !:m=tani
External and internal work is given by
W. =S·b.uext
W. t=N·b.v+P'b.u/cosi=N·(b.v/b.u)·~u+(N+S·tani)·tan~·b.uw ' U
Equating external to internal work done i.e.W. t=W t,we have,~n ex
For an irregular surface we put tani=v,i.e. an average dilation rate.
(2.68)
(2.72)
where ¢f an average friction angle for different irregularity orientations.
It can be taken ¢f~¢U~ 300±50. For high loads complete shearing of the
asperities occurs,and the strength of the discontinuity becomes the same
with the strength of the rock(brittle ductile transition,Mogi(I966)).
The strength of the rock is given by Fairhurst(I964),
1'P=q (i.O-n-o/q )0.5·(m_LO)/n , n=q /(-TO)
, m=(n+1.0)0·~(2.69)u u u
However the two modes occur simultaneously before the transition point
is reached , the total peak being partially due to overriding and parti-
ally due to shearing of asperities.If As the portion of surface A over
which the asperities are sheared off and as=As/A,a linear interpolation
between the two modes a =0 and a =i.O is performed, i.e.s s
S=(S1+S2+S3)' (1.0-as)+S4· as (2.,70)
o·O\O-a Hv+tan¢ )+a 'q .(1.0+n.o/qu)0.5 ·(m-1.0)/nSUs u (2.71)
1.0-(1.0-a )·v.tan¢s f
For a and v an exponential interpolation is made between two extremess
o/q =0 a =0 v=taniu s
o/q >1.0 a =1.0 v=Ou s
a =1.0-(1.0-o/q )k1 kl=1.5s u
v =(1.0-o!q )k2 k2=4.0u
In all the above formulae,qu has been used instead of 0T as suggested
-53-
( / ) 0. 75 . (1 0 (/ '(}.25)1075·t .by Goodman.The new formulae for a = 0 q ,V= 0 - 0 q ) ° anas u u
suggested by Ladanyi and Archambault(I980) have not been implemented.
The degree of interlocking factor n might be useful to be incorporated
in the program.It affects qu and 0T by modifying them to become qu·n,
0Ton respectively.
Peak shear strength of joint 2.(Figo2.IO)
This is a mixed Mohr-Coulomb,Patton failure criterion.
The Mohr-Coulomb failure criterion is TP=S +O.tan$o jJ
The Patton law is : TP=ootan($ +i) for -o<-qjJ u
TP=q °tan($ +i)+(O-q )otan$ ,u jJ u r
The mixed criterion then becomes:
-o>-qu
TP=sO+ootan($r+i) for -o~-qu
TP=SO+q otan($ +i)+(O-q )otan$ for -o>-qu r u r u
This model is characterized by 4 parameters so,q ,$ ,iou r
rResidual shear strength T •
(2.75)
Very little is known on the variation of residual shear strength
Tr
with ° .It is known that for high confining pressure,peak strengthn
equals to residual,i.e. for -o>-q +Lr=TP•u
At very low confining pressure Tr/TP=B ,where O<B <1.0. A linear inter-o 0
polation between the two extremes is used,
O<o/q <1.0u
o/q >1.0u
(2.76)
If the peak shear strength has been attained and shearing continuous,
some asperities will break;depending on the normal stress and shear di-
splacement.Thus for a new load the peak shear strength cannot again be
given by the previous relations unless modified.In the program the new
peak shear strength is taken to be given by the following relation,
TPk=percoTP+(1.0-perc)oTr , perc=(Tpk_Tr)/(TP_Tr) (2.77)
where TPklies on the raIling part of the shear stress vs shear strain
-54-
curve.It must be acknowledged that the incomple.teness of the model might
be misleading in some cases.Say for example a small normal stress is
first applied on the joint with a small shear stress that causes
slip to occur.Asperities are overriden but very few broken.Nevertheless
perc might become zero and for higher normal stresses strength w.ill be
moving on the residual envelope which underestimates the strength being
uneconomical-On the other hand,if at very high normal loads some slip
occurs so that perc say becomes 0.9 , almost all high stepped asperities
will have been broken.If then the normal load is reduced, the prototype
would be able to attain only residual stresses,whereas the model would
predict much higher stresses. In Fig.2.II. the effect of load history
due to strain softening is illustrated,as conceived by the author,as
a multiple S shaped curve, that should be modelled in terms of distribu-
tion of asperity steepness;i.e. in the functions of a and v.s
Curve 1 is due to partial shearing at stress LeveI o.s Cur-ve 2 is due to
further partial shearing at stress level 0'2.This has not been programmed
but might be an important point for further investigation.
For good behaviour of the model,elastic behaviour of the joint is
required if reversal of the load is expected as for initial consolida-
tion and then excavation.The model for monotonic loading would under-
estimate the strength of the prototype,whereas for reversed loading
might overestimate its strength. These problems cease to exist if strain
softening is not occuring before the final load step.It might sometimes
be reasonable to work with residual values for shear strength,which are
the long term values for shear strength for soft rocks as is suggested
for fissured clays.By reducing TP so that TP_Tr becomes small, the error
is also becoming smaller.
-55-
I
residual s·tre th
eak stren th
Curve 2I
qu a
Expected prototype peak shear strength- - - --
model
Figure 2.11 Load history effect on current peak shear strength.
-56-
2.3.2.2. stress vs strain.
Normal stress vs normal strain. (Fig.2.12)
Joint I
The law is a hyperbolic curve used first by Goodman(1975).
v =-V • ["vom mc 0(2.78)
["t is a negative very small stress
V is the maximum strain(positive) that can be attained from 0=[,,1.mc
00 is the negative initial stress.
V is the negative minimum closing strain for 0=00•m
The tangent stiffness is given then by,
k -=dO/d£ =OO~V ~(V ...s .)-2·C.,.1.0)'(_1-.0)=02/(Q,'V )n nn . m m nn' me
Joint 2
(2.79)
The law is a trilinear compression curve used by Goodman et al.(1978)
The normal stress vs normal strain space has been divided into three
regions,
£ <.V -V + 0=nn m me
V-V <£ <V + 0=00+kn·• £nnm mc nn m (2.80)
V ~ £ + 0=0m nn
Vm is defined here as the positive strain from 0= 00 to o=O,and is
given by,
V =-0 /km 0 n
Shear stress vs shear strain. Fig.2.13.
(2.81)
An elastoplastic multilinear relationship has been adopted between
shear stress and shear strain.Strain softening and hysteresis loops are
simulated with this law. The shear strain axis has been divided into 5
regions determined by the strains £rn'£pn'£pp'£rp that are the points
at which negative or positive,peak or residual strains are first attai-
ned. They are defined from the formulae below.
-57-
O.Vm(-'
Region II
a =Vm..'OnO/ (Vm..£.nn)
Vm=-Vmc"'~rh. on
.Jol:.----+Vmc( +)-)
Region I infinite stress
Region III zero stress
JointI - Hyperbolic compression curve
aVmc(+) ;=-t
Vm(+
Region II
a :OnO+kn.*e: nn
Vm=-OnO/k. n
e:nn
I II
,I......7, III
Joint2 - Trilinear compression curve
Figure 2.I2 Normal stress vs normal strain law
-58-
e pk=tP k/ks
rPk=TP.'perc+ Tr • (i-perc)
Shear stress vs shear strain Law.
The shear stiffness is taken zero if T is outside the elastic range.
Figure 2.13. Shear strain vs shear stress.
-59-
E =TP/k peak, positivepp s
E =M'E residual,positiverp pp M=4.0
E: =-E:pn pp peak, negative
E =-E residual, negativern rp
E =TPk/k reduced peak positive due to strain softeningpkp s
E =-E reduced peak negative due to strain softeningpkn pkp
Table 2.1. Shear strain regions
I 2 3 4 5 6
Region Min. strain Max.strain Name al.ope Id s) shear stress (T)
I - E negative 0 -Trn residual r
II E Epkn negative (T -T )/(E: -E ) -T +d '(E -E )rnfalling p r rn pn pk s s pkn
III Epkn Epkp elastic k k 'ES S s
IV Epkp E positive (T -T )/(E -E ) T +d '(E -E )rp falling p r pp rp pk s s pkp
V E - positive 0 Trpresidual r
_..- -- ~ ~
During the simulation of excavation or loading,having arrived at a
stable position (E ,Tl),after a number of iterations,this point willSl
represent the end of the load step.A new load step then will be applied
and a movement from (E ,T l) will occur which must be compatible withSl
the constitutive law illustrated in the figure.A number of displacements
will be tried then through iterations,that. will always refer to (E ,T l)Sl
till a new stable position is arrived.As can be seen, u/. is the plastic~
strain plus the initial strain due to residual stresses and uD
is the
elastic strain at the end of a load step.In Fig.2.I4 a 3-D view of Es'
T and 0 is illustrated, for a linear shear strength envelope and no strain
softening.If E lies in the elastic range,we move on the elastic plane froms
o to A.Otherwise we move on the plastic plane from
of the elastic origin.
B .Note the shift
-60-
T
o
",,//,-
"/ ,-,/ ,,- .
/ , /
" //
" /// /
"" /
//
,"
"
plastic plane
T=k s,,·£~
T P= ;\·0
defines elastic plane
defines plastic plane
Figure 2.14 Three diQensiona1 sketch for £s,T,O •
-61-
2.3.2.3. Normal strain vs shear strain (dilatancy).Fig.2.I5.
The normal strain is assumed to be comprised of two components,£nn
being independent of shear strain and dependent on normal stress and
£ dependent on shear strain and secondarily and indirectly throughns
shear strain to shear stress. The normal strain may thus be written
e =£ +£n nn ns
Dilation refers always to the initial shear strain which for reversed
loading might give unrealistic behaviour.It depends on normal stress
and shear strain. The equation is as follows
£ =a.·tani·£ns s
e =a. ·t.ani·£ns r
for £ <£s r
for £ ~£S r
The two models have different functions for a. •
Joint 1 assumes variation for a. such that dilation is the same as the
dilation used in the derivation of the failure criterion, i.e.
a.=(i.O-o/q )4 for o/q <1.0u u
a.=0 for o/q ~i.ou
Joint 2 assumes for a.
a.=1.0-o/q for o/q <1.0u u
a.=0 for o/q ~1.0u
(2.85a)
(2.85b)
This is not consistent with the failure equation assumption of constant
iJfor o/q <1.0 and i=O}for o/q >1.0.Nevertheless the bilinear failureu u
envelope is an approximation to a curved one and the variation of a.
used would be both physically as well as numerically more suitable.
Dilation}as a normal strain depending on normal and shear stres~
should introduce cross terms in the stiffness matrix.
st.John(Goodman and St.John, (197 7)) ,suggested a diagonal constitutive
matrix and correction for dilation in the next iteration to avoid asym-
metry of the matrix.Physically this can be explained as preventing all
the dilatancy on the adjacent elements applying external compressive
a-tani
-62-
Ens
dilation
£p=TP /k. se r=4. 0-£ p
max£ns=£r" a.··tani
£sshear strain
1.0
a =v/taniO. a L----- ..:::::::====_L =_
0.0
Joint1
La o /qu
a1..0
O. a '--------------=:::::..I:-....--_e-0.0
Joint2
La o /qu
Figure 2.15 Dilation - shear strain law for the two models.
-63-
stresses to the joint and in the next iteration withdrawing these stres
ses.In the approach used the idea of preserving the diagonality of the
constitutive matrix has been kept, but the diagonal components will be
modified as will be explained in the next section.
2.3.3. Iterative solution.
Application of the loads is through a sequence of load phases,hen
ceforth called activities and which correspond to an actual work phase,
as an excavation of a hole,or the installation of rock bolts.
The load corresponding to each activity is applied proportionally in a
number of steps,which have been chosen to be between 3 and 5,as sug
gested by Hittinger et al.(loc.cit.). Within each step an acceptable
solution, i.e. a displacement field compatible with the stress field,is
found,through a number of iterations.Hittingeret al.(loc.cit) sugges
ted 5 iterations per step.In the problems run we allowed sometimes a
maximum of 16 iterations per step for convergence to be achieved.
A check is made at the end of each iteration,whether all the unbalanced
loads, i.e. the difference between the loads at each node,found by the
elastic solution and those that can actually be carried by the struc
ture for the same displacement, are less than a certain threshold which
has been chosen to be between 1 and 5% of the expected load at the no
des.If the answer is positive,we proceed to the next step;otherwise we
proceed to the next iteration.If the number of iterations exceeds the
maximum number of iterations allowed, the analysis stops,and the dia
gnostic "The solution does not converge" is printed, indicating failure
of the structure.
The notion of iteration,step,activity and load sequence may be
written in set terminology as,
iteration~ step ~ activity C load sequence
where ~ is the symbol of 11 is .a subset or identical of".
-64-
A variable stiffness approach is used.The analysis is path independent
for iterations,but path dependent for load steps.After each iteration
the incremental displacements,strains,and stresses corresponding to
the elastic solution are added to the total displacements,strains and
stresses.Any unbalanced loads are added to the load vector for the
next iteration.
2.3.3.1. Normal stress vs normal strain.(Fig.2.I6,2.I7)
A new shooting point is sought for the next iteration and the un
balanced loads are added to the load vector.If no dilation exists the
approach is:
i.Joint closing and normal stress compressive.
The new shooting point is defined as a point with the same displace
mentJbut with an applied stress such that the tangent from the point
to the constitutive law curve touches the curve at the existing stress
level. The slope of the tangent is the new normal stiffness.The unbalan
ced normal stress then becomes,
~N=-(DELV-VREAL)·D22
ii.Joint opening or normal stress not compressive.
The new shooting point is defined as the point on the constitutive law
curve with the same strain.The normal stiffness then is defined as the
tangent at this point.The unbalanced stress then becomes,
~N=SIGMA-SIGM
The different definitions of the new shooting point in the above
cases are chosen to ensure the existence of that point.
2.3.3.2. Dilation.
If dilation is not zero,then the constitutive law curve will be
the one discussed previously augmented by the dilation,which will cause
a shift of the curve to the right(fig.2.I8 and 2.19).
shootin
(O,resid2)
-65-
o
J oint closing and 0< 0
\\ Elastic system\\ \ ,.,
(SIGMA,DELV) ....,---- (SIGM,VREAL)\ /,
/'/ DfLN=(VREAL-DELV).D2
(STRESS,DELV) ~ - - - Jrnew shooting point VREAL=VM.(SIGM-RESID2)/SIGM
I
oint.
c,
- - - - -~(SIGMA,DELV)
point on curvenew shooting point .
,,
, Elastic system.,SIGM=RESID2-VM/(VM- Erti
J oint opening or 0> 0
Figure 2.16 Iterative process(for compression)- Joint~,no dilation.
-66-
J oint closing and 0< 0
"-"
'\(SIGMA,DELV) 'f" rJ GM, VREAL)
II' DELNi-1~ Elastic system(STRESS,DELV)new shooting point
o
Elastic system(SIGM,VREAL)(STRESS,DELV)
~ - ..LSIGMA,DELV)DELN /'1 - - _ _ Enn
OLD SHOOTING POINTJoint opening or 0>0
o
Note :The stiffness in zone I is taken 10 4 -xksIThe stiffness in zone III is taken 10 -4-:'-xks i
Figure 2.17 Iterative process(for compression)- Joint2~no dilation.
>-%j
1-"
(J"Q ~ CD l\) . H 00 H c+ CD Ii Il' c+ 1-" < CD "0 Ii a o CD fJl
fJl
1;' a Ii o a .a Ii CD fJl
fJl
1-" a ::l ........
C-j a 1-" ::l c+ H ~ 1-" c+ ::r'
~ ..... I-'
II' c+ 1-" a ::l .
Join
tclo
sin
g.
"<
,
<,
o
..kDILAT~
I
, "I
Join
to
pen
ing
.,
Kla
s.ti
csy
stem
.
Eon
-,. -I I I
DIL
AT
--
--.,..
\
\ \ \ \E
lasti
csy
stem
.I 0'
-J
\I
---,-,~
I 0'
co I
(SlG
M,V
REA
L)Enn
ILN
IlR
sr'frO
~D
o--
-+:-
-(S
TRES
SDE
LV)I
~DI
LAT
-*'
Ela
stic
system~(C
-r--
~IGMA,DELV)
DELN
I
Join
top
enin
g
DIL
ATE
last
icsy
stem
I"'
Ell
at"
csy
stem
~s1
.(S
IGH
,.
Ii
DIL
N(S]J;i~'L~D]LP
__V
-r
__'
___~
~,
I(STRESSi
DEL~
DELN
---.1...
.....
--
-"
~D
IAT
;r-
I<K
1I
DIL
NI
~
DILA
T....
.....,
(SIG
M,V
REA
L)
--\/- /
//
/
/
1/
-;' I I
DILN
I I/'
I1
/+
-.J
/(STRESS,DEL~)
Jo
int
clo
sin
g
Ela
stic
syst
em
(SIG
MA
,DEL
vJ,.-
--D
EL
N*
~ o ~" ::s c+ I\) ~ c+ ::r p.,
~. I-'
P'
c+ ~" o ::s .I-xj~"
(JQ ~ (0 I\) . H <o H c+ (0 I-j P'
c+ ~" <;
(0 '"d I-j o a (0 til
til
,.-.
.H
) o I-j a o 3 '"d I-j
(0 til
til
~" o ::s .......... I
-69-
Joint 1..
The new shooting point is found now in a similar way on the augmented
curve. The stiffness matrix will now correspond to this augmented curve.
The total normal strain is the sum of one strain associated with normal
stress and one strain associated with shear displacement.
e =£ +£ , e =(o/q -1.0)4. t an i .£ £ = min( IE 1,£ ) (2.86)n nn ns ns us) S S r
The differential strains are calculated from,
d£ =d£ +d£n un ns (2.87)
d£ =(d£ /do)·donn nn
d£ =(a£ /aO)·dO+(a£ /a£ )·(a£ /aT)·dTns ns ns s s
The derivatives of strain are given by,
(a£ns/ao)=(4/qu)·«o/qu)-i.0)3.tan i ·£s=Fdi l
de /do=-f,,-eV /o2='1.0/Dnn me nn
d£ =(du/dT)·dT + F ·dOs ns
The cross flexibility F is given by,sn
(2.90)
(2.91)
(2.92)
F =(a£ /a£ )·(a£ faT)sn ns s s
and is zero for 1£ 1>£ •s r
TThe tangential stiffness matrix relating the strain vector (d£ ,d£ )s n
D =tang (2.94)
Fnsdu/dT
to the stress vector (dT,do)T is the following:-1.
a£ /do+d£ /dons un
Diagonality as described by St.John is attained by putting F = F =0 insn ns
the augmented stiffness matrix.The diagonal stiffness termS are given
by the inverses of the diagonal terms of equation 2.94.They are:
D =D /(1.0+D .Fd"l)nnn nn J.
D =dT/dus
where Dun,Fdi l are defined in eqs.2.91,2.90. _
-70-
Joint 2.
The total normal strain is the sum of one strain associated with normal
stress and one strain associated with shear displaeement.
E =E +En nn ns, E =a·tan i .£ =(1.0-0/q )-tan ins s u
--ES
E =(0-0 )/knn 0 n
Working similarly as for joint i we derive the stiffnesses as
dE Ido=i.O/k -tan i .£ /q =i.Ojk +Fd"ln n sun ~
D =do/dE = k /(1.0+k -Fd"l)n n n n ~
Fdil=-tan i .Es/qu
E is given by eq. 2.87s
The shear stiffness D =dT/du is unchanged.s
2.3.3.3. Shear stress vs shear strain.
(2.96)
(2.97)
(2.98)
At the end of each step a shooting point is defined.If the shear
strain in the next step is in the elastic region then the stiffness
is ks;if not the stiffness is taken to be zero.This has been found be
neficial for the problems analysed. The unbalanced shear stress is de-
fined by ~S=TAU(computed)-TOR(on the constitutive law curve).
Two simple illustrations of iteratively approaching an acceptable
solution are shown in Fig.2.20.The problem becomes more complex and
numerically slower if 0 and T vary simultaneously and the cross terms
become important.Very stiff joints might also create numerical pro-
blems.
-71-
T
To
-- do -
Block
Shear strain iterations under constant 0
o
-- 00 D
k2
Block Joir..t
Normal strain iterations under constant T
Figure 2.20 Iterations for simple examples.
-72-
2.3.4 Examples.
The two following examples are to test the convergence process of
strain softening joint elements with or without dilation. The strain
softening parameter BO
has always been taken 0.5 and the residual fri
Oction angle 10 .Plane strain elements have also been used.Young's mo-
dulus has been chosen to be 100 and Poisson's ratio O.
a.Twoplane strain and two joint elements.
The discretization is shown in Fig.2.2la.The two plane strain ele-
ments Pl,P2 are prestressed, the former in the horizontal direction with
an initial stress -1 and the latter with a vertical stress.Joint ele-
ments j 1,j2 have been prestressed by an initial normal stress -1.
A stable situation was sought through a number of iterations for vari-
ous material parameters of the joints and vertical stresses.Both joint
element models have been used.In table 2.2 the total displacements and
the maximum unbalanced 10ads(U.L) have been followed through the itera-
tions.In rows 1 to 4 the joint element model 1 has been used,whereas
in row 5 the joint element model 2 has been used.In row 1 it can be
seen that although shear displacements are within the elastic range,
they fluctuate after the second iteration without further convergence.
This is due to the lack of cross stiffness components of the joint e-
lement stiffness matrix and the dominance of the shear displacements.
In row 2 a larger initial vertical stress a and a lower strength qv u
have been chosen so that the total displacements lie within the plastic
range of the shear stress vs shear strain curve, but the normal stress
does not exceed q .The same phenomenon observed in row l,was observedu
also here.In row 3 q has been chosen small so that the normal stressJ u
exceeds q and hence dilation and strain softening do not occur. Theu
problem converged in two iterations.In row 4 the dilation has been cho-
sen to be zero but normal stresses did not exceed q .Shear displacementsu
-73-
%/ ""..- #/
fxr
13 14 15
P216
12l.0
+, 7(a) 7 6 5
j P1 j2
1+8
L 1 2 39
'" ~L 2.0 ;;f
(b)
Figure 2.21 Strain softening joints (examples)
-74-
Table 2.2 Two plane strain and two joint elements.0E=100,v=0,BO=0.5,¢=10 ,OH=-l,k =1.0,V =10.0,q /TO=10.0,~l=0.025.s mc u
row description Iteration Ver.disp. Max U.L. Comments
1 Model 1 1 0.487 0.438 Within elastic
q =100 2 0.486 0.065 range.Fluctua-u
3 0.509 0.067. 50l=
4 0.481 0.068tion of displa-
o =-1 5 0.507 0.061 cements due tov
6 0.481 0.068 lack of k terms.7 0.507 0.061
sn
2 Modell 1 0.973 0.547 In plastic range.
q =10 2 1.310 0.079 Displacementsu3 1.270 0.287
. 50l=4 1.190 0.314 fluctuate due to
° =-2 5 0.969 0.411 zero cross stif-v6 1.000 0.243 fnesses k ,k •7 0.846 0.329 sn ns
8 0.867 0.222
9 0.746 0.293
10 0.799 0.084
3 Model.l 0 1 0.973 0.153 In plastic range,q =1,l=5
2 1.200 0.001 and 0- beyond q •u n u(J =-2 I Iv
4 M!jdel ~ 0 1 0.973 0.278 In strain softeningq =10,l=0
2 1.370 0.047u° =-2 range.v3 1.450 0.005
5 Model 2 0 1 0.485 0.176 In strain softeningq =100,i=5 2 0.756 0.027 range.u
uP=0.267,ur=1.071o =-1 k =1v ' n •3 0.794 0.003
-75-
were in the falling part of the shear stress strainoCUI"ve.Convergence
was achieved within three iterations.In row 5 the shear displacements
were in the falling part of the shear stress strain curve.Convergence
was achieved within three iterations.
This analysis showed that the lack of cross stiffness components might
cause apart from slowing down the convergence process,divergence of an
actually stable problem. As far as strain softening is concerned there
were divergence problems when we chose negative stiffness for tangent
stiffness.By defining the tangent stiffness to be zero when negative,
these problems were overcome.
b.One plane strain and three. joint elements.
The discretization is shown in Fig.2.2lb.The plane strain element
Pl was initially prestressed horizontally with a horizontal stress
0H=-l.The three joint elements jl,j2,j3were also prestressed with a
normal stress a =-l.A stable position was sought through a number ofn
iterations for various material parameters for both types of the joints.
In table 2.3 the vertical displacement and the maximum unbalanced
load within the top joint and the two side joints was followed through
the iterations.In rows 1 and 2 the joint model 1 was used with diffe-
rent values of ~l so that the shear displacements of j oint elements jl
and j2 lie within or outside the elastic range. The number of iterations
needed for convergence is about the same showing at least for that e-
xample that dilation rather than plasticity or strain softening was the
main factor slowing down convergence.In row 3 the joint element model
2 has been used and convergence has been achieved within three itera-
tions,illustrating the quicker convergence of joint model 2 compared
with joint model 1 due to the linear nature of the normal stress strain
and dilation laws of the former. This is reversed when the normal dis-
placements are near the corners of the normal stress strain constituti-
ve law of joint model 2.
-76-
Table 2.3 One plane strain and three joint elements.
E=100,V=0,BO=O.5,$=100'OH=-1,O =-l,k =l.O,V =lO.O,q /T O=10.O,i=5?n s me u
row description Iter. Ver.displ. Maximum Unbal.load Comments
jl&j2 j3
1 Model 1 1 0.108 0.132 1.122 Within elastic
q =10 2 0.235 0.155 0.259 range.u
t,;l=0.025 3 0.289 0.045 0.018
4 0.291 0.0097 0.000
!
2 Model 1 1 0.827 0.222 0.066 Always out of
q =10 2 2.140 0.258 0.093 the elasticu
t,;l=l 3 4.460 0.114 0.149 range_the side
4 6.990 0.037 0.078 joints.
5 7.720 0.000 0.004
3 Model 2 1 0.658 0.603 0.000 Side joints
q =10 2 1.570 0.078 0.000 always inuk =1.0 3 1.690 0.003 0.000 plastic range.n
-77-
2.4 Change, of the geometry.
Change of the geometry results from excavation,i.e. removal of
structural material,and construction, i.e. the addition of structural
material.Simulation in the program is achieved through addition or
subtraction of elements. The method used is similar to the one used by
Hittinger et al.(1978).
2.4.1 Excavation.
Previous work includes,Clough and Duncan(1969),and Christian and
Wong(1973).They used interpolation functions for the stress field with
higher continuity than those for the displacement field,to take account
of stress concentrations.In the program this has not been implemented
and results might not be satisfactory}if such stress concentrations
occur near the excavation surface. The quadratic nature of the model
and a finer mesh would compensate for this.
The air elements suggested by Desai give ill conditioned stif-
fness matrices.
The method used is as follows:
a sNode s within the excavation area become inactive and f'Lxedv'I'hus
the number of d.o.f. is reduced and an identification array containing
the new number of each d.o.f. is formed.
b.Elements excavated but not lying on the excavation surface cease
to be active.
c.Elements excavated and lying on. the excavation surface,get zero
stiffness but continue to exist until the end of the activity,to unlo
ad their stresses on the excavation surface. This is achieved by calcu
lating the equivalent nodal forces along the surface bounding the ex
cavation,and applying them in the opposite direction to create a stress
free excavated surface.
-78-
These forces are calculated from,
nGP=-fB:a·ds~- E (B~a·J·w)~ - ~ p=l - ~ P
(2.99)
d.Add the incremental displacements, strains and stresses to the to-
tal ones.
e.lf there are unbalanced forces in the remaining elements iterate
with these forces as the load vector until convergence is reached.
All three types of finite elements,i.e. membrane, plane strain~and
joint,may be excavated by this method. The boundary element region dis-
cussed in the next chapter has not been programmed to be excavated al-
though it might be convenient sometimes.
Stress path dependency necessitates a number of steps of excavation in
order to obtain real deformation paths and to avoid numerical instabi-
lity,i.e. in one activity (excavation) several load steps are used to
impose the load •.The physical meaning of it is, "an excavation progres ...
ses in the direction of the tunnel axis;step by step more unloading
occurs as the problem transforms from three dimensional to two dimen-
sional".
The condition of an element i.e. it exists or it is excavated,or it
lies on the excavation surface,is characterized by a flag IFLAG.The
information stored in the files for an element depends on the value of
that f'Lags Lf' IFLAG is 0 the element exists.Tf it is I the element is
excavated but lies on the excavation surface.If it is 2 the element is
excavated. during this activity and lies not on the excavation surface.
If it~greater than or equal to 3 the element was excavated during a
previous activity.Some information always. remains for an excavated
element.
-79-
2.4.2. Construction.
The method is as follows:
a.Nodes within the excavation become freed and/or new nodes are ad
ded.Thus the number of d.c.f. is increasedpan identification array con
taining the new number of each d.o.f. is formed and the displacement
vector is lengthened. The first option to free a fixed node is prefera
ble as nodes can be numbered more efficiently.If the program is run
interactively and the eventual development of the mesh cannot be for
seen then new nodes have to be added.
b.Elements will be added.If they are of the same type as the previ
ous ones, they will be put in the same order of element type as the e
xisting ones.New types will be added at the end. The small displacement
theory is used and the added elements will be computed with dimensions
corresponding to the undeformed states.Stresses will be computed for
incremental displacements after their placement.For joints, strains are
total relative displacements between opposite sides of the element.
Hence if a joint is half connected to previously fixed nodes(new mesh)
and now freed,and half to the old mesh,it will inherit an initial strain
which can be removed by adding an equal and opposite initial stress.
-80-
2.5. Types of activities.
The aformentioned changes in geometry constitute important types
of activities treated in the program. Other types considered are gravi
tational loading,residual stresses ,pressure , concentrated forces and
quasi-static earthquake loading.Water flow drag and thermal effects
have been not considered. The loads considered have been combined in
four different ways to form four types of activities. They are:
a.Gravity,residual stresses and pressure,applied on each element.
b.Concentrated forces,applied on the nodes.
c.Activities a and b together.
d.Quas,istatic earthquake load,applied on each element.
Activity a is usually applied to consolidate the space,i.e. to
arrive at a situation where the stress field is the premining one.Care
must be taken during that stage that no plastic strains occur,or at
least that no strain softening occurs,as this stage is artificial and
plastic strains would alter the material properties.
When applying concentrated forces,it must be remembered that nodes
belong to quadratic elements and the stress distribution in the neigh
bourhood of the node would depend upon whether the node is midside or
corner.
Quasi-static earthquake load can be applied to all or to selected
plane strain or membrane elements. This type of loading is suitable for
limit equilibrium analyses also.
Due to path dependency(plastic behaviour) ,several steps are needed
for each activity,in order to obtain deformations approaching reality
and avoid such numerical problems as slow convergence and ill condi
tioning.
-81-
CHAPTER 3 - ELASTIC REGION
3.0 General,
The rock that lies at some distance from the excavation will
not undergo plastic strains ,neither it will exhibit large strains.
This region we intend to model as continuous and elastic.It must be
acknowledged that variability in initial stresses, orientation of
discontinuities,as well as other physical factors would cause this
region to be far from homogeneous,resulting in an increase of stif
fness with depth.As a first approximation we assume each elastic
region to be linear and homogeneous,with elastic properties those
of the rock mass near the excavation. The apparent elastic proper
ties of a large volume of rock, containing discontinuity features
such as joints, schistosity planes,cleavage or bedding,will hence
forth be called the equivalent elastic properties of the rock mass,
and will be dealt in Section 3.1.
The coefficients of a matrix that relates tractions to displa
cements for each such region is computed using a boundary element
program,discussed in Section 3.2.
The problems we intend to solve ,we assume to satisfy plane
strain conditions, the plane being perpendicular to axis 3.
-82-
3.1 Equivalent elastic properties of a jointed rock mass
We assume that the anisotropic rock material can be described
as orthotropic elastic. Two classes of discontinuity patterns can
be described this way. The first pertains to three orthogonal joint
sets with any deformational characteristics for each discontinuity
(Gerrard and Harrison(1970),Wardle and Gerrard (1972)~Ei$sa (1980),
HarrisonandGerrard (1972),Gerrard (1982 a,b,c,d),Pande and Gerrard
(1983) ).
The second pertains to two inclined sets of discontinuities
with the same material properties (Bray (1976)).
Other invesigators who have worked on the subject are Salamon
(1968),who assumed the rock mass to be transversely isotropic,and
Singh(1973) who dealt with discontinuous (staggered) joints,and in-
vestigated the problem in the context of composite material theory.
3.1.1 Three orthogonal sets of joints.
The equivalent properties of a rock mass crossed by three or-
thogonal sets of joints (fig. 3.1) b,d,f,that have normals parallel
to the axes 3,1,2, respectively,are given by the following formulae
(Gerrard (1982b):
Ela/El=l+Flld
E2a/E2=1+F22f
E3a/E3=1+F33b
~21a/v21=1+F22f
v31a/v31=1+F33b
v32a/v32=1+F33b
G12a/G12=1+F12d+F12f
-83-
material r;.~r::ff===1
material
,,I./Fr f
'-, ,.....'material b
3
I
Figure 3.1. Three orthogonal sets of joints.
2
n
joint 2
I
Figure 3.2. Two oblique sets of joints.
joint I
-84-
where subscript a denotes intact rock,and the denominators of the
left hand side of the equations are equivalent properties. The ad-
=ditional non-dimensional compliances F..k due to the joints are de
lJ
fined from the following formulae:
Flld=(Ela/knd)·Frd-prd
F22f=(E2a/knf)-Frf-prf
F33b=(E3a/knb)-Frb-prb
F12d=(G12a/ksd)-Frd-Prd
F12f=(G12a/ksf)-Frf-Prf
where k,k are the normal and shear stiffnesses of the joints d,n s
f,b, relating tractions to displacements,and Fr is the frequency of
the joints.We have introduced another multiplier to the~terms,Pr,
to take account of the persistence of the joints. This parameter
lies between 1 and 0 and is co~pletely empirical.
3.1.2 Two oblique sets of joints.
Two sets of joints are assumed,intersecting at an angle 8.J
(fig. 3_2)_ Bray(1976) showed that if the intact rock material is
isotropic and
cos 28.J
1/Knl-l/Kn2+1/Ksl-l/Ks2=
::"'1/~K::"'n-l--::"'1/~K::"'n-2-+=Kn-2-1""(~K-nl-·~K~s-2~)1:~K~n-1""/""(K~n-2-·~K-s--C-l )
where K=k/Fr, and Fr the frequency of the joint, then the material
may be represented as an equivalent orthotropic elastic continuum.
-85-
If F .. ,F~. are the components of the compliance matrix for planeJ.J J.J
strain for the,equiva.lent ortfio-tropic material and the isotropic
intact rock respectively,then according to Eissa (1980),
1 [(l+COS 2n)2 sin22n (I-cos 2a)2 sin22a ] ,F = - - + + + + F11
11 4 Kn2
Ks2
Knl
Ksl
F =!- f(....L - -L)-sin22n + (-.L - ---.l.-)-sin22a J +Fl'2 = F2l12 4 t Kn2 Ks2 Knl Ksl .
F =!.[(1-COS2n) 2 + sin22n (l+cos 2a)2 sin22a
]++ + F;222 4 K Ks2 Knl Ks ln2
sin22a cos 22n cos 22a sin22a
F33=+ + I (3.4)+ + F33K Ks2 Ksl Knln2
where
The derivation of the formulae shows that they hold also for
orthotropic intact rock material,whose principal axes coincide with
the principal axes of the joint system. In that case the intact rock
compliances F~. will take their orthotropic values.J.J
A special case arises if Ksl=Ks2 and Knl=Kn2• From equation
3.3 we observe that cos 28. can take any value,that is the equivaJ
lent continuum is orthotropic for any angle 8 .• This can be concludedJ
also directly,due to the existence of three orthogonal planes of
symmetry. These planes are the two planes bisecting the joints and
the plane of plane strain. Thus,
a=8./2 , n=n/2 -8. +a = n/2 -a, 2-n = n - 2-aJ J
SUbstituting to equations 3.4 we get,
Fl l=0_5 ·{(l-cos 2 rJI /K +sin22a/K }+F'n s 11
F12=0.5- {(l/K -l/K )- sin22 a }+ F'n s 12
. 2F22=0.5-{(1+cos 2a)IKn+sin22a/Ks + F~2
F33= (sin22a/Kn+cos22a/Ks) + F33
<3.6)
-86-
3.2 Implementation of the direct boundary integral method
The direct formulation of the boundary integral method is used
to develop a program for plane strain linear orthotropic elasticity
with quadratic boundary elements. The displacement and stress field
is divided into two components, the first called the complementary
function, that satisfies the homogeneous differential equation of
elasticity,and the second called the particular integral,which is
a particular solution of the differential equation and satisfies
the boundary conditions at infinity,if such conditions are imposed.
The total solution then would be the sum of the complementary func-
tion,the particular integral and the initial conditions. This may be
written as,
t cpou =u + u + u(3.7)
where u and t are the displacements and the tractions,and superscr-
ipts t,c,p,o stand for total,complementary,particular and initial
respectively.
The differential equation for elastostatics in terms of dis-
placements are those due to Navier for isotropy.
(3.8a)
where b is the body force,and A and ware the Lame constants. L*
is a linear differential operator. For general anisotropy the
equation would be of the form,
L*u=-b
L* is now a more general linear differential operator.
(3.8b)
-87-
3.2.1 The integral equation for the complementary function.
The complementary function satisfies the homogeneous equation
L*u=O
The boundary integral equation for that function may be obtained
from Betti's theore~ and the divergence theorem (Watson (1979)),or
distribution theory (Lachat (1975)) ,and is given by
J (U.. (x,y).t~(y)J.J J
S+s(x!E)
where
- T.. (x,y).u~(y)).dS = 0J.J J Y
x: the position vector of a point of the region and not at the
boundary.
y: the position vector of a point at the boundary.
S: the boundary surface.
S(XIE): the surface of an infinitesimal sphere around x with
radius E.
i,j: the directions of the first and second arguments of the
kernels respectively.
U.. : the singu1ar solution,that is the displacement at'y' on theJ.J
boundary in direction 'j',dueto a unit force acting at 'x'
in direct~on 'ire Note that U.. (x,y) is symmetric both inJ.J
arguments and indices for orthotropy,whereas for general ani-
sotropy it is symmetric only in arguments,i.e U.. (x,y);:U .. (x,y).J J. J.J
T.. : the traction at 'y' on the boundary in direction 'j',due toJ.J
a unit force at 'x' in direction 'it.
The definition of U and T can be written in an algebraic form,
u~(y)=U.. (x,y).e.(x) , L*U=oJ J.J J.
t':(y)=T .. (x,y).e.(x) T=O(U).~ = T(n).UJ J.J J.
(3.10)
(3.l1a)
-88-
where
e. (x)1
6
a(U)
n
a force acting at 'x' in the direction tit
is the Dirac's delta function
the stress field due to displacement field U
the normal, unit vector,to the boundary
a differential operator which for isotropy is given by,
T(n) =A.B.~. + w~.2 + ]l.!;.yT (3.llb)
From equation 3.9 we get the following equation, that is known
as Somigliana's identity.
u~(x)=fSU.. (x,y)·t?(y)·dS - fST . . (x,y)·u?(y)·dS1 lJ J Y lJ J Y
(3.12)
This equation may be used if the displacements at points within
the region are to be evaluated,after the values of t and u at the
boundary have been determined.
If 'x' defines a point on the boundary,equation 3.9 may be used
but integration has to be performed on the surface S +s(xls) - s(xls),
where
sex Is) is the part of the surface of the sphere with centre at
'x',and radius s ,that is contained within the region,
s(xls) : is that part of S contained within the sphere.
The integral equation is given by,
c .. (x).u?(x) + lim fT .. (x,y).U:(y).dS =lim Ju .. (x,y).t?(y).dSlJ J s+O lJ J Y s+O lJ J y
s-s(xls) s-s(xls) (3.13)
The integrals are Cauchy principal values.The coefficients c .. (x)lJ
are given by,
c .. (x) = lim f T.. (x,y).dSlJ s+O ( I )lJ Y. s x S
For a continuous tangent plane,
c .. =1/2·0 ..lJ J-J
where 0 .. the Kronecker delta.lJ
(3.l4a)
(3.l4b)
-89-
3.2.2 Kernels U and T.
These kernels are defined by equations 3.10 and 3.11. The
convention for position of the indices and arguments is arbitrary,
and in this work the convention used by Watson(1979) has been re-
tained. Banerjee and Butterfield (1980) have the indices and argu-
ments of the kernels interchanged. These definitions are shown in
fig. 3.3. Kernel U being symmetric will be the same in both cases.
For isotropy and plane strain the kernels are given by Lachat (1975):
(l+v) 1U.. (x,y)= ·{C3-4·v)·o ..• log(-) +
1J 4.'!T.E.(1-v) 1J r
1 x.-y.T.. (x,y)= .{(1-2.v).(n.(y).·J J
1J 4.'!T.(1-v).r 1 r
(x. -y . ) • (x. -y . )1 1 J J}
r 2
x.-y.n.(y). 1 1) +
J r
(x.-y.).(x.-y.) x -y+((1-2.v).o .. +2. 1 1 J .J)en (y). s s}
1J r2 s r
r is the distance between points x and y. Kernel U can be seen
from the formula to be symmetric in its arguments and indices.
The diagonal terms Eo log(l/r) ,whereas the off-diagonal terms are
pnoduct.s of'
Kernel TEo l/r
the direction cosines of ;.
The first bracket is antisymmetric in its subscr-
ipts and vanishes for the diagonal terms. The second bracket is
symmetric.
The formulae and their derivation for orthotropy for kernels
U and T are shown in Appendix 3,section A3.l.
-90-
send
Watson(I979) convention
U.• (x.y)1J
T.. (x.y)1J
Banerjee and Butterfield(I980) convention
G •• (y.x)J1
F .. (y.x)J1
U.. (x.y)=G .. (y.x)1J J 1
T.. (x.y)=F .. (y.x)1J J1
Figure 3.3 Conventions for kernel arguments.
-91...
3.2.3 Isoparametric element
The boundary element is a three node isoparametric one
(fig. 3.4). The tractions as well as the displacements are defined. ,
by the same shape functions as the geometry. This is written as
x . (I;) = Na(l;) -x~1 1
u. (I;) = Na(I;)_u~ (3.16)1 1
t. (I;) = Na(I;)_t~1 1
where subscripts i are the directions of orthotropy 1 and 2 and
superscripts a are the nodes 1,2,3 of the element.
The shape functions are given by,
N1(1;)= (1/2)-1;-(1;+1)
N2(1;)= (1/2)-1;-(1;-1) (3.17)
N3(1;) = 1-1;2
The derivatives D of the shape functions with respect to I; are
given by,
D1(1;) = I; + 0.5
D2 (1; ) = I; 0.5
D3(1;) = -2-1;
The Jacobian is given by
(3.18)
J(I;)=dS/dl;~{(dxl/dl;)2+ (dx2/dl;)2} = I{ (Da_x~)2+ (Da_x~)2}
(3.19)
mapping of line S to line ~1•~=-I
-92-
3•~=o
Element
2•
~=+I
Shape functions.
1.0
1.0
Figure 3.4 Isoparametric boundary element.
2
H
I
Figure 3.5 Coordinate systems H,V and-l,2.
-93-
3.2.4 Nodal collocation.
A system of simultaneous equations is written in terms of dis-
placements and tractions at nodes of boundary elements,that apprcoci-
mates the boundary integral equation. We assume that the directions
for displacement and traction at all the points are parallel with
a global cartesian coordinate system,this being chosen to coincide
with the axes of orthotropy,if any such axes exist. The equations
of nodal collocation then become
b=l e=l
p( a ) c( a) tc .. x ·u. x + L
1J J
3
L c( d(b,e))u. x •
J
p
= Lb=l
where
(3.20)
a is the collocation point number,i.e a node number that can
take values in the closed interval (l,q).
q the total number of nodes.
p the total number of elements.
b an element number.
e a node of element b,i.e l,or 2,or 3.
d(b,e): is the global node number of local node number e of ele-
ment b (l<d(b,e)<q).
The above equations can provide 2'q equations. The unknown quan-
tities may be more than 2'q if there are unknown tractions with dif-
ferent values either side of the nodes,that is when sharp corners
exist at the nodes.Many workers have investigated the problem of
providing the additional equations needed and a treatise on the
subject can be found in Banerjee and Butterfield (1980).For the
-94-
purpose of this work we considered it was sufficient to assume that
non-specified tractions were equal either side of the nodes. Errors
introduced due to this assumption are discussed in Chapter 4.
In accordance with established practice,the units of stress
and distance for the purpose of construction of the system of equa-
tions are taken to be a modulus of elasticity and the greatest di-
mension of the mesh. The choice of distance ensures that matrix U
of equation 3.22 is non-singular.
b=l e=lL sc
p
+ I3 C( d(b,e)) (I=")u. x a e J sI .J • J (T.. (x y(t;:)tL )·N (t;:).-·dt;:=
lJ' SC LS sc
b
p 3 t?(xd(b,e)) a e J(t;:)= I I J E • J (u .. (x ,y(t;:))·E )·N (t;:).--.dt;:
lJ sc Lscb=l e=l S sc
b
(3.21)
. c cwhere the complementary functl0ns u and t have been scaled in the above
equation by L ,that is the largest distance between nodes and E ,sc sc
that is the first Young's modulus. The above equation may be written
in matrix form as,
uc'"'"T·--Lsc
tC
= U·'::::"- E
sc
where uC and t C the displacement and traction vectors for all nodes- .....,
respectively, and T and U the matrices that contain the coefficients
of u: and t~ in equation 3.21-J J
-95-
3.2.5 Numerical integration.
The integrals of the kernel - shape function products of equa-
tion 3.21 can be evaluated by using Gauss-Legendre quadra-
ture formulae (G.L.Q.F),provideddue precautions are taken where the
integrand tends to infinity.If node xa does not belong to the element
over which integration . is performedJthe G.L.Q.F can be used direc-
tly for kernels U and T. Kernel components U12=U21 never tend to
infinity as they are independent of r and hence again G.L.Q.F may
be used directly. Table 3.1 shows the way the integration of the ker-
anels over an element is performedJif the node x belongs to the ele-
mente
Table 3.1 Integration of kernel - shape function products over anelement containing the first argument
Kernel order a is middle node; a is extreme node;Integration over two Integration over twosubelements adjacent elements
a=d{b,e) afd{b,e) a=d{b,e) ald{b,e)
T l/r spiral G.L.Q.F spiral G.L.Q.F
Ul l,U22 log{l/r) analytic analytic* analytic analytic*+ G.L.Q.F + G.L.Q.F + G.L.Q.F + G.L.Q.F
U12=U21 1 G.L.Q.F G.L.Q.F G.L.Q.F G.L.Q.F
Nd{b,g) ,........-- I ---..... t>-... ....<'1 c=::meea ! a a a .... db;ed{b,e) d{b,e) d{b,e) d'{b,e)
order ofNd{b,e)
at xa 1 1. r r
Note: An asterisk indicates that the use of only a G.L.Q.F suffices
The method of the spiral used to evaluate the cofficients of T and
the analytic method used to calculate coefficients of U,are shown in
Appendix 3, Section 3.2. A four point G.L.Q.F is used.
-96-
3.2.6 Rotation of axes.
The coefficients of matrices U and T have been calculated in the
principal directions of orthotropy 1 and 2. A rotation of axes is ne-
cessary for tractions and displacements to be related in the horizon-
tal,vertical coordinate system H , V.In the following we suppose va-
lues of u and t are scaled. Also quantities within parenthesis
with subscripts 12 or HV denote that their components are refer-
ring to the 1,2 or H,V coordinate system. Then,
where
-1S = U ·T 0.25)
A second order tensor coordinate transformation is needed in order
to rotate the axes 1,2 by an angle -8 ,to H,V_(fig. 3.5),
that is,
0.26)
where TR the vector coordinate transformation matrix. Then,
0.27)
Substituting 3.7 to 3.27 we get,
0.28)
-97-
3.2.7 Particular integral.
The particular integral for a distributed body force within the
region could be chosen to be,
u~(x) = f U.. (x,y)·b.(y)·dVol ,i,j = H,V1 Vol lJ J
This would entail the evaluation of volume integrals.¥or a uni-
form vertical body force _peg ,
- pegeh + p )o 0
(3.30)
where Po is the vertical stress at level hoJand KA
the ratio of
horizontal to vertical stresseTh;j.s-satisfies the equations of equili-
brium and the boundary conditions in the far field,and so the displa-
cements in the H,V system may be chosen to be,
(3.31)
where parameters a to E are independent of position. The evaluati-
on of volume in~rals is therefore avoided e
The values of the parameters and their derivation are shown in
Appendix 3,Section 3.3.
-98-
3.2.8 Infinite domain.
vfuen we deal with boundary element regions that extend to infi-
nity,it would be convinient if the inl1gration of the kernels U and
T with first arguments within the near field, over the remote
boundary could be avoided. Let us prove that the integrals over infi-
nity,
I l= lim f U.. (xa,y).t~(y).dsl.J J
r-klof° r
I 2= lim f T.. (xa,y).u~(y).dsl.J J
r-klof° r
i=1,2 j=1,2 •(3.32)
ain an infinite elastic body are independent of position of x. Let
us consider two points ax and xO (fig. 3.6),within a circle of ra-
dius R ,and an outer boundary f at a distancer °r O from x 0 We
denote by Ta , TO,Ua , UO th t b t t· th t t· the wo y wo ma r1.ces a con a1.n e com-
ponents T..l.J
and U..l.J
a,with first argument of the kernels at x or
°and second argument of the kernel at r Then,x .r
Ta_ TOd
d ° °/::,T = = (/::'ro- + t::,wo-- )T(r +81°/::,r , w +82o/::,w)
Ua_ uO drdW ° ° (3033)
/::,u '" = (/::'ro.Q. + /::,wo~ )U(r +83
o/::,r , w +84
o/::,w)dr dW
where we have assumed T and U to be functions of wand r,and
a °/::'r=r -r
a °t::,w=w -w
and
The orders of the individual terms of /::,T and /::,U are,
-99-
Figure 3.6 Integration over remote boundary.
>1
-100-
a 1-T[o-aw r O
But also !1w E 0 (l/r0)' hence
a 1l1re-U E o~-
ar rO
.1.. U E. o law
U.34a)
Hence l1T~ 0 (l/r~)
a,liwe - U E' (l/r0)
awU.34b)
(J.34c)
If tC
, uC
are equal to the singular solutions T and U,then
hence,
lim I l1T • •eu:eds = 0 ,l.J J
r ~ ro r
lim I liU .. et:edS = 0l.J J
U.36)
Now let us consider the entire, elastic space within r and ar
force field acting within rR • The total solution is the superposi~
tion of the effects of the individual forces. The displacements and
tractions due to forces em ,acting at points m, for rO~' are
given by,
U.37)
Hence the integrals II and 12 become,
\ 10m mL U.. eTk.eekeds
l.J Jm r
r
Noting that,
Um = U
O + liUm
m
1 0 m mT.. eUk.eekeds
l.J Jrr
and from equation 3.36
-101-
that the integrals of ~T_UO and ~U_TO are
zero, the expressions for the intgrals 11, 12 become,
I = I U O_T 0 -I em-ds1 ij kj k "
r mr
For an equilibrating force field,
I em = 0m
and the integrals become zero.
o 0IT . .-Uk'·J.J Jr
r
I e~-dsm
(3.40)
If the force field is non-equilibrating , then let us consider
the difference,
where
- \' eme = L
m
. 000 0-= Li.m I (U .. -Tk.-T .. -Uk.)-ekedsJ.J J J.J J
ro~ rr
Matrix A is antisymmetric.The diagonal terms hence are zero and the
non-diagonal terms in expanded form are given by,
We apply Betti's theorem for the fields due to unit force in 1 di-
rection and 2 direction. We note that,
) I (Tl l-U2l + T12-U22)eds
rrare the displacements due to field 1 multiplied to the tractions
due to field 2, and the displacements due to field 2 multiplied
to the tractions of field 1 ,on the remote
lye Hence,
boundary respectiva-
J A12
- d Sr
r
-102-
=
ou ..
J.Jis the displacement in j direction at xO due to a unitwhere
force in i direction at ox _
For orthotropy o au12 = u2l = 0 and hence the integral I is zero_00
If now the infinite body contains holes on which distributed non-
equilibrating tractions exist,this body may be considered without
the holes,by filling them and applying additional tractions to the
infillJso that the displacements on the hole boundaries remain the
sameo(this is in accordance with the indirect boundary element formula-
tion). Thus we arrive in the previous problem of a resultant force
within an infinite elastic body without holes for which the integral
lover a boundary approaching infinity is zero.00
-103-
3.3 Exampleo
A brick crossed by two oblique joints is subjected to self weight.
The displacements are calculated in two different ways.In the first
the body is assumed to be discontinuous and we model the discontinui-
ties with joint elements and the intact rock with plane strain ele-
mentso(fig. 3.7). In the second the brick'is modelled as an equivalent
continuum,with boundary elements (fig. 3.7).
In both configurations,nodes 5,6,7 are fixed in both directions,and
the vertical faces are fixed in the horizontal direction. The materi-
al parameters are:
Intact rock
E.=lOO , v=O , p·g=l.Ol.
Joints
k =20 , k =2.0 , Fr=0.57n s
The deformed shapes and flow field of the brick are shown in
figures 3.8 and 3.9.In the last figure the stresses at the centres
of the plane strain elements are also shown.
In table 3.2,values for the displacements at nodes are shown.
Table 3.2 Displacements at the nodes of a brick
Node Discontinuum Equivalent continuum
'\r uv '11: uv
1 0 -0.126 0 -0.146
9 0.007 -0.111 - -2 0 -0.143 0 -0.155
8 0 -0.104 0 -0.100
3
4
5
-104;"
2
6
1
8
7
~N~~~~eA~~S~LEMENT REGION SUBJECTED TOL~Ng~~~ITATIONAL ~E~Di.o_ID-1 UNITSACTIVITY 0
3
g 4.C\l
17 2 19
8
16 24.,::....-------------------_...:...-_-~
567
f------- 4. 00 - - - - - - - 1PLANE STRAIN AND JOINT ELEMENTS SUBJECTED TO A GRAVITATIONAL FIELD.INITIAL MESH LENGTHS _ = 1.0-10-1 UNITSACTIVITY 0
Figure 3.7 Initial meshes for the examples of section 3.3
-105-
SE~g~~~SR~E~~E"ENT REGION SUBJECTED TOL~Ng~~~ITATIONAL ~E~Di.0_l0-1 UNITSACTIVITY 1 LOAD STEP I ITERATION 1 DISPLACE"ENTS _= 1.0-10-1 UNITS
A BOUNDARY ELE"ENT REGION SUBJECTED TO A GRAVITATIONAL FIELD.FLOW FIELD LENGTHS . _ = 1.0-10-1 UNITSACTIVITY 1 LOAD STEP 1 ITERATION 1 DISPLACE"ENTS __ 1.0_10-1 UNITS
Figure 3.8 Boundary element region subjected to
gravitational field.
-106-
~~~~~M~~R~l~HAND JOINT ELEMENTS SUBJECl~2G~2SA GRAVITATION~LI~A~~g:1 UNITSACTIVITY I LOAD STEP I ITERATION I DISPLACEMENTS _ = 0.5-10-1 UNITS
l
J
1
\
~tS=EF~~~SIN AND JOINT ELEMENTS SUBJECl~2G~2SA GRAVITATION~LI~A~~g:1 UNITSACTIVITY I LOAD STEP I ITERATION I DISPLACEMENTS _ = 0.5-10-1 UNITS
•
t
PLANE STRAIN AND JOINT ELEMENTS SUBJECTED TO A GRAVITATIONAL FIELD.STRESS FIELD LENGTHS _ = 1.0-10-1 UNITSACTIVITY I LOAD STEP I ITERATION I STRESSES _ = 0.5_100 UNITS
Figure 3.9 Plane strain and joint elements subjected to
gravitational field.
-107-
CHAPTER 4 - COUPLING REGIONS WITH CONTINUOUS AND DISCONTINUOUS
DISPLACEMENT FIELDS
4.0 General
Finite elements, boundary elements as well as finite differences
have been identified for sometime now to have a common basis, and should
be used by engineers as allied tools rather than distinctly separate
methods.High stress concentrations or potential gradients,anisotropy,
infinite space,or large volume to surface ratio are areas where the
boundary integral method can be successful. On the other hand inhomo
geneities,non-linearities and plasticity are areas where the finite
element method can be successful.
The idea of coupling boundary and finite elements is attributed
to Wexler in 1972Jwho used integral equation solutions to represent
the unbounded field problem, the advantage being that this allowed for
the use of appropriate conditions to represent the infinite domain.
The first combination of the two methods in elastostatics is by Osias
in 1977,although for wave propagation problems the method was used by
Mei in 1975 by employing variational techniques. The idea developed by
Lachat(1975) of using interpolation functions to define the variables
along the elements allows for the combination of finite and boundary
elements without any loss of continuity.Shaw in 1978 used a weighted
residual procedure,so that a finite difference or finite element sys
tem of equations was obtained to describe the inner region of an infi
nite body,that was non-linear and inhomogeneous.He approximated the
outer region, that was linear and elastic,by deriving a boundary inte
gral equation around the interface boundary of the inner and outer re
gions in terms of the dependent variable and its derivatives. This in
tegral relation was a suitable boundary condition,with which to link
the finite difference or finite element approximation.
-108-
Brady(1981) was the first to use a combined method with finite and
boundary elements in rock mechanics. Beer(1982),by using his coupled
finite-boundary element algorithmJs~owed that although central proces-
sor time for using either only finite elements or coupled finite and
boundary elements was comparable, a significant gain was made by the
latter in the real time spent by the user of the program.Lorig(1982)
coupled discrete and boundary elements to simulate the behaviour of
excavations within jointed rock.
The combination of finite elements and boundary elements may be
achieved in two ways.In the first, that will be used in this chapter,
the boundary element region is treated as a"finite element region and
can thus be easily incorporated into existing finite element computer
packages (Zienkiewicz,Kelly,Bettess (1977) ; Kelly,Mustoe,Zienkiewicz
(1979) ; Mustoe (1979) ; Zienkiewicz (1975». In the second, the fini-
te element region is considered as a boundary element region.
4.1 SymmetriC coupling
The equivalent nodal forces for distributed tractions t on a fi-
nite element are given by,
T - T~=-fN .t.ds = -J! .~.ds.t = -C·t
TQ=J~ .!I.ds
This formula may be applied also for the boundary element region to
derive the equivalent nodal forces from the nodal tractions. By premu-
ltiplying then equation 3.28 by Q we get,
c·S-u = C.t
or
where
K =C·S-1 --
-109-
This stiffness matrix is derived without the use of a variational
principle and is non-symmetric.From energy considerations this is
inconsistent,as for various load paths leading to the same final load,
different energy requirements exist.Also finite element software usu-
ally assumes symmetric stiffness matrices.To generate a symmetric sys-
tem of equations from the direct boundary integral procedure,we requi-
re the minimization of an energy functional (Kelly et al.(1979) of
the form,
IT = 0.5.I uC.tc.dr - I uC.~.drr rwhere t C are the prescribed tractions on the boundary.
The physical meaning of IT is total potential energy,the first term on
the right handside is strain energy and the second term is work done
by the prescribed tractions. Discretization gives,
u = N.(~)·u.~ ~
t = N. (~).t.~ ~
, i=1,2,3
From Chapter 3 equation 3.27 we have,
c ci = §.~
Substituting 4.4 and 4.5 into 4.3 we get,
where in equation 4.6 we have dropped the superscript c.
By minimizing IT with respect to u we get,
Kt.£c + pc = 0
where
c= -C·t
Substituting for tC,uc from,~ ...-
(4.7)
(4.8a)
(4.8b)
-110-
we arrive at,
where,
Equation 4.10 is of the same form as the finite element stiffness
equation, and therefore Kt and pt can be assembled into the standard- ,..,
finite element system of Chapter 2,as contributions from a new ele-
mente
4.2 Validation
Validation of the program. is achieved through the analysis of
two series of problems .• The first series consists of problems analysed
also by Mustoe(1979), so that a direct comparison with a similar pro-
gram would be possible. In the second series the program is validated
further,by comparing the results with the known analytical solutions.
Series 1
a. Square block in tension
The block shown in fig.4.1 has been modelled as a boundary ele-
ment region,.and was subjected to tension. Young's modulus is 2.6 and
Poisson's ratio is 0.3. The expected displacements at node 5 are gi-
ven by,
I-v2 -v· (l+v) LUx=T .L.tx=1.4 , uy= E ."2. t x=- 0. 3
The displacements given by the analytical
(L the side of thesquare)
solution, the program and
by Mustoe(1979) are shown in table 4.1.
-lll~
6
8
2
4.0
Figure 4.1 Square block in tension
5
4
t =1.0x
x
a. Hole within infinite rock mass
surrounding rock
1Pe
b. Thick cylinder
Figure 4.2 A circular hole under pressure.
-112-
Table 4.1 Square block in tension
Analytical Program AJROCK Mustoe(1979)
Node u u u . u u ux y x y x Y1 0 0.3 0 0.332 0 0.336
2 0.7 0.3 0.716 0.284 0.720 0.283
3 1.4 0.3 1.420 0.325 1.417 0.324
4 1.4 0 1.460 0 1.457 0
b. Circular hole within infinite space and internal pressure.
The solution for a thick cylinder subjected to internal and ex-
ternal pressures p. and p (Fig.4.2b) is given by Obert and DuvallJ. e
(1967). Assuming plane. strain, infinite domain and external pressure
zero,we get the solution for the problem shown in Fig.4.2a. The dis-
placements and stresses at any point r are given by,
Ro = -0 = _p .• (_)2rr ee J. r
For the particular example the Young's modulus is 2.6 and the Pois-
son's ratio is 0.3.The radius of the tunnel is 2.0 and the internal
pressure is 1. O.
First we model the problem with only boundary elements as shown in
fig.4.3 using 12 elements.The displacements at nodes 1,2,3,4 are
shown in table 4.2.
We then model the same problem with both boundary and finite ele-
ments as shown in fig.4.4. The displacements at various nodes are
shown in table 4.3.
In table 4.4 the stresses at the Gauss points and the middle point
of the plane strain elements are shown.
-113-
surrounding rock.
3
2
1 >H
Figure 4.3 Hole within infinite rock mass modelled byboundary elements only. Vt
rock.
;>
H
8
...... .... 3 7__~~
Figure 4.4 Hole within infinite rock mass modelled byboundary and finite elements.
-114-
Table 4 .. 2 Circular hole modelled by boundary elements only ..
Analytical Program AJROCK Mustoe(1979)
Node u u u u u ux y x y x Y
1 1 .. 000 0.. 000 0.. 950 0.. 000 0.. 950 0.. 000
2 0.. 966 0.. 259 1 .. 000 0.. 268 1 .. 000 0.. 268
3 0..866 0.. 500 0.. 823 0.. 475 0.. 823 0.. 475
4 0.. 707 0.. 707 0.. 733 0.. 733 0.. 733 0.. 733
Table 4 .. 3 Circular hole modelled by boundary and finiteelements.- Displacements at nodes ..
Analytical Program AJROCK Mustoe (1979)
Node u u u u u ux y x y x Y
1 1..000 0.. 000 1..010 0.. 000 1..003 0.. 000
2 0.. 966 0.. 259 0.. 962 0 .. 258 0.. 965 0.. 259
3 0.. 866 0.. 500 0.. 876 0 .. 506 0.. 867 0.. 501
4 0.. 707 0.. 707 0.. 704 0 .. 704 0.. 707 0.. 707
25 0.. 800 0.. 000 0 .. 797 0.. 000 0.. 794 0.. 000
26 0.. 693 0.. 400 0.. 690 0.. 398 0.. 688 0.. 397
37 0.. 667 0.. 000 0.. 650 0.. 000 0.. 653 0.. 000
38 0.. 644 0.. 173 0.. 654 0.. 175 0.. 651 0 .. 174
39 0.. 578 0.. 333 0.. 563 0.. 325 0.. 566 0.. 327
40 0.. 472 0.. 472 0.. 479 0.. 479 0.. 477 0.. 477
Table 4.. 4 Circular hole modelled by boundary and finiteelements - Stresses within plane strain elements ..
Distance Analytical Program AJROCK
r a =-0 a aeerr ee rr
2.2113 -0 .. 818 -0 .. 820 0.822
2.. 7887 -0 .. 514 -0 .. 513 0.. 515
2.. 5000 -0 .. 640 -0 .. 638 0.. 637
-115-
c. Tension of a long plate.
A plate of 64 units length(fig.4.5a) and 16 units width is sub-
jected to uniform tension applied on face 2. Young's modulus is 2.6
and Poisson's ratio is 0.3. The movement in the x direction of face
4 is constrained. In fig. 4.5b the plate has been modelled by two
boundary element regions. In fig. 4.5c the plate has been modelled by
a finite element and a boundary element region. Finally in fig. 4.5d
the plate has been modelled with two boundary element regions that
form sharp corners on the interface at nodes A,D,and E. In table 4.5,
displacements for discretizations b and c calculated-by the program
are compared with the analytical solution to the example.
Table 4.5 Tension of a long plate modelled by sy.mmetric mesh
Analytical discretization b discretization c
Node u u u u u ux y x y x Y
A 11.2 -1.20 11.1 -1.21 11.2 -1.19
B 11.2 -0.90 11.1 -0.90 11.2 -0.91
C 11.2 -0.60 11.1 -0.59 11.2 -0.60
D 11.2 -0.30 11.2 -0.30 11.2 -0.30
E 11.2 0.00 11.1 0.00 11.2 0.00
F 22.4 -1.20 22.2 -1.23 22.2 -1.23
G 22.4 -0.90 22.3 -0.78 22.3 -0.78
H 22.4 -0.60 22.2 -0.56 22.2 -0.56
I 22.4 -0.30 22.5 -0.29 22.5 -0.29
J 22.4 0.00 22.2 0.00 22.2 0.00
-116-
yjtx
I (3)a.
16.0 (4) (2)
J-1
(1)
32.0 II 32.0 J1\
64.0
I A FI I I I
B Gb. ..c H
D I.. r-E -J
r-
I II I I I
c.
A
BC","",
1)
;;'
F
GHIJ
d.
E
Figure 4.5 Tension of a long plate.
D
F
G
H
I
J
K
L
-117-
In table 4.6 the displacements for discretization (d) ,calculated by
the program for various positions of node. D in the x direction (YD=-4),
are compared with results calculated by Mustoe(1979) and to the ana-
lytical solution.
Table 4.6 Tension of a long plate modelled by asymmetric mesh
Analytical Program AJROCK Mustoe (1979)
x =40 x =40 x =36 x =34 x =40D D D D D
Node u u u u u u u ux y x y y y x YA 11.2 -1.2 11.4 -2.29 -1.80 -1.15 11.18 -1.19
B 12.13 -0.6 12.3 -1.67 -1.21 -0.52 12.12 -0.56
C 13.07 0.0 13.3 -1.00 -0.65 0.07 13.04 -0.05
D 14.00 0.6 14.3 -0.17 -0.09 0.66 13.98 0.64
E 11.20 1.2 11.4 0.13 0.61 1.27 11.16 1.24
F 16.80 -1.2 16.8 -1 •.82 -2.15 -1.08
G 16.80 1.2 17.2 0.61 0.26 1.34
H 22.40 -1.2 22.4 -1.48 -2.57 -1.03
I 22.4 -0.6 22.5 -0.80 -0.19 -0.37
J 22.4 0.0 22.6 -0.24 -1.32 0.19
K 22.4 0.6 22.7 0.32 -0.76 0.75
L 22.4 1.2 22.7 0.98 -0.11 1.42
It can be seentha.t progra.mAJROCK rotates the plate,which can be
seen by the non-zero values for u at the nodes C and J.This errory
does not exist in Mustoe's results.If this rotation term is subtrac-
ted from the values of the other nodes on the same vertical line,the
results are reasonable.Nevertheless a vertical displacement of 1 at
node J,corresponds to a distributed vertical traction on face 2 equal
to 0.0025,that is an error in the applied traction of 0.25%.
-118-
Series 2
a. Lined circular tunnel within infinite space.
A lining rock system is examined. The rock medium is assumed to
be homogeneousisotrop~c and linearly elastic,and the lining material
is also assumed to be linear elastic with zero flexural stiffness.Two
extreme cases of interface conditions are considered.In the first ca-
se no slip is permitted and the lining is assumed perfectly bonded
to the rock mass.In the second case free slip is allowed between the
lining and the rock mass. The material properties are ,
Rock mass:
E=lO.O
v=0.20
Lining:
E ·t =20.0c c
v =0.00c
Interface (only second case):
k =1000.0n
k =1.0s
¢ =0.0
The diameter of the tunnel is 2 and the thickness t of the liningc
is negligible.Two discretizationsare used as shown in fig. 4.6a and
4.6b. The rock mass has been discretized with boundary and plane
strain elements in fig. 4.6a or with only boundary elements in fig.
4.6b.The lining has. been discretized with membrane elements.In the
case of perfect bond, the lining is acting directly on the rock mass.
In the case of free slip at the interface,the interface is modelled
with joint elements.
-119-
y f surrounding rock mass
(a)
surrounding ~ock mass
x
(b)
Figure 4.6 Lined opening.
-120-
The rock is assumed to be weightless and the overburden pressure p=
-1 and lateral pressure KA-p are applied in one load step)after the
hole has been excavated and the lining installed. In table 4.7 the
displacements for perfect bond at interface and three KA ratios are
compared at three points on the liningJas computed by the program
for the two discretizationsJand as given by the analytical solution
(Poulos and Davies (1974)).
In table 4.8 the displacements at three points on the rock mass
at the interface are shown for free slip between lining and rock
mass.Poisson's ratio was chosen to be 0.20,0.00,' and 0.333.The other
material properties and dimensions were the same as for the fully
bonded case. The stress ratio KA is chosen to be 0.5 and O. An ana
lysis for KA=1,V=0.20 is carried out also,and as no slip isoccu
ring,displacements are almost identical with those shown in table
4.7. For the discretization of fig. 4.6a displacements are shown
only f'or v =0.20.The analytical solution is given by the following
formulae:
u =0.5.(p/M).{(1+K ).(l-V).(l+( C-l) )-4.(1-K)· (1-V)2.cos 28}r Al-2·v -C+l A l-2-v
where
M=E·(I-V)/{(ltv)-(1-2·V)}
(D is the diameter of the tunnel)
These formulae do not agree with the formulae 11.38 and 11.39 given
in Poulos and Davies (1968) ,derived originally by Hoeg(1968) and
which are not correct.
An important conclusion is drawn from the analysis of this pro-
blem,involving all the types of elements available to the program.
The order of the Gauss formulae used initially in the calculations
-121-
of this example is 4 for the boundary elements,2 for the plane strain
elements,5 for the joint elements and three for the membrane elements.
This causes the computed stresses within. the joints to fluctuate a-
round a mean value within an element. and the analysis not to conver-
ge.Subsequently the Gauss formulae of the elements neighbouring the
joint were chosen to be of similar order and the analysis converged
within two iterations.
For the discretization of fig. 4.6a, we reduced the order of the
Gauss formula for the joint elements to 3. For the discretization
of fig. 4.6b a four point Gauss formula was used for all the ele-
ments. In the former case the fluctuation almost disappeared, where-
as in the latter it completely disappeared.
c. Excavation of a circular tunnel.
The displacements at the surface of a tunnel are given by Obert
and Duvall(1967) as,
Young's modulus and Poisson's ratio are chosen to be 10 and 0.0.
The radius of the tunnel R is 1.The vertical stress field p is takeny
to be -1.0, the horizontal stress field p being zero. The rock to bex
excavated is modelled by either 7 plane strain elements (fig. 4.7),
or 8 plane strain elements and a boundary element region (fig. 4.8)
The displacements obtained by the program for the two idealizations
before and after excavation, and those given by the analytical solu-
tion are shown for comparison in table 4.9.
-122-
surrounding rock.
B.E.
x
Figure 4.7 Excavation of a CirCU1;r'ttunnel(B.E.and 7 plane strain el.)
surrounding rock
B.E. x
Figure 4.8 Excavation of circular tunnel(2 B.E. regions and 8 P.S.el.)
-123-
Table 4.7 Lined circular tunnel with full adhesion on interface
Analytical Discretization a Discretization b
KA e u u u u u ux y x y x y
0 0.0347 0 0.0379 0 0.0343 0
,0.5 rr/L 0.0019 -0.0619 0.0013 -0.0608 0.0017 -0.0619
rr/~ 0 -0.1200 0 -0.1220 0 -0.1190
0 0.0565 0 -0.0561 0 -0.0563 0
1.0 rr/4 0.0399 -0.0399 -0.0397 -0.0397 -0.0398 -0.0398
rrt: 0 -0.0565 0 -0.0561 0 -0.0564
0 0.1257 0 0.1320 0 0.1250 0
0.0 rr/4 0.0438 -0.0837 0.0422 -0.0819 0.0436 -0.0837
rr/2 0 -0.1821 0 -0.1880 0 -0.1820
Table 4.8 Lined circular tunnel with free slip on interface
Analytical Discretization a Discretization b
V KA e u u u u u ux y x y x y
0 0.0500 0 0.0491 0
0.5 rr/4 0.0350 -0.1060 0.0338 -0.1060
rr/2 0 -0.1500 0 -0.15100.0
0 0.1670 0 0.1660 0
0.0 rr/4 0.1180 -0.1650 0.1160 -0.1640
rr/2 0 -0.2330 0 -0.2340
0 0.0537 0 0.0512 0 0.0531 0
0.5 rr/4 0.0380 -0.0978 0.0356 -0.0959 0.0368 -0.0977
rr/2 0 -0.1383 0 -0.1370 0 -0.13900.2
0 0.1638 0 0.1590 0 0.1640 0
0.0 rr/4 0.1158 -0.1557 0.1110 -0.1510 0.1140 -0.1550
rr/2 0 -0.2202 0 -0.2160 0 -0.2210
continued
-124-
Table 4.8 Continued.
Analytical Discr. a Discretization b
v KA 8 u u u u u uyx y x y x
0 0.0528 0 0.0525 0
0.5 rr/4 0.0402 0.0919 0.0364 -0.0884
rr/2 0 -0.1260 0 -0.1260
•.33~
0 0.1543 0 0.1540 0
0.0 rr/4 0.1080 -0.1430 0.1070 -0.1420
rr/2 0 -0.2031 0 -0.2030
Table 4.9 Excavation ofa circular tunnel
Before excavation After excavation
Analyt. a &'b Analytical a b
8 u u u u u u u u u ux y x y x y x y x y
0 o 0.0 0 0.0 0.1000 0.0000 0.,0988 0.000 0.0988 0.000
rr/8 0 0.0383 0 0.0383 0.0918 -0.115 0.0932 -0.116 0.0932 -0.116-
rr/4 0 0.0707 0 0.0707 0.0707 -0.212 0.0698 -0.207 0.0698 -0.207
3rr/f: o 0.0923 0 0.0924 0.0381 -0.277 0.0386 -0.281 0.0386 -0.281
rr/2 0 0.1000 0 0.1000 0.0000 -0.300 0.0000 -0.293 0.0000 -0.293
Note:
a corresponds to the discretization of figure 4.7
b corresponds to the discretization of figure 4.8
-125-
4.3 Inherent errors.
4.3.1 Causes of errors.
Some inherent errors have been determined to exist in the coup-
led model due to
i. Discontinuous tractions at nodes
ii. Symmetrization
iii. Dependent interpolants for u and t
iv. Error in numerical integration
i. Discontinuous tractions at nodes
If discontinuous tractions exist at nodesJthen the U matrix
is no longer square,the number of columns being greater than the
number of rows by a number equal to the additional tractions due to
the discontinuities. To make U square a number of equations equal
to the number of the additional unknowns needs to be added to the
system. A treatise on the subject has been written by Mustoe (19791
in which six ways to make U square are proposed as follows.
a. Take the tractions to be equal (fig. 4.9a) either side of the
node. This has been implemented in the program.We assumed that the
error tends to zero as the distance from the node increases.
b. Use the boundary interpolants on the two neighbouring elements
for the displacements to derive approximate expressions for the two
traction vectors at the corner in terms of the adjacent nodal dis
placements.This relation between t and u can be evaluated by propo
sing a corner finite element (fig.4.9b)
c. Alter the position of collocation so that extra equations may
be generated (fig. 4.9c). This method produces discontinuous tracti
ons at the nodes,but does not account for the continuity of the
-126-
e1. (r-)
a. Continuous traction assumption.
b. Corner finite element.
x are collocation points
c. Collocation points not at corner.
kk d. Gallerkin method weight functions.
Shape functions for geometryand displacements
e. Chaudonneret.
Shape functions for tractions
f. Differing shape functions for geometryand tractions.
Figure 4.9 Various methods to determine the limiting values of tractions
at the two sides of a corner.
-127-
stress tensor.It has been used successfully by Mustoe(1979).
d. Do a Galerkin weighting formulation. This method was suggested
by Mustoe.At the nodes of discontinuous tractions he used the two
parts of the shape function of that node over each adjacent element
as two separate weighting functions,to create the additional equati-
ons (fig.4.9d). This method is similar to method c.
e. Formulate two extra equations fom the continuity and symmetry
of the stress tensor, the invariance of the trace of the strain tensor
and an assumption for the variation of the displacements along the
boundary as found by Chaudonneret(1978). This method is similar in
concept with method b.
f. Use different order interpolants for tractions and displace-
ments.Mustoe using the shape functions shown in fig. 4.9f obtained
a singular matrix U.
A formula that evaluates the error due to the assumption of
continuous tractions at nodes of known discontinuous tractions is
derived in APpendix 4.
ii. Symmetrization
Let us take ~'and!l t o ' be the symmetrized and the non-
symmetrized stiffness matrices respectively. Then
where
TK = C • S-1 - -
Subtracting the first equation from the second and noting that
K' = (1/2)·(~1+~i)
we arrive at
( T) c c c c(1/2)· K -K ·u =fF =F -F-1 -1 .... ... ... 1 ....T
that is the error in the nodal forces found is proportional to ~l-~l'
-128-
which shows how far from symmetric matrix ~l is,and proportional to
the displacements uC•
N
The violation of the convergence criteria must also be examined. The
first criterion states "no straining of an element is permitted to
occur when the nodal displacements are caused by a rigid body dis-
placement".In order to ensure that the rows and columns of the matrix
K' for a finite region will sumtDzero,the following equations should
be satisfied.
.L: K/•• 0 A=1,2, ••• ,N/2 i=1,2, •••• ,N.j=2A-l lJ{Q.}= =
1 N is the number of degrees of freedom
L: 'K'.. 0 of the boundary element region.j=2A lJ
Another equation requiring that rigid body rotations do not cause
strains should also be satisfied.For the case of infinite regions
these sums are not zero. Care must be taken if the error is to be
spread over the stiffness terms,that symmetry is retained.
The simplest way to achieve this might be to subtract the error from
the diagonal terms.
iii. Independent interpolants
The direct boundary integral procedure assumes independent
interpolants for u and t.This is not physically consistent (Kelly
et al.(1979»,as definition of the boundary variation of onecomple-
tely defines the other through the solution of the boundary value
problem. Therefore the resultant energy distribution modelled by
(1/2)·f uC• t C
• ar cannot be correct.From that follows that the der- -rivativescHI/dl; are not calculated accurately and equation 4.7
contains an inherent source of error.In finite elements a similar
-129-
error exists,when we assume known the distribution of displacements
within the element. That is also not physically consistent as defini-
tion of displacements along the boundary completely defines the dis-
placements in the interior of the element through the solution of
the boundary value problem. This may be summarized in table 4.10.
Table 4.10 Prescribed values in finite and boundary elements..
~Displ. at bound. traction at bound. Displ. intO'
Method
Boundary integral prescribed prescribed -Finite element prescribed - prescribed
This error is probably a cause of the asymmetry of the directly
evaluated stiffness matrix K1•
iv. Error in numerical integration.
A four point Gauss-Legendre quadrature formula has been used in
the boundary integral module.As the functions to be integrated are
not simple polynomials,errors might exist due to incorrect integra-
tion.This error becomes particularly important if neighbouring ele-
ments are of different lengths (e.g. 1:4),in which case results might
be very inaccurate.
-130-
4.3.2 Examples.
Five example problems have been analysed to illustrate the mag-
nitude of errors. The first two examples have no corners,so that er-
rors are not due to corner effects. Examples c and d have corners,
but the shape functions are theoretically capable of modelling the
exact solution. The last example illustrates the errors due to great
discrepancy between paticular and total solutions.
a. Two boundary element regions
An infinite rock mass is divided by an imaginary circular con-
tour into an outer and an inner region (fig. 4.10).A uniform verti-
cal stress field of unit intensity is considered.Young's modulus is
taken to be 10 and Poisson's ratio O.A particular solution satisfy-
ing this field is applied to the external region.Two choices are
considered for the particular solution of the internal region.
Case 1 : The particular solution of the internal region is the
correct solution.
Case 2 The particular solution (uP,tP) of the internal region
is zero.
Theoretically both cases must give the same results. In table 4.11
the equivalent nodal forces and computed displacements for the two
cases are compared.The former are given by K'.uP-Pp•- ,.... ,....
The displacements given by case 1 are exact.Displacements given by
case 2 are in error by less than 2%.The equivalent nodal forces for
the internal region should be zero as no body forces exist. Never-
theless the correct displacements correspond to non-zero equiva-
lent nodal forces.
-131-
region
2
outer region
6
5 >H
1
Figure 4.10 Two boundary element regions.
j
'\
:>H
1
9
7Figure 4.11 Circular disc.
-132-
Table 4.11 Equivalent nodal forces and displacements forvarious particular solutions
Node 1 Node 2 Node 3 Node 4
Function H V H V H V H V
Eq.nod.for.extvregf.on o -0.486 0.063 -0.759 0.069 -0.344 0.152 -0.315both cases
Eq.nod.for.int.region o -0.030 ~0.002 0.027 0.004 -0.021 ~0.006 0.011case 1
Eq.nod.for.int.region 0 0 0 0 0 0 0 0case 2
Displacem.case 1 0 0.100 0 0.0924 0 0.0707 0 0.0383
Displacem.case 2 0 0.0981 0.0003 0.0935 0.0003 0.0694 0.0002 0.0387
Table 4.12 Equivalent nodalTforces given by use of stiffnessmatrices !.' '~l'~l.
Case and Node 1 Node 2 Node 3 Node 4
st.matrix H V H V H V H V
Case l,orcase 2 & K
l0 -2.743 1. 956 -4.715 1.940 -1. 940 4.715 -1. 956
Case 2 & KT 0 -3.442 1.688 -4.070 2.434 -2.434 4.070 -1.6881
Case 2 & K' 0 -3.093 1. 822 -4.393 2.187 -2.187 4.393 -1.822
-133-
b. Circular disc.
A circular disc (fig. 4.11) in a state of plane strain of radi-
us l~Young's modulus 1, and Poisson's ratio 0, is subjected to the
two following types of loading.
Case 1
Case 2
Uniform unit pressure normal to the circumference
Initial unit displacement normal to the circumference
The pressure is related to the radial displacement by the equation,
Inserting the known values for E,r,v we find p=uo at the circumfere-
nce,that is the equivalent nodal forces for both cases should be
equal. These nodal forces are given by ~,.~o - f .
In case 1 the equivalent nodal forces are independent of K'
and are termed here the correct equivalent nodal forces. The equiva-
lent nodal forces calculated for case 2,by using as stiffness matrix
Kl is calculated to be equal to the correct one.By using as stiffness
matrix~' or ~i the equivalent nodal forces differ from the correct
one. In the former case the discrepancy is about ±lO%.These results
may be seen in table 4.12. The displacements calculated in case 1
have a ±2% error when compared with the analytical solution.
c. Square block modelled with 32 boundary elements
A square block (fig. 4.12) of side 8,Young's modulus 10,Pois-
son's ratio 0 ,and unit weight equal to 1 ,is subjected to various
particular solutions and boundary conditions. Two parameters of the
particular solution are varied. The height at which zero stresses
occur is taken to be 1 and 8,the latter value corresponding to the
correct vertical stress distribution. The ratio KA of the particular
solution is taken to be 0,0.5 and l,the first value corresponding
to the correct lateral stresses.Face 1 is always restrained in the
-1.34-
..(2)
8.0
(I)>0-
f r H8.0
If iFigure4.12 Square block modelled with 32 boundary elements.
)' 8.0 ¥
H
8.0
/////
l- f-
~ ~ ~ a- o:7
Figure 4.13 Square block modelled with B.E. and P.S. elements.
-135-
vertical direction.Faces 2 and 4 are taken either free or restrained
in the horizontal direction.Face 3 is always free. The node at the
lower left corner is always restrained in both directions.
In table 4.13 the maximum and minimum displacements of face 3
in both directions are shown for the various cases. The correct ans-
wer is 3.20 for the vertical displacement and zero for the horizon-
tal displacement.
Table 4.13 Square block modelled by 32 boundary elements
Height KA uv . Uymax ~min ~maxFace 2 Face 4mJ.n
1 0.5 3.16 3.23 -0.0018 0.0100 fixed in H fixed in H
1 0.5 3.16 3.23 -0.0079 0.0106 free fixed in H
8 0.5 3.20 3.20 -0.0130 0.0248 free fixed in H
8 0.5 3.20 3.20 -0.0100 0.0141 fixed in H fixed in H
8 loG 3.20 3.20 -0.0100 0.0100 free fixed in H
1 0.0 3.16 3.23 -0.0100 0.0100 free fixed in H
1 0.0 3.16 3.23 -0.0100 0.0100 fixed in H fixed in H
8 0.0 3.20 3.20 0.0000 0.0000 free free
The error in the vertical displacements for incorrectly chosen
particular solution is about ±l% .The same block with face 2 fixed
in the H direction and face 4 free is subjected to unit tension on
face 3.In that case the displacements in the vertical direction
fluctuate between 0,805 and 0.795,the correct answer being 0.8,that
is an error of ±0.6%.
-136-
d. Square block modelled with boundary and finite elements.
A block (fig. 4.13) with the same dimensions and material pro-
perties as the block of example c is discretized with 8 plane strain
elements and one boundary element region. The block is subjected to
gravitational loading. The ratio KA
of the particular solution is
taken to be zero. The height of the particular solution,at which zero
stress occurs is taken to be 8 and 0, the former corresponding to the
correct stress and displacement distribution.
The computed displacements are shown in table 4.14.
Table 4.14 Square block modelled by finite and boundary elements
Height~m~
uv .~m~ ~minmln
8 3.20 3.20 0 0
0 3.36 3.23 0.1 -0.1
The large error in the .case in which the height is 0 is due to
the high value of the ratio of lengths of neighbouring elements,
especially near a corner, such as those on the top corners of the
boundary element region,where the ratio of the lengths of the
boundary elements is 2:l.These errors become of the order of the
previous example as soon as this ratio becomes close to l,and the
number of boundary elements increases.
-137-
e. Large problem.
The accuracy of the program is further tested by the analysis
of the large problem shown in fig. 4.14. The exterior infinite rock
mass is modelled as a boundary element region numbered BEl.The inte
rior intact rock is modelled by three boundary element regions num-
bered BE2,BE3,BE4,and 26 plane strain elements.
Young's modulus and Poisson's ratio are taken everywhere 100000 and
o respectively. The unit weight is 0.27-10-4 and the depth of point
o is 100000. The ratio of horizontal to vertical stress KA,takes
the values 0 and 1.In the case KA=l,no errors are expected due to
corners,whereas for KA=O the error due to corners is expected to be
very close to maximum.
The particular solution of the internal boundary element regi-
ons BE2,BE3,BE4is varied,by varying the values of KA
and hO
of equa-
tion 3.30 .Theoretically the results should always be the same, as
the sum of the particular solution and the complementary function
should give always the same total solution.
In tables 4.15 and 4.16 the calculated stresses at the centres of
the plane strain elements for various particular solutions of the
internal boundary element regions are compared with the analytical
solution.In table 4.15 the ratio KA of the infinite field (exterior
boundary element region BEl),and therefore of the total solution
is O.In columns 3 to 7 the particular solutiomof the internal
boundary element regions are varied,by varying KA and hO.In column
8,the calculated stresses at the same points are shown for the same
configuration, but by now modelling lines AA and BB with very strong
(¢=800) and very stiff (k =k =1000) joint elements.s n
-138-
(
Figure 4.14 Large problem with boundary and finite elements.
Tab
le4
.15
Str
esse
sat
the
cen
tres
of
the
pla
ne
stra
inel
emen
tso
fla
rge
prob
1em
.KA
ofth
eto
t.so
1.i
sO
.
Part
icu
lar
solu
tio
nin
pu
tfo
rin~ior
boun
dary
elem
ent
reg
ion
s
12
34
56
78
K=
0,K A
=0.
5K
=0
h=
106
K=
0h
=10
7K A
=0.
5Jo
ints
,k=
k=
103
hA=1
05
h=
5·10
LA
'0A
'0h
=10
7K
=0
h=
105
S
aa
aA
'0P
oin
tA
nal
yti
c-0
.-0
.-0
.-0
-0.
-0-0
.-0
.m
J.nn
uri
mJ.n
.m
axID
J.nm
axm
J.nm
J.n
C2
.34
92.
341
2.31
52.
725
2.04
56.
561
2.24
36.
617
2.3
47
D2.
402
2.39
62.
394
2.42
30.
077
3.09
70.
440
3.8
16
2.4
01
E2
.44
92.
448
2.4
50
2.4
30
0.03
82
.24
80.
413
2.26
12
.45
3
F2.
505
2.50
62
.50
72.
493
0.02
32
.35
90.
255
2.33
72
.51
0
G2
.56
02.
563
2.5
65
2.5
36
0.01
82.
262
0.19
42.
229
2.5
66
H2.
616
2.61
92
.62
02.
600
0.01
62.
413
0.16
32.
409
2.6
19
I2
.67
22.
673
2.6
74
2.6
36
0.00
72
.27
90.
795
2.19
42
.67
4
J2
.62
02.
619
2.6
18
2.65
30.
077
3.5
67
0.27
04.
743
2.6
19
K2
.45
92.
457
2.4
55
2.5
18
0.05
83
.22
00.
548
3.79
22
.45
6
L2
.29
92.
295
2.2
93
2.34
50.
067
3.05
50.
534
3.7
22
2.2
94
H2
.13
82.
136
2.1
36
2.12
20.
054
2.32
10.
256
2.77
12
.13
6
I I-'
VJ
'-0 I
Tab
le4
.16
Str
esse
sat
cen
tres
of
pla
ne
stra
inel
emen
tso
fla
rge
prob
lem
.K A
of
the
tota
lso
l.is
1
Part
icu
lar
solu
tio
nin
pu
tfo
rin
teri
or
boun
dary
elem
ent
reg
ion
s1
2'1
1.t)
67
K A=
1,h O
=100
000
K A=
0.5,
h O=5
0000
K A=
0.5,
h O=1
0000
000
K=
l,h
=10
6K
=l,
h=
107
A0
Aa
Po
int
An
aly
tic
-0-0
.-0
-0
.-0
-0
.-0
.-0
."m
axm
lnm
axm
lnm
axm
lnID
lnID
ln
C2.
349
2.34
02
.34
52.
273
2.3
38
5.16
08.
883
3.3
36
13
.24
8
D2.
402
2.39
52
.39
62.
363
2.3
99
1.9
15
5.5
67
2.84
57
.34
0
E2
.44
92.
449
2.4
52
2.4
40
2.4
50
2.26
13
.51
42.
608
4.1
77
F2.
505
2.50
62
.50
72
.50
12.
507
2.33
13
.09
52.
591
3.4
28
G2
.56
02.
562
2.5
64
2.5
59
2.5
66
2.27
82
.93
12.
613
3.1
14
H2.
616
2.61
82
.61
92
.61
42.
622
2.22
42
.99
82.
667
3.1
62
I2.
672
2.67
32
.67
32.
672
2.67
82.
057
2.6
99
2.68
02
.75
5
J2
.62
02.
619
2.6
21
2.58
22.
624
1.8
57
6.3
59
3.1
36
8.2
97
K2
.45
92.
456
2.4
61
2.43
22.
453
2.93
75
.08
62.
848
6.7
22
L2
.29
92.
294
2.3
01
2.27
32.
292
2.58
55
.06
02.
691
6.6
00
M2
.13
82.
135
2.1
39
2.11
52
.14
01
.54
54.
533
2.49
36
.03
5
I f-J
-l' o I
-141-
In table 4.16,the ratio KA of the infinite field is 1.In co
lumns 3 to 7 the particular solution of the internal boundary ele
ment regions is varied by varying KA and hO' as in table 4.15.
It can be seen that the stresses for particular solutions of
the interior boundary element regions near to the total solutions
are very satisfactory (columns 3 and 4). As the difference between
the particular solution and the total solution increases (columns
5 to 7),the results become less accurate,especially near sharp
corners.Hence care must be taken that the stresses due to theparti
cular solution and the total one are of similar order.
-142-
CHAPTER 5 - STABILITY OF AN OVERHANGING ROCK ~~DGE IN AN EXCAVATION
5.0 General
The behaviour of a wedge in a roof of a tunnel is governed by
its geometry , the mechanical parameters of the joints forming the we-
dge,the stresses in the rock mass,and the flexibility of the rock
mass.In section 5.1,the main parameters that govern the mechanism
of failure are identified~and the forces that create limit conditions
are evaluated.In section 5.2 the above mentioned forces are computed
by the use of numerical models,which can take account of a greater
number of parameUrs,and tables are produced in which closed form
solutions are compared with more sophisticated numerical ones.
5.1 Idealized behaviour (Fig. 5.1)
The logic adopted in this section is due to Bray (1975),and al-
lows calculation of the factor of safety of a rock wedge against fa-
ilure,and the reinforcement required. The assumptions rnadeare:
• The weight of the wedge does not act until after the excava-
tion has been completed.
• Blasting does not influence the forces acting on the joint.
• No vertical forces act,other than the weight and the support
force (i.e. no initial vertical force ).
The procedure is as follows:
• Assume that joints are initially infinitely stiff,so that the
rock mass may be regarded as continuous elastic and homogeneous.
• Carry out an elastic analysis,to determine the stress in the
crown of the excavation.Follow the usual procedure,whereby the weight
of the rock in the immediate vicinity of the opening is ignored.
• Take the stiffness of the joints to be reduced to values k ,s
k and take both of these to ben
small by comparison with the stif-
fness of the rest of the ro£k mass,so that the intact rock including
-143-
h
surrounding rock
Flexible joints
L
w
Rigid wedge
t vExcavation roor 11----------------)1
pA
A W
Figure 5. I Wedge ideal ization.
-144-
the wedge may be regarded as rigid.
• Apart from the forces at the joints,the wedge will be acted upon by
its own weight WJ and another applied force A (e.g. rock bolt in tension).
Let P be the resultant of Wand A ,i.e. P=W-A. (5.1)
• Find the magnitude of the force P acting in the same direction aso
P and replacing it,which will cause the wedge to be in a state of limit
equlibrium.
-• Assume two independent failure criteria for the joints.
i. The tensile strength of the joints is zero.
ii. The shear strength of the joints is purely frictional.
The factor of safety may be defined here as;
FS=I+(A-A )/W=l+c-co 0
where
c=A/W, CO=AO/W, AO=W-PO
5.1.1. Symmetric wedge (Fig. 5.2)
Due to symmetrY,only the half wedge needsto be considered.The force
with which the surrounding rock acts on the half wedge,before softening of
the joint starts and Po is applied (stage l),is HO.lt is assumed at this
stage that elastic behaviour is exhibited.This necessitates that a<¢.
After the joint deforms and Po is applied (stage 2),the force with which
the surrounding rock acts upon the half wedge is J.
The initial conditions require;
NO=H(!cOS a , SO= HO-san a
For a.j oint without dilation, the following system of equations needs to
be solved.
1 0 0 -k .cos a o l S sin as
0 I 0 k .sin a 0 N cos an
1 -tan ¢ 0 0 0 • H = HO• 0 (5.4)
-sin a cos a 1 0 0 d 0
-cos a sin a 0 0 I PO/2 0
-145-
~- L
Geometry and forces
without dilation
Force components
d
Roughness
d
Figure 5.2 Symmetric wedge - <j»a.
or in abbreviated form
-146-
CoX= H .y_..... 0 ...
where rows 1,2 are the constitutive equations for the joint, row 3 is the
failure criterion and rows 4,5 the equation of equilibriumo
The solution of the matrix equation gives;
PO/2=MoHO
H=PO/2ocot (~-a)
d=H osin (~-a)/(Dok )a n
where
D=cos aocos ~ok /k +sin aosin ~s n
M=(cos2aok/k +sin2a)osin (~-a)/Ds n
For a joint with dilationJthe two first rows of the matrix
(5.5)
(5.6)
equation 5.4
need to be modified. The normal displacement component is assumed to be
the sum of two components va and vd ,where vd is a function of shear
displacement,that causes dilation due to roughness,
(5.7)
va is a function of the applied normal stress only.Its relation to the
vertical movement is given by
(5.8)
Thus in equation 5.4 we put instead of C24=knosina
the value
C24=knsin (a-i)/cosi and resolve the system. This gives
P0/2=Mo HO
H=P0/2 °cot (~-a)
d=sin (~-a).HO/(kn.Docos i) (5.9)
N=(cos'aocos iok /k +sin (a-L) -s i,n a) -cos ~.HO/(Docos i)s n
S=(cosaa.cos i.k /k +sin (a-i).sin a)osin ~.HO/(D.cos i)s n
-147-
where,
D=cos a-cos ¢-k /k +sin ¢-sin (a-i)/cos is n(5.10)
M=(cos2a-cos i-k /k +sin (a-i)-sin a)-sin (¢-a)/(D-cos i)s n
For k <<Ie ,S n
P =2-H -sin (¢-a)-sin a/sin ¢a a (5.11)
i.e Po is independent of the dilation angle i.Note that ¢ is the total
friction angle. The case with dilation includes the case i=O, i.e a joint
without dilation. The nondimensional ratio M=PO/ (2HO) depends only on the
mechanical properties of the joint and its geometry,and it will be used
frequently in this chapter.
The mean horizontal stress is given by
The weight of the wedge is given by,
The resultant force Po is given by,
PO=W-AO=(1-c O)-0.5.p-g-L-horP =2-H -M=2-h-o ·Ma 0 HO
Equating the last two equations
By defining the stress concentration factor Scf as,
z is the depth of the excavation, equation 5.15 becomes,
l-c =4- (a/L) -S f-Ma c
Scf depends on the geometry of the excavation and the
(5.14a)
(5.14b)
(5.15)
-148-
stress ratio KA ' and may be assumed independent of depth for z/L > 3,
where we have assumed L of similar magnitude to the largest vertical di-
mension of the opening. This factor can be obtained directly from tables
(e.g Eissa (1980),page 139), for various shapes,positions,and KA•
Equation 5.16 uncouples the various contributions to the carrying capacity
of the joint.
Friction angle p <a (Fig. 5.3)
If the previous equations are used,then a negative carrying capacity will
be required for limit equilibrium. This is because at the end of stage 1
we have passed the limit equilibrium and a negative force Po is required
to keep the force acting on the wedge by the surrounding rock on the fai-
lure envelope.If slip is allowed to occur,then row 1 in equation 5.4 is
no longer applicable making the system of equations indeterminate.If we
define N=NO
' then no total movement occur-a, but plastic shear movenent of
the joint occurs •. This will corespond to force P 12 in the figure.omax
For N= 0 ,PO will. become zero, i.e for ¢ < a at least the whole weight
of the wedge must be supported.
Behaviour of. resistance force P~
Graphs relating M to the angle a for various friction angles, stiffness
ratios,and dilation angles are shown in Appendix 5.They indicate that
the most important of the above mentioned parameters is the stiffness
ratio,whichis also the most difficult to determine.For very low values
of k/kn,M is no liDnger. a monotone decreasing function of angle a.This
means that if wedges of various angles are considered,an angle may be
found, below which the resistance will become smaller.As an example let
us consider a wedge with constant base length L=lOxl03,k Ik =O.Ol,i=O,s n
¢=400,pg=27xlO-6.The required horizontal stress at limit equilibrium be-
comes: a =peg.L/(4eM)=O.0675 eM.HO
d
-149-
1h Wedge
~P 1,1t L ;(
Force components
N
Displacement VB normal force VB horizontal force
Figure 5.3 Symmetric wedge - ¢<a
-150-
Let us take the angle a to be 10~20~30~Then M'OHO ~ and FS vary as
follows:
a M °Ho FS(OHo=0.355)
10 0 0.167 0.404 0.878
20 0 0.190 0.355 1.000
30 0 0.136 0.496 0.715
The factor of safety if no support force exists is given by)
FS=l-co =Po/w =4M.oHO /(pg~) = °HO·M/0.0675
In column 4 in the table above the factors of safety for 0HO=0.355 are
shown. In the diagram~Fig 5.4~the behaviour of these three wedges is
shown. This has also been confirmed by the program and the results can be
seen in table 5.1 •
For k /k ::::0 IS n
M=sina ·sin(¢-a)/sin¢
dM/da=(cosa • sin (¢-a) - sin a·cos (¢-a»/sin ¢
For obtaining an extremum we put dM/da=O.This results in
tana=tan(¢-a) 'or a=¢/2
This can be seen geometrically(Fig. 5.5)as the path of point C on the
circumference of a, circle)at which the line OH subtends an angle 180-¢.o
The maximum is achieved when the triangle COH becomes isoceles ,o
i.e a=¢-a.Another extremum (minimum)can be identified in the graphs at
very small angles a.The curves tend to smoothen as k /k becomes largers n
or ¢ becomes smaller. The distance between points C and B depends on the
sngLe e where) for i=O,
zero.M isand .a=O
e~arctan(k /k ·cot a)s n
k /k #0- and a=O + e=rr/2 and M=P / (2· H )=tan ¢s n 0 0 •
k !k =0s n
For
For
This is because it is imp.ossible to achieve
-151-
k. /k =0'.01s ne=arctan((k /k )·cota)
o s n"i=3.2459
oe =1. 5738
m 0e
h=O.3308
H: Wedge with 0a=30
M: Wedge with 0a=20
L: Wedge with a=IOO
are lines for k /k =0S !l
Figure 5.4 Examples for very low stiffness ratio joints.
Low
Medium
circle
o
e=arctan«k /k ) ·cota)s n
-152-.-: -------- --,.,IPomax/2 ~ " /;I .r: ,
" II
I
I/
II
I_ -- _ Po/2 I
/I
II
/
Figure 5.5 Behaviour of symmetric rigid wedge.
-153-
a shear for~e S through movement,and if there were non zero S lit wouldo
be impossible to change it.Also No due to a=O would not change.
llexibility of the wedge
If the flexibility of the ' ..irltact 'rock of the wedge is significant with
respect to the flexibility of the joint,then the stability of the wedge
is affected.Flexibility in the direction of the joint causes partial yield
of the joint surface before failure, decreasing thus the final load that
can be sustained.On the other hand flexibility in the horizontal direction
increases the stability of the wedge ,because it acts to increase the
apparent normal flexibility of the joint and hence increase the angle e.
This is illustrated in Fig. 5.6 .The simplest model that could reproduce
flexible behaviour of a wedge is shown in the upper part of figure 5.7.
A system of 10 simultaneous equations must be solved in the 10 unknown
quantities,ioe two values for each N,S,H and one value for each R,dx'd y'
P • The determination of k,k needs engineering jud~ent.If the wedgeo r n
is assumed to fail simultaneously on the whole face,then the simpler
model shown in the lower part of the figure 5.7 may be appropriate.
The system of equations to solve the problem if dilation is excluded is,
1 a 0 k ·sin a -k ·cos a a S sin as s
a 1 a k ·cos a k ·sin a a N os an na a 1 -k a a H
H •1 (5.18)h
1 -tan <l> a a a a d 0 ax-sin a -cos a 1 a a a d ay-cos a sin a a a a 1 P /2 a
0
or in abbreviated form, C·X=H .y- - 0-
where the first two rows are the constitutive equations of the joint.
In row 3 is,the constitutive equation of the wedge.Row 4 is the failure
criterion.Rows 5 and 6 are the equilibrium equations.
-154-
Increarein wedge stability due to horizontal flexibility.
Decrease in wedge stability due to flexibility parallel to the joints.
Figure 5.6 Effect of intact rock flexibility.
-155-
Variation of stress along the joint.
Kinematiks
AH
Constant stress along the joint.
Figure 5.7 Models for elastic wedge.
-156-
Performing Gaussian elimination in Eq. 5.18 ,we get
a' b' a d r 5x
c'+~ d' 0 - d = H - 0Y 0
-d' f' 1 P /2 00
Solving 5.19 we get
d d'x
=dY
P /2o
d /dx Y
-~~+c')
-d ,2 -c' -f' -f' -~_
= -d' / (~+c t)
(5.20)
(5.21)
where the parameters , b' c' d' f'a, , , , ,r5 have the following values:
a'=k-(-(k /k )-sin a + cos a-tan ~)n . s n
a-cos a
b'=k -((k /k )·cos an s n
c'=k -((k /k )·sin2an s n
d'=k -(l-k /k )-sinn s n
f'=-k -(k /k cos2an s n
+ sin a·tan ~)
c'-k -ks n
(5.22)
For ~=00 , we get
P /2=H -f'-r /b'005
d /d =d =0x Y xand
which is the same equation as the first of equations 5.5 ,for the
stiff wedge.The enhancement in stability due to flexibility may be
shown by the ratio P /f5 ,o 0
d,2/f'+c'+~
-a'-d'/b'+c'+~
=
a'-d'-b'-c'-b'-~ f'---------P /p =
o 0
-157-
The function,
a'-f' + b'-d' = -k /k -(tan $-COS a - sin a)-k2s n n
is negative for ¢>a.
Hence
-f'<
b'(5.26)
Also f' is negative, and a',b',c',d' are positive for
k /k < 1 • Hence from 5.24 and 5.26 we find that,s n
p j'p > 1o 0
i.e. the stability of the wedge increases due to the flexibility
of the wedge , if this model is appropriate.
-158-
Stress redistribution Fig. 5.8
Let us assume a hydrostatic stress field not varying with depth and exa-
mine the half. portion of a symmetric wedge.OA is the force on the joint
before excavation, which for the hydrostatic stress field would be normal
to the joint.We may propose that the stiffness of the excavated material
is reduced to zero but the stresses on the excavation face remain.
Thus no movement will have occured till now.Then,let us propose that a
traction is applied slowly on the.·8xcavation surface in the opposite
sense to the existing tractions and propor~ional to them.If the joint
flexibility is very large by comparison with the flexibility of the rock
mass, the stress app.Lf.ed.: on the j oint will depend on the wedge movement
as stresses due to excavation will be redistributed outside the wedge
(path AD).On the other hand if the flexibility of the joints is zero
then an elastic solution without taking into account the. JOints would be
appropriate (path AB).In the case where the stiffnesses of the joint and
intact rock are of similar magnitude we may suggest that an intermediate
path is followed from A to C which would be a straight line if the
stiffnesses remain constant.
Let us consider the case ¢ > a.By applying to the excavation face the
tractions that will make the force on the joint horizontal,we equilibra-
te the forces acting on the joint due to the vertical stress field, but
not the weight of the wedge.H in Bray)s theory corresponds to OB ,o
whereas fora very flexible joint,H corresponds to OD.Bray)s theory aco
cepts that the joint now softens and from B we arrive at E' on the fai-
lure envelope at an angle e,specified in Fig. 5.4. ,from the normal to
the joint.For the very flexible joint from D we continue to D' at the
same angle e.
If the joint stiffness is of some significance, then we presume that we
will follow an intermediate path and meet the failure env.81ope at a point
envelope
o
o
<1>< a
*D
-159-
*C
A
Failure
..... ....... .....
*E *F B
Figure 5.8 Stress redistribution.
-160-
such as C' .the (positive)carrying capacity being between the extreme
values D~D* and E'E*.
Thus we see that the joint carrying capacity depends on the relative
stiffness of the rock mass and joint. because this determines the relati-
ve position of the path.whereas e determines the limiting paths. As e
is always positive by definition. from point B we can not arrive at a
point such as C'. by specifYing appropriate values for k /ks n
Nevertheless the assumption made by Bray seems very reasonable because
joint stiffness is not constant but reduces continuously.this being taken
into account implicitly by Bray using two joint stiffness values.
In this respect. the introduction of a loosening factor LF which would
multiply H could be suggested,to bring point B at a point such as E.o
It must be noticed here that in some cases,e.g. a small wedge in a large
opening,wedges may be subjected to much reduced horizontal stresses, that
might be also tensile, due to bending ,and point B might lie even to the
left of point D.
Let us now consider the case 4>< a.In this case we meet the failure en-
velope before the force becomes horizontal at E' for infinitely stiff"
joints in the direction towards B,) corresponding to the elastic solution...
or D' for infinitely flexible joints.Again it is presumed that for joint
stiffness of the same order as the rock stiffness an intermediate path
AC will be followed,that would be linear for constant stiffness.Note
here that F'F*corresponds to P / 2 in Fig. 5.3 .This value we see re-max
duces now even. for infinite.ly stiff joints to E'E* which shows the signi-
ficance of the load path. (Points E',F' correspond to different distribu-
- tions of tractions around the excavation face.)
According to the stated logic, the plastic movement will start after we
\ve reached the line E'D' .In all cases we can propose for design pu~~
~s a zero resistance of the joint,requiring only full support of the
-161-
wedge 10ad.We expect less plastic movement as we move from F' to D',
so this could be an indicator of potential explosive failure,especi-
ally if strain softening is anticipated.
A verification of these thoughts might be given by experimental test
results obtained by Crawford and Bray(1982),that show failure stresses
substantially lower than predicted by Bray's theory.
5.1.2 Asymmetric wedge.
Let us consider a rigid non-symmetric wedge in a horizontal tunnel
roof (Fig.5.9),within infinitely stiff rock.The forces acting on the
wedge faces are not equal and displacement will no longer be vertical.
A rotation that might also occur has not been considered. Two critical
stages exist; first yield and failure.
Before first yield, all deformations are elastic and at first yield
one face reaches yield.At failure both faces exhibit yield,but befo-
re failure one unyielded face exists.In the force component diagram
of Fig.5.9 the mirror image of the forces acting on face 2 is shown
together· with..the forces acting on face 1.Vectors HoDl and HoD2 are
the differences between the forces acting initially and those acting
at first yield.Joint shear and normal displacements are defined in
terms of wedge displacements as follows:
-sin
cos
sin
-cos ::}[::J(5.27)
Let us suppose that face 2 yields first,as shown in the figure,while
face 1 remains elastic. The system of equations to solve is
-162-
Force diagram.
--
Displacement components.
II
I
I
1
//
dxFigure 5.9 Asymmetric rigid wedge.
o 1 0 0
o 0 1 aa 0 0 1
o 0 1 -tan ¢2
sin a1 cos a1 -sin a2 -cos a2cos aI-sin a1 cos a2 -sin a2
1 a a a
-163-
-k -cosslkn1-sin
-k -coss2kn2-sinooo
a1 ks1-sin a1a1 kn1-cos a1a2 -ks2-sin a2a2 -kn2-cos a2
ooo
o Slo N
1a S2
a - N2o d
Yo dx
-1 Po
cos
sin
=H - cosooao
(5.28)
where the first four rows are the constitutive laws for the two joints,
row 5 is the failure condition on face 2,and the last two rows are
the equilibrium equations of the whole wedge.
Performing Gaussian elimination the system of equations becomes,
aY bY 0 d r YY 5
cY dY 0 - d = H - 0x 0
eY fY -1 P 00
Solving this system we get,
dY
dx
Po
=(5.30)
d /d =-cY/dYx Y
SY=arc tan (-cY/dY)
where the parameters aY to r Y are given by,5
aY=kS2-Cos a2+kn2-sin a2-tan ¢2
bY=ks2-sin a2-kn2-cos a2-tan ¢2
cY= -(kn1-ks1)-sin aI-COS a1 + (kn2-ks2)-sin a2-cos a2
dY=-ks1-sin~a1-kn1-cos2a1-ks2-sin2a2-kn2-cos2a2
eY=k -cos 2a +k < -sin~a +k· -cos 2a·+k -sin2a ::'::L:k+dYsl 1 n1 1 s2 < 2n2 2.
fY=(k -k )-sin a -cos a -(k -k )-sin a -cos a =-cYn1 sIll n2 s2 2 2
r~= sin (¢2-a2)/cOS ¢2
(5.31)
-164-
The angles Y Y are given by,el ' e2
eY = arc tan ((ksl/knl)·cot (al+~))1 (5.32)eY ~= arc tan ((ks2/kn2)·cot (a2-S))2
These angles are independent of the friction angles and can be used
to find graphically or analytically the yield force,if the failure
envelope is multilinear or non-linear.
The equations for al=a2'¢1=¢2,ksl=ks2,knl=kn2 give cY=O and
d =d /d =0 d =H ·rY/aY P =H .eY.rY/aY M=eY.rY/aY (5.33)x x Y , Y 0 5 ' 0 0 5' 5
which are the same as equation 5.5 for the symmetric wedge.
The ratio cY/dY is a measure of the asymmetry of the wedge.If this is
small)then equation 5.33 may be applicable.
By interchanging the indices 1 and 2 we get values for P coro
responding to first yield on face 1.The lower value is to be chosen.
Usually the flatter face,that is the face with larger angle a,will
yield first.
Let us suppose now that face 2 has yielded first.It is assumed that
shear deformations on face 2 are plastic, but normal deformations con-
tinue to be elastic,so that their resultant lies on the failure en-
velope.Vectors DlCl,D2e2 are the changes in forces acting in the two
faces from yield until failure.To form the new system of equations,
corresponding to failure,we have to replace in equations 5.28 the
third row that corresponds to elasticity in the shear direction of
face 2 by an equation that specifies yield on face 1 at failure.The
system of equations then becomes:
-165-
1 a a a -k °cos al kslosin al a Sl sin alsl
0 1 a 0 knlosin al knlocos al a Nl cos al
a a 1 -tan <1>2 0 0 a S2 a
0 0 0 1 kn20sin a2 -k °cos a2o 0 N2
=H 0 cos a2n2 0
1 -tan <1>1 0 0 0 a a d 0Y
sin al cos aI-sin a2-cos a2 a a a d ax
cos al -sin al cos a2-sin a2 0 0 -1 P a0
(5.35)
Performing Gaussian elimination,
f bf a d fa r?yf df
0 d = H 0 f (5.36)c 0 r6x 0
f ff -1 P fe r 70
Solving this system we get
a
o
f f f f f f fa of -e °b c °b -a °d
Ho
df
f= 0 -c
afodf_cfobf f f f fe od -c of
dY
dx
Po
f f f f "f f f fdx/dy=(-c or5+a or6)/(d or5-b or6)
Sf=arc tan (d /d )x y
wh~re the parameters _af until T~ are "given by,
fa =kslocos al+knlosin alotan ~
bf=_k osin a +k °cos a otan <I>sl 1 nl 1 1
cf=_(k -k )osin a °cos a +k osin ao(cos a2+sin a20tan <1>2)nl sIll n2 2
df=-kslosin2al-knlocos2al-kn2°cos a2o(cos a2+sin a2
0tan <1>2)
ef=kslocos2al+knloSin2al+kn2oSin a2~Sin a2-cos a20tan <1>2) (5.38)
ff=(knl-ksl)osin alocos al-kn2 ocos a2o(sin a2-cos a20tan <1>2)
r~=Sin (<I>l-al)/cos <1>1
r~=sin a20sin (<I>2-a2)/cOS <1>2
r~=-cos a20sin (<I>2-a2)/cOS <1>2
-166-
f f fThe angles el ' e2 are dependent on both friction angles. el is given
by,
The value P for failure might sometimes be smaller than its valueo
for yield. This happens if the line joining the middle points of DID2
and CIC2 is sloping downwards from D to C.This willcause a brittle ty
pe of failure when first yield occurs.If this line is sloping upwards
a ductile type behaviour happens, the horizontal distance between D and
C being a measure of the ductility. This ductility for a symmetric we-
dge is zero and hence we conclude that symmetric wedges always exhibit
brittle failure.
For an oblique wedge(Fig.5.10),two force components P and P mustp n
be considered.P is parallel to the wedge face and except for veryp
shallow excavations,is small by comparison with H and may be ignored.o
P is normal to the wedge face.Calculations may now proceed as for then
wedge with the face horizontal, but with al,a2 as shown in the figure.
This analysis has not been validated.
Apart from the symmetric wedge,the wedges might have an important
rotational component, which so far has not been taken into account.
The simplest idealization of the problem is the one shown.in fig.5.11,
where we assume contact points between the wedge and the surrounding
rock at nodes A,B,C,D •.For failure to occur,yield must occur at all
four points. The ultimate load and displacement can be found by follo-
wing successive yield happening at the nodes. The procedure is similar
to the one used for analysing the asymmetric wedge.A system of 12 equ-
ations in 12 unknowns must be solved each time.The unknowns are the
-167-
dy
Figure 5.10 Oblique wedge.
·(tan az-tan al)/2
h·(tan az-tan al)/6
h-(tan az-tan al)/3
Figure 5.1 I Wedge with rotation.
h
-168-
three displacement components d ,d ,d and two force components forx y w
each of the four nodes. The equations are the three equilibrium equati-
ons of the wedge,the four constitutive equations relating normal force
to normal displacement and 5 equations from the 4 yield conditions
and the four constitutive laws relating shear force to shear displa-
cement.
-169-
5.2 Numerical solution
A series of numerical analyses has been carried out to validate
and to give an insight of the applicability of the simplified solutions
proposed in the previous section. Various wedges with ¢>a and i=O we-
re subjected to progressively increasing distributed pullout loads.
This load was applied in steps of magnitude ~W=~c·w.
The total load applied before failure has occured is P =(l-c )·W.o 0
From now on we drop the subscripts (0). The following subscripts are
used:
min Value corresponds to the last converging step of the numerical
computation.
max Value corresponds to the first non-converging step of the
numerical computation.
gr Value is calculated analytically (from graphs of appendix 5)
c Value is calculated numerically by the program.
th Value has been calculated analytically from theoretical solu-
tion. (asymmetric wedge)
y Value corresponds to yield (asymmetric wedge )
f Value corresponds to failure.
BS The value of H has been calculated by an elastic solution,o
in order to do the rest of the calculations analytically.
Ho The value of H has been calculated by the program in ordero
to do the rest of the calculations analytically.
-170-
5.2.1 Symmetric wedge.
Two types of pullout test analyses were carried out. The first ty-
pe of analysis consisted of wedges of varying flexibility, within an
infinitely stiff rock maas, subjected to a horizontal stress field. In
the second type of analysis the same wedges were considered to be em-
bedded in a flexible rock mass. The stiffness of the intact rock varies
from very stiff to very soft thus introducing a new parameter not ta-
ken into account in the formulae of section 5.1.
Infinitely stiff outer rock mass.
The surrounding rock mass (Fig.5.12) is represented by the fixed nodes
1 to 17.The elastic wedge is modelled by 7 plane strain elements PI
to P7.The joints are modelled by eight joint elements jl to js' one
face of each of which is fixed to the surrounding rigid rock mass.
An initial horizontal stress is applied to the system together with
the weight of the wedge. Then additional distributed load proportional
to the weight of the wedge is applied in the vertical direction until
failure occurs. The horizontal stress is applied through an initial
stress within the joint and plane strain elements.
50 ° ° 400,The angle a is taken to be ,20 ,35 , the. friction angle ¢ is
and the dilation angle i is 0°.
In table 5.1 numerically cal.culated results for 9 almost rigid
wedges (E=5000 GPa) surrounded by rigid rock are shown and the cal
culated value of M in column 9,M =P /(2.H ) is compared with thatc c 0
predicted by the simplified theory of section 5.1,in column 10.
The stiffness ratio is taken to be 0.001,0.01,0.1 for each value of a.
The total force P is applied in up to 10 steps. The last converging
step is n,and P is the corresponding load. M. corresponds to Pn m~n n
and M to P +1. M is given bymax n
-171-
P2 42
45
Js
iVb
P4
P5
46
40
4136
PI
4
2
I 18 35 44 34 17
Figure 5.12 Symmetric flexible wedge within rigid rock.(WDG series)
Tab
le5
.1S
ymm
etri
cal
mo
stri
gid
wed
gew
ith
inin
fin
itely
sti
ffro
ck.
E=
50oI
05,
$=40
0,i
=0
0,
k n=IO
,L=
IOoI
03,0
HO
=0
.42
/co
sa,k
R=
0.O
I,0
.IO
,I.0
0.
Seri
es
WDG
___
•
I2
3a3b
45
67
89
10II
ak
.Ik
aT
PB
t .10
4M
min
Mm
ax~c
Mc
Mgr
Com
men
tss
nn
050.
001
0.15
60.
040
3000
0.20
7514
0.06
20
.09
30.
025
0.08
70.
087
050
.01
00
.II9
0.07
169
00"
0.14
40
.16
70.
014
0.15
70
.15
8F
ail
ure
was
050
.10
00.
277
0.19
719
800
"0
.4II
0.4
79
0.05
20.
463
0.46
3si
mu
ltan
eou
sin
200.
001
0.17
80.
145
2180
0.81
4334
0.1
79
0.20
30.
005
0.18
30.
183
all
join
tele
-
200.
010
0.19
30.
150
2170
"0
.17
80
.20
20.
012
0.18
90.
190
men
ts.
200.
100
0.22
60.
190
2940
"0.
239
0.2
74
0.00
10.
240
0.24
1
350.
001
0.30
40.
241
400
1.3
65
65
80.
055
0.0
82
0.02
30
.07
80
.07
8
350.
010
0.29
10.
242
540
"0.
075
0.0
86
0.00
30.
077
0.0
78
350.
100
0.29
80.
247
560
"0
.07
60
.08
60.
004
0.08
00.
080
Not
eth
at
the
mea
ning
of
the
ind
ices
of
that
tab
leis
asfo
llo
ws:
min
:V
alue
corr
esp
on
ds
toth
ela
st
con
ver
gin
gst
epo
fth
en
um
eric
alco
mp
uta
tio
nm
ax:
Val
ueco
rres
po
nd
sto
the
firs
tn
on
-co
nv
erg
ing
step
of
the
nu
mer
ical
com
pu
tati
on
c:
Val
ueex
trap
ola
ted
from
nu
mer
ical
resu
ltg
r:
Val
ueca
lcu
late
dan
aly
ticall
y
I I--'
....::l
I\) I
The initial
-173-
horizontal stress is
The value of M lies between M. and M and can be found by extra-c m1n max
polation from.M .• The stresses within the joint at the end of loadm1n
step n O(n) and L(n) do not vary along its length/due to the high ri
gidity of the intact rock. These values are known and shown in column
3 of the table. Then if ~d is the additional movement needed to cause
failure,
Solving for k ·~d,n
The additional load then provided would be
The additional value for M would be given by,
The value of M is calculated fromc
M = M. + ~~c m1n c
As can be seen from the table , computed (column 9) and theoretically
predicted (column 10) values of M are almost identical.
Validation for Mmi n can be made from the known o(n) and L(n) as,
Mmi n = p~tan a/(oHo·L) = (T·COS a-o·sin a)/(oHo·cos a) (5.47)
From the table we may observe that for values of k /k <0.01 the coms n
puted values of M are not monotone decreasing functions of a,
which was also predicted by the theory of section 5.1 (also example
of figure 5.4).
-174-
The same three wedge geometries are analysed by the program again,
now taking the stiffness of the intact rock to have the realistic
value 100 GPa. The joint shear stiffnesses are taken to be 0.2,1.0,
and 2.0. The normal joint stiffness is taken to be 20. The rest of
the material parameters are taken to be the same as for the very rigid
wedge.
A horizontal stress field of -2.7 MPa is applied through initial
stresses within the plane strain and joint elements. Then a load
is applied in steps of magnitude 6c·W. The value of 6c is 1.
In table 5.2 the values of (I-c) characterizing the failure capacity
of the wedge as computed by the program (column 4) and as predicted
by the simplified theory (column 5) are compared. The value in column 4
cor-reaponds-, to the last converging step of the program. Thus the
correct value for (I-c) lies between (I-c) . and (I-c). + c. Duem1n m1n
to the lower flexibility of the intact rock, the values of stresses
along the joints were not constant,so that no extrapolation could be
performed. The relation between M and (I-c) is
(1-c)=40·M
Elastic surrounding rock mass
The surrounding rock mass (Fig.5.13) is modelled by boundary ele-
ments,the nodes of which are numbered 32 to 59. The angle a is taken
oto be 20 • All combinations of three stiffness ratios and three Young's
moduli are examined.Two types of problems were considered.
In the first (Fig.5.13a) the hole already exists,and horizontal stress
is applied through initial stresses in the 8 joint elements and the
8 plane strain elements and through far field stresses in the bounda-
ry element region. A load proportional to the weight of the wedge is
then applied downwards until failure occurs. In table 5.3 the values
of (I-c) for the last converging step are shown in column 6 for com-
a.N
oex
cav
atio
nis
per
form
ed.
b.E
xca
vat
ion
sequ
ence
isp
erfo
rmed
.
I ..... -..J
VI
I
13
73
7.
b1
ja
58
PI
P2
,P
3,
P4
b2
j7
5II
32
59
bli
t
jIO
,
55
Pa
P1
0
P1
2
.....
63 '"
60
3I
It
62
40
P72a
53
54
2
61
P9 Pll
b6 j3
~b
12
L..~10000.~
49 51
50
4a
ba
jI
b7 j2
b9 j9
blO
Bou
ndar
yel
emen
t
reg
ion
.
Ela
stic
spac
e;
Ho
le.
-1--10
000y
ll' II /'V aaI; f-J.
c+ ::r f-J. =' CD I-'
ll' til c+ f-J.
Q Ii o Q ~
"%j
f-J.
O'Q ~ Ii CD VI . H v
.J
......
.C
/)
~~
3:
St"
"'CD c+
tilIi
CDf-
J.Ii
Qf-
J. CDCD
tilI-
'll' til c+ f-
J.Q ~ P
.O'
Q CD til
Tab
le5
.2S
ymm
etri
cela
stic
wed
gew
ith
inri
gid
surr
ou
nd
ing
rock
.
o=-2.7,E=100.103,p.g=27.10-~~c=I.0,¢=400,i=0.0,k
=20
,L=
1000
0,1.
0-c=
40.M
HOn
-6
I2
34
5
ang
lek
k/k
(I.O
-c)
.(I
.O-c
)ex
Ms
sn
mJ.n
gr
0.2
0.01
66
.32
0.15
8
05°
1.0
0.0
513
13
.92
0.3
48
2.0
0.1
018
18
.52
0.46
3
0.2
0.01
77
.60
0.19
0
20°
1.0
0.0
58
8.6
00.
215
2.0
0.1
09
9.6
40.
241
0.2
0.01
33
.12
0.07
8
35°
1.0
0.0
53
3.1
60.
079
2.0
0.1
03
3.2
00.
080
Not
eth
at
the
mea
ning
of
the
ind
ices
of
that
tab
leis
asfo
llo
ws:
min
:V
alue
corr
esp
on
ds
tola
st
con
ver
gin
gst
ep
of
the
nu
mer
ical
com
pu
tati
on
gr
:V
alue
isca
lcu
late
dan
aly
tica
lly
<(g
rap
hs)
.
I I-'
-:l
0'
I
Tab
le5
.3.
Sym
met
ric
ela
stic
wed
gew
ith
inela
stic
rock
,w
ith
ou
tex
cav
atio
nse
quen
ce.
O.
-60
i=O
O,
kn=
20,
h=I3
737,
L=IO
OO
O,
W=I
854
-S
erie
sTA
D-T
CF.
a=20
,p·
g=27
·IO
,~c=I,
¢=40
,
I2
34
56
78
9
Tes
tna
me
ksk
s/k
nE
4IO
-3°H
O(I
-c)m
inM
gr(I
-c)g
rF
ailu
rest
art
sat
TAD
0.2
O.O
IIO
O.
2.8
77
0.I
90
8.0
8at
the
top
TAE
1.0
0.05
IOO
.3
.I5
II0
.2I5
IO.0
3at
the
top
TAF
2.0
O.I
OIO
O.
3.2
7I3
0.2
4I
II.6
8at
low
elem
.
TBD
0.2
O.O
IIO
.O3
.36
II0
.I9
09
.46
at
the
top
.
TBE
1.0
0.0
5IO
.O3
.60
I30
.2I5
II.4
6at
bott
om.
TBF
2.0
O.I
OIO
.O3
.64
>I3
0.2
4I
I3.0
0at
low
elem
.
TCD
0.2
o.or
I.O
O3
.65
>II
0.I
90
IO.2
7at
low
elem
.
TCE
1.0
0.05
1.0
03
.72
I30
.2I5
II.8
5at
low
elem
.
TCF
2.0
O.I
O1
.00
3.7
2>
I20
.24
II3
.28
at
low
eLe
m,
Not
e:In
dic
es"g
r"in
dic
ate
that
val
ues
are
com
pute
dfr
oman
aly
tical
solu
tio
n(g
rap
hs)
Ind
ices
"min
"in
dic
ate
that
val
ues
are
com
pute
dn
um
eric
ally
and
corr
esp
on
dto
the
last
con
ver
gin
gst
ep
.
I I--'
...:J
...:J I
-178-
parison with the values predicted by the simplified theory,shown in
column 8.The horizontal stress acting before the vertical load is ap-
plied is shown in column 5. The value of (I-c) is computed fromgr
l4.8l.(OH -M )o gr
In the second type of problem (Fig. 5.l3b) a nearly hydrostatic stress
field is created initially by applying far field stresses as well as
initial stresses in the 12 plane strain elements and the 10 joint e-
lements. An excavation is then performed by removing elements P9 to
P12 and j9 to jlO so that a free face of the wedge is exposed.
Then a downward load proportional to the weight of the wedge is appli-
ed until failure occurs. In table 5.4 computed results are compared
with the results predicted by the simplified theory.
In column 5 are shown the average horizontal and vertical tractions
in the joints before excavation. In column 6 are average horizontal
tractions on the joint after excavation,assuming infinitely rigid jo-
ints and using elastic theory. In column 7 are the average horizontal
tractions acting on the joint after excavation, as calculated by the
program(the difference between columns 6 and 7 being due to the dif-
ferent flexibilities of the joints).In columns 8,9 and 10 are shown
the non-dimensional value.s of the failure capacity of the wedge as
calculated by the simplified theory(corresponds to column 6),as cal-
culated by the simplified theory but with horizontal stress calcula-
ted from column 7(correct value of initial horizontal force),and as
calculated by the program respectively.
The values in columns 8 and 9 are calculated from the corresponding
tractions t in columns 6 and 7 and from values of M from column 7 in
table 5.3,from the formula
(I-c) = l5.77-(t.M )gr
Tab
le5
.4.
Sym
met
ric
ela
stic
wed
gew
ith
inela
stic
rock
,w
ith
exca
vat
ion
seq
uen
ce.
o.
-60
_.,,-
.00
h=I3
737,
1=10
000,
W=
I854
,~c=I.O
-Seri
es
WA
D-W
CF.
a=20
,po
g=27
·IO
,¢=
40,
1=,
kn=
20,
.-
I2
34
56
78
910
II
Tes
tna
me
kk·
;jk
EOIO
-3tH
,tV
t BStH
O(I
-c)B
S(I
-c)H
O(r
-e)
Fai
lure
sta
rts
ss
nc
at
WAD
0.2
0.01
10
0.
3.2
3,1
.17
3.8
20
.75
IIo
44
2.2
62
.0to
pel
emen
t
WAE
1.0
0.0
51
00
.3
.03
,1.2
83
.58
2.I
I1
2.1
37
.15
7.0
top
elem
ent
WAF
2.0
0.1
010
0.2
.98
,1.4
73
.52
2.6
31
3.4
01
0.0
09
.0bo
ttom
elem
ent
WED
0.2
0.01
10
.03
.12
,1.6
73
.69
2.5
71
1.0
57
.70
8.0
top
elem
ent
WBE
1.0
0.0
51
0.0
2.9
7,1
.68
3.5
13
.34
II.9
01
1.3
31
1.0
bott
omel
emen
t
WBE
2.0
0.1
01
0.0
2.9
3,1
.68
3.4
63
.49
13
.15
13
.26
12
.0b
ott
om
elem
ent
WCD
0.2
0.01
1.0
02
.93
,1.7
33
.46
3.4
81
0.3
61
0.4
21
2.0
bo
tto
mel
emen
t
WCE
1.0
0.0
51
.00
2.8
9,1
.72
3.4
23
.58
II.5
912
",14
13
.0b
ott
om
elem
ent
WCF
2.0
0.1
01
.00
2.8
8,1
.72
3.4
23
.59
13
.00
13
.60
13
.0bo
ttom
elem
ent
Not
eth
at
ind
ices
,BS
ind
icat
eth
at
val
ues
corr
esp
on
dto
Hca
lcu
late
dby
ela
stic
solu
tio
n.
The
an
aly
tic
solu
tio
nis
use
d.
Ho
ind
icat
eth
at
val
ues
corr
esp
on
dto
HOca
lcu
late
dn
um
eric
ally
.T
hean
aly
tic
solu
tio
nis
use
d.
cin
dic
ate
sth
at
val
ues
have
been
calc
Hla
ted
com
ple
tely
nu
mer
ical
ly.
I f-'
-...J
<o I
-180-
There is reasonable agreement between columns 9 and 10,but great dis-
crepancy between columns 8 and 10 ,for high values of E.
In fig. 5.14, the change of the force vector acting on the joint
of the wedge is shown for test run WAE referred to in table 5.4.
This may be compared with fig.5.8a. As we see, the path fromA to G
is linear and corresponds to the loading applied to create a stress
free surface.We continue loading,by apply:i.ng in steps load proportio-
nal to the weight of the wedge.We are moving again on a straight line
at 16° to the normal on the joint,that is greater than e=7.820
(e=arc tan «k /k )·cot a) ,which as discussed earlier in section 5.1s n
is due to the flexibility of the rock.
The last converging step is at (1.52,0.45).Until that point minor
yielding occurs which does not deviate the line from linearity.After
that point partial and subsequently total yielding occurs which does
not allow the analysis to converge at G'.In fact the line bends,be-
coming concave downwards and thus meeting the failure envelope at a
lower point and bringing the solution(column 10 of table 5.4) closer
to the predicted value(column 9).
Tests WAD,liBD,WGD,WGE when run at ten times the stress level
(at 1000 m depth) fail a.t ten times the pullout load (b,c=l,Q), indi
cating a paralle'l shift of the stress. path ccAGG' ,(Figs. :5.8 and 5.14).
-181-
,....o
..coU"I.
I'II
'I/1
I II 0
I ' 0N
/~I e--:
II
II
I
a
,....l/"t\D.
..\Dl/"t.
,....coN.
..Mo.
Figure 5.14 Example of stress redistribution in a symmetric wedge(WAE)
horizontal stress
-182-
5.2.2 Asymmetric wedge.
The program is used to calculate the yield and failure loads of
various wedges within infinitely rigid surrounding rock.All possible
asymmetric permutations for a=50,200,350; E=lOO,lO,l GPa; k =0.2,1.00,s
2.00 are studied. The other parameters are taken to be
¢=400 , i=OO , k =20 , L=lO m.n
An initial horizontal stress 0Ho=-2.7 is applied in the plane strain
elements. The initial stress applied in the joint elements to create a
of -2.7 MPa is calculated from O=OH ·cos2a,o 0
T=OH -cos a- sin a, and is tabulated for the three angles a below.o 0
a T o0 0
50 0.234 -2.679
200 0.868 -2.384
350 1.269 -1.812
The wedge is modelled by 8 plane strain elements PI to P8 and the jo
ints by 8 joint elements jl to j8' one side of each joint element
obeing fixed.In figure 5.15, the discretization of a wedge with al=35o
and a2=5 is shown.
In tables 5.5,5.6,5.7 the non-dimensional resultant forces at first
yield and failure,calculated from the numerical solution,are compared
with the predicted values calculated from the formulae developed in
section 5.1.2. Each table refers to one geometry,that is one pair of
angles a. In columns 4 and 5 the analytically calculated M parameters
for each angle are' shQ1Nll.ln column 6, the non-dimensional failure
capacity of the wedge calculated from an average M assuming two
wedge parts separated by a vertical line and moving in the vertical
direction is shown. Then
(I-c) =(Ml
+M2 )e2eOH /(pege L)=20.(Ml +M2 )gr gr gr a gr gr(5.50)
-183-
48 I
~-
31 7 34
P429 9 612740
12700
j~.)
6 33
8 35
39
10 37
p6
P8
3
10000
8890
P7
2
15500
Figure 5.15 Asymmetric Wedge (WAS series)
Tab
le5
.5
Asy
mm
etri
cw
edge
surr
ound
edby
rig
idro
ck;
al=
350 ,a
2=
05
0;
-6po
g=27
0IO
,l:!.
c=0
.20
.
E=
IOO
·I0
3,1
0°1
03,I
0103
,<1
>=40
0,i=
OO
,k=
20,L
=IO
ol03
,aH
O=2
.7,k
=0
.2,1
.0,2
.0,h
=I2
70
0,W
=I7
I4.
Ser
ies
WAS
•-
ns
-
I2
34
56
78
910
II
-
kk
/kE
oI0
3M
M(I
-c)g
r(I
-cy
)th
(I-c
f)th
(I-C
v)(I
-cf)
Com
men
tss
sn
19
r2
gr
0.2
0.01
100
0.07
70
.15
84
.70
1.8
22
.18
1.6
01
.80
3.6
83
.60
rFaf
.Lur
est
art
s1
.00
.05
100
0.0
79
0.3
48
8.5
42
.04
1.8
0
2.0
0.1
010
00.
080
0.46
31
0.8
62
.26
5.1
72
.00
5.0
0in
up
per
m.id
dIr
join
t,el
emen
ts
0.2
o.or
10
.0.
077
0.1
58
4.7
01
.82
2.1
81
.80
2.4
0
1.0
0.0
51
0.
0.0
79
0.3
48
8.5
42
.04
3.6
82
.00
5.0
0F
ail
ure
star-
2.0
0.1
01
0.
0.08
00.
463
10
.86
2.2
65
.17
2.0
07
.00
ted
inlo
wer
mid
dle
join
ts.
0.2
o.or
1.0
0.07
70
.15
84
.70
1.8
22
.18
-4
.60
1.0
0.05
1.0
0.07
90
.34
88
.54
2.0
43
.68
-7
.80
-2
.00
.10
1.0
0.08
00.
463
10
.86
2.2
65
.17
-8
.20
Not
eth
efo
llo
win
gm
eani
ngo
fth
ein
dic
esus
edin
this
and
the
foll
ow
ing
tab
les
of
this
chap
ter:
gr:
Val
ueh
asbe
enca
lcu
late
dan
aly
ticall
y(f
rom
the
gra
ph
s)th
:V
alue
has
been
calc
ula
ted
an
aly
ticall
yfr
omas
ym
etri
cw
edge
solu
tio
ny
:V
alue
corr
esp
on
ds
toy
ield
f:
Val
ueco
rres
po
nd
sto
fail
ure
.
I I--' co -l'- I
Tab
le5
.6A
sym
met
ric
wed
gesu
rro
un
ded
byri
gid
rock
.a 1
=2
00
,a2=
050
;-6
p.g
=2
7-I
O,~c=0.25.
E=
IOO
°·I0
3,1
001
03,1
0103
,<I>
=40
0,i=
OO
,kn
=20
,L=
IO01
03,
0HO
=2.
7,k
s=0
.2,1
.0,2
.0,h
=2
2I5
0,W
=29
9fl.
Seri
es
WA
S_
.
I2
34
56
78
91
0II
kk
/kE
'I0
3M
M(I
-c)g
r(I
-c)t
h(I
-cr)
th(I
-c)
(I-c
r)C
omm
ents
ss
nIg
r2
gr
yy
0.2
0.0
11
00
0.1
90
0.1
58
6.9
65
.13
.48
.10
5.0
0
1.0
0.0
51
00
0.2
15
0.3
48
11
.26
7.1
6.6
7.7
5-
-2
.00
.10
10
00
.24
10
.46
31
4.8
08
.79
.5-
10
.25
0.2
0.0
11
0.
0.1
90
0.1
58
6.9
6-
3.4
-6
.75
1.0
0.0
51
0.
0.2
15
0.3
48
11
.26
-6
.6-
11
.25
-2
.00
.10
10
.0
.24
10
.46
31
4.0
8-
9.5
->
12
.75
0.2
0.0
11
.00
.19
00
.15
86
.96
-3
.4-
11
.75
1.0
0.0
51
.00
.21
50
.34
81
1.2
6-
6.6
->
12
.75
-
2.0
0.1
01
.00
.24
10
.46
31
4.0
8-
9.5
->
12
.75
I I-'
00
\.J1 I
Tab
le5
.7A
sym
met
ric
wed
gesu
rrou
nded
byri
gid
rock
;a 1=
350 ,
a2=20~;
-6p
°g=
27°I
O,
b.c=
0.25
.
E=I
OO
oI03
,Io
oI0
3 .r-I
03
,<I>
=400 ,i=
OO
,k=2
0,L
=IO
°I0
3,
aH
O=2
.7,k
=0
.2,L
0,2.
O,h
=94
00,W
=I26
9.S
erie
sW
AS•
ns
-
I2
34
56
78
9IO
II
kk
/kK
OIO
)M
M(I
-c)g
r(r
-e)t
h(I
-cr)
th(I
-c)
(I-c
r)C
omm
ents
ss
n1
9r
2g
ry
y
0.2
O.O
IIO
O0.
078
0.I
58
4.72
2.40
4.60
-4
.00
LO
0.05
IOO
0.07
90.
348
8.54
2.50
5.IO
4.5
0-
-2
.0O
.IO
IOO
0.08
00.
463
IO.8
62.
605.
50-
5.00
0.2
O.O
IIO
.0.
078
0.I
58
4.7
22.
404.
60-
3.75
LO
0.05
IO.
0.07
90.
348
8.54
2.50
5.IO
-5.
00-
2.0
O.I
OIO
.0.
080
0.46
3IO
.86
2..6
05.
50-
5.50
0.2
O.O
II.
O0.
078
0.I
58
4.72
2.40
4.60
-4.
75
I.O
0.05
LO
0.07
90.
348
8.54
2.50
5.IO
-6.
00-
2.0
O.I
OL
O0.
080
0.46
3IO
.86
2.60
5.50
-6.
00
I I--'
(X) o I
-187-
Equation ,5.50 has. been used by Goodman,Shi and Boyle(1982).In column
7 is shown the value for (I-c) when first yield occurs calcula-
ted analytically from equation 5.30.In column 8 is< thevallle for O.-c)
when failure occurs as calculated analytically from equation 5.37.
In column 9 the value for (I-c) at which first yield occurs is shown.
In fact first yield occurs between (l-c )-~c and (I-c) as the loady y
is applied in steps of magnitude ~c·W.
In column 10 the value for (I-c) at which the program last converges
is shown.Failure actually occurs between (l-cf) and (l-cf)+~c.
As pointed out in section 5.1.2, (I-c) for failure might sometimes be
less than for first yield.In this case a brittle type failure is ex-
pected and first yield and failure are expected to occur simultaneou-
sly.In column 11 are comments on where failure started first.
There is a tendency for the factor of safety to increase with flexi-
bility.Rotation of the wedge is an additional cause for the discre-
pancy between the simplified and the numerically calculated solutions.
First yield values for stiff wedges are a bit lower than predicted by
the simplified solutions, because yield does not occur simultaneously
at every point on one face. The flatter face of the wedge always yie
ld first and then fails.
-188-
CHAPTER 6 - APPLICATION OF THE PROGRAM TO ORE STOPING
In this chapter the computer program previously described is
applied to a stoping problem. The geometry is shown in Fig. 6.1. The
shaded area to the right of the figure represents the ore to be
excavated. The shaded area to the left of the figure represents the
position of a drive to be excavated before stoping starts.which is
used for access. Two joints (joint A and joint B) intersect over the
drive. Additional information on relative distances may be obtained
from Fig. 4.14.
The existing stress field is hydrostatic. The unit weight of
the rock mass is 27 KN/mm3 and the depth below ground level of the
floor of the drive or the stope is 100 m.
When the drive is excavated,the two intersecting joints form a
wedge, the stability of which depends on the material properties and
the horizontal stress field. Excavation of the ore then proceeds in
the stope from lower to higher levels,thus reducing the horizontal
stress field acting on the wedge. Thus the installation of struts
might be necessary in order to support the wedge.
The following set of units is used~
Quantity
Length
Force
Stress
Unit
mm
N
MPa
The material properties assumed in the analyses are shown in
Table 6.1
-189-
otoIIIen
o...oCD
r
",-~-10~
\
\\
'\10~\J 20-FI
" "
Joint B.JI ---
Joint A
II
I+1
I+1...
~L
Figure 6.1 Stope and drive georeetry.
2723171393
60
5 39
Figure 6.2 Stope,drive and surrounding rock diseretizatiQn.
-190-
Table 6.1 Material properties
Structure Parameter Unit Value
MPa3 3
E. 100x10 ,10x101.
Rock mass Vi - 0
N/mm3 -4P.xg 0.27x10
1.
Joint qu MPa 20
k N/mm3 0.2,0.1s
general BO - 1.0,0.5
V mm 1.0mc
cPr degrees 40 , 30
i degrees 0 , 5
------- - ------ ------- -------------
model 1 -q /T - 10u 0
~l MPa 0.05,1.0
------- - --- - - -- -----------------------model 2 So MPa 0
k N/mm3 20
n
E MPa 200x103s
2 -3strut A mm /mm 50,50x10
s
N/mm3 10-5
P xgs
Whenever two values appear for the same parameter,the first is
used unless otherwise specified. The discretization of the various
structures is summarized in Table 6.2 and shown in Fig. 6.2. There
is a total of 187 nodes.
-191-
Table 6.2 Discretization
Model Prototype Symbol
1 Exterior boundary element region Rock mass BE
3 Interior boundary element regions Intact rock BE
26 Plane strain elements Intact rock p
20 joint elements Discontinuities j
6 membrane elements Struts m
The activities being simulated are shown in order of occurence
in Table 6.3.
Table 6.3 Activities
No Activity Geometry change No of loadsteps
1 Gravitational loading - 1
2 Excavation of drive P9 to P14 and 3
ju and j18
2a Installation of struts mI to m6 1
3 First level ore excavation P19 and P20 3
4 Second level ore excavation P2I and P22 3
5 Third level ore excavation P23 and P24 2
6 Fourth level ore excavation P25 and P26 2
7 Doubling the weight of the wedge - 2
The number of load steps is chosen in an empirical way , within
limits suggested by Goodman (Hittinger and Goodman (1978)), larger
for activities that are intuitively predicted to cause large stress
redistribution. Results for problems ii. and iv. described below are
similar, when the number of load steps is doubled. These analyses
are not sensitive to the number of load steps,due to the brittle
-192-
type of failure (almost elastic behaviour until failure).
Nevertheless,due to path dependency of any other systems analysed,it
is suggested that the number of load steps of each activity be
varied, so that the sensitivity of the system with respect to the
number of load steps may be determined.
Activity 2a is included in the following analyses , only where
specifically mentioned. A set of analyses with joints with dilation
was unsuccessful,as even for a maximum number of iterations equal to
30 the computation did not converge. This is attributed to the lack
of cross stiffness terms in the joint element.
In analysing the problem with joint model 1,the wedge does not
fail for any combination of ~r' k s ' BO ' and ~l • In subsequent
analyses of the problem,Joint model 2 is used. The results are
summarized below.
i. ks=O.l
, BO
=1 . 0 or 0.5. The wedge failed during the third
step of activity 2 (excavation of drive)
ii. ks=0.2
, BO
=1 . 0 or 0.5. The wedge failed during the first
step of activity 6 (fourth level ore excavation). The case BO
=1 . 0 is
illustrated in Figures 6.3 to 6.8.
iii. Installation of struts. For very flexible initially
-3unstressed struts (A =50x10 , corresponding to rock bolts)
s
failure, that is separation of the wedge and the surrounding rock
mass occurs as in case ii., during the first step of activity 6,and
prior to failure the struts remained practically unstressed. Similar
behaviour is observed for stiff struts (A =50) without or withs
initial prestressing, the initial prestressing force being equal to
60 percent of the weight of the wedge.
iv. E =10 , BO
=0 . 5 or 1.0. The wedge did not fail.i
-193-
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.INITIAL MESH LENGTHSACTIVITY 0
Figure 6.3 Initial mesh.
UNITS
-194-
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.DEFORMED MESH LENGTHSACTIVITY I LOAD STEP I ITERATION I DISPLACEHENTS
__ = 2.0-10' UNITS__ = 0.5_10° UNITS
---
\ \ \ \ \ \ \111) 1111 I / / I\ 1 f / / /\ III' / I / / /-, ~ / / /'-, '~, /////<, ,'I ,,, ./ /' .."...-<, \'11 ,', ,,_ ............. ~
\ . , ~
, \ I I ,... . - .... ---- -- --_ _--- - - - --------.--
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.FLOW FIELD LENGTHSACTIVITY I LOAD STEP I ITERATION I DISPLACEHENTS
++++XX++
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.STRESS FIELD LENGTHSACTIVITY I LOAD STEP I ITERATION I STRESSES
__ = 2.0-10' UNITS__ = 1.0_10° UNITS
Figure 6.4 Gravitational loading.
-195-
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.DEFORMED MESH LENGTHSRCTIVITY 2 LOAD STEP 3 ITERATION 4 DISPLACEMENTS
UNITSUNITS
,/
---//
A- -c _ _ ..----_ .... ,., ...... _-
-,
\\\\\ \ '11/1111/ I 11/, I
\, \ I I / /\ I\ I\I
'"
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.FLOW FIELD LENGTHSACTIVITY 2 LOAD STEP 3 ITERATION 4 DISPLACEMENTS
Xx\~
++-f--f-
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.STRESS FIELD LENGTHSACTIVITY 2 LOAD STEP 3 ITERATION 4 STRESSES UNITS
UNITS
Figure 6.5 Excavation of the drive.
-196-
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.DEFORMED MESH LENGTHSACTIVITY 3 LOAD STEP 3 ITERATION 1 DISPLACEMENTS
UNITSUNITS
- ........................ -
\1111111' I 11/, I" I / /I ,I' /////
i ",' //// //...A ...-////- - ...-4 _ .. too---_ ... ,;~,." ..... _-
....
"
\ \ \ \ \ \
""
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.FLOW FIELD LENGTHSACTIVITY 3 LOAD STEP 3 ITERATION 1 DISPLACEMENTS
UNITSUNITS
~+
f-+-/-
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.STRESS FIELD LENGTHSACTIVITY 3 LOAD STEP 3 ITERATION 1 STRESSES UNITS
UNITS
Figure 6.6 First level ore excavation.
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.DEFORMED MESH LENGTHSACTIVITY 4 LOAD STEP 3 ITERATION 1 DISPLACEMENTS
UNITSUNITS
\ \ \ \ \ \ 1111/1111 I I I /I I\ , I I I /, I,
, 1 / / / / /
"\1
/ /
A/
.... "" / / / /'
.... /'
...-- .... ...-.- . ,____ ,"',1" ...... ... .. .... .... - -
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.FLOW FIELD LENGTHSACTIVITY 4 LOAD STEP 3 ITERATION 1 DISPLACEMENTS
X-t~-
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.STRESS FIELD LENGTHSACTIVITY 4 LOAD STEP 3 ITERATION 1 STRESSES
Figure 6.7 Second level ore excavation.
UNITSUNITS
-198-
EffECT Of SHADOWING ON A WEDGE IN A TUNNEL ROOf.DEfORHED HESH LENGTHSACTIVITY 5 LOAD STEP 2 ITERATION. DISPLACEHENTS
UNITSUNITS
\ \ \ \ \ \ 111111111 I / / I\ I-, \ I I I /
\ I-, \ , / / 1/ /"- ' I ,/
..... A ,/
.... .,/
.,/- ,,' .....-~, .,
- __ .- ...... ',"'11 .. . , , ... .... .... --
EffECT Of SHADOWING ON A WEDGE IN A TUNNEL ROOf.fLOW fIELD LENGTHSACTIVITY 5 LOAD STEP 2 ITERATION. DISPLACEHENTS
UNITSUNITS
EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF.STRESS FIELD LENGTHSACTIVITY 5 LOAD STEP 2 ITERATION. STRESSES
Figure 6.8 Third level are excavation.
__ = 2.0-10' UNITS__ = 1.0_10° UNITS
-199-
The aim of these analyses is to demonstrate the
program. Nevertheless the following conclusions may be drawn:
a. In joint model 1, k decreases rapidly,while k remainsn s
constant. Thus for similar initial parameters of joint models 1 and
2, the former estimates the wedge to be more stable.
b. Due to the brittle type of failure of the wedge, BO
is
irrelevant to its stability.
c. The lack of cross stiffness terms for the joints may cause
divergence of the iterative calculation in large problems with rough
joints.
-200-
CHAPTER 7 - SUMMARY AND CONCLUSIONS
A computer program that simulates the behaviour of fractured rock
near underground openings is developed. Based on Goodman's original
joint element model t the following additional features are
incorporated.
(a) Quadratic joint element
(b) Exterior and interior regions modelled with boundary
elements
The an~lytical and numerical formulation of (a) and (b) is
described and comparisons are made with other solutions in order to
check the accuracy and to validate the program. Finally the
developed codes are applied to determine the stability of a wedge in
a tunnel roof subjected to a horizontal stress field. Analytical
solutions are derived for this problem and graphs are drawn,relating
geometry and material properties for factor of safety equal to 1.
It is found that:
- The joint element is capable of modelling the behaviour of
discontinuities within rock. Problems of computed stresses
oscillating along the joints are encounteredtwhen the order of
integration used in the joint and adjacent elements is different.
The way strain softening is implemented does not pose any serious
numerical problem. On the contrary the way dilation is programmed
poses serious problems in convergence, the reason being the lack of
cross stiffness terms in the joint elements t even though the effect
of dilation in the diagonal stiffness terms is added. For small
problems with dilationtusually convergence is achieved. This is not
so for the large problem of Chapter 6.
-201-
- The boundary element method is suitable in representing
realistically the boundary conditions of the near field.It is found
that no integration needs to be performed over the boundary
approaching infinitY,even if the forces acting on the rock are not
self equilibrating,as in the case of excavations.
Numerical techniques are developed,in order to evaluate the kernel
shape function product integrals over elements containing the first
argument of the kernel.
- The way symmetric coupling is implemented in the program causes
some error in the neighbourhood of sharp corners and at points of
known discontinuous tractions. Some other minor causes for error are
identified such as, symmetrization, assumption of independent
interpolants for displacements and tractions,ratio of length between
neighbouring elements large, (e.g. >4) , and particular solution for
the interior boundary element region in a higher order of magnitude
than the total solution.
- The stability of a wedge in a tunnel roof due to a horizontal
stress field is investigated analytically. This analysis is based on
the assumption that,before self weight and any additional support
force are taken to be active,the joints are infinitely stiff. Also
the wedge and surrounding rock is assumed to be rigid and the two
joint faces are assumed to fail simultaneously. The factors of
safety calculated by the program and by the analytical solution for
various wedge configurations are in reasonable agreement. For deeper
excavations,the factor of safety of the wedge is calculated to be
proportional to the depth. Nevertheless at higher depths,the
parameters will change,especially the stiffnesses of the
joints ,which will become much larger. Thus for great depths,it seems
-202-
more suitable to assume the joints infinitely stiff, and calculate
the factor of safety of the wedge,on the basis of stresses
calculated by an elastic solution. This will overcome the difficulty
in determining the joint stiffnesses and their variation.
The effect of the flexibility of the rock mass and the wedge on
the factor of safetyis found not to be consistently higher or lower
than that calculated by the analytical solution.Nevertheless all the
problems analysed showed an increase in the factor of safety with
flexibility. The analytical solution can prove useful in
understanding the mechanism of failure, and identifying the range
within which the factor of safety lies, whereas the numerical one
may determine the factor of safety sufficiently accurately for
engineering purposes,if the properties of the joints are known.
Suggestions for further work
Further developments of this work are proposed here in three groups.
The first refers to work that needs to be done in order to make the
existing code more general. The second pertains to alternative
features that would improve the existing program. The last includes
some further suggestions.
Enhancement of generality
a. Infinite boundary elements: For very shallow excavations the
free surface of the ground must be modelled. If it suffices to model
this surface as a straight line free of traction, then the half
elastic space singular solution (e.g.Gerrard and Wardle (1973)) may
be used,for the derivation of the kernels for the exterior boundary
element region. Otherwise infinite boundary elements (Watson (1979))
should be used. These allow as well as having a ground surface of
arbitrary geometry and loading, more than one exterior boundary
-203-
element region,the interfaces of which extend to infinity. The
halfspace kernels are more expensive to compute.
b. Fluid elements: These are line elements,that simulate the
water flow through the fractures and the interaction with the rock.
Such elements have been used by Vargas(1982) to simulate the fluid
rock structure interaction problem. This model might be useful in
solving problems such as seismic induced activity in reservoirs.
c. Heat deformation: This can be easily incorporated in the
existing finite elements,if the temperature within each element is
given.If the source of the heat is given, then the heat conduction
system of equations may be formulated,using the existing
descretization, to find the temperature within each element.
d. Non-linear or plastic plane strain elements: The use of such
elements may be found in Owen and Hinton(1980). An appropriate
failure criterion for rocks, such as the Hoek and Brown (Hoek and
Brown,(1980»,and flow rule must be incorporated.
e. General anisotropy and extension to three dimensions: There
is no closed form singular solution for general anisotropy. The way
the boundary integral equation is formulated in this case may be
found in Wilson and Cruse (1978). This might be useful in
conjunction with an extension of the program to three dimensions.
The author's personal opinion is that this extension is necessary,
as underground excavations are rarely two dimensional.
Improvement of existing program
a. Refinement of the joint element: Cross stiffness terms must
be introduced in the joint elements,if rough joints are to be
modelled. The present method in which are assumed zero value cross
stiffness terms is unsuccessful in solving large problems. This
amendment will create non-symmetric stiffness matrices during
-204-
iterations, that must be symmetrized, if a symmetric solver (e.g.
Wilson et al.(1974» is to be retained. Also the constitutive
relations for the joint element should be posed in terms of
plasticity theory, were dilation is plastic strain in the normal
direction and slip is plastic shear strain. Strain softening must be
modelled in the same context.
b. Boundary element formulation: The system of equations is
formulated in terms of displacements and tractions. A non-symmetric
solver must be used and no symmetrization of the joint element
stiffness matrices needs to be performed during iterations, which
might result in fewer iterations. Improvements can be made in the
cost of solution of the system of equations,by performing
elimination of the unknowns at nodes that do not belong to the
interface with the non-linear region, prior to the simulation of the
activities.
c. Hermitian cubic elements: They may be used ,especially for
the three dimensional extension, as they are believed to be more
efficient than the quadratic ones. Hermitian joint and membrane-beam
elements may be developed and incorporated in the system.The
performance of Hermitian cubic plane strain finite elements needs to
be investigated.
d. Variable shear stiffness: The shear stiffness is known to
vary with stress level. This becomes very important for the analysis
of the wedge,discussed in Chapter 5,where the normal stress varies
continuously. Goodman (1974) has developed a constant peak shear
displacement model,which might be suitable.It is suggested that the
behaviour of the wedge be examined by using a model in which both
stiffnesses vary.
-205-
Further suggestions
a. Dynamic solution. This extension would permit the modelling
of fault propagation and attenuation of waves with distance. Three
dimensional modelling is needed for realistic results.
b. Fracture initiation or propagation.
c. Experimental data. There are insufficient or no data on the
following phenomena or properties.
Cross stiffness terms: Determination of these parameters would
require special strain controlled direct shear test machines.
Reversed loading: Celestino (1979) has performed a number of
tests on artificial specimens.
Model tests on a wedge in a tunnel roof: Crawford and Bray (1983)
conducted a series of experiments on artificial specimens. More
results are needed in order to understand the mechanism of failure
for the wedge.
Restart code: 0
-206-
APPENDIX 1
DESCRIPTION OP INPUT POR PROGRAM AJROCK
A1.1 Control cards and ordering of input deck
1. Heading card(A80)
Columns 1 - 80 Title card for program identification
2.Control cards
First card(4I5)
Columns 1 - 5 number of nodal points
6 - 10 number of element types
11 - 15 restart code
16 - 20 save code
initial problem
problem restarted
2 problem restarted and displacement reset to
zero
Save code o if saving of the results not required
control cards for restart will be saved in
disc
Second card(10I5)
Columns 1 - 5 Execution code
6 - 10 displacement printing code
11 - 15 blank
16 - 20 equation data printing code
21 - 25 graphical output code
26 - 30 mesh drawing code
-207-
31 - 35 fields drawing code
36 - 40 blank
41 - 45 frequency of graphical output
46 - 50 scaling code
Notes on second control card
Execution code : use 1 for a data checking run
Displacement printing code
o incremental and total displacements are printed
only total displacements are printed
2 no displacements are printed
Equation data printing code :
o print equation numbers and storage requirements
suppress printing
Graphical output code :
o no graphical output
plots every load case
2 plots every 'n' steps
3 plots every 'n' iterations at every step
Note 'n' is the frequency given in col.41 - 45.
Mesh drawing code:
o no mesh drawn
initial mesh only
2 initial and deformed meshes
Field drawing code:
o no field drawn
stress field only
2 stress and flow field
-208-
Frequency of graphical output·
o every step/iteration
n every nth step/iteration
Scaling code:
o scales adjusted every plot
scales adjusted to first plot
2 scales given by user
Third card (2F10.0) (only if scaling code 2)
Columns 1 - 10 Displacement scale
(1 plot cm = scale x 1 length unit)
11 - 20 Stress scale
(1 plot cm = scale x 1 stress unit)
3. Nodal point information
See section A1 .2
4. Element information
Columns 1 - 5 Keyword
Keyword may be 'BELEM', 'FELEM', 'ENDEL' .
'BELEM' indicates following information pertains one boundary
element region. Read following information for the boundary element
region according to section A1.3· For each boundary element region a
new card 'BELEM' must be read.
'FELEM' indicates following information pertains to a finite
element region. Read following information according to section
A1.4·
'ENDEL' indicates end of input pertinent to elements.
-209-
5. Activity parameters(A5.2I5,2F10.0)
Columns 1 - 5 code for load type
'GRAV ' : Gravity.residual stresses and pressure
loads only.
'NOD Nodal point loading.
'EXC Excavation.
'GRNOD': Gravity.residual stresses. pressure and
nodal point loads.
'CON Construction.
'EQ quasistatic earthquake load.
6 - 10 number of load steps (default is 1)
11 - 15 maximum number of iterations allowed(default 1)
16 - 25 convergence criterion(force units per unit width)
(default is 10-9)
26 - 35 upper limit on unbalance (divergence criterion)
6(default is 10 x convergence criterion)
6. Load information
A sequence of cards headed by an activi ty parameter card is
required for each activity. (The form of these data for each type of
activity is described in section A1.5.)
7. End of problem
Column 1 - 5 keyword
keyword may be :
'END 'for stopping the execution of the program
'NDATA' for allowing input for a new problem.
-210-
A1.2 Explanation of nodal point and boundary condition cards
Columns 1 - 5 keyword
keyword may be 'CAR r for cartesian coordinates to be read
'POL 'for polar coordinates to be read
'ORIG ' for origin shift from initial (0.0)
'FIX 'for fixing or freeing nodes
'ASSOC' for associating displacements of nodes
'ENDND' for ending this type of information
For keywords 'CAR,
and 'POL
Columns 6 -. 10 number of last node of the group
11 - 20 H or R coordinate
21 - 30 V or phi coordinate
31 - 35 generator index KN
36 - 40 number of first node of the group
41 - 45 H or R coordinate
51 - 60 V or phi coordinate
61 65 H direction boundary condition
65 - 70 V direction boundary condition
If the index KN is zero no automatic generation of nodes is
performed and the last node is the node being specified.If KN is not
zero but the first node is zero.automatic generation proceeds with
first node the last node of the previous card. Automatic generation
creates intermediate nodes evenly distributed between the first and
the last node.with numbering incremented by KN.
The boundary condition codes are
-Zero or blank. to indicate that the nodal point is free to move
in that direction
-One,to indicate that the nodal point is fixed from displacing in
the indicated directions.
-211-
For keyword 'ORIG '
Columns 11
21
20 H coordinate of temporary origin
30 V coordinate of temporary origin
For keyword 'FIX
Columns 6 10 number of first node of the series
11 15 last node of the series
16 - 20 increment to the node numbers (default 1)
21 - 25 H direction boundary conditions
26 - 30 V direction boundary conditions
If the last node is blank or zero only the first node is
processed.
For keyword 'ASSOC'
Columns 6 - 10 number of first node of group A
11 - 15 number of first node of another group B
16 - 20 number of last node of group A
21 - 25 number of last node of group B
26 30 increment to the node numbers of series A
(default is 1)
If the last node of group A is zero only the two first nodes will
be associated.The sequence of keywords is suggested to be:
'CAR ','POL ','ORIG'
'ASSOC'
'ENDND'
The keyword may be left blank.In this case the previously defined
keyword is applicable.
-212-
A1·3 Boundary element region
Element card(A5.6I5)
column .- 5 keyword
6 - 10 number of first element in a group
11 - 15 number of last element in the group
16 20 increment of node numbering (optional)
21 - 25 node of first element (Fig.A1 .1a)
26 .. 30 node 2 of first element
31 - 35 node 3 of first element(intermediate).
The keyword is 'ELE ' or 'ELE C' .In the latter case it is presumed
that the elements read, form a closed contour;thus the last node of
the group is given the number of the first node of the group. If the
last element in a group is zero only one element is processed .If
node 3 is zero or blank, node 3 is calculated to be the mean value of
the extreme node numbers 1 and 2.
Material card (A5,7F10.0)
Column -- 5 keyword
6 - 10 material type
If material type is 1 then material type is isotropic
If material type is O.then material type is orthotropic
For isotropic material'
Column 11 - 20 Young's modulus
21 - 30 Poisson"s ratio
For orthotropic material'
Columns 11
21
31
20 Young's modulus in direction
30 Young's modulus in direction 2
40 Young's modulus in direction 3
41 - 50 Shear modulus 12
-213-
(a) Element numbering
n>outwardnormal
(-)
(b) Orientation of element
<noutwardnormal
(t)
material
2
H
(c) Elasticity and global axes
Figure AI.I Boundary element convention •
-214-
51 - 60 Poisson's ratio 21
61 - 70 Poisson's ratio 32
71 - 80 Poisson's ratio 31
Oblique set of joints(A5,5F10.0)
Column -. 5 'JOL ' (keyword)
6 - 15 angle 'a' of joints (Fig.3.2)
16 - 25 normal stiffness of joints
26- 35 shear stiffness of joints
36 - 45 frequency of joints
46 -. 55 factor accounting for the persistence or
staggering(default is 1)
The joints are assumed to have the same mechanical properties and
to be symmetrically inclined to the axes 1 and 2.
Orthogonal set of joints (A5.I5.4F10.0)
Column - 5 'JOI I (keyword)
6 - 10 direction of the normal to the joint plane(1 .2.or
3. The out of plane direction is 3)
11 -- 20 normal stiffness of joint. (If omitted this is
assumed to be infinite.)
21 .- 30 shear stiffness of joint. (for the direction 3, this
value is irrelevant)
31- 40 frequency of joint spacing.
41 -. 50 factor that takes into account the effect of trace
length,persistence,etc.)
Orientation (A5.13I5.F10.0)
Column -- 5 I CUE ' (keyword)
6 - 10 cue elements for
" each closed BE region with sign
-215-
65 70 (see for sign convention in Fig.A1 .1b)
70 - 80 angle in degrees from direction H to direction 1 .
(see Fig.A1.1 c)
Elements with initial conditions(A5.15I5)
Column .- 5 keyword
5 - 10 number of elements
" with given initial
75 - 80 conditions
The keyword is either,
'DDE ' for given initial displacements.
or.
'TTE ' for given initial tractions.
Comment card (A5.A75).(obligatory)
Column - 5 . L , (keyword)
6 -- 80 comment
Initial loading
Columns 1 -- 5 keyword
If the keyword is 'DDQ ',corresponding to given initial
displacements,or 'TTQ ',corresponding to given initial
tractions. both varying parabolically ,then
Column 10 ., 20 displacement or traction at node 1 in the
horizontal direction
21 - 30 displacement or traction at node 1 in the vertical
direction
31 - 40 displacement or traction at node 2 in the
horizontal direction
41 -. 50 displacement or traction at node 2 in the vertical
direction
51 - 60 displacement or traction at node 3 in the
-216-
horizontal direction
61 - 70 displacement or traction at node 3 in the vertical
direction
71 - 75 element number
76 .- 80 element number
If the keyword is 'DDU '.corresponding to initial displacements.or
'TTU '.corresponding to initial tractions being constant within an
element. then
Columns 10 .. 20 displacement or traction in the horizontal
direction
21 30 displacement or traction in the vertical direction
31 .- 35 element number
" element number
75 - 80 element number
Particular solution(A5.5x.4F10.0)
Columns - 5 'PRT ' (keyword)
11 .- 20 height to the free surface (+)
21 - 30 ratio of horizontal to vertical stress (KA)
31 - 40 unit weight of the rock mass
41 - 50 pressure at the free surface
End of that boundary element region data(A5)
Columns 1 - 5 'ENDB I (keyword)
-217-
A1.4 Finite element region
A1.4.1 Membrane elements
Control card(A5,3I5)
Columns - 5 'BAR ' (keyword)
6 - 10 number of elements
11 - 15 number of different material properties
16 -- 20 number of integration points (default is 3)
Member properties(I5.3F10.0)
Columns - 5 material identification number
6 - 15 Young's modulus
16 - 25 cross sectional area
26 .- 35 uni t weight
Member data cards(6I5,F10.0)
Columns - 5 member number
6 - 10 nodal point
11 - 15 nodal point 2
16 - 20 nodal point 3
21 - 25 number of material of the member
26 - 30 optional parameter K causing automatic generation
of member data(default is 2)
31 - 40 initial stress in the membrane.
--Element data generation. Element cards must be in element number
sequence.If cards are omitted}data for the omitted elements will be
generated.The nodal numbers will be generated with respect to the
first card in the series by incrementing node numbers by K. All
other information will be set equal to the information on the last
card .The mesh generation parameter K is also specified in the last
card.
-218-
A1 .4·2 Plane strain elements
Control card(A5.2I5)
Columns .- 5 'PLANE' (keyword)
6 - 10 number of elements
11 - 15 number of different materials
Material property cards(2I5,7F10.0)
Columns - 5 material identification number
6 - 10 material type
11 - 20 unit weight
If material type is 1 ,the material is isotropic,
if the material type is O.the material is transversely isotropic.
For isotropic material,
Columns 20 .- 30 Young's modulus
31 - 40 Poisson's ratio
For transversely isotropic material,
Columns 21
31
30 Young's modulus in
40 Young's modulus in
, s '
. ,n
direction
direction
41 - 50 Shear modulus in I sn ' plane
51 60 Poisson's ratio giving the strain in the 'n'
direction due to stress in the's' direction(v )sn
61 - 70 Poisson's ratio giving the strain in the 't'
direction due to stress in the's' direction(v st)
71 - 80 Direction of's' axis in degrees (Fig.A1.2b)
Element card
First card(11I5,2A5)
Columns _. 5 element number
6 - 10 node 1 (Fig.A1.2a)
-219-
tP3
67 5
P4 8
nL.4
P2
< :>
(a) Element
I 2 3
n
(b) ElasticH
axes
Figure A1.2 Plane strain element convention.
nt2 4 6
:;.-~
I 3 5
(a) Node numbering
nt°n'tt L sn:1. ~
r------------,I I
L sn ton
(b) Positive displacements and stresses
Figure AI.3 Joint element convention.
-
-220-
11 15 node 2
16 - 20 node 3
21 - 25 node 4
26 - 30 node 5
31 - 35 node 6
36 ... 40 node 7
41 - 45 node 8
46 50 material identification number(default is 1)
51 - 55 element data generation K
56 .- 60 keyword
61 - 65 keyword 2
keyword 1 may be·
'SAMEB' :all node numbers are automatically incremented by K
'NOD48':all node numbers are automatically incremented by K,
except for nodes 4 and 8 which are incremented by K/2.
'NOD26':all node numbers are automatically incremented by K.
except for nodes 2 and 6 which are incremented by K/2.
If keyword 1 is blank the previously defined keyword 1 is
applicable.Automatic generation proceeds as described for the
membrane element.
Keyword 2 may be either,
'NEXT ': next card is second card for this element
blank: second card for this element does not exist.
Second card(7F10.0)
Columns .0 10 pressure on face
11 - 20 pressure on face 2
21 - 30 pressure on face 3
31 - 40 pressure on face 4
41 - 50 residual stress a xo
-221-
51· 60 residual stress
61 - 70 residual stress
A1 .4.3 Joint elements
Control cards(A5.4I5)
ayo
Txyo
Columns - 5 'JOINT' (keyword)
6 -. 10 number of joint elements
11 - 15 number of different materials(less or equal to 7)
16 - 20 number of integration points (2 to 5;default is 2)
21 - 25 code for law of behaviour
Law of behaviour'
'1' ,Hyperbolic closure with Ladanyi and Archambault shear
failure criterion
'2',Trilinear closure approximation with Mohr-Coulomb.Patton
shear failure criterion.
Material property cards(I5,8F10.0)
Columns - 5 material identification number
6 - 15 q .the unconfined compressive strength of the wallu
rock
16 _. 25 quiTO for model 1, or the shear strength
intercept of the joint for model 2
26 -. 35 shear stiffness ks
36 .. 45 BO.the ratio of residual to peak shear strength at
very low normal pressure
46 -. 55 Vmc(+)
56- 65 ';1 (.) for model 1 .or k for model 2n
66 - 75 the friction angle for a smooth joint
76 - 80 the dilatancy angle at zero(model 1 ),or
-222-
low(model 2) normal pressure
Element cards(9I5.2F10.0)
Columns 1 - 5 Element number (must start from 1)
6 - 10 node (Fig. A1 . 3a)
11 - 15 node 2
16 - 20 node 3
21 - 25 node 4
26 - 30 node 5
31 - 35 node 6
36 40 material identification number(default is 1)
40 - 45 element data generation K.
as described for membrane element.
46 - 55 initial shear stress
56 .- 65 initial normal stress 0'0n
(Fig.A1.3b)
-223-
A1 .5 Activities
A1.5.1 Activity 'GRAV I .Gravity, residual stresses and pressure
loading(SF10.0)
Columns - 10 percent of total loading for the first step
n
71 - SO percent of total loading for the eighth step
For gravity,residual stresses and pressure load,the equivalent
forces of the three loadings are computed.added,and multiplied by
the percentage for this step. The percentage may be specified for up
to the first eight steps. Blank will generate equal steps for the
remaining percentage of the total load.This activity is usually
applied to consolidate the field.
A1 .5.2 Activity 'NOD' .Nodal point loading(I5,2F10.0)
Columns - 5 Nodal point number
6 15 load in the H direction
16 25 load in the V direction
The sequence must be terminated with a blank card.
A1 .5.3 Activity 'EXC '.Excavation.
Nodal point cards(A5.5I5)
Columns _. 5 keyword
6 - 10 number of last node of the group
11 15 boundary condition in H direction
16 20 boundary condition in V direction
21 - 25 generator index KN
26 - 30 number of first node of the group
The keyword may be.
'BNMNP' ,or blank for input of nodes
'ENDND' for ending nodal point input.
-224-
Nodes must be given in order.
Element cards
First card (15)
Column - 5 Non zero only if at least one element to be
excavated is not on the excavation surface of this activity. If a
zero exists in the first card then skip to the third card of the
group
Second card group(1615).
Columns - 5.6 - 10.etc.List here all elements excavated that do
not have a nodal point on the excavation surface of this
activity.There must be a separate list for each element type in the
same order as in the original input deck. (Element types are
membrane. plane strain. and joint elements of type 1 and type 2) A
blank card must be provided for any element type that has no
individual elements being excavated.Always end each list(for each
element type) with a blank space.
Third card group(1615).
Columns 1 - 5.6 - 10.etc.List all elements excavated that do have
at least one nodal point on the excavation surface of this
activity.AII information of the paragraph above pertaining to the
second card group applies here as well.
A1 ·5.4 Activity 'GRNOD' .Gravity.residual stresses. pressure. and nodal
point loading. Follow the same format as for 'NOD' . shown in section
A1 ·5·2.The percentage of load or unload applied each step is
automatically 100% divided by the
A1 .5.5 Activity 'CON' .Construction .
Nodal point cards
-225-
First card(I5)
Column 1 - 5 new total number of nodal points.
Default is old total number of nodal points.
Second card
Freed nodal points are input as described in activity
'EXC'.section A1.5·3.New nodal points are input as described in
section A1.2. 'ASSOC' and 'FIX' keywords apply only to the new
nodal points.In both types of nodal points end sequence of cards
with keyword 'ENDND' .Freed nodal points are those that were input in
the original mesh as dummies.unattached to any element and
originally fixed. Freeing such nodes and connecting them to new
elements is an alternate method of adding new material.This approach
permits one to optimize the bandwidth.
Element cards
All elements to be added,must be described by a new element
deck, following the formats of section A1 ·4. The control parameters
will refer only to the new elements.For example if 3 new plane
strain elements all of a single material type are added to a 100
element mesh. the first control card demanded in section A1 .4.2 will
have 3 in column 10 and 1 in column 15. Assemble the element types
in the same order as in the initial data deck. End the sequence with
a card containing the keyword 'ENDEL' in columns 1 to 5.
A1 .5.6 Activity 'EQ '.Earthquake loading.
Acceleration cards(4F10.0.A5)
Columns
11
10 minus acceleration in H direction(in g units)
20 minus acceleration in V direction(in g units)
21 - 30 unit weight to be used for plane strain elements.
If zero. the originally defined unit weight for
each plane strain element is applicable.
-226-
31- 40 unit weight to be used for membrane elements.
If zero. the originally defined unit weight for
each membrane element is applicable.
41 - 45 keyword
If the keyword is 'ALL ',then the loading is applied to all
elements of the finite element region,and the element cards are not
read. If the keyword is blank,then the loading is applied to
selected membrane and plane strain elements. which are given in the
next cards.
Elements cards.
List here all element types(membrane,or plane strain only) ,on
which earthquake loading is applied.There must be a separate list
for each element type in the same order as in the original input
deck.A blank card must be provided corresponding to any element
type,on which elements, earthquake loading is not applied. Note that
element types in this case are considered only the membrane and
plane strain elements.For the joint elements no blank card should be
provided.Always end each list(for each element type) with a blank
space.
Example of input data
The following is the input data for the problem shown in
Figures 6.3 to 6.8 in Chapter 6 .
-227-
IJIIUJ DATA FCIl JIEIUM. Iff IWfIlE ~QI) PAGE 1......................................................***••***...............~... efFECT (E StMDCIRJC (If A lEDGE Itt A 1\It£l ROCF•... 117 2... 0 2 1 2 2 2... CM 5 -12200. o. 1 3 -17200. 0.0... 7 -1200. o. 1... 9 -5110. O• :1... tt -JOOO. o. 1... 15 3000. o. 1... 17 5110. o. 1... 19 1300. O. 1... 21 13000• o. 1... 23 19000. o. 1... Z7 31500• o. 1... 35 22800. 23100• 1... 39 10't00. 23100. 1... '13 3500. 23100. 1... '17 -3500. 23100• 1... 53 -17200. 23800• 1... 61 -17200. o. 1... 68 10300. 23800. . 1 61 17912. 2975.... 73 o. 1'12'10. 1 61 2916.7 22206.7... 11 3000. 6000 • 1... 17 5110• o. 1... 92 o. 1'12'10. 1 11 -2916.7 22206.7... 100 -3000• 6000. 1... 106 -5110. o. 1... ttl o. 1'12'10. 1 to' -2916.7 22206.7... tt7 3500. 23800. 1... 122 3000. 6000. 1 117 '1816.7 1000.... 130 o. 1'12'10. 1•u 138 -3000. 6000 • 1... 1'1'1 -5180. o. 1... 1'18 -3000. o. 1 1'f't -3800. 'tOOO.... 151 -1900. '1000. 1 1'18 -l5O. 10350.... 153 -1500• o. t... 163 o. O. 1 153 O. 10350.... 166 1900. 'tOOO. 1 163 l5O. 10350.... 161 1500. O. 1... 171 3000. O. 1 161 3550. 3000.... 171 3900• 2000.... 1'1'1 -3CJOO• 2000.... 172 o. l2'tOO.... 11'1 350. tt'lOO. 1 173 -350• 11'100.... 2 1 1 1 1... 112 165S0. 23100. 1 175 2'1163. 2975.... 11'1 15'100. 11000. 1 112 19800• 6000.... 117 21600. 11000• 1 185 26000. 6000.... EJIII)... IIEl.EM... B..E C 1 29 2 3 5 'I
-228-
JIIlVf MTA Fell PERUSIL rtf RCMDtE ~QI) MGE 2.............................................................................*a. MT I 100.£+03 0.0- ~ -1- La. MT 100.£+03 1• 27.£-06- ..E"r I 3 2 ·3 5 'I
'I 9 10'1 1055 12 -2 10'1 102 103
13 a '11 17a. U 1'1 19 2 '11 '19 'IIa. 20 S9 3 60a. I 100.£+03 0.0a. ClE Ia. La. PRJ 100.£+03 I. 27.£-06a. EJDa. IEUM**-' EL£ 1 3 2 11 19 IIa. EL£ 'I 23 62 61a. EL£ S 6 2 62 6'f 63a. EL£ 1 66 39 61a. EL£ I 9 2 39 '11 'to... ELE 10 '13 69 61... ELE II II 2 69 11 10... ELE 19 IS 11 86... MT I 100.£+03 0.0... CU 1... L... PRJ 100.£+03 1. 27.£-06... BD:au IElBt... ELE 1 2 2 '13 '15 'I't... ELE 3 '11 101 106... ELE 'I 1 2 101 109 101... ELE 1 ll5 '13 tt6... MT 1 loo.E+03 0.0... CU 1... L... PRJ 1~.E+03 1. 27.E-06... BDau FElEM... PUlE 26 1... 1 1 27.E-06 100.£+03 0.0... 1 130 131 132 112 121 129 130 130 1... 2 153 11'1 121 112 132 113 153 153 1... 3 132 133 lJ1 I'll lS3 113 132 132 1... 'I 121 11'1 153 163 126 127 121 121 1... 5 lS3 I'll lJ1 135 136 1'19 155 15'1 1... 6 lS3 15'1 155 16'1 12'1 125 126 163 1
-229-
JIIIUI' MfA,. IIIRUM. rtt ReM. ~QI) MiE J
.................................................................................~au 1 ISS 1'19 II 137 1. ISO 151 156 1... I tSS tS6 t65 t22 til t2'l 1M 1au 9 151 ISO 1. 1ft 1'10 1 1 159 lSI 1au 10 151 lSi 159 166 120 121 122 165 1... It 159 tSI t'lO 1'15 t'16 t52 16t 160 1au 12 159 160 161 161 16" 161 120 166 1au 13 161 152 1'16 1'tl II 12 13 162 1au 1'1 161 162 13 1'1 IS 1'70 169 161 1... 15 1'10 I'll 1'12 1'ft 1'16 1'15 1'10 1'10 1au 16 120 161 169 111 III lI9 120 120 1... 11 1'12 1'13 9 10 II 1'17 1'16 1'ft 1au II 169 110 IS 16 11 III III 111 1au 19 2J 2'1 2S 111 I"" 112 62 61 1au 20 2S a Z7 21 29 lIS I"" 111 1au 21 62 112 I"" In 111 113 M 63 1... 22 I"" lIS 29 30 31 I. 111 In 1... 23 M 10 111 119 110 11'1 66 65 1au 2'1 111 1. 31 32 33 117 110 119 1... 25 66 In 110 111 37 • ft 61 1au a 110 117 33 lit 35 36 n 111 1... ,"00 it 2 3 2lSI I 20. 00. 0.2 1.0 1. 20. '10.... 2 20. 00• 0.2 1.0 1. 20. '10.... 1 '11 '11 11 106 a 107 2 0.0 -2.'13lSI 2 a 101 19 101 90 109 2 0.0 -2.'13... 'I '13 '13 lI6 61 liS 69 2 2 0.0 -2.'13lSI 5 itS 69 It'l 10 lI3 11 2 0.0 -2.'13lSI 6 lI3 11 1t2 12 III 13 2 0.0 -2.'13... 1 92 130 93 131 crt 132 1 0.0 -2.'13lSI 13 10'1 1'12 105 1'13 9 9 1 2 0.0 -2.'13... 1'1 130 13 129 7'1 121 75 1 0.0 -2.'13... 15 121 75 127 "" 121 n 1 0.0 -2.'13lSI 16 121 n 125 11 12'1 19 1 0.0 -2.'13... 11 12'1 19 123 10 122 11 1 0.0 -2.'13... 11 122 11 121 12 120 83 1 0.0 -2.'13lSI 19 120 13 lI9 n ttl 15 1 0.0 -2.'13... 20 lI8 15 III 16 11 17 1 0.0 -2.'13... 9IIELlSI fiIIJN 1 5 20. 20.£+06lSI 100.... E>E 3 5 20. 20.£+06... II". 139 1 1 18 121... II"• 152 1 I 1 151... 162 1 1 1 158... 161 1 1 1 166... BIlelSI 0lSI 9 10 II 12 13 1'1lSI II 18lSI E)O; 3 5 20. 20.£+06
-230-
...................•.....•......................................................,JIIIU1' MFA Felt PElUSM. rtf AM'DE ~QI) PAGE ..
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.- .... 111 1 1.- 8ae.- 0.- 19 20.- 0... £)E 3 S 20• 2O.E+06.- .... In 1 I I 116.- 115 I I 3 112.- eae.- 0
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-231-
APPENDIX 2
OVERALL STRUCTURE OF THE PROGRAM
The program has been subdivided into a root segment and seven
primary overlays.
Overlay 1 copies the input data from file 4 to file 5 and echoes
them to the output file.
Overlay 2 is further subdivided into five secondary overlays.
These calculate the information needed for the formulation of the
equations, that is the individual finite element stiffness matrices
and load vectors.
Overlay 3 assembles the system of equations.
Overlay 4 solves the system of equations and calculates the
displacements stresses and unbalanced loads.
Overlay 5 contains the graphics routines.
Overlay 6 contains various routines seldom used.
Overlay 7 constructs the stiffness matrix and load vector of the
boundary element regions. It is subdivided into seven secondary
overlays.
-232-
FLOW CHART OF PROGRAM 'AJROCK'
Zeroise oounters and flags. Read general information.Read ooordinate and boundary oondition data.
Finite element
Read information for boundary
element region.
Construct stiffness matrix and
load vector for it.
(Overlay 7
zo
Read information for finite
elements.
Construct the stiffness matrix
and the load vectors.
(Overlay 2 ,2.1 .2.3 ,2.4
y
Draw initial mesh(Overlay 5)
Zeroise displacements. Calculate parameters for the solver. (Overlay 6
Read type of activ1ty.
y
N
+ 1
Calculate applied load this step.
Add any unbalanced loads (U.L) to the load vector. Assemble stiffness matrix and load
vector (Overlay 3). Solve the system of equations to find the incremental displacements.
Calculate the total displacements the total and incremental stresses,and any unba-lanced loads. Zeroise load vector. (Overlay 4)
y
Draw appropriate drawings.
y
N
y
Figure A2.l Flow chart of program AJROCK •
-233-
APPENDIX 3
ADDITIONAL INFORMATION,RELEVANT TO CHAPTER 3
A3.l Orthotropic kernels
The generalized Hooke's law for orthotropy may be written as
£11 l/El -v2l/E2 -v3l/E30 0 0 °Il
£22 -v12/E l 1/E2 -v32/E30 0 0 °22
£33 -v13/El -V2/ E31/E
30 0 0 °33
=1/(2.G12)
•£12 0 0 0 0 0 °12
£23 0 0 0 0 1/(2.G13) 0 °23
£31 0 0 0 0 0 1/(2.G3l) °31
or
£ = F·o (A3.l)<f'J - ....
fromdefines
taken in the notation of v ..• If,~J
i=1,2,3 (no sum double index convention)
Care must be3
e .. = LF . . ·0 ..11 j=l ~J JJ
then Lekhnitski (1963)
F.. =-(v . .lEi) =-(v . .IE.)~J J~ ~J J
whereas Tomlin and Butterfield (1974) and .Gerrard (1982b) define
v.. from~J
F .. =- (v . .IE. ) =- (v .. /E . ) , i -F j •~J ~J a J~ J
In the following the latter convention is used.
Due to the symmetry of matrix F ,
or
v .. /E . =v.. /E .~J ~ J~ J
i¢j (A3.2a)
-234-
The following inequalities must also be satisfied:
El,E2,E3,G12,G13,G31 > a
(1-V12eV21) , (1-V23eV32) , (1-V31eV13) > a
I-V12eV21-V23~1-V21eV13-2eV21eV32eV13 > a
For plane strain E33
= E23
= E31
= a , hence
F.. are the compliances of equation A3.1. ThuslJ
Ell Fl l F12 a °Il
E22 = F21 F22 0 e 022
E12 0 0 F33 °12
where
(A3.2b)
i,j=1,2 (A3e5a)
Substituting for F.. we getlJ
Fl l = l/El - V31/E3> a
F12 = -V21/E2 - V31eV32/E3 = F21
F22 = 1/E2- V3/ E3 ~ > a
F33
= 1/(2eG12) > 0
(A3.5b)
Tomlin and. Butterfield (1974) give the equations for stresses
and displacements due to a point load in an infinite orthotropic con-
tinuum. Let us define first the parameters used.
Parameters independent of position
.1. .!.a = «FlleF22)2 - F12- F33)2
(real or imaginary) (A3.6a)J...!.
b = «FlleF22)2 + F12+ F33)2(real) (A3.6b)
n = (b 2_ F33)ea2 - (a2 + F33)eF33 (A3.6c)
-235-
A = arbitrary = 1
Parameters dependent on position
m = Fll·zi + 2.F12coz~oz~ + F22°z~ =1 1
= «F )2oZ2 + (F )2oZ2) 2_ 2oa2oz2oz211 1 22 2 1 2
~ll= boz~ / (F22°z~ + (F12 + F33 )oz~ )J_ 1.
~12=~2l=/2ozloz2/«Fll)2oz~ + (F22)2oZ~ )
~22= boz~ / (Flloz~ + (F12 + F33 )oz~ )
where
(A3.6d)
(A3.6e)
(A3.7a)
(A3.7b)
(A3.7c)
(A3.7d)
z , =x. -yo111
i=1,2 (A3.B)
b is always real. This may be proven by substituting in equation
A3.6b for F.. from A3.5bo Then due to the last inequality oflJ
A3.2b we have that b2 - F33
> a and hence b is always real.
m is always positive. This is proven as follows:
If z2*O equation A3.7a becomes,
m/z~ = Fli(zl/z2)4+ 2oF12Co(zl/z2)2+ F22
The discriminant of the right hand side of the equation is,
Di = F2- FlloF22 (A3.l0)l2c
If Di < a , then m > a
If Di ~ a , then for m>O it is sufficient that
(zl/z2)2> (-F12c + (F~2c - FlloF22)! )/Fl l (A3.ll)
b2> 1.from a we get (FlloF22)2 > -F12c
from Di> a get FlloF22 < 2we F12c
From the last two inequalities we conclude that F12c > a
Hence the right hand side of inequality A3.ll is always negative,
and the inequality is always satisfied.
We may be proceed similarly for z2 = a , zl ¢ O. m becomes zero
for zl = z2 = O.
The kernels U
-236-
are the displace~ents due to a unit force and
are given directly by Tomlin and Butterfie1d(1974). The functions are
different according to whether a is zero,rea1 but not zero,or imagi-
nary. They are given by the following formulae:
a zero
Ul l = U'+ (F33-t11)/(4-/2-n-/F22)
U12 = F33-t12/(4-n-b) = F33-t21/(4-n-b) = U21
U22 = U'+ (F33-t 22)/(4-/2-n-/Fll)
a real
(A3.12)
U = U' - n/(4-/2-n-a-/F )-arctan (a-t )11 22 11
U = (a2_b2~F2 )/(a-n-a-b)-log«l + a-t )/(1 - a-t »= U12 - 33 12 12 21(A3.B)
U = U' - n/(4-/2-n-a-/F )-arctan (a-t )22 11 22
Note that arctan is to be specified in the interval [D,n [
a imaginary
U = U' - n/(a-/2-n-(1-a)-/F )-log«l + l-a-t )/(1-11 22 11
U = (a2_b2+ F2 )/(4-n-(1-a)-b)-arctan(1-a-t ) = U12 33 12 21
U = U' - n/(a-/2-n-(pa)-/F )-log«l + l-a-t )/(1-22 11 22
pa-tl l)
)
(A3.14)
pa-t22)
)
Note that arctan is to be specified in the interval ]-n/2,n/2}
U' = {(b2_ F )-F + (a2+ F )_b 2} l(a-/2-n-b-/F )-log(F -A'+ 1m)33 33 33 11 11
(A3.15)
1 =1(-1)
-237-
The kernels T are derived from the stress field equations given
by Tomlin and Butterfield (1974).
=222= z2·{1F11o(b2- F33)ozi + 1F22·(b
2+ F33)oz~} /(2o/2 oTIomob)
-122= zl/z2°=222 (AJ.16)
=111= zl·{1F22o(b2-F33)oz~ + 1F11o(b2+ F33)ozi}~ /(2o/2oTI omob)
-121= z2/z1o=111
where =..k is the stress 0.. due to a unit force in the directionlJ lJ
k • The relation between _ and T becomes
t.(y) = T.. (x,y)oe.(x) = 0J.k(y)onk(y) = =.k.(x,y)oe.(x)onk(y)J lJ 1 J 1 1
or
T.. (x, y ) == .k. (x , y ) 0 nk(y )lJ J 1
e.(x) is a load at x in the i direction.1
The full expression of kernels T then becomes
(A3.17)
T =-(1/(2o/2oTIomob))o{((b2+ F )olF - 2o(a2+ F )ob 2/1F )oz2 oz on +12 33 11 33 22 2 1 2
+(Fl l/1F22)0 (b2_ F33)ozi
on2 - 1F22o(b2- F33)onloz~ -
-1F11o(b2+ F33)ozioz2onl }
-238-
T2l=-(1/(2e/2enemeb))e{((b2t F33)e1F22 - 2e(a2t F33)eb2/1F11)ez~ez2enlt
t (F22/1F11)e(b2.;. F33)ez~enl - 1F11e(b2- F33)ez~en2---
- 1F22e(b2t F33)ez~ezlen2 }
where s takes the values 1,2. (A3.l8)
This is equivalent to
Singular solutions for line loads applied within half or whole ortho-
tropic space are given also by Gerrard and Wardle (1973,1980).
Behaviour of kernels U
We assume that zl and z2 are not simultaneously zero.
If a is real then,1
F12c = F12 t F33 < (FlleF22)2
Prove that (ltae~2l)/(1-ae~2l) is greater than zero and finite.
It suffices to prove that
l-ae~2l > 0 71+ae~2l > 0 )
~{ae/2ezleZ2/(1F11ezf + 1F22ez~ )}2 < 1
~ 2ea2ezfez~-< (lFllezf + 1F22ez~)2
which holds because m > O.
The function arctan is discontinuous and multivalued. In order to
perform numerical integration of this function over an element, this
function must be defined to be single valued and continuous in the
entire region.
-239-
Define the functions arctan(aetll) , arctan(aet22)e
Let us first examine the arguments aetll,aet22'
.....
a-tIl -------==---F22ez~ + Fl2cez~
a- t22
-------'""Fllez~ + Fl2cez~
If El 2c > 0
and
then
, then let us define the angle w from,
ae.Q,11 = ±ro
aet22 = ±ro
For the function arctan to be continuous,we define the angle to
lie within the interval [O,n [,and
z2/z1 = ±tan wI + arctan(aet l l) = n/2
zl/z2 = ±tan w2 + arctan(aet22) = n/2
In figure A3el the lines defined by z2/zl= ±tan wI ' z2/zl=±tan w2
are showneThe sign of the functions ae~l' ae~2 changes as point
(zl,z2) passes these lines respectively,their sign being shown in the
figure within parenthesise
From b2 > 0 and Fl 2c < 0 we get,~ ~
1{(FlleF22)} >-F I 2c + (Fll/(-FI2c))2 > ((-FI2c)/F22)2,
or
or
wI + w2 < n/2
This proves that the line defined by ~ is always lower than the
line defined by w2 in the first quarter of the coordinate axes.
arctan(a-i11)=n/2(+)
(-)
(-)
(+)
(-)
(+)
(+ )
arctan(a-i2~)= n/2(-)
-240-
(+)
(+)
F < 012c
a-i =11
(-)
(-)
(-)
± 00
(+)
(+)
Figure A3.1 Lines on which the orthotropic kernel U is undefined.
-241-
If a is imaginary then,
Prove that (1 + leae~22)/(l - leae~22) is always positive and fi
nite,so that the function log (1 + pae~22)/(l - paeQ,22) has mea
ning. It suffices to prove that,
and1 + paeQ,22 > 0
this is equivalent to
1 - leaeQ, > 022
Substituting for· t 22 we get
_a2eb2ez~ /(Fllezi + F12cez~ )2 <1
or
F ez4 + F ez 4 + 2eF ez2ez 2 > 011 1 22 2 12c 1 2
The last inequality holds , as m > O.
Similarly we may prove that (1 + leaeQ,ll)/(l - leaeQ,ll) is positive
and finite.
Define the function arctan(leaeQ,21) to be continuous and single
valued. We proceed as follows:
le aeQ,21 never becomes infinite,and is continuous in the zl,z2 plane.
Hence arctan(leaeQ,21) must be defined to lie within the interval
]-n/2,n/2 t ,to be continuous.
In table A3 el the function arctan for a real or imaginary is
defined
Table A3el Determination of function arctan
argument aeQ,.. 0J.J
arctan(aeQ, .. ),i=j 0J.J
arctan(leaeQ, .. ),i~j 0J.J
positive
JO,n/2[
JO,n/2[
±oo
n/2
negative
J n/2,n [
J-n/2, 0 I
-242-
Relation between parameters of orthotropy and isotropy
The compliances when the orthotropy degenerates to isotropy become,
Fl l = F22 = (1 - V 2 )/E
F = F =-V-(l + V)/E1221
F33 = 1/(2-G) = (1 + V)/E
Substituting into A3.6 from A3.20 we get
a =0
lb = {2-(1 - v2 )/ E}2
n = - F 2 = -(1 + v)2/E233
Substituting from A3.20 and A3.2l into A3.7 we get,
l£12 = £21 = zl-z2/r 2-{2-E/(1 - V
2 ) 2
£22 = (z2/r)2-/{2-E/(1 - v2) }
£11 = (zl/r)2-/{2-E/(1 - v 2) }
(A3.20)
(A).2l)
(A3.22)
By substituting these values to A3.l2,A3.l5,A3.l8 we get the
formulae 3.15 of chapter 3 for isotropy.
-243-
A3.2 Integration of kernel-shane function products over an element
containing the first ill:gument
Kernel 1I
Instead of using a logarithmic Gaussian quadrature formula in
order to integrate the logarithmic term of a kernel U over an
element,we may integrate analytically over a straight line element
tangent to the actual one, whLch in general is curved, and then add
the difference between the two integrals, which may be evaluated
numerically, using the Gauss-Legendre quadrature formula (G.L.Q.F).
If the intermediate node of a straight line element lies at
the middle of the element, the Jacobian is constant and equal to
the half length of the element, and r ~ ~ (Fig. A3.2a).
Hence the shape functions are second order polynomials of r. For
isotropy the logarithmic term is Ug = Al·log(l/r), where Al is
a constant and the kernel U - shape f~~ction product integral isg
a sum of terms Al- J ug-(r/al)n- dr • Let us evaluate these terms.
alJ U -(r/al)n-dro g
al
= A - J log(l/r)-(r/al)n-dr =1 0
In Table A3.2 the values of I for n= 0,1,2 are shovm.n
Table A3.2 Integral I for isotropyn
Integral value
10 -a -(log a - l)-A111
11 -(a/2)-(log al - 1/2)-A1
12 -(al/3)-(log al - 1/3) - Al
spiral
a-2
<1 separates elements
(a) a is extreme node
a at Ix (b)
I 3 2
;fa 1
.-.- r-a at 2x (c)I 3 2
1/ 1/I
a 1 /
r r
ax at 3 (d)
I 3 2
al 4- al -iFigure A3.2 Analytical integration of a log-polynomial product
over a straight line element.
(b) a is intermediate node
a-l
subelement........
spiral
Figure A3.3 Spiral method used for the determination of the diagonal
terms of matrix T.
-245-
For orthotropy the logarithmic term is in the form
terms_
tegral is the sum of terms
Ug
= A2-1og(F11/m)
where A2 is a constant_ The kernel U ,shape function product ing
J ug-(r/al)n- dr .Let us evaluate these
elementa
la
lIn = ~ ug-(r/al)n-dr = A2-~ log(Fll/m)-(r/al)n-dr =
= -al-A2/(ntl)-{4-(log al - l/(ntl))t Cal
where
In table A3.3 the values of I for n=O,1,2n
are shown.
Table A3_3 Integral I for orthotropyn
Integral Value
10 -a -A -{4-(log a - 1) t C }1 2 1 a
11 -a -A /2-{4-(log a - 1/2) t C }1 2 1 a
12 -a -A /3-{4-(log a ..- 1/3) t C }121 a
Let s = rial • In table A3.4 the analytically evaluated integrals
J U (xa,y)_Ne(y) edr are evaluated. In column 1 the internal nog
de number of the first argument of the kernel a and the internal
node number e of the shape function are shown.In column 2 the
relation between s and ~ is shown. In column 3 the shape func-
tions are shown as functions of s. In column 4 the value of the
analytically evaluated integral is shown. In column 5 the relative
length k of the tangent element to the Jacobian at the node aa
is shown_ In column 6 the appropriate figure for the tangent ele-
ment is given_
-246-
\Vhen a=3 the integration is performed over two straight line ele-
ments and the total integral is the sum of the two integrals. As Ug
is symmetric with respect to a lit suffices to add the shape func-
tions in each half of the element and perform the integration over
one half only.
Having calculated analytically the logarithmic parts of the ke-
rnels U over the tangent straight line elements,we need to add an
extra term to account for the deviation of the real element from the
linearity_ This term is given by
R = f N-U odS - f N-U -dS (A3_25)es Sr r r St gt t
where subscript r denotes the real element and t the tangent one.
Expressing the functions in terms of intrinsic coordinates ~,
dS = J-d~r
dS = J -k -d~t a a
R becomes,es 1
R = f N-(U -J - U -k -J )-d~es ':'1 r gt a a
This last integration may be evaluated numerically.
(A3.26)
Table A3.4 Analytically evaluated integrals of kernels Ug
1 2 3 4 5 ~6
a , e ~(d N'" (s ) f U -N"'(s)-dS ka=al/JaFig.gt t
1 1 - 3-s + 2- s'" I O-3 I l+2 I 21 2 1 - 2- s -s + 2- s 2 -I +2-I 2 A3.2b1 2
3 4-s -4-s 2 4-I - 4-I1 2
1 -s + 2- s 2 -I +2-I1 22 2 2- s - 1 1 - 3- s + 2- S
2 I -3-I +2-I 2 ~3.2c0123 4-s -4- S
2 4-I - 4-I1 2
1 S2 I 23 2 s S2 I 2 1 A3.2d
3 2 - 2- S2 2-I - 2-Io 2
--247-
Kernel T
The principal value of the integral
for nodes a and d(b,e) coinciding, over one subelement (Fig.A3.3b),
or over an element (Fig. A3.3a) , for node a middle or corner node
respectively,does not exist. This principal value exists only if the
two subelements or two adjacent elements are taken together. In other
treatises on the subject the total integrals are determined, either
by considering a rigid body translation of the region, or by direct
evaluation of the c .. terms and the Cauchy princiual values of thelJ •
integrals. In the program we have extended a method developed by
Watson (1981,1982) for potential flow problems,to elasticity. It
relies on the rigid body translation method. Each node is taken to
be separated from the rest of the body by an imaginary spiral contour
having end points the two neighbouring corner nodes and passing
through an intermediate point inside the body, called the halfway
node. A rigid body translation in each of two directions is assumed
for each node, together with the portion of the body within the spiral.
(A3.27)IT .. (xa,y(~)) -ds
r lJs
If d(b,e) = a, and a is an intermediate node (Fig. A3.3b),then
c .. (xa) + IS T.. (xa,y(~)).J(tJ.d~ =
lJ b lJ
- - IS T.. (xa,y(~)).(1_N3).J(~).d~b lJ
where r is the spiral contour.s
If d(b,e) = a, and a is an extreme node (Fig. A3.3a), then
(A3.28)
where now So is the sum of Sol and Sbr.
The integrals on the right hand side of the equations may be
evaluated numerically using the G.L.Q.F.
-248-
A3.3 Particular integral
The following stress field satisfies the equations of equili-
brium,hence it is a particular solution.
(0) =[:H] =[ ~A1-(p-g-x - peg-h. + p )HV V V 0 0
0HV 0
The boundary conditions at infinity are assumed to be also satisfied
by this equation.
is the vertical stress at level h,o
KA is the ratio of horizontal to vertical stress,
H,V are the horizontal and vertical axes of the coordinate system.
If 1,2 are the principal axes for orthotropy,then
(£)12 = M-(g)HV
M is the stress vector transformation matrix.
The constitutive law that relates stresses to strains is
(A3.30)
(A3.31)
F is the compliance matrix.in the principal directions 1 and 2.
We take the displacements to be given by a second order polynomial
Then the strains are given by,
where
(A3.32)
2-a Yl1
A = Y2 2-S2
0.2 +Yl/2 Sl + y2/ 2
-249-
By equating .E: from equations A3.3l,A3.33 we get",
Also from A3029,
where
T,!S* = (KA,1,0)
Substituting Q in A3.34 we get
T Tpogo~o~o~*o(sin 8,cos 8,-hO+Po/(pog) )o(xl,x2,1) =~o(xl,x2,1)
(A3.35)
where
(Fllocos28 +F12osin28)oKA+Fll-sin28
(Fl2-cos2 8 +F22- sin28) -KA+F12- sin28
F33-sin 8-cos 8-(1-KA)
+F -cos2812
+F -cos2822
(A3036)
For A3.35 to hold for any pair of (xl,x2),all nine coefficients
Tmultiplying the vector (xl,x2,1) must be identical on both sides
of equation A3035, that is
2-0. =C -sin 81 1
Y =C -sin 82 2
aZ+Yl/2=C3-sin8
Yl=Cl-cos 8
2-S2=C2-cos 8
Sl+y2/2=C3-cos 8
o =C -(-h +p /(p-g))1 1 0 0
E: =C -4-h +p /(p-g))2 2 0 0
(o2+~~)/Z=G3-(-ho+Po/(p-g))
(A3.37)
Three parameters may be chosen arbitrarily,to allow for a rigid body
motion o We define sl=s2=0 which imposes zero translation at the o
rigin of the coordinate axes .~.
-250-
We arbitrarily also specify d~/dxH=O at the origin which gives the
following relation,
Thus the parameters a to £ become
(A3.38)
al=1/2.Cl-sin 8
Sl=C3-cos 8- C2/2-sin 8
yl=Cl-cos 8
°l=Cl-(-hO+po/(p-g))
£1=2-C3-(-hO+po/(p-g))-02
a2=C3-sin8- Cl/2-cos 8
S2=1/2-C2-cos 8
If the displacements in the HV system are given by
uH=aH-x~ + SH-X~ + yH-xl-x2 + 0H-xl + £H- x2
~=~-x~ + SV-x~ + YV-xl-x2 + 0V-xl + EV-X2 (A3.40)
the parameters~ to £V are related to the parameters al to £2
as follows
[~SH YH °H ::J ~~::
8-sin ~t Sl Yl °1 oj
Sv YV °v 8 cos 8 a2 S2 Y2 °2 £2
(A3.4l)
The tractions are related to the stresses as follows
t H = 0H-(cOS 8-nl sin 8-n2)
tv = 0V-(sin 8-nl + cos 8-n2) (A3.42)
-251-
Special cases
Directions 1,2 and R,V coincide, that is 8=0_
Then
Cl = P-g-(Fll-KA + F12)
C2 = P-g-(F12-KA + F22)
C = 03
Isotropy
the angle 8 may be taken zero,hence
C = P-g-(l+V)/E-((l-V)-K - v)1 A
C = p-g-(l+v)/E-(-v-K + 1 - v)2 A
C = 03
Isotropy and KA=V/(l-v) (corresponds to lateral constraint)
~ = 0
"v = p-g-(l+v)/E - (1-2-V)/(1-v)-x2-(0_5-x2-hO+p/(p-g))
OR = V/(l-v)-p-g-(xV - hO + po/(p-g))
-252-
APPENDIX 4
Estimate of error due to the assumption of continuous tractions
at nodes.
The simultaneous e~uations approximating the integral equation
may be written in matrix form as
u ·1 = 1.-,Y,
If t is discontinuous at nodesJthen the equation may be rewritten
as
where Ur , Ul are the matrices that have components equal to the
integrals of the kernels U times the shape functions to the right
r 1or left of the node respectively, and t ,t are the tractions to the- ~
right and left of the nodes.
Le~ us say now that
where
and t' is arbitrary.Then
a - S = I
I is the unit matrix. Equation A4.2 then becomes
(:~(+Ul)_ i' + (Ur_a + ~-S )-4.-t = ~ ~
or
or
C-t' + C_U-1_(Ur_a + Ul-S)-t.t = K-u- ""- - ,..., - - - - ""- -1 --
(A4.7)
(A4.8)
where the non-symmetrized stiffness matrix has been used for the
sake- of clarity.
-253-
Let us define ~' from the following relation:
-P=C-t' = J N-t-ds = J N_(Nr_tr + Nl_tl)_ds = Cr_tr + Cl_tl~ (A4.10)",....., - r- r- f- - ~ -,....., - ~ - ~
t are ,the actually applied tractions on the boundary r.Substituting fortr,t l from A4.4 in equation A4.10 we get
~ ,-..J
Cr_a + cl-S =0 (A4.11)- - - - -
Also
(A4.12)
From equations A4.,,6,A4.11,and A4.12 we get
a = C-l_Cl ~ = _.2.--1_Cr
From equation A4.9 it can be seen that an additional term I1P is
needed for the correct answer to be obtained.Substituting fora
and.@. from A4.13 into A4.9 we get,
-11P=C-U-1_(Ur-C .1_Cl_Ul_C-1_Cr)_l1t= (Cl_C_U-1_Ul)_l1t =~-- - - - "- ---
=_(Cr -C-U -l_Ur) -l1t=- ---
=(1/2)_{(Cl_Cr)_C_U-1_(Ul_Ur)}_~t
If the elements are all of equal length then
Cr=Cl=0.5_C- - -
_t.P=(1/2)-C-U -1_ (Ur _Ul) -l1t- ........
The error in displacements is found from
(A4.15)
(A4.16)
(A4.17)
-254-
APPENDIX 2
GRAPHS FOR ESTIMATING THE STABILITY OF A WEDGE IN A TUNNEL ROOF
In this appendix,diagrams are drawn, that relate the non-dimensi-
onal parameters M ,to the angle a (ALFA),for various friction an-
gles ¢ (PHI),stiffness ratios k /k , and dilation angless n i (IOTA).
The value of the friction angle of each curve can be read as the va-
lue of ALFA ,at which the curve meets the ALFA axis.
CURV
ESFO
FVA
RIOU
SPH
I-VA
LUE
OFPH
ION
ALFA
AX
IS
.. .. ~KS/KN=Oo0009IOTR=O
.. ": o
CURV
ESfO
R"V~RIOUS
PHI-V
ALU
EO
fPH
ION
ALf
AA
xiS
!:: -'K5
/KN=
0·0
01
9lO
TR
=0Cl
t E: 1
.. .. 0' ~ <'f .. '? ~,
_A
LF
A_
I 0' I D I D N r- D .. • D :I
tD
E:
•1
.. 0' 0 .. 0 ::I D' .. N D
-AL
fA_
I 1\)
\J1
\J1 I
VR
iHo
us
PH
I-V
RlU
EOF
-PHC
D"R
LF
R-A
xl
.. -'KS
/K
N=
0D0
02
9lO
TR
=0
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