NUMERICAL SIMULATION OF EXCAVATIONS WITHIN JOINTED ROCK OF INFINITE EXTENT
by
ALEXANDROS I . SOFIANOS (M.Sc.,D.I.C.)
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 -2-
A bstract
The subject of the thesis is the development of a program to study i 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,construetion 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 identif i ed. 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. * 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 * G reece.
Furthermore thanks are due to: * Dr. J. 0-Watson, for his supervision of this study,and for introducing me to the boundary element method. Dr. J.W.Bray,for discussions. 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.
*
* TABLE OF CONTENTS
Page ABSTRACT 2 ACKNOWLEDGEMENTS 3 TABLE OF CONTENTS 4 LIST OF FIGURES 7 LIST OF TABLES 11 NOTATION AND CONVENTIONS 14
CHAPTER 1 - INTRODUCTION 20
CHAPTER 2 - NUMERICAL MODELLING OF JOINTED ROCK 23 2.0 Distribution of stresses and displacements 23 2.1 The continuum 24 2.1.1 Mechanical properties 24 2.1.2 Simulation 26 2.2 Discontinuities - A literature survey 34 2.2.1 Mechanical properties 34 2.2.2 Sim ulation 39 2.3 The joint element 47 2.3.1 The element 47 2.3.2 The constitutive law 50 2.3.3 Iterative solution 63 2.3.4 Examples 72 2.4 Change of the geometry 77 2.4.1 Excavation 77 2.4.2 Construction 79 2.5 Types of activities 80 - 5 -
Page CHAPTER 3 - THE ELASTIC REGION 81 3.0 General 81 3.1 Equivalent elastic properties of a jointed rock mass 82 3.1.1 Three orthogonal sets of joints - 82 3.1.2 Two oblique sets of joints 84 3.2 Implementation of the direct boundary integral method 86 3.2.1 The integral equation for the complementary function 87 3.2.2 Kernels U and T 89 3.2.3 Isoparametric element 91 * 3.2.4 Nodal collocation 93 3.2.5 Numerical integration 95 3.2.6 Rotation of axes 96 3.2.7 Particular integral 97 3.2.8 Infinite domain 98 3.3 Example 103
CHAPTER 4 - COUPLING REGIONS WITH CONTINUOUS AND DISCONTINUOUS DISPLACEMENT FIELDS 107 4.0 General 107 4.1 Symmetric coupling 108 * 4.2 V alidation 110 4.3 Inherent erro rs 125 4.3.1 Causes of erro rs 125 ** 4.3.2 Examples 130 -o-/
Page CHAPTER 5 - STABILITY OF AN OVERHANGING ROCK WEDGE IN AN EXCAVATION 142 5.0 General 142 5.1 Idealised behaviour 142 5.1.1 Symmetric wedge 144 5.1.2 Asymmetric wedge 161 5.2 Numerical solution 169 5.2.1 Symmetric wedge 170 5.2.2 Asymmetric wedge 182
CHAPTER 6 - APPLICATION OF THE PROGRAM TO ORE STOPING 188
CHAPTER 7 - SUMMARY AND CONCLUSIONS 200
APPENDIX 1 - Description of input for program AJROCK 206
APPENDIX 2 - Overall structure of the program 231
APPENDIX 3 - Additional information relevant to Chapter 3 233 A3.1 O rthotropic kernels 233 A3.2 Integration of kernel - shape function products over an element containing the first argument 243 A3.3 Particular integral 248
APPENDIX 4 - Estimate of error due to the assumption of continuous tractions at nodes 252
APPENDIX 5 - Graphs for estimating the stability of a wedge in a tunnel roof. 254
REFERENCES 279 - 7 -
LIST OF FIGURES
Page
CHAPTER 2_ % Fig. 2.1 Eight node serendipity element 27 Fig. 2.2 Axes for transverse isotropy and global cartesian system 29 * Fig. 2.3 Sign convention for internal forces 29 Fig. 2.4 Isoparametric three node membrane element 31 Fig. 2.5 Peak shear strength 36 Fig. 2.6 Peak shear strength 37 Fig. 2.7 First joint element 43 Fig. 2.8 Three dimensional joint elements 43 Fig. 2.9 Isoparametric quadratic joint element 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 s ,a , T . 60 Fig. 2.15 Dilation vs shear strain law for the two models 62 Fig. 2.16 Iterative process (for compression) - Joint 1, ♦ no dilation 65 Fig. 2.17 Iterative process - Joint 2 without dilation 66 Fig. 2.18 Iterative process - Joint 1 with dilation 67 * Fig. 2.19 Iterative process - Joint 2 with dilation 68 Fig. 2.20 Iterations for simple examples 71 Fig. 2.21 Strain softening joints (examples) 73 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 fie ld 105 Fig. 3.9 Plane strain and joint elements subjected to gravitational field 106
CHAPTER 4_ Fig. 4.1 Square block in tension 111 Fig. 4.2 A circular hole under pressure 111 Fig. 4.3 Hole within infinite rock mass modelled by boundary elem ents only 113 Fig. 4.4 Hole within infinite rock mass modelled
by boundary and finite elements 113
Fig. 4.5 Tension of a long plate 116
Fig. 4.6 Lined opening 119
Fig. 4.7 Excavation of a circular tunnel 122
Fig. 4.8 Excavation of a circular tunnel 122
Fig. 4.9 Various methods to determine the limiting values
of tractions at the two sides of a corner 126
Fig- 4.10 Two boundary element regions 131
Fig. 4.11 Circular disc 131 Fig. 4.12 Square block modelled by 32 boundary elements 134 Fig. 4.13 Square block modelled by boundary and plane strain elements 134 Fig. 4.14 Large problem with boundary and finite elements 138
CHAPTER _5 Fig. 5.1 Wedge id e a liz a tio n 143 Fig. 5.2 Symmetric wedge - Friction angle greater than a 145 Fig. 5.3 Symmetric wedge - Friction angle less than a 149 Fig. 5.4 Examples for very low stiffness ratio joints 151 Fig- 5.5 Behaviour of a symmetric rigid wedge 152 Fig- 5.6 Effect of intact rock flexibility 154 Fig. 5.7 Models for elastic wedge 155 Fig. 5.8 Stress redistribution 159 Fig. 5.9 Asymmetric rigid wedge 162 Fig. 5.10 Oblique wedge 167 F ig. 5.11 Wedge with ro tatio n 167 Fig. 5.12 Symmetric flexible wedge within rigid rock 171 Fig. 5.13 Symmetric elastic wedges within elastic rock ; a=20^ 175 Fig. 5.14 Example of stress redistribution in a symmetric wedge 181 Fig. 5.15 Asymmetric wedge 183 - 1 0 -
Page CHAPTER 6 Fig. 6.1 Stope and drive geometry 189 % Fig. 6.2 Stope.drive,and surrounding rock discretization 189 Fig. 6.3 I n itia l 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. A l.l Boundary element convention 213 * Fig. A1.2 Plane strain element convention 219 F ig . 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 244 Fig. A3.3 Spiral method used for the determination of the diagonal terms of matrix T 244
* - 11 -
LIST OF TABLES
Page
CHAPTER 2 Table 2.1 Shear strain regions 59 Table 2.2 Two plane strain and two joint elements 74 Table 2.3 One plane strain and three joint elements 76
CHAPTER .3 Table 3.1 Integration of kernel shape function products
over an element containing the first argument 95
Table 3.2 Displacements at the nodes of a brick (example) 103
CHAPTER 4_
Table 4.1 Square block in tension 112
Table 4.2 Circular hole modelled by boundary elements only 114 Table 4.3 Circular hole modelled by boundary and finite elements - Displacements at nodes 114 Table 4.4 Circular hole modelled by boundary and finite
elements - Stresses within plane strain elements 114
Table 4.5 Tension of a long plate modelled by symmetric mesh 115 Table 4.6 Tension of a long plate modelled by asymmetric mesh 117 Table 4.7 Lined circular tunnel with full adhesion
on interface 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 - 12 -
Page Table 4.11 Equivalent nodal forces and displacements for various particular solutions 132 Table 4.12 Equivalent nodal forces given by use of stiffness matrices K',K^,(K^) T 132 Table 4.13 Square block modelled by 32 boundary elements 135 Table 4.14 Square block modelled by finite and boundary elements 136 Table 4.15 Stresses at centres of plane strain elements of large problem for K^=0 139 Table 4.16 Stresses at centres of plane strain elements of large problem for K =1 140 i l CHAPTER 5 Table 5.1 Symmetric almost rigid wedge within infinitely stiff rock 172 Table 5.2 Symmetric elastic wedge within rigid surrounding rock 176 Table 5.3 Symmetric elastic wedge within elastic rock, without excavation sequence 177 Table 5.4 Symmetric elastic wedge within elastic rock, with excavation sequence 179
Table 5.5 Asymmetric wedge surrounded by rigid rock; 3^=35°,a2= 5° 184
Table 5.6 Asymmetric wedge surrounded by rigid rock; a^=20^,a2= 5° 185
Table 5.7 Asymmetric wedge surrounded by rigid rock; a^=35^,a2=20^ 186 - 13 -
CHAPTER 6 Table 6.1 Material properties 190 Table 6.2 Discretization 191 4 Table 6.3 A ctiv itie s 191
APPENDICES * Table A3.1 Determination of function arctan 241 Table A3.2 Integrals I for isotropy 243 Table A3.3 Integrals I for orthotropy 245 A3.4 * Table Analytically evaluated integral's of kernels U s 246
*
*
* NOTATION AND CONVENTIONS The following notation is used unless defined otherwise A : Support force in Chapter 5 A s : Cross sectional area 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
H Horizontal coordinate In Chapter 5 horizontal force In Identity matrix of order n J Jacobian matrix In Chapter 5 force
JCS Joint compressive strength (same as q )
JRC Joint roughness coefficient (similar quantity to
K S tiffn ess
Ratio of horizontal to vertical stress KA L Length ; In Chapter 5 base length of wedge
M In Chapter 5 strength parameter - 15 -
N Shape function
In Chapter 5 normal force
P 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
Stress concentration factor Scf T Kernel function
Tensile strength of wall rock * To TR Coordinate transformation matrix
U Kernel function
Vertical coordinate * V V Maximum closure for a joint me W Work
In Chapter 5 Weight ¥■ X Solution vector
Y Load vector
a Nodal displacement * In Chapter 5 wedge angle
b body force
b. Boundary element i 1 * c In Chapter 5 non-dimensional support force
c . . Coefficient of the free term ij d Total displacement
d(b,e) External node number of node e of boundary element b
e Internal node number for boundary elements - 16 -
In Chapter 5 angle
f Internal finite element nodal force
g Acceleration of gravity-
h height
* In Chapter 5 the height of the wedge
height to free surface ho i Dilation angle
Joint element i
k Joint normal stiffness n k Joint shear stiffness s m. Membrane element i 1 ♦ Normal direction to a surface
nG Number of Gauss points
Pressure at the free surface po Plane strain element i ♦ Pi qu Unconfined compressive wall strength r Distance between the first and the second argument of
the kernel * s Tangential direction to a surface
Cohesion so t Traction on a surface
* t Given tractions on a boundary
u Displacement component(in shear direction if direction
not specified)
* Displacement component in normal direction V Dilation rate
w Weight function
X First argument of the kernel (or coordinate)
y Second argument of the kernel (or coordinate) - 17 -
z Depth of excavation
r Boundary-
A Difference operator
Stress component ij due to a unit force in direction k 4 _iJk n Total potential energy
£ Addition operator
ft Domain
* a Acceleration
3 Angle
Y Engineering shear strain
♦ 6. . Kronecker delta ij 6 Virtual displacement operator
6(x) Dirac’s delta
e m Strain C Scaled coordinate
U Second curvilinear coordinate
0 Angle between axes H and 1 # D C
“3 l“ • Angle between two joint sets
1 Square root of (-1)
X Lame' constant
* y Lame/ constant
V Poisson’s ratio
First curvilinear coordinate
Very low stress *
7T Ratio of length of circle circumference to diameter
P Density
PS Unit weight (p«g)
0 Stress ; in Chapters 2 and 5 , normal stress -18-
o ’ Effective stress
0 c Unconfined compressive strength of unweathered rock
T Shear stress
t ? Peak shear strength , (same as )
* TP Residual shear strength , (same as )
4) Friction angle
Basic friction angle of unweathered dry smooth surface *h
4) Residual friction angle r r
4) Angle of frictional sliding resistance along the y contact surfaces of the teeth
CjO : angle
♦
Superscripts
c Complementary function
P Particular integral ■4*
0 Initial
T Transpose
*
* - 19-
CONVENTIONS
If A is any alphabetical symbol,then
A : Matrix A
A : Vector A
(A) Components of A in HV coordinate system 4 HV (A) Components of A in 12 coordinate system 12 (A) Components of A in sn coordinate system sn (A) Components of A in xy coordinate system xy
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
4k
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 Unit
Length mm
Force N
Stress MPa -20-
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 4 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. 4 The constitutive law parameters for the various m aterials 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 4 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.
4 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 -21-
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. -22-
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. -23-
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 Ergiin(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(l97l) 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 m ass,till the material becomes a me chanism and collapses. 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.I.I. Mechanical properties. 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 Poissons ratio 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 fc ( Peres , (1966)) was found to be adequately represented by the formula x^/A^+y^/B^+Z^/G^=I, E=x^+y^+z^, 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 * order equation of the form;x^/l^+y^/m^+z^/n^+x?z^/p^+y?z^/q^=I ,which is. a quartic. Structural material * The mechanical characteristics of the various materials are well de fined and may be found in appropriate codes as GPIIO or_DIN-IQ4.5::forocon- crete,BS 4.4.9 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 2li7 GPa (Hoek and Brown,(1980)) with Poisson's ratio 0.25 . The Young's modulus for steel sets or rock bolts is 207 GPa . -26-
2.1.2.Simulation. Intact rock will be simulated with plane strain elements,and concre te lining and rock bolts with membrane elements. 2,1.2.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:
9/9x 0 — __ . — * 9/9x 9£/9x 9ri/9x 3/35 3/35 0 9/9y N. N. = H. ^J-1 N, B. = . 1 9/9y 3£/3y 9n/3y 3/3n 3/3n 9/9y9/9x L-. -
N= (l2HltI2N2, ----- •i 2H8) (2.2) (& ,B 2, ...... ,Bg) (2.3) The Jacobian matrix is
9x/9£ 3y/35 9N./3C-X. 3 1 ^ / 3 5 ^ J= 11 ,i=I,2,..,8 (2.4-) 9x/9r) 9N./3t t x . 3y/3n __ 1 13 ^ / 3 ^ The inverse Jacobian becomes
35/9x 9ri/9x 3y/9n -3y/3£ r 2* = (1 /d etj) * 35/3y 3n/3y -9x/9ri 9x/9£ (2.5) The constitutive law is transverse isotropy.In three dimensions the law is given by: * es l/Ej - V E2 -V1/E1 0 0 0 °s en -V2/E2 !/E 2 -v 2/e 2 0 0 0 0 n ez -V1/E1 ”V2//E2 0 0 0 Oz * Y'nz 0 0 0 1/G2 0 0 Tnz ^sz 0 0 0 0 i ./g -l 0 Tsz ^sn 0 0 0 0 0 1/G2 Tsn — — - where the axis of symmetry is n,subscripts 1 describe behaviour within the isotropy plane sz,subscripts 2 describe behaviour within the aniso- -27-
Node coordinates in plane. Node €i Hi I -I -I 2 0 -I 3 i -I k i 0 5 i I 6 0 I 7 -I I 8 -I 0
Shape functions Corner nodes N?=0.25(1+5's^)•(l+n*n^)*(5’C^+rrn^-l) ,i=l,3,5,7 Midside nodes N?=0.50( (£??(l+^vf^) * (l-o2)+n|' (l+n*riiM.l-52)) >i?=2,4.,6,8.
¥
Figure 2.1 Eight node serendipity element (Owen & Hinton 1980) -28- tropy planes sn,zn and G^ is dependent i.e. G^=E^/(2 '(1+V^)).If axis z coincides with the axis of the excavation,the relation for plane strain becomes (Zienkiewicz,I977) : n(l-i>v2)/(l+V^) »v2 0 (e) = E2 n*vr 1-V ~ sn n (l-V1-2-n'-vp n Ysn 0 m-(l-V1 -2nA>p sn
(e)~ sn =(D) - sn -(a) ~ sn ,n=E./E0,m=G0/Eo 1' 2 2 2 (2.7) If axis s.'is inclined to x at an angle a (Fig.2.2) .then a rotation of axes is needed for D. (D)- xy =Ti— *(D) - sn *T— (2.8) cos 2 a / sm. 2 a* sina«cosa T= sm. 2 a* cos 2 a * -sm a-cosa (2.9) -2-sinacosa 2*sina*cosa cos^a-sin^a where T the engineering strain coordinate transformation matrix i-e- K-a+f=Q (2 .10 ) ~ ~ nG nG K. .= J 1 / 1B ;' D *B. *detJ • d£d0 =£ I t'(S ,n )..-w -w (2 .II) ij -l -l -l - li P q ij P q p q T'.i j .=B. ~ i * D----- *B .*j detJ The integration above has been performed numerically using a Gaussian quadrature formula.The internal forces f are given by (Fig. 2.3) : f=f,+fb s +fn 0 (2. 12) where b denotes body forces,s surface tractions,and 0 initial stresses. j q f,bi .=(P xi* . ,P yi . )^=b -l-ii J^N. • b-detJ-d£dr| =ZE11Y1.(£ p ,0 q )*w p -w q (2.13) . p q Q.=N.•b-detjl l . For gravitational loading b is given by b=-cg-(0,l)T (2 .14 ) For quasistatic earthquake loading with -a ,-a accelerations^b is x y Pg-(“X>“y) (2.15) -29- # ' * s axis on isotropy plane n axis of symmetry for transverse isotropy )► Figure 2.2 Axes for transverse isotropy and global cartesian system Figure 2.3 Sign convention for internal forces. -30- The nodal force at node i due to distributed tractions on surface f is 1, t " 3 x/35 f s i.=(P xi .,P yi .)= s f /N.-1 t n (2.16) - t n t, t 3 y/35 The nodal force due to residual stresses is « f„ , = (P .,,P .)!= )) b1 ( .T «) -j-d? dn (2.17) 0i x i y i 0 -l-i i xO y0’ xyO The external loads at node i are ,T Q.=(Pl x i.,P y i.)' (2.18) All integrations are implemented numerically.A two point Gauss fo rmula is used,i.e. 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(1977) 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. 2.1.2.2.Membrane element. The characteristic of a membrane is, it is thin and stresses do not ■ t 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 Poisson's ratio is neglected). The element is three node isoparametric (Fig. 2.4-). Geometry. Three nodes define the position of the element as -31- # =x/N.1 e =3 ug/ 9s o=E*e 5=-i 5=0 5=+i 0---- —e- -o Mapping on line £ % 3 N^.=0.5* 5* (5-1) *■ N2=l-52 * « Figure 2.4- Isoparametric three node membrane element. -32- x=(x,y)T=Ni»xi=Ni*(x±,yi)T (2.19) The shape functions JI. ..are defined on line K in Fig.2.4-. The first derivatives of the shape functions are N.|=£-0.5 ,N2=-2*£ ,N^=5+0.5 (2.20) The displacements are defined from u=(u x ,u y )T=N.*a.=N.*(a l l l xi.,a yi.)T, i=l,2,3. (2 ,21) The mapping of line s of the element to line £ is through the Jacobian j(?)=ds/dS=((dx/d5)2+(dy/d5)2)1/2 (2.22) (d/d£) x=(d/d£) !L*x^=N^*x^ (2.23) The direction of line s is defined from dx/ds=(dx/d£)•(d^/ds)=N^*x^,j “1 (2-24) The strain is defined as the change in length per unit length of an infinitesimal element. £=(d/ds) x^*(d/ds) u= J~2«(xT*N.)*N.*a. (2.25) The strain displacement matrix _B is defined from C—B*a , B=(B^,B2»B^) (2.26) B.=J"2*(x T*n !)*n ! 1 J J 1 (2.27) a—_/ v a^, a^»3^ xT / (2.28) The constitutive law is given by G=E*£+GU =3«B«a+anr* •+ U (2.29) Imposing static equilibrium 6aT*Q+/6uT*b*dV = J6eT*G*dV (2.30) V a* Substituting for u,e ,G from 2.21,2.26,2.29 we get 6aT*Q+/6aT*NT«b*dVr-* ^ e** = J\5aT*BT-(E*B*a+Gn)-dV A/ Aj -V U (2.31) K*a+f=Q — A/ r* (2.32) The stiffness matrix component K^. relating nodes i and j of one element is given by, -33- K.iJ . =E* A5• 5 /BT*BT*ds=E*A;/B,F*BT*J'd i J 5 i J £=E*A .^(B?* |,= i i bT* J J* w) P - (2.33) The internal forces due to body forces b and initial stresses are f= fb+f0 (2.3 i) f, =A-/HT-b-J-d5 = b‘A-Z(NT-J-w) (2.35) * ~b s ^ ~ p T T b=-pg(0,l) or b=pg(a x ,a y) T nG t fn=A* /B-an-J-dC = an-AyZ(B-J-w) (2.36) A^is the uniform- thickness of the membrane.For rock bolts A5is 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.All solid elements may be made plastic by changing the appropriate module 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 4 ntinuous, if not it is termed discontinuous.Continuous strength anisotropy for rock has been investigated by Jaeger (i960), McLamore and Gray(l967) and others.A special case of discontinuous anisotropy,is a rock isotropic in strength,cut by a continuous joint set (jaeger (i960)).His theory was 4 named the plane of weakness theory.Bray (1967) investigated discontinuous strength anisotropy with one, two, six and multiple joint sets.In the last case ». if the strength of the joints is the same or continuously varied 4 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 4 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. 4 2.2.1. Mechanical properties. It is argued that experimental results obtained on isolated rock joints IT 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 4 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.I.I. 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 fa ilu re . 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 model,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.In 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 (I980)having more results from laboratory tests adjusted some constants into their failure equation (power constants for a ands v). * They derived also a varying i Patton law (Fig. 2.5).The prediction of stre ngth in biaxial tests was also investigated.Jaeger (1971) suggested-a va rying cohesion law (Fig. 2.5).A general criterion for rough joints was ♦ developed by Barton (1971,1973»1974-) .It is in the form (Fig. 2.6) x/a^=tan((jRC) -logI0 (JCS/a^)+ * Patton's Law low stress x p =otan((h,ti) high stress t =s0+o*an<|>0 # * Figure 2.5 Peak shear strength. to Xp =(o( I-a )-(v+tan(l) )+as-T0)/(l-(l-a s>v.tan(|)r ) v=(l-a/(rpT) )^tani0> a s=I-(l-a/(r>oT) )L K=4-,L=I. 5 v=0 , a =1 for a/a^I.O’ri #■ * B arton’s Law. in Region II V °n =tan( JRCi°gi0(JCS/^ n^ ^ * in Region III n=tan(jRC-logI0(oi -a3)/anH«t> fe) in Region I straight line q and JCS are the same O axial stress at failure O confinement Figure 2.6 Peak shear strength. -38- be taken 30^ if not known.The joint wall compressive strength(JCS=qu) varies from Gfor unweathered joints to 0.25*c?c for weathered ones. If the dilation angle at peak i p and the maximum dilation angle i at o extremely low normal stress are known,then the formula might be written T/an =tan((90°-i o )• (i p /i o )+i o ) (2.38) No significant scale effect exists for peak shear strength of tension joints.Scale effect is significant for displacements. For a/JCS<0.0I Barton(I976) suggested a straight line to zero.A curve with normal tangent at 0^=0 seemed also appropriate.For high stresses i.e. a/JCS>I.0 he assumed that confinement was of importance and hence JCS was no longer appropriate.He introduced in his formula instead of JCS,a^-a^ where is the axial stress at failure and G^ the confinement. This formula reduces to the previous one for G^=0 in which case 0^=JCS. Barton and Bandis(l982) 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(l967) 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 parameter i n=T / t^^. 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(l976). 2.2.1.2 Deformability. The deformability of joints has been discussed by Goodman(1970,1974-> 1977).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(l979) performed cycled tests on artificial specimen joints with, very regular geometric form.Hungr and Coates(1978) studied the relation of deformability of joints to rock foundation settlements.Walsh and Grosenbauch (1979) model led the compressibility of fractures.Swan (1983) showed the functional re lationship between .normal stress,normal 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 al (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..I. No tension and laminar elements. Zienkiewicz,Valliapan 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 element(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 -4.0- 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, (1981)) 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.) 4 The method is especially suitable for dealing with cracks (Roberts and Einstein, (1979)) and slit like openings.lt 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 /ft n each element. Dds m dn f V 2a "k - + D,dn =u**-u+n n , D, ds =u s -u+ s (2.4.2a) If tractions are prescribed* t^ A ^ . Dl (2.39) I.f displacements are prescribed* (2.40) if mixed conditions exist,the above equations may be rearranged to form a system of linear equations with b^ the known quantities, * bd=cdj-Dd (2-^> The displacement discontinuities are defined in the sketch above and can be written in vector form Dd=(Dds ’Ddn)T (2 ' i2) where s,n is the local coordinate system;-/+ indicates the side of the element;i,j are nodal points;and ,B^ are boundary influence coefficients for the tractions and the displacements respectively. *• D^must be computed first.Tractions and displacements will be computed by substitution of 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(1968).Their model shown in Fig.2.7 is afour noded two dimensional element.Two independent com ponents for stress and strain exist i.e an>T and e , zQ»Applying stan dard finite element procedures,the stiffness matrix becomes, 2k 0 0 -k 0 0 s CO s "2ks 0 2kn . 0 kn 0 -kn 0 “2kn k S 0 2ks 0 "2ks 0 -ks 0 0 kn 0 2kn 0 ■“2kn 0 -kn k = l (2.43) - ks 0 - 2ks 0 2ks 0 ks 0 0 "kn 0 -"2kn 0 2kn 0 kn 1 to 0 0 0 0 to "ks ks 2ks 0 "2kn 0 0 kn 0 2kn n and k s are functions of e n ’e s , and the ; 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 shown in Fig.2.8.The no tension element technique has been compared to the joint element one by Heuze' et al.(l97l) for the case of borehole jack deformability tests.Goodman and Dubois(1971,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 _D has become full.The constitutive law for dilatancy was formulated by transforming D for an assumed smooth plane parallel to the direction of the asperities.For a smooth plane J^is diagonal and given by, ks 0 ■ So" (2.44) _0 kn - -43- ♦ »• * Figure 2.7 First joint element (Goodman et al.I9&8) w Figure 2.8 Three dimensional joint elements (Mahtab & Goodman 1970) - u - By rotating the coordinate system by an angle i,D becomes ,T k s 0 k s -cos^it k n-sin^i sini-cosi * (k -k s ) n D=TR *T 2 2 0 kn J sini*cosi*(k -ks ) n k • n -cos i+k - ssin i "k-to CQ k * sn k -k (2.45) k ... ns sn L ns tin. 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.lt 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 would 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 nsk and k sn exist and the variable stiffness 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 ss =(3 t / 3 u ) , k nn =(3a/3v) , k sn = (3 t / 3 v ) , k_ ns =(3a/3u) (2.46) 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, (1972)),, showed the suitability of the method for parametric study.Ghabousi,Wilson and Isenberg(l973) 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. X ■AAAAAr »t W W ^-A /W - k:S * ke kj ke The stiffness matrix is ke + kj -kj K= (2.4-7) . - kj ke + kj ♦ For large kj/ke^this matrix becomes ill conditioned.If now the unknowns are changed to be u^ and Au=u^-u^,and a= F~1 if ,then Lo i j u=(u^,u^)^=a*u =a*(Au,u*))'1' and ♦ K*u=K*a-u =(K*a)*u =P , (a^«K-a)*u =a^* P (2.48) The new stiffness matrix avoids the previous problem.lt is given by ke+kj ke K'=aT* K•a= (2.49) ke 2ke For the two dimensional case they considered they defined a as I*i 0*f lit where I if and Of* are the 4X4 unit and zero J)n Iif Otf * matrices respectively.The housekeeping of the global stiffness matrix becomes more complicated. Goodman(1975) included a two dimensional joint element program in his #> book.Desai,(Desai,(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, -4 6 - - T ks 0 0 Au T = a = 0 kn 0 Av (2.50) M0 0 0 k0) Aw - a 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).M q is the moment about the centre caused by the linear variation of a ,Aw is the rotation of the element caused by the linear variation of Av.The rotational stiffness k =k0) n*L3/4>where L the length of the element.Hittinger and Goodman(1978) 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(1978),that can treat the entire behavioural history of a rock discontinuity without dilation.Ke Hsujun .(1979*1981) used joint elements,non linear material properties and load cycling.Heuze(1979) 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(1981,I982) presented a new model for axisymmetric in terfaces, such as found in shaft and footing design.A new model for di- latational effects was also included.VanDillen and Ewing(I981) discussed * a new version of BMINES,a s ta tic three dim ensional non-linear program with joint elements,whose constitutive relations are posed in terms of plasticity theoryji.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 et al (1983,1934) presented a thin layer element. Carol and Alonso(1983) presented an isoparametric quadratic joint element using a constant peak shear displacement law and dilation. -47- 2.3.The .joint element. 2.3.I.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^O^x^+x^) , x 2=0.5(x ^ + x ^ ) , x 3=0.5(x ^ + x ^ ) (2.51) The coordinates at any point in the middle line may be calculated as x(?)=Ni (?).x1 , 1=1,2,3. (2.52) 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 i in the xy plane is given by, Aai=(Aui,Avi)T=(u^-u^ ^ ,v^-v^ ^) T= -1 0 1 0 ( u ^ . v ^ u ^ v W ) 1- T *• a i (2.53) _ 0 -1 0 1 1= [-v j (2.54) For all three nodes T T Aa=(Aai, A§2, Aa3) , a=(ai,a2,a.3) (2.55) 6X1 12X1 The relative displacements at any point are 1 _ Au =(Au ,Au )T;=N. •Aa.=] (2.56) ~xy X y 1 - 1 -N. 0 N. 0 1 1 N.= (2.57) 1 0 -N. 0 N. 1 1 The Jacobian determinant is given by J=ds/dg= /((dx/d£)2+(dy/d£)2 )= /( (2 .58) 1 where N is the derivative of N with respect to E, The direction of s is defined from dx/ds=(dx/d£)* (d£/ds)= J"V n ! ^ , i=l,2,3. (2.59) -4 8 - * x(a)=x:lE(5) i=l,2,3. ♦ ( 1) * Element Shape functions N1=0.5-^-(C-X) ♦ n 2= ^ 2 1.0 ♦ N3=0.5* £•(£+!) 5 = -l 5=0 5 =+l * f x t e s ’ t' t✓ i e : S Sign conventions for stresses ^77777777777777 and strains. Figure 2.9 Isoparametric quadratic joint element. -49- The coordinate transformation matrix is dx/ds dy/ds R= (2 .60 ) -dy/ds dx/ds The strain is defined as e= (eQ ,e^) = (i. 0/t) (Au, Av^= (1.0 /t)•£-(Au,Av)^ = (1.0 /t)'lla±= =B.- l • a ~i . (2.61) 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=1.0 • The constitutive law can be written in the form, dg=d(T,a) =D*de=D/13 *da , (2.62a) D= ss sn k k. . *- ns nn for constant JD the relationship becomes, g=g0+D*B* a (2 .62b) Imposing static equilibrium by equating the virtual work of the external forces to the work of the internal forces, 6a^,#Q=/6g^* q*ds = /6a^* B^* (ao+D_#B_*a)*ds= /6aT-_BT* g0*ds + /6aT*BT-D-B-a*ds + Q=f0+K*a (2.63) f o=/BT* g0-ds , K=/BT*_D*B *ds where fo the initial loads and K the stiffness matrix. Integration is performed numerically as, nG nG f 0= Z (B*a0'J*w) , K= Z (B -D-B-J-w) (2.6,0 - p-i - - p - p-i ----- 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 D 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 imbalanced stresses at the Gauss points after the state determina tion has been computed,are transformed to unbalanced nodal forces as, m nG T f =f B *(ae-ar)«ds = E (B •(ae-ar)*w'-J) (2.65) ~unb e — ~ ~ p P=* 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 stress vs normal strain,shear . stress vs shear strain.lt 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.I. Shear strength (Fig.2.10) Peak shear strength of joint I. Ladanyi and Archanbault*s(l970) failure criterion is used.The shear strength of the joint is assumed to be the sum of four separate strength components,ST1 ,S0 2 ,SQ,S..The 3 4 three first components assume no shearing occurs. From static and limit equilibrium we have, N* cosi+S*sini=V -51- y» a r 1 a — t J o in tl Ladanyj & Archambault dilation rate v=dy/dx Jo in t2 . Mohr Coulomb-Patton Residual to peak re la tio n Figure 2.10 failure criteria and param eters. -52- From kinematic consideration we get Av/Au=tani Ebcternal and internal work is given by W .=S‘Au ext * W^n^=N* Av+P• Au/cosi=N* (Av/Au) * Au+ (N+S* tani) * tanc{)^Au Equating external to internal work done i.e^int^ext^6 kave> S=N'tani+S*tani*tan S=N* v+S' v % tanct^+N * tancj)^ (2.68) where (J)^, an average friction angle for different irregularity orientations, It can be taken 30^±5^. 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(l966)). The strength of the rock is given by Fairhurst(l964j, TP=qu(1.0-n'a/qu)°*^*(m-1.0)/n , n=qu/(-TQ) , m=(n+1.0)°*?(2.69) 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 ag=As/A,a linear interpolation between the two modes a =0 and a =1.0 is performed,i.e. s s (1.0-a a (2.70) S=(S,+S0+S0)1 2 3 * s )+S U • s 0 5 S a*(l'.0-a )*(v+tanc{) )+a »q •(1.0+n«a/qu) * *(m-1.0)/n S ]J s u TP= _ = (2.71) 1.0-(1.0-a )-v*tan4)„ S X For ag and v an exponential interpolation is made between two extremes ♦ a =0 , v=tani a/q =0 s a/qu>l.o a =1.0 , v=0 a =1.0-(1.0-a/q„) kl , ld=1.5 s u (2.72) v =(1.0-a/qu)k2 k2=4-.0 In all the above formulae,qu has been used instead of as suggested -53- by Goodman.The new formulae for a =(a/q )^’l^,v=(d..O-(a/qu f’^'*)^‘*'^,'tani suggested by Ladanyi and Archambault(l980) have not been implemented. The degree of interlocking factor r| might be useful to be incorporated m in the program.lt affects q^ and Cf^ by modifying them to become Gu*n» aT* n respectively. Peak shear strength of .joint 2 . (Fig.2.10) This is a mixed Mohr-Coulomb,Patton failure criterion. The Mohr-Coulomb failure criterion is T^=SQ+CJ«tan(})^ (2.73) The Patton law is : TP=G*tan(<}> +i) for -G<-q U U (2.74) T -q *tan((J) +i)+(G-q )’ tancj) , -0>-q Mu u Hu r Hu The mixed criterion then becomes: TP=sn+o*tan( Very little is known on the variation of residual shear strength * T with a .It is known that for high confining pressure,peak strength r p equals to residual,i.e. for -0>-qu +T =T . At very low confining pressure Tr/TP=B ,where 0 polation between the two extremes is used, tr/TP=BQ+ (4.0-BQ) • 0/qu , 01.0 nu 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, ^^perc'x^+Cl.O-perc) »xr , perc=(Tp^-Tr )/(TP_Tr ) (2.77) where lies on the falling part of the shear stress vs shear strain -54- curve.lt must be acknowledged that the incompleteness of the model might be misleading in some cases.Say for example a small normal stress is first applied oh 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 will 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 level Gi* Curve 2 is due to further partial shearing at stress level O2 .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 so that becomes small,the error is also becoming smaller. -55- Expected prototype peak shear strength model Figure 2.II 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(I975). « o= a -V /(v -e ) , V =-V -5/on (2.78) Om mnn m me 0 is a negative very small stress V is the maximum strain(positive) that can be attained from Cf=£l. ♦ C q is the negative initial stress. is the negative minimum closing strain for 0=(Jq » The tangent stiffness is given then by, * (2*79) Joint 2 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 4 regions, £ V is defined here as the positive strain from 0= O q to a=0,and is given by, V =-an/k (2.81) % m 0 n Shear stress vs shear strain. Fig.2.13. 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 £ ,£ ,£ ,£ that are the points ° rn pn pp rp at which negative or positive,peak or residual strains are first attai ned. They are defined from the formulae below. -57- 4 Region II O. =Vm»0nO/( Vra-£nn) Vm=-Vmc*5rfo ' nO ♦ 4 JointI - Hyperbolic compression curve Region I infinite stress Region III zero stress ♦ Region II O =OnO+k^.>enn 4 Vm=-(7nO/kn Figure 2.12 Normal stress vs normal strain law -5 8 - * The shear stiffness is taken zero if t is outside the elastic range. % Figure 2.13.Shear strain vs shear stress. -59- peak,positive £ =M •£ residual,positive ; M=4.0 rp pp £ =-£ peak,negative pn pp (2.82) £ =-£ residual,negative rn rp * =TPk/k reduced peak positive due to strain softening £pkp s reduced peak negative due to strain softening Gpkn £pkp Table 2.1. Shear strain regions I 2 3 k 5 6 Region Min.strain Max.strain Name slope (d g) shear stress (x) I - £ negative 0 -T rn r residual II negative (t -T )/(£ -£ ) Ern £pkn p r rn pn falling - V ^ v W III elastic k k • £ Gpkn Gpkp s s s * IV positive (t -T )/(£ -£ ) Gpkp Grp p r pp rp falling V d s( v W V positive 0 T r Erp residual During the simulation of excavation or loading,having arrived at a * stable position (£ jT1),after a number of iterations,this point will s 1 represent the end of the load step.A new load step then will be applied and a movement from (£g ,T X) will occur which must be compatible with * the constitutive law illustrated in the figure.A number of displacements will be tried then through iterations,that, will always refer to (e ,t x) s 1 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 u^ is the elastic strain at the end of a load step.In Fig.2.14- a 3-D view of £g, T and a 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 from s 0 to A.Otherwise we move on the plastic plane from B .Note the shift of the elastic origin. - 6 0 - T=k^'£^ defines elastic plane p T = A*o defines plastic plane Figure 2.14. Three dimensional sketch for e ,T, 0 . * c s -6l- 2.3.2.3. Normal strain vs shear strain (dilatancy).Fig.2.15. 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 through ♦ shear strain to shear stress.The normal strain may thus be written z =z +£ (2.83) n nn ns Dilation refers always to the initial shear strain which for reversed Ik loading might give unrealistic behaviour.lt depends on normal stress and shear strain.The equation is as follows z =a* tani* z for z 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=(1.0-a/qii)4’ for Cf/q^l.O lu Lu (2.85a) a=0 for a/q >1.0 Joint 2 assumes for a I a=1.0-a/q for a/q <1.0 hi ui (2.85b) a=0 for a/q >1.0 This is not consistent with the failure equation assumption of constant i.for a/q <1.0 and i=0,for a/q > 1 .©.Nevertheless the bilinear failure 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 stress^ 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 - 6 2 - Jointl Joint2 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.Hittinger et 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 W 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 c step C activity Q. load sequence where c is the symbol of "is a subset or identical of". 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.2.16,2.17) 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 ment, but 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, AN=-(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, AN=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). -65- Joint closing and q< Q nn \ * (0,resid2) Elastic system \ previous shooting point. (SIGMA,DELV) ^ ---- SIGM, VREAL) ' . - * // DELN=(VREAL-DELV )* D22 L/ V_ _ L (STRESS,DELV)^ ^ new shooting point VREAL=VM*(SIGM-RESID2)/SIGM * -M------*( SIGMA, DELV) enn —*------7^"^------DELN=SIGMA-SIGM/ j N /'“^ ^ ( s i G M ,VREAL) point on curve ♦ / (STRESS,DELV) new shooting point r \ '\old shooting point \ Elastic system. \ SIGM=RESID2*VM/(VM- (0,RESID2y Joint opening or q>Q * ♦ Figure 2.16 Iterative process (for compression)- Joint.^no dilation. - 6 6 - * * * * * 4 Note:The stiffness in zone I is taken I O'* »xksi The stiffness in zone III is taken 10 “^xksi Figure 2.17 Iterative process(for compression)- Joint2^no dilation. 4 « * * 4 0 Figure 2.18 Iterative process (for compression)- JointI with Dilation. * # 0 0 0 * * Figure 2.19 Iterative process(for compression)- Joint2 with dilation. -69- J oint 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 + e , e =(a/q -1.0)^«tan i •£ f i o= min(|e |,e ) (2.86) n nn ns ns u s * s s r The differential strains are calculated from, ♦ de =de +de (2.87) n nn ns (2.88) de nn =(de nn /da)*da (2.89) de ns =(9e ns /9a)-da+(9e ns /9ee)•(9e s s /9x)*dx The derivatives of strain are given by, 2 90 (9ens^3a^ = ^/ qu^* ((aAlu)"^#0)3*tan 1 *^s=Fdil ( . ) o (2.91) de nn /do=-5fV me / 2=1.0/D nn * de =(du/dx)*dT + F *dO ( . ) s ns 2 92 The cross flexibility F is given by, SH F =(8e /3e )•(3e /3t ) (2.93) sn ns s s and is zero for Ie I>e * 's' r The tangential stiffness matrix relating the strain vector (de .de ) n T to the stress vector (dx,da) is the following: - 1 . du/dx F ■# ns D, tang (2.94) sn ae/ao+dens nn /da Diagonality as described by St.John is attained by putting F = F =0 in sn ns 4 the augmented stiffness matrix,..The diagonal stiffness terms are given by the inverses of the diagonal terms of equation 2.9/-.They are: D =D /(1.0+D *F,.J (2.95) n nn nn dil D =dx/du s where D .F.,.- are defined in eqs.2.9I»2.90. nn dil -70- Joint 2. The total normal strain is the sum of one strain associated with normal stress and one strain associated with shear displacement. £ =£ +£ , £ =a«tan i *E = (d.0-a/q )*tan i *E n nn ns ns s hi s £ =(cJ-Gn)/k nn U n Working similarly as for joint i we derive the stiffnesses as de n'/da=1.0/k n -tan i «E s ^u /q =i.0/k n +Fj., dil (2.96 ) ♦ Dn =da/d£ n = k n /(i.0 + k n *F dil ) (2 .9 7 ) F,d il =-tan i »E s /q ni (2.98 ) is given by eq. 2.87 * The shear stiffness Ds=dx/du is unchanged. 2.3.3.3. Shear stress vs shear strain. 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 k g;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 AS=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 Q and t vary simultaneously and the cross terms become important.Very stiff joints might also create numerical pro blems. 4 -71- Shear strain iterations under constant a * 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 Bq has always been taken 0.5 and the residual fri ■* ction angle 10^.Plane strain elements have also been used.Young’s mo dulus has been chosen to be 100 and.Poisson’s ratio 0. a.Two plane, strain, and two joint elements. ♦ The discretization is shown in Fig.2.21a.The two plane strain ele ments Pj_»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 2 »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 loads(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 0 and a lower strength q^ 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 observed also here.In row 3^qu has been chosen small so that the normal stress exceeds qu and hence dilation and strain softening do not occur.The 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 displacements -73- * * * ♦ % ♦ Figure 2.21 Strain softening joints(examples) -74- Table 2.2 Two plane strain and two joint elements. E=100,V =0,B0=0.5, 1 Model 1 1 0.487 0.438 Within elastic 2 0.486 0.065 q =100 range.Fluctua- u 3 0.509 0.067 i=5 tion of displa- 4 0.481 0.068 H II 1 5 0.507 0.061 cements due to 6 0.068 0.481 lack of k terms. sn 7 0.507 0.061 2 Model 1 1 0.973 0.547 In plastic range. 2 1.310 0.079 q =10 Displacements 3 1.270 0.287 i=5° fluctuate due to 4 1.190 0.314 av=-2 5 0.969 0.4H zero cross stif- 6 1.000 0.243 fnesses k ,k sn ns 7 0.846 0.329 8 0.867 0.222 9 0.746 0.293 10 0.799 0.084 3 Model 1 1 0.973 0.153 In plastic range, and on beyond q^. V 1’- 5 2 1.200 0.001 i V 2 4 Model 1 n 1 0.973 0.278 In strain softening q =10,i=0 2 1.370 0.047 au =-2 o range. 3 1.450 0.005 5 Model 2 n 1 0.485 0.176 In strain softening q =100,i=5U range. 2 0.756 0.027 up=0.267,ur=1.071 a„=-l,kv n=1. 3 0.794 0.003 -75- were in the falling part of the shear stress strain curve.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.2.21b.The.plane strain element p^ was initially prestressed horizontally with a horizontal stress a^=-l.The three joint elements were also prestressed with a normal stress On=-l.A stable position was sought through a number of 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 so that the shear displacements of joint elements j^ and j^ lie within or outside the elastic range.The number of iterations 4 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=ioo,v=o,BQ=o.5,(})=io0,aH=-i,an=-i,ks=i.o,vmc=io.o,qu/T0=io.o,i=50. row description Iter. Ver.displ. Maximum Unbal.load Comments 4k ji&j2 h 1 Model 1 1 0.108 0.132 1.122 Within elastic q =10 2 0.235 0.155 0.259 range. 0.018 ♦ Ci=0.025 3 0.289 0.04.5 A 0.291 0.0097 0.000 2 Model 1 1 0.827 0.222 0.066 Always out of * q =10 2 2. HO 0.258 0.093 the elastic 5i=l 3 4..4-60 O.IH 0.14-9 range the side A 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 in * k =1.0 1.690 0.000 n 3 0.003 plastic range. ♦ 2 .k 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.(l978). 2.4-.1 Excavation. Previous work includes,Clough and Duncan(19&9)>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. Nodes within the excavation area become inactive and fixed.Thus 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 lyjjig 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, r nG P=-/B*o*ds=- E (B^a-J-w) (2.99) ~ --- , — ^ P P=1 % d. Add the incremental displacements,strains and stresses to the to tal ones. e. If 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,nan 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 m information stored in the files for an element depends on the value of that flag.If IFLAG is 0 the element exists.If it is 1 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 itwgreater than or equal to 3 the element was excavated during a previous activity.Some information always remains for an excavated element 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.o.f. is increased0an 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 aforraentioned 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. Quasistatic 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.lt 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,resuiting 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 (l972),Eissa (1980), Harrison and Gerrard (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(l973) who dealt \ 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=1+Flld E2a^E2=1+E22f E3a^E3=1+E33b V21a^V21=1+E22f (3.1) V31a^V31=1+E33b V32a//v32=1+E33b G12a^G12 1+F12d+F12f -83- material f~ ~==t material 1 3 b a Figure 3.1. Three orthogonal sets of joints. 2 1 joint 1 Figure 3.2. Two oblique sets of joints. 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.due to the joints are de fined from the following formulae: (3.2) where k ,k are the normal and shear stiffnesses of the -joints d, f,b, relating tractions to displacements,and Fr is the frequency of the joints.We have introduced another multiplier to these terms,Pr, to take account of the persistence of the joints.This parameter lies between 1 and 0 and is completely empirical. 3.1.2 Two oblique sets of joints. Two sets of joints are assumed,intersecting at an angle 0. (fig. 3.2). Bray(l976) showed that if the intact rock material is isotropic and (3.3) 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 plane strain for the equivalent orthotr.opic material and the isotropic intact rock respectively,then according to Eissa (1980), (l+cos 2rj)2 sin22ri (l-cos 2a)2 sin22a F = — < ------+ ------+ ------+ ------+ F11 11 4 K K K K n2 s2 nl si F = — < • ~ — ) • sin22ri + (r— - --^-)*sin22a +F = F 12 12 21 4 n2 K s 2 nl si (l-cos 2q)2 sin22n (l+cos 2a)2 sin22a F = — 1 ------+ ------+ ------+ ------+ F' 22 22 4 , K K K K n2 s2 nl si sin22a cos22ri cos22a sin22a F = + F (3.4) 33 33 K K K n2 s2 si nl where f ^1=F22=(1-V2)/e , f ^2=-V*(1+v )/E , F ' =1/G 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. A special case arises if K s-^=Ks2 and ^nq=^n2 * From equation 3.3 we observe that cos 29. can take any value,that is the equiva- J lent continuum is orthotropic for any angle 0..This can be concluded J 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=0./2 , ti=tt/2 -0. +a = 7t/ 2 -a, 2«n = tt - 2-a (3.5) J J Substituting to equations 3.4 we get, f 1;l=0.5 - { ( 1 - c o s 2 F12_0,^ * ^ 1^Kn“1/Ks^* sin22 a ) + F'12 (3.6) F99-0.5«{(l+cos 2a)/K +sin22a/K + F' ^ n s 22 F^^= (sin22a/Kn+cos22a/Ks) + F ^ -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 m 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, «*=«“+ uP+ u° (3.7) t V t tp+1° ♦ 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. L*u=A-V* (V. u)+u* V# (V 'U )+yV. (V-u)T=-b (3.8a) * where b is the body force,and X and u are the Lame constants. L* is a linear differential operator. For general anisotropy the equation would be of the form, L*u=-b (3.8b) L* is now a more general linear differential operator. -87- 3.2.1 The integral equation for the complementary function. The complementary function satisfies the homogeneous equation L*u=0 The boundary integral equation for that function may be obtained * from Betti's theorem and the divergence theorem (Watson (1979))»or distribution theory (Lachat (1975)),and is given by (U (x,y)*t^(y) - T ..(x,y)-u°(y))*dS = 0 (3.9) -*-j j j y S+s(x|e) where 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(x|e): 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 singular solution,that is the displacement at *yT on the boundary in direction ’j’jdue to a unit force acting at tx l in direction 'i1. Note that U..(x,y) is symmetric both in arguments and indices for orthotropy,whereas for general ani sotropy it is symmetric only in arguments,i.e U ..(x,y)#J..(x,y). J ^ T..: the traction at !y ’ on the boundary in direction ’j ’,due to a unit force at Tx' in direction ’i1. # The definition of U and T can be written in an algebraic form, u?(y)=U .(x,y)*e (x) , L*U=6 (3.10) J J ^ t?(y)=T..(x,y)*e.(x) , T=a(U)-n = T(n)-U (3.11a) J J -88- where e_^(x) : a force acting at ’x’ in the direction fi' 6 : is the Dirac's delta function a(U) : the stress field due to displacement field U * n : the normal,unit vector,to the boundary (n) : a differential operator which for isotropy is given by, T^=X*n*V.+ u-n.V + U«n.VT (3.11b) * From equation 3.9 we get the following equation,that is known as Somigliana's identity. u°(x)=; D (x,y)-t°(y)-dS - /qT (x,y)-u°(y)-dS (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(x|e) - S(x|e), where s(x|e) : is the part of the surface of the sphere with centre at ♦ ’x ’jand radius £ ,that is contained within the region. S(x|£) : is that part of S contained within the sphere. The integral equation is given by, * c . . (x)*u?(x) + lim T. . (x,y) *u?(y) *dS =lim U . .(x,y)*t?(y)*dS £->0 J e->0 j y (3.13) S-S(x|£) S-S(x|£) The integrals are Cauchy principal values.The coefficients c_(x) * are given by, c..(x) = lim / T..(x,y)*dS (3.14-a) £-m d / | \ y S(X |£) For a continuous tangent plane, c..=1/2*6.. (3.lib) where 6.. the Kronecker delta, ij -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) (1+v) -i (x.-y-Mx -y ) U. .(x,y) = ------{ (3-^-v)-6ij.‘log(— ) + ------— J-} J 4-7T*E*(1-v ) T..(x,y)=--- ±----- •{ (1-2*v)• (n (y)•—*1—— *1- - n .(y)- — — -■) + J 4-• tt- (l-v) *r J *v* (x. -y.) • (x .-y.) x -y +((1-2* v) *6. .+2* — ^---- J-lJ-Jnfyy-S-lSj (3. !5) * ij 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 £o log(l/r),whereas the off-diagonal terms are products the direction cosines of r. Kernel T € o l/r . 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 A3.1. ♦ - 90- send ♦ Watson(1979) convention Ui0.(x,y) ^ ( x . y ) Banerjee and Butterfield(I980) convention G.-ty.x) Fji(y,x) uij(x »y)=Gji(y.x ) Tij.(x»y)=Fj.i(y,x) 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^ S ) = Na (5)‘xa u± (5) = Na (5)*ua (3.16) m t.(5) = Na (5)-ta 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(S) = (1/2)(5+1) N2 (5) = (l/2)«5*(5-l) (3.17) N3 (C) = 1-52 The derivatives D of the shape functions with respect to E, are given by, p - dMS) =5 + 0.5 D2(5) =5-0.5 (3.18) D 3(5) = -2-5 * The Jacobian is given by j(£)=ds/d£=i/{ (dx^/d£)2+ (dx2/d^)2} = /{ (Da*x^)2+ (Da *xa )2} (3.19) - 9 2 - * Figure 3.4 Isoparametric boundary element. It A 2 V * H Figure 3.5 Coordinate systems H,V and 1,2. -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 approxi ♦ 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 P 3 e (xa ).u®(xa ) + l l u«(x*<*-e>). / T (xa,y(£)>Ne(£)*j(£)«d£= J J b=l e=l b P 3 = I l t^(xd(b,e))-/ U (xa,y(£))-Ne(0-J(S)-d5 (3.20) J b=l e=l S. b where ♦ 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. t d(b,e) : is the global node number of local node number e of ele ment b (l 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. c/ as p 3 cr d(b,e)\ ur(x ) ^ u.(x ) „ _ J(5) (Xa).-J----- + J l / (T (xa ,y(€)>L > N ® ( 5 ) ----- d£ 13 1 j sc T sc sc sc * b=l e=l b p 3 tc(xd(b,e)) J(?) - z Z - ^ ------/ (Uij(xa ,y(5))-Eso)>K®(5)- ■d? (3.21) sc b=l e=l SC * where the complementary functions u and t have been scaled in the above equation by L ,that is the largest distance between nodes and E , SO so that is the first Young's modulus. The above equation may be written 4 in matrix form as, c , c u t - T = B'e " (3.22) SC ""sc ♦ c c where u and t the displacement and traction vectors for all nodes respectively,and T_and_Q the matrices that contain the coefficients C 0 of u. and t. in equation 3.21. 4 -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),provided due precautions are taken where the integrand tends to infinity.If node x does not belong to the element over which integration is performed,the G.L.Q.F can be used direc tly for kernels U and T. Kernel components never ^enc^ 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- £L nels over an element is performed,if the node x belongs to the ele ment. Table 3.1 Integration of kernel - shape function products over an element containing the first argument Kernel order a is middle node; a is extreme node; ♦ Integration over two Integration over two subelements adjacent elements a=d(b,e) a^d(b,e) a=d(b,e) a^d(b,e) T 1/r spiral G.L.Q.F spiral G.L.Q.F * log(l/r) analytic analytic* analytic analytic* U11’U22 + G.L.Q.F + G.L.Q.F + G.L.Q.F + G.L.Q.F 1 G.L.Q.F G.L.Q.F G.L.Q.F G.L.Q.F U12=U21 ^ - 1 ---- ><^| -* Nd(b,e) . a i a a a d(b,ej d(b,e) d(b,e) d(b,e) d(b, e) order of Md(b,e) at xa . 1 r 1 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 JJ,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 JJ 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 K , 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, (3*23) 4 (S*u°)12=(t°)12 (3.2A) where S = U -1*T (3.25) * A second order tensor coordinate transformation is needed in order to rotate the axes 1,2 by an angle -9 ,to H,V (fig. 3.5) > that is, •* ( S ) ^ = 'raT-(S)12*TR (3.26) where TR the vector coordinate transformation matrix. Then, ^HV = ^HV (3.27) Substituting 3*7 to 3.27 we get, (t^-t^-t0)^ = (S-(^-^-u0) )HV (3.28) * -97- 3.2.7 Particular integral. The particular integral for a distributed body force within the region could be chosen to be, * u?(x) = / U ..(x,y)-b (y)*dVol ,i,j = H,V (3.29) 1 Vol J This would entail the evaluation of volume integrals.Por a uni form vertical body force -p*g , ♦ /—s V h o II > = 1 •(P‘g'Xv - P'g'hQ+ pQ) (3.30) °v > 0 * where p is the vertical stress at level h .and K. the ratio of oJ A horizontal to vertical stress.This 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, * "h = V X1 +iV X2 + V Xl‘X2 + V * 1 + eH-X2 (3.31) < = where parameters a to £ are independent of position.The evaluati on of volume integrals is therefore avoided. The values of the parameters and their derivation are shown in Appendix 3,Section 3.3. ♦ -98- 3.2.8 Infinite domain. When we deal with boundary element regions that extend to infi nity, it would be convinient if the integration 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-j_= lim / U _ (xa ,y) «t^(y)«ds * rn-*°° 0 r r i=l,2 ; j=l,2 (3.32) I = lim f T (x ,y)*u?(y)*ds ^ J-J J r n-x» r 0 r 3. in an infinite elastic body are independent of position of x . Let us consider two points x and x (fig. 3.6),within a circle of ra dius R ,and an outer boundary at a distance rq from x^*. We 3 0 3. 0 * denote by T ,T ,U ,U the two by two matrices that contain the com ponents T . . and U . . ,with first argument of the kernels at x or x and second argument of the kernel at Then, « 9 3 n o % AT = Ta- T = (Ar— + Aw — )T(r +e -Ar,w +0 -Aw) n 9r 9w n n (3.33) AU = U - U = (Ar*9_ + Aw 9 )U(r +9 »Ar,u) +9. »Aoj) 9r 9co * 4 where we have assumed T and U to be functions of w and r,and * a a 0 Ar=r -r Au)=U)a a — (jl) 0 and * ^)<®1,®2,®3,®4. <'L The orders ofthe individual terms of AT and AU are, - 99- % Figure 3.6 Integration over renote boundary. -100- Ar— T C o — Ar« — U C — 9r r* 3r r 0 (3.34a) — T £ o — -^-U £o 1 9w rQ 9u) ♦ But also Am £ o (l/ip, hence Am —T £ (1/r*) ,Au— U £ (l/rn) (3.34b) 3m 3m * Hence AT £ ° (l/ri) AU € o (l/r0) (3.34c) c c If t , u are equal to the singular solutions T and U,then t C o (l/rQ) , u° £ o log (l/rQ) (3.35) ♦ tJ J hence, lim / AT..*u?»dsiJ J = 0 , lira / AU. iJ.*t force field acting within T_ . 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 in, for r,--*00 , are 0 given by, , c „ m m c TT in m t. = T. .»e, , u. = U. .*e. (3.37) J kj k j kj k * Hence the integrals 1^ and 12 become, r r TT 0 in m , In = ) J U. . «T, . *e, *ds , I = y Jf T.m 0. «U,TT ^. *e, ® 1 L ij kj k * 2 L ij ^ k m r in r * Noting that, Um = U° + AUm Tm = T° + ATm (3.39) -101- and from equation 3.36 that the integrals of AT*U^ and AU*T^ are zero, the expressions for the intgrals 1^ , become, I-. = / e?/ds i / T.9-U, ?• I e“*ds (3.4-0) * 1 ij kj ^ k X2 = ij kj ^ k T m r m r r For an equilibrating force field, l em = 0 (3.a ) m and the integrals become zero. If the force field is non-equilibrating,then let us consider the difference, I = I.-I. = lira / (U.?«T. ?-T.?«U. ?)*e. *ds Aik*ds (3.^2) 00 1 2 ij kj lj kj; k ey mf r n -x» r where 0 r — p i e = L e m ♦ Matrix A is antisymmetric.The diagonal terms hence are zero and the non-diagonal terms in expanded form are given by, A =_A =U *T +U «T -T «U -T • U (3.4-3) 12 21 11 21 12 22 11 21 12 22 % We apply Betti’s theorem for the fields due to unit force in 1 di rection and 2 direction. We note that, % * *'Ull‘T21 + U12'T22^‘dS •> ^ ^Tll‘U21 + T12‘U22^"ds W Fr are 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 boundary respective ly. Hence, -102- 0 0 (3 .U) {. A 1 2 ' iS U12 " U21 r where is the displacement in j direction at x due to a unit w force in i direction at x^. For orthotropy u ^ = UgJ = 0 and hence the integral 1^ is zero. If noii the infinite body contains holes on which distributed non w- equilibrating tractions exist,this body may be considered without the holes,by filling them and applying additional tractions to the infill,so that the displacements on the hole boundaries remain the same,(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 I over a boundary approaching infinity is zero. 4 * * 4 -103- 3.3 Example, 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 ments, (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 3^=100 , V—0 , p* g=l.0 Joints k =20 , k =2.0 , Fr=0.57 n 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 " h uv uv 1 0 -0.126 0 -0.146 9 0.007 -0.111 - - 2 0 -0.H3 0 -0.155 8 0 -0.104 0 -0.100 7 6 5 * 7fT3 6 1 . i Figure 3.7 Initial meshes for the examples of section 3.3 section of examples the for meshes Initial 3.7 Figure BUDR EEET EIN UJCE T A RVTTOA FIELD. GRAVITATIONAL A TO 0 SUBJECTED REGION ACTIVITY ELEMENT BOUNDARY A NTA MS LNTS --O' UNITS l-O-lO*' = _ 2.00 LENGTHS MESH INITIAL CIIY 0 ACTIVITY LN SRI AD ON EEET SBETD O GAIAINL FIELD. GRAVITATIONAL A TO SUBJECTED ELEMENTS JOINT AND STRAIN PLANE 3 INITIAL MESH LENGTHS LENGTHS MESH INITIAL 7 1 2 17 h -104- 4.00 2 6 __ l0I- UNITS l-0»I0-1 = 7 1 9 8 24 -105- * A BOUNDARY ELEMENT REGION SUBJECTED TO A GRAVITATIONAL FIELD. % DEFORMED MESH LENGTHS _= 1.0»10* UNITS ACTIVITY 1 LORD STEP I ITERATION 1 DISPLACEMENTS _ = 1.0-10*' UNITS * * R BOUNDARY ELEMENT REGION SUBJECTED TO R GRAVITATIONAL FIELD. FLOW FIELO LENGTHS __= 1-0-10* UNITS ACTIVITY 1 LOAD STEP 1 ITERATION 1 DISPLACEMENTS _ = 1.0-10'1 UNITS Figure 3.8 Boundary element region subjected to gravitational field -10 6 - ♦ PLANE STRAIN AND JOINT ELEMENTS SUBJECTED TO A GRAVITATIONAL FIELD. DEFORMED MESH LENGTHS _= 1.0*10* UNITS ACTIVITY 1 LOAD STEP 1 ITERATION 1 DISPLACEMENTS _ = 0.5*10*‘ UNITS l I * II ♦ I 1 * PLANE STRAIN AND JOINT ELEMENTS SUBJECTED TO A GRAVITATIONAL FIELD. FLOW FIELD LENGTHS — = 1.0*10* UNITS ACTIVITY 1 LOAD STEP I ITERATION 1 DISPLACEMENTS _ = 0.5*10*' UNITS * % PLANE STRAIN AND JOINT ELEMENTS SUBJECTED TO A GRAVITATIONAL FIELD. STRESS FIELD LENGTHS __ = l .O-IO"1 UNITS ACTIVITY 1 LOAD STEP 1 ITERATION 1 STRESSES _ = 0.5*10° UNITS Figure 3.9 Plane strain and joint elements subjected to gravitational field -107- CHAPTER U - COUPLING REGIONS WITH CONTINUOUS AND DISCONTINUOUS DISPLACEMENT FIELDS J+.O General Finite elements,boundary elements as well as finite differences ♦ have been identified for some time 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 1972, who 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(l975) 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(l98l) was the first to use a combined method with finite and boundary elements in rock mechanics. Beer(l982),by using his coupled finite-boundary element algorithm,showed 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, P=-/NT .t.ds = -/NT .N.ds.t = -C-t U.l) C=/NT -N-ds This formula may be applied also for the boundary element region to derive the equivalent nodal forces from the nodal tractions. By premu- M ltiplying then equation 3.28 by (3 we get, C-S-u = C-t (4.2a) or j^.u = -P (4.2b) where K^C-S U . 2c) -109- This stiffness matrix is derived without the use of a variational principle and is non-symmetrie.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.(l979) of •#> the form, n = o.5>/ru°*tc*dr - /ruc*t°-dr (a .3) — c where t are the prescribed tractions on the boundary. The physical meaning of II 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 i(5)-ui , t = Ni(5)-ti , i=l,2,3 (4.4) * From Chapter 3 equation 3.27 we have, tC = S«uC (4.. 5) Substituting 4-.4- and 4-*5 into 4-.3 we get, * II = 0.5#u^«S^*/pN^,N*dr,u - uJ*/j,N^*t*dr U.6) where in equation 4-. 6 we have dropped the superscript c. By minimizing II with respect to u we get, * K'*uC + PC = 0 U.7) where K' = 0.5*(C*S + (C*S)T) = 0.5*(K + K b (4-.8a) * P° = - C t ° (4.8b) c c Substituting for t ,u from, 4.0 ft * p . 0 c t p 0 t = t - t - t , u = u -u -u (4.9) -110- we arrive at, K ,*ut+(Pt-(K,*(uP+u°)-(PP+P°))) = K ,«ut+P,=0 U.10) M V -V/ _ ~ /v A/ 1— ^ A/ ^ where, pt = c-t5 , PP = C*tP , P° = C-t° U.ll) Equation 4-»10 is of the same form as the finite element stiffness equation,and therefore K ’ and P 1 can be assembled into the standard finite element system of Chapter 2,as contributions from a new ele ment . * 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-.l 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, u =^r^ *L*t =1.4- » u = V *p#t =-0.3 (L the side of the square) The displacements given by the analytical solution,the program and by Mustoe(1979) are shown in table 4-«l» - 1 1 1 - * Figure 4..I Square block in tension Figure 4.2 A circular hole under pressure. -112- Table 4.1 Square block in tension Analytical Program AJR0CK Mustoe(1979) Node u u u . u u u X y X y X y 1 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 1.4-20 0.325 1.4-17 0.324 4 1.4 0 1.4-60 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 pg (Fig.4-.2b) is given by Obert and Duvall (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, / R \ 2 U = —r1+v r - p. •— R‘ r E *1 r rr = -Oeeqo = -Pi’(7 )2 * 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 1.0. 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. - 1 1 3 - 4 Figure 4.3 Hole within infinite rock mass modelled by boundary elements only. y ^ Figure U*U Hole within infinite rock mass modelled by boundary and finite elements. Table 4-.2 Circular hole modelled by boundary elements only. Analytical Program AJROCK Mustoe(1979) Node u u u u u u X 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 0.4-75 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 finite elements - Displacements at nodes. Analytical Program AJROCK Mustoe(1979) Node u u u u u u X 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 finite elements - Stresses within plane strain elements. Distance Analytical Program AJROCK D r a =-aQQ a CD CD rr 00 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 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 symmetric mesh Analytical discretization b discretization c Node u u u u u u X 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 -0.39 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 -0.36 22.2 -0.56 I 22.4 -0.30 22.3 -0.29 22.5 -0.29 J 22.4 0.00 22.2 0.00 22.2 0.00 -1 1 6 - A y t X ♦ > X * m w q _, * c CnMWO^xj * Figure 4.5 Tension of a long plate In table 4.6 the displacements for discretization(d),calculated by the program for various positions of node D in the x direction (y^=-4), are compared with results calculated by Mustoe(l979) and to the ana lytical solution. Table 4*6 Tension of a long plate modelled by asymmetric mesh Analytical Program AJROCK Mustoe (1979) o K ii n Q o -?>> xd =36 x d=34 XD=40 Node u u U u u u u u X y X y y y X y A 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 1 1 .1 6 1.24 F 16.80 -1.2 1 6 .8 -1.82 -2.15 -1.08 G 16.80 1.2 17.2 0 .6 1 0 .2 6 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 seen that program AJROCK rotates the plate,which can be seen by the non-zero values for u y at the nodes C and J.This error does not exist in Mustoe’s resuits.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- S eries 2 a. Lined circular tunnel within infinite space. A lining rock system is examined.The rock medium is assumed to % be homogeneous isotropic 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: m E=10.0 v= 0.20 L ining: E c -t c =20.0 v c=0.00 Interface (only second case): kn =1000.0 k s =1.0 d> =o.o The diameter of the tunnel is 2 and the thickness t of the lining % is negligible.Two discretizations are 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. -1 1 9 - * at #■ ♦ * (b) * Figure 4-. 6 Lined opening -120- The rock is assumed to be weightless and the overburden pressure p= -1 and lateral pressure K.• 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 ratios are ♦ compared at three points on the lining^ as computed by the program for the two discretizations, and 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 is chosen to be 0.5 and 0. An ana lysis for K^=l,y=0.20 is carried out also,and as no slip is occu- ring,displacements are almost identical with those shown in table 4.7. For the discretization of fig. 4.6a displacements are shown only fory =0.20.The analytical solution is given by the following form ulae : ur^0.5.(p/M)-{(l+KA).(l-v)-(l+(1J ;^ .m )-^(l-KA)-{ ^ -2-cos 28} u0=2*(p/M)-~—^ (l-K A)*sin 28 where M=E*(l-v)/{(1+v)•(1-2-v)} 0=0.5*7^--- tM«D t ;—I" (D is the diameter of the tunnel) (l-v)«Ec »t c These formulae do not agree with the formulae 11.38 and 11.39 given in Poulos and Davies (1968) ,derived originally by Hoeg(l968) 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, * "I _ V )2 u ="rr_ *R*{(p +p )+2«(p -p )«cos 20} r E *x *x uQ=-—=r~1-V)2 •R«2*(p -p )*sin 20 0 E x -^y Young’s modulus and Poisson’s ratio are chosen to be 10 and 0.0. The radius of the tunnel R is l.The vertical stress field py is taken to be -1.0, the horizontal stress field p^ being zero.The rock to be 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. -1 2 2 - 4 *• * % * Figure 4-7 Excavation of a circulary f tunnel(B.E.and 7 plane strain el.) * * 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 A nalytical Discretization a Discretization b * K. 0 u u u u u u A X y X y X y 0 0.034-7 0 0.0379 0 0.034-3 0 0.5 tt/4 0.0019 -0.0619 0.0013 -0.0608 0.0017 -0.0619 tt/2 0 -0.1200 0 -0.1220 0 -0.1190 * 0 -0.0565 0 -0.0561 0 -0.0563 0 1.0 tt/4 -0.0399 -0.0399 -0.0397 -0.0397 -0.0398 -0.0398 ir/ 2 0 -0.0565 0 -0.0561 0 -0.0564. * 0 0.1257 0 0.1320 0 0.1250 0 0.0 7r/4 0.04-38 -0.0837 0 . 04-22 -0.0819 0.04-36 -0.0837 tt/ 2 0 -0.1821 0 -0.1880 0 -0.1820 * Table 4*8 Lined circular tunnel with free slip on interface A nalytical Discretization a Discretization b V 6 u u u u u u ka X y X y x y 0 0.0500 0 0.04-91 0 0.5 7T/4- 0.0350 -0.1060 0.0338 -0.1060 tt/ 2 0 -0.1500 0 -0.1510 0.0 ♦ 0 0.1670 0 0.1660 0 0.0 7r/4- 0.1180 -0.1650 0.1160 -0.1640 tt/ 2 0 -0.2330 0 - 0 .2 3 4 0 * 0 0.0537 0 0.0512 0 0.0531 0 0.5 tt/4- 0.0380 -0.0978 0.0356 -0.0959 0.0368 -0.0977 tt/ 2 0 -0.1383 0 -0.1370 0 - 0 .1 3 9 0 0.2 0 0.1638 0 0.1590 0 0 .1 6 4 0 0 0.0 tt/4- 0.1158 -0.1557 0.1110 -0.1510 0.1140 -0.1550 tt/ 2 0 -0.2202 0 -0.2160 0 -0.2210 continued -124- Table 4.8 Continued. A nalytical D iscr. a Discretization b V K. 0 u u u u u u A X y x y X y <* 0 0 .0 5 2 8 0 0.0525 0 0.5 tt/4 0 .0 4 0 2 0.0919 0.0364 -0.0884 tt/2 0 - 0 .1 2 6 0 0 - 0 .1 2 6 0 .333 * 0 0.1543 0 0.1540 0 0.0 tt/4 0.1080 -0.1430 0.1070 - 0 .1 4 2 0 7T/2 0 -0.2031 0 - 0 .2 0 3 0 * Table 4*9 Excavation of a circular tunnel Before excavation After excavation Analyt. a & b A nalytical a b * 0 u u u u u u u u x y X y , x y x y u x u y 0 0 0.0 0 0 .0 0.1000 0.0000 0.0988 0.000 0.0988 0.000 tt/8 0 0.0383 0 0.0383 0.0918 -0.115 0.0932 - 0 .1 1 6 0.0932 -O.II 6 * tt/4 0 0.0707 0 0.0707 0.0707 -0.212 0.0698 -0.207 0.0698 -0.207 3tr/£ 0 0 .0 9 2 3 0 0 .0 9 2 4 0.0381 -0.277 0.0386 -0.281 0.0386 -0.281 tt/2 0 0.1000 0 0.1000 0 .0 0 0 0 - 0 .3 0 0 0.0000 -0.293 0 .0 0 0 0 - 0 .2 9 3 N ote: 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 nodes,then the IJ 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 it 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 (1979), * 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 - V 1)=V r) a. Continuous traction assumption. * * x are collocation points c. Collocation points not at corner. * * m e . Chaudonneret. * Shape functions for tractions Shape functions for geometry f. Differing shape functions for geometry and displacements and tractions. Figure A.9 Various methods to determine the limiting values of tractions at the two sides of a corner. -127- stress tensor.lt has been used successfully by Mustoe(l979). 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 m 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(l978). 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 K1 and to be the symmetrized and the non- * symmetrized stiffness matrices respectively.Then —K*.uC ~ = F° ~ , K “1 • u°=~ F° ~»-L where * Kx = G T• S Subtracting the first equation from the second and noting that K' = (1/2) • (K-^+K^) we arrive at (1/2) • (Kn—l -K?’)*uC=AFC=F?-FC —i — ^ ~ that is the error in the nodal forces found is proportional to T -128- which shows how far from symmetric matrix is,and proportional to Q the displacements u/V . 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 sum to zero,the following equations should be satisfied. z k '. . 0 j=2X-l A=l,2,...,N/2 ; i=l,2,....,N. {Q.} = = N is the number of degrees of freedom ♦ I 'K' . 0 H C_i of the boundary element region. II ro e_i. __ I— 1 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 a l.(1979)) ,as definition of the boundary variation of one comple tely defines the other through the solution of the boundary value problem.Therefore the resultant energy distribution modelled by (l/2)*/puC»tC*dT cannot be correct.From that follows that the de rivatives 9H/9u 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 * Function Displ. at bound. traction at bound. Displ. int. Method Boundary integral prescribed prescribed - Finite element prescribed - prescribed * This error is probably a cause of the asymmetry of the directly evaluated stiffness matrix K_^. * 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 corners7 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 0.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 (u^,t^) 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'-u^-P*5. 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. Figure 4.. 10 Two boundary element regions. Figure 4..II Circular disc. -132- Table 4.11 Equivalent nodal forces and displacements for various particular solutions Node 1 Node 2 Node 3 Node 4 Function H V H V H V H V Eq.nod.for. ext.region 0 -0.4-86 0.063 - 0.759 0.069 -0.344 0.152 -0.315 both cases Sq.nod.for. int.region 0 -0.030 -0.002 0.027 0.004 -0.021 -0.006 0.011 case 1 Eq.nod.for. int.region 0 0 0 0 0 0 0 0 case 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 nodal^forces given by use of stiffness m atrice s K1,. Case and Node 1 Node 2 Node 3 Node 4 st.m atrix H V H V H V H V Case l,o r case 2 & 0 -2.743 1.956 -4.715 1.940 -1.940 4.715 -1.956 Case 2 & k J 0 -3.442 1.688 -4.070 2.434 -2.434 4.070 -1.688 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-*ll) in a state of plane strain of radi us 1,Young’s modulus 1, and Poisson’s ratio 0, is subjected to the two following types of loading. Case 1 : Uniform unit pressure normal to the circumference Case 2 : Initial unit displacement normal to the circumference The pressure is related to the radial displacement by the equation. (u°/p)=(l+v)•(l-2*v)*r/E Inserting the known values for E,r,v we find p=u° at the circumfere nce, that is the equivalent nodal forces for both cases should be equal.These nodal forces are given by K,#u0-P,. 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 * is calculated to be equal to the correct one.By using as stiffness matrix K_’ or T the equivalent nodal forces differ from the correct one. In the former case the discrepancy is about ±10$.These results * may be seen in table b.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 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 3 4 - 8.0 Figure 4»I2 Square block modelled with 32 boundary elements, A 8.0 - x y & 8.0 ... ZS------A ------ZS------A /- & / 77777- //'//> ///// 7-77TS Figure 4-.13 Square block modelled with B.E. and P.S. elements. -135- vertical direction.Faces 2 and k 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 tab le 4-. 13 the maximum and minimum displacem ents 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 ^m in ^Vmax UHmin UHmax Face 2 Face 4- ♦ 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 0.024-8 free fixed in H 8 0.5 3.20 3.20 -0.0100 0.014.1 fixed in H fixed in H * 8 1.0 3.20 3.20 - -0.0100 0 .0 1 0 0 free fixed in H 1 0.0 3.16 3.23 -0.0100 0.0100 free fixed in H 1 0.0 3 .1 6 3.23 -0.0100 0 .0 1 0 0 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 ±1% .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 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 displacem ents are shown in tab le 4-.14-. Table 4--14- Square block modelled by finite and boundary elements Height UVmax UVmin ^m ax ^m in 8 3.20 3.20 0 0 0 3 .3 6 3.23 0.1 -0.1 V 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 2:1.These errors become of the order of the ft previous example as soon as this ratio becomes close to l,and the number of boundary elements increases. ft -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 e x te rio r in fin ite rock mass is modelled as a boundary element region numbered BE^.The inte rior intact rock is modelled by three boundary element regions num bered BE^jBE^BE^, and 26 plane strain elements. Young’s modulus and Poisson’s ratio are taken everywhere 100000 and 0 respectively.The unit weight is 0.27*10 ^ and the depth of point 0 is 100000. The ratio of horizontal to vertical stress K.,takesA the values 0 and l.In the case K^=l,no errors are expected due to ♦ corners,whereas for K^=0 the error due to corners is expected to be very close to maximum. The particular solution of the internal boundary element regi ons BE9,BE^,BE^ is varied,by varying the values of and h^ of equa 4 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 tab les A .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-»l5 the ratio of the infinite field (exterior boundary element region BE^),and therefore of the total solution is O.In columns 3 to 7 the particular solutions of the internal * boundary element regions are varied,by varying and h^.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 ((J)=80^) and very stiff (k s =k n=1000) joint elements. -138- Figure 4-»I4- Large problem with boundary and finite elements. * * * * ♦ * * Table 1.15 Stresses at the centres of the plane strain elements of large problem.K of the tot.sol.is 0. Particular solution input for interior boundary element regions 1 2 3 i 5 6 7 8 6 11 0 K =0.5 S' =107 Jo in ts,k =k =103 0 KA=0’ 5 KA=0’V 10 > KA=0-57 h"=105 h^S -lO " h0= 10 ka =0’V 105S Point Analytic -amm . -a min . -a min . -a■ max -amm . -amax amin ~amin C 2,34-9 2.3a 2.315 2.725 2.045 6.561 2.243 6.617 2.347 D 2.4-02 2.396 2.394 2.423* 0.077 3.097 0. 448 3.816 2.401 E 2 .U 9 2.4-48 2.450 2.430 0.038 2.248 0.413 2.261 2.453 F 2.305 2.506 2.507 2.493 0.023 2.359 0.255 2.337 2.510 -139- G 2.560 2.563 2.565 2.536 0.018 2.262 0.194 2.229 2.566 H 2.6l6 2.619 2.620 2.600 0.016 2.413 0.163 2.409 2.619 I 2.672 2.673 2.674 2.636 0.007 2.279 0.795 2.194 2.674 J 2.620 2.619 2.618 2.653 0.077 3.567 0.270 4.743 2.619 K 2.4-59 2.457 2.455 2.518 0.058 3.220 0.548 3.792 2.456 L 2.299 2.295 2.293 2.345 0.067 3.055 0.534 3.722 2.294 M 2.138 2.136 2.136 2.122 0.054 2.321 0.256 2.771 2.136 » « • * -0 0 0 Table 4.16 Stresses at centres of plane strain elements of large problem. K of the total sol. is 1 Particular solution input for interior boundary element regions 5 7 5 O H 1 2 3 4______rH II II < KA"1,h0=100000 ka=0-5’V =50000 V ° . 5 . hQ=10000000 KA=1' V 10b O Point A nalytic -a. max -O min . -0max -amm . -amax -amin . amin -amin . C 2.34-9 2.34-0 2.34-5 2.273 2.338 5 .1 6 0 8.883 3 .3 3 6 13.248 D 2.4-02 2.395 2 .3 9 6 2 .3 6 3 2.399 1.915 5 .5 6 7 2.845 7.340 E 2.4-4-9 2.4-4-9 2.452 2.440 2.450 2 .2 6 1 3.514 2.608 4.177 F 2.503 2.506 2.507 2.501 2.507 2.331 3.095 2.591 3.428 G 2.560 2.562 2 .5 6 4 2.559 > 2.566 2.278 2.931 2.613 3.114 H 2 .6 1 6 2.618 2.619 2.614 2 .6 2 2 2.224 2.998 2.667 3.162 I 2.672 2.673 2.673 2.672 2.678 2.057 2 .6 9 9 2.680 2.755 -on- J 2 .6 2 0 2.619 2 .6 2 1 2.582 2.624 1.857 6 .3 5 9 3 .1 3 6 8.297 K 2.4-59 2.4-56 2.461 2.432 2 .4 5 3 2.937 5.086 2.848 6.722, L 2.299 2.294 2.301 2.273 2.292 2.585 5.060 2.691 6 .6 0 0 M 2.138 2.135 2.139 2.115 2.140 1.545 4.533 2.493 6.035 - l a in table 4-.16,the ratio of the infinite field is l.In co lumns 3 to 7 the particular solution of the internal boundary ele ment regions is varied by varying and h^, 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 ^ $ to 7),the results become less accurate,especially near sharp corners.Hence care must be taken that the stresses due to the parti cular solution and the total one are of similar order. ♦ * P CHAPTER 5 - STABILITY OF AN OVERHANGING ROCK WEDGE 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 parameters,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 made are: . 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 kn and take both of these to be small by comparison with the stif fness of the rest of the rock mass,so that the intact rock including Flexible joints ♦ » * Figure 5.1 Wedge id e a liz a tio n . - l u the wedge nay be regarded as rigid. . Apart from the forces at the joints,the wedge will be acted upon by its own weight Wa and another applied force A (e.g. rock bolt in tension). Let P be the resultant of tat and A ,i.e. P=W-A . (5.1) l l • Find the magnitude of the force Pq acting in the same direction as 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=l+(A-A0)/W=l+c-c0 (5.2) where c=A/W , =0=AQ/W , A0=W-PQ (5.3) * 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 Pq is applied (stage l),is HQ.lt is assumed at this stage that elastic behaviour is exhibited.This necessitates that a<<{) . After the joint deforms and Pq is applied (stage 2),the force with which * the surrounding rock acts upon the half wedge is J. The initial conditions require; Nq =H^cos a , Sq= Hq «sin a For a joint without dilation,the following system of equations needs to * be solved. — - 1 0 0 -k s »cos a 0 S sin a 0 1 0 kn «sin a 0 N cos a n . 1 -tan 0 0 0 • H EC 0 (5.4) O -sin a cos a 1 0 0 d 0 -cos a sin a 0 0 1 V 2 0 _ — — h Geometry and forces v / Figure 5.2 Symmetric wedge - or in abbreviated form C«X= Hq »Y where rows 1,2 are the constitutive equations for the joint, row 3 is the failure criterion and rows 4»5 the equation of equilibrium. The solution of the matrix equation gives; V2=M-Ho H=P0/2*cot (<|>-a) d=Hg*sin (cl>-a)/(D-kn) (5.5) N=(cos2a*k s /k n +sin2a) *cos where D=cos a*cos v^= u*tan i (5.7) Vq is a function of the applied normal stress only.Its relation to the vertical movement is given by v0 =v-vd=d*sin (a-i)/cos i (5.8) Thus in equation 5.4 we put instead of C,=k *sina the value <.l{. n C5/=k*sin (a-i)/cosi and resolve the system.This gives P0/2=M* H0 H=Pq/2 *cot ( d=sin (({)-a)«HQ/(kn -D*cos i) (5.9) N=(cosaa»cos i «ks /k n tsin (a-i) .sin a) .cos ^ H~/(D l) .cos i) S=(cosaa«cos i*k s /k n +sin (a-i).sin a).sin where, D=cos a*cos (b-k /ks +sin n (J)-sin (a-i)/cos i m ( 5.10) M=(cos2a*cos i*k /ks +sin n (a-i)»sin a)*sin ( For k s«k n , ♦ PQ=2*HQ*sin ( i.e Pq is independent of the dilation angle i.Note that 4) is the total friction angle.The case with dilation includes the case i=0, i.e a joint * without dilation.The nondimensional ratio M=Pq/(2Hq ) 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 * aH0=HQ/h ' (5.12) The weight of the wedge is given by, ♦ W=0.5*p*g*L«h=0.25*p*g*L2/tan a (5.13) The resultant force Pq is given by, Pq =W-Aq =(1-Cq )« 0.5*p*g*L*h (5.14a) or * P0=2.H0.M=2*h«aH0-M (5.14b) Equating the last two equations l-c0=4-aH0*M/(p«g«L) (5.15) By defining the stress concentration factor SC£ as, * ^cf=CJHc/^*^*Z^ where z is the depth of the excavation,equation 5.15 becomes, l-c0=4-(z/L)-Sof*M (5.16) S . depends on the geometry of the excavation and the -1 4 8 - stress ratio , 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 K^. * Equation 5.16 uncouples the various contributions to the carrying capacity of the joint. Friction angle 6 < 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 P q 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=Nq , then no total movement occurs, but plastic shear movenent of it* the joint occurs ..This will corespond to force P /2 in the figure. For N= 0 ,PQ 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,which is also the most difficult to determine.For very low values of k /k ,M is no longer, a monotone decreasing function of angle a.This s n w 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 3 us consider a wedge with constant base length L=10xl0 ,k /k =0.01,i=0, s n (J) =40 0 ,pg=27xl0 -6 .The required horizontal stress at limit equilibrium be comes: a^Q=p*g*L/(4,M)=0.0675,M. -1 4 9 - Figure 5.3 Symmetric wedge — ct> Let us take the angle a to be 10°20?30°Then M,aun and FS vary as H U 1 follows : a M a„ FS(a„ =0.355) Ho Ho O O i ♦ —1 0.167 0.1,0/, 0.878 o w o 0.190 0.355 1.000 o cr\ O 0.136 0.4.96 0.715 % The factor of safety if no support force exists is given by, FS=l-co =PQ/W =4M.Oh o /(pg-L) = aH 0 .M/0.0675 In column 4- in the table above the factors of safety for our=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 0/kn-O , w M=sina • sin( dM/da=(cosa • sin ( For obtaining an extremum we put dM/da=0.This results in 4 tana=tan(({)-a) or a=c})/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 0Ho subtends an angle 180-(J). The maximum is achieved when the triangle C0Hq becomes isoceles , 4 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^ becomes larger or (J) becomes smaller.The distance between points C and B depends on the t angle e where, for i=0, e=arctan(k /k «cot a) (5.17) s n For k /k £0 and a=0 -*-e=7i/2 and M=P /(2*H )=tan ♦ ♦ * 4 * M: Wedge with a=20^ L: Wedge with a=IO^ 4 --— — — — are lines for k s/k n =0 Figure 5.4. Examplesfor very low stiffness ratio joints. -152- ^Pomax/2 I Figure 5.5 Behaviour of symmetric rigid wedge. -153- a shear force S through movement,and if there were non zero SQ ,it would be impossible to change it.Also NQ due to a=0 would not change. Flexibility of the wedge If the flexibility of the 'intact 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,i.e two values for each N,S,H and one value for each R,d .d ; 4 y Pq. The determination of k ^ k ^ needs engineering judgement.If the wedge 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 0 0 k *sin a k *cos a 0 S sin a s s 0 1 0 k «cos a k *sin a 0 N cos a n n 0 0 1 0 0 H 1 =H • (5.18) o 1 -tan (J) 0 0 0 0 d 0 X sin a -cos a 1 0 0 0 d 0 y cos a sin a 0 0 0 1 P / 2 0 o - or in abbreviated form, C«X=H «Y , ---o ~ 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. -1 5 4 - IncreaaB in 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. -1 5 5 - * A H p ^ AR Figure 5.7 Models for elastic wedge. -156- Performing Gaussian elimination in Eq. 5.18 ,we get a 1 b* 0 d X r5 d» 0 • d = H • 0 (5.19) * °,+kh y o -d' f ' 1 P n /2 0 Solving 5.19 we get d d 1 * X d = ~(k^+c') •r^*Ho/(a’*d1-b'•c ’-b!'k^) (5 .20 ) y _P 0 /2 -d,2-c''f*-f ' - k b d x /d y = -d'/O^+c’) (5 .21) where the parameters a*,b»,c*,d»,f'»r^ have the following values a , =k *(-(k /k )*sin a + cos a«tan <$>) n s n m b , =k *((k /k )*cos a + sin a*tan({>) n s n c'=k *((k /k )«sin2a + cos2a) (5.22) n s n d ’=k •(l-k /k )*sin a*cos a n s n f ’=-k *(k /k cos2a + sin2a) = c*-k -k n s n s n rr=sin (cJ)-a)/cos For k^= °o ,we get dx/d^=dx=0 , and * P /2=H »f1• v c / b t (5.23) 0 0 o 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 /P , o o d t2+c»*f»+f'-k. -b* d ,2/f'-fc’+k^ P /P = ______h « ___ (5.24) ° 0 a»*d»-b»«c,-b,*k1_ f» -a' ’d'/b'+cHk The function a ^ f 1 + b ’^d1 = -k /k *(tan (j)*cos a - sin a)*k2 (5.25) s n n is negative for (j)>a . Hence d'2 a«*d» ------< ------(5.26) -f» b» Also f' is negative, and aSb'jcSd' are positive for k /k < 1 . Hence from 5.24. and 5.26 we find that, s n P /P > 1 o o i.e. the stability of the wedge increases due to the flexibility of the wedge , if this model is appropriate. 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. excavation surface in the opposite sense to the existing tractions and proportional to them.If the joint flexibility is very large by comparison with the flexibility of the rock mass,the stress applied on the joint 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 G which would be a straight line if the stiffnesses remain constant. Let us consider the case 4) > 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.Hq in Bray’s theory corresponds to OB , whereas for a very flexible joint,Hq corresponds to OD.Bray’s theory ac 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 envelope at a point Failure enveL nfinitely stiff joint * * # * D C 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 /k .1 ’, s 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 q could be suggested,to bring point B at a point such as E. 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 $ < 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- 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 have reached the line E'D’ .In all cases we can propose for design purr- poses a zero resistance of the joint,requiring only full support of the -161- wedge load.We expect less plastic movement as we move from F* to D*, so this could be an indicator of potential explosive failure^speci 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 HqD^ and 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: * - ' _ U-. cos a^ -sin a^ d u „ cos a„ sin a_ d 1 y 2 2 2 y * 9 • V-. sin an cos a., d v_ sin a_ -cos a_ d 1 1 1 X 2 2 2 X __ , _ _ - (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 - 1 6 2 - -163- —- — 1 0 0 0 -k •cos k • sin 0 sin a.. si ai si ai 1 i 0 1 0 0 k .. *sin k n•cos 0 Nn cos an nl ai nl ai 1 1 0 0 1 0 -k *cos -k *sin 0 sin a» S/£ a2 s d. a2 So2 2 0 0 0 1 k -sin -k «cos 0 • =H • cos a_ n2 a2 •n2 a2 No2 o 2 0 0 1 -tan (J>2 0 0 0 d 0 y ■* sin a^ cos a^ -sin a2 -cos a2 0 0 0 d X 0 cos a^-sin a^ cos a2 -sin a2 0 0 -1 P 0 0_ - (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, i i ---- ____ % V by 0 d U y cy dy 0 • = H • 0 d X 0 (5.29) ey fy -1 0 p o _ % Solving this system we get, i i — __ d L y H T y ° 5 cy d X cy-by-ay-dy (5.30) P cy.fy.ey.dy ♦ o J d /d =-cy/dy x y $y=arc tan (-cy/dy) y y « where the parameters aJ to r^ are given by. ay=k s2 „«cos a0+k 2 n2 0*sin a_«tan 2 v2 d)_ by=k s2 0»sin a0-k2 n2 „«cos 2 a0»tan ^2 The angles are given by, e^ = arc tan ((k ../k -,)*cot (an+^)) l si nl 1 (5.32) ©2 = arc tan ((kg2/kn2) • cot (a,^)) 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 a.. =a_,(t>.. = The ratio c^/d^ 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 Pq cor responding to first yield'on face l.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.lt 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 ^^-[>^2^2 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- — cti 1 0 0 0 -k - cos k , *sin 1—1 0 sin a. si ai si S1 0 1 0 0 k _*sin k ,*cos 0 cos a . nl ai nl ai N1 0 0 1 -tan (J>2 0 . 0 0 S2 O 0 0 l—1 k -sin -k *cos 0 • =H • cos a. n2 a2 n2 a2 N2 0 O 1 -tan O 0 0 0 d H y sin a^ cos a-^-sin a2-cos a2 0 0 0 d X cos a^ -sin a^ cos a2~sin a2 0 0 -1 p 0 (5.35) Performing Gaussian elimination, f ,f f a b 0 d rt y 5 f f c d 0 • d = H • r, (5.36) X 0 6 f S f e f -1 P r„ 0 _ L 7 J Solving this system we get 1 ___ <+H d r, -bf 0 H f f -c~ a 0 (5.37) f ,f f , f a *d -c *b f ,f f -f f _f f ,f f , f e *d -c *f a *f -e *b c *b -a , /, , f f, f fx //,f f , f fx dx/dy=(_c *r5+a 9r6 " ' d *r5"b ' r 6 ' S"^=arc tan (d^/d ) f f where the parameters a until r^ are given by, f a =k , -cos a..+k , .sin a, .tan d>. si 1 nl 1 ^1 f b =-k , *sin an +k .. *cos a, *tan (L si 1 nl 1 1 £ c =-(k ,-k .,) *sin a *cos an+k «sin a -(cos a0+sin a »tan £ r^=sin (cJ)^-a^)/cos e?= arc tan ((k ,/k -,)*cot (an+ 3^)) (5.39) 1 si nl 1 The value PQfor failure might sometimes be smaller than its value for yield.This happens if the line joining the middle points of D-^D^ and 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 Pn must be considered.P is parallel to the wedge face and except for very P shallow excavations,is small by comparison with Hq and may be ignored. Pn is normal to the wedge face.Calculations may now proceed as for the wedge with the face horizontal,but with a^»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- Figure 5.IO Oblique wedge. h -(tan a2-tan al)/2 h-(tan a2-tan al)/6 h-(tan a2-tan al)/3 dy Figure 5.I I Wedge wi th rotation. -168- three displacement components d ,d ,d and two force components for x 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 calculat.ed numerically by the program. th Value has been calculated analytically from theoretical solu- tion. (asymmetric wedge) y Value corresponds to yield (asymQetric 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 order o 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 var.ying flexibility, within an infinitely stiff rock mass, 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 l7.The elastic wedge is modelled by 7 plane strain elements PI to P7_ The joints are modelled by eight joint elements jl to j8' 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. The angle a is taken to be 50 ,20 °,35 °, the friction angle ¢ is 400, and the dilation angle i is 0°. In table 5.1 numerically calculated 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 that c c 0 predicted by the simplified theory of section 5.l,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 P n mln n and M to P +1. M is given by m~ n -171- 7 46 4 P2 42 P5 3 36 41 45 2 40 PI P4 I 18 35 44 34 17 Figure 5.12 Symmetric flexible wedge within rigid rock. (WDG series) * * ♦ ♦ * Table 5.1 Symmetric almost rigid wedge within infinitely stiff rock. E=50‘I05, a k /k a T P Bt , Mmin Mmax AMc Me Mgr Comments s n n •io4 05 0.001 0.156 0.040 3000 0.207514 0.062 0.093 0.025 0.087 0.087 05 0 .0 1 0 0.II9 0.071 6900 it 0.144 0.167 0.014 0.157 0.158 Failure was 05 0 .1 0 0 0.277 0.197 19800 it 0.4II 0.479 0.052 0.463 0.463 simultaneous in all joint ele- 20 0.001 0.178 0.145 2180 0.814334 0.179 0.203 0.005 0.183 0.183 it 20 0 .0 1 0 0.193 0.150 2170 0.178 0.202 0.012 0.189 0.190 ments. -172- 20 0 .1 0 0 0.226 0.190 2940 ti 0.239 0.274 0.001 0.240 0.241 35 0.001 0.304 0.241 400 1.365658 0.055 0.082 0.023 0.078 0.078 35 0 .0 1 0 0.291 0.242 540 it 0.075 0.086 0.003 0.077 0.078 35 0 .1 0 0 0.298 0.247 560 11 0.076 0.086 0.004 0.080 0.080 Note that the meaning of the indices of that table is as follows: 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 c : Value extrapolated from numerical result gr : Value calculated analytically -173- The initial horizontal stress is Ojjo= 0.4-2/cos a (5.4-1) The value of M lies between M . and M and can be found by extra- c m m max J m polation from M . . The stresses within the joint at the end of load step n and 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 Ad is the additional movement needed to cause * failure, (O/ x-k -Ad-sin a)-tan (}) = T/ x+k -Ad-cos a (5.4-2) (n) n (n) s Solving for kn »Ad, k -Ad = (O/ x-tan (J)-T/ x)/(sin a-tan (J)+cos a-k /k ) (5.4-3) n vn; s n The additional load then provided would be AP = k -Ad-(sin2a+cos2a-k /k )*L/sin a (5.4-4-) n s n * The additional value for M would be given by, AM = AP*tan a/(oTT *L) = k «Ad* (sin2a+cos2a*k /k )/(cos a*aTT ) c Ho n s n Ho (5.4-5) The value of M is calculated from c M = M . + AM (5.4.6) c m m c As can be seen from the table , computed (column 9) and theoretically predicted (column 10) values of M are almost identical. * Validation for M . can be made from the known o , s and t , s as, m m (n) (n) M . = p «tan a/(oTT *L) = ( t * c o s a-o*sin a)/(aTT *cos a) (5.4-7) m m n Ho Ho From the table we may observe that for values of k /k <0.01 the com- 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-) • 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 Ac*W. The value of Ac is 1. In table 5.2 the values of (l-c) characterizing the failure capacity of the wedge as computed by the program (column U) and as predicted by the simplified theory (column 5) are compared. The value in column 4- corresponds to the last converging step of the program. Thus the correct value for (l-c) lies between (l-c) min . and (l-c) mm . + c. Due 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 (l-c) is (l-c)=10*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 to 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 (l-c) for the last converging step are shown in column 6 for com- -175- i+o b.Excavation sequence performed. is b.Excavation sequence a.No excavation is performed. excavation is a.No * * f * • * * Figure 5.13 Symmetric elastic wedges within elastic rock - a=20 o (WHML series) * 0 # 0*0 0 Table 5.2 Symmetric elastic wedge within rigid surrounding rock. •g=27‘I0"?Ac=I .0,4>=40° ,i=0.0,kn=20,L=I0000,I . -c= - #M 0h o =“2,7,e =i o o *i o 3,p 0 40 I 2 3 4- 5 6 angle k k /k (1 .0-c) . (1 .0-c) M s s n min gr gr 0.2 0.01 6 6.32 0.158 05° 1.0 0.05 13 13.92 0.34-8 2.0 0.10 18 18.52 0.4-63 0.2 0.01 7 7.60 0.190 -176- o o 1.0 0.05 8 8.60 0.215 2.0 0.10 9 9.64 0.24-1 0.2 0.01 3 3.12 0.078 35 o 1.0 0.05 3 3.16 0.079 2.0 0.10 3 3.20 0.080 Note that the meaning of the indices of that table is as follows: min : Value corresponds to last converging step of the numerical computation gr : Value is calculated analytically.(graphs). # ♦ 0 * 4 * 0 Table 5.3. Symmetric elastic wedge within elastic rock , without excavation sequence. a=20°, p*g=27*I0"6, Ac=I. I 2 3 4 5 6 7 8 9 Test name ks ks/kn E'I0‘3 (i-c)min Mgr °H0 (i-c)gr Failure starts at TAD 0.2 0.01 100. 2.87 7 0.190 8.08 at the top TAE 1.0 0.05 100. 3.15 II 0.215 10.03 at the top TAP 2.0 0.10 100. 3.27 13 0.241 11.68 at low elem. TBD 0.2 0.01 10.0 3.36 II 0.190 9.46 at the top. TBE 1.0 0.05 10.0 3.60 13 0.215 11.46 at bottom. TBF 2.0 0.10 10.0 3.64 >13 0.241 13.00 at low elem. -177- TCD 0.2 0.01 1.00 3.65 > II 0.190 10.27 at low elem. TCE 1.0 0.05 1.00 3.72 13 0.215 11.85 at low elem. TGF 2.0 0.10 1.00 3.72 > 12 0.241 13.28 at low elem. t Note: Indices "gr" indicate that values are computed from analytical solution (graphs) Indices 11 min" indicate that values are computed numerically and correspond to the last converging step. -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 -(l-c) is computed from * (l-c) = 4-*0go*M/(p-g-L) = H . 81.(0Ho'M ) (5.4.8) In the second type of problem (Fig. 5-13b) 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 p^ to p^2 and j^ to j^^ 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 values 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 (l-c) = 15.77*(t»M ) (5.49) Sr 0 * ■ 0 # # ♦ Table 5.4. Symmetric elastic wedge within elastic rock , with excavation sequence. a=20°, P-g=27-I0, <>=40°, i=0°, kn=20 , h=I3737, L=I0000, W=I854, Ac=I. 0 Series WAD-WCF. I 2 3 4 5 6 7 8 9 10 II Test name k k /k E ’lO"3 t , t (l-o)o Failure starts H -V "b S "ho (I-0)BS ^I_C^H0 s s n at WAD 0.2 0.01 100. 3.23,1.17 3.82 0.75 11.44 2.26 2.0 top element WAE 1.0 0.05 100. 3.03,1.28 3.58 2.II 12.13 7.15 7.0 top element WAF 2.0 0.10 100. 2.98,1.4-7 3.52 2.63 13.40 10.00 9.0 bottom element WBD 0.2 0.01 10.0 3.12,1.67 3.69 2.57 11.05 7.70 8.0 top element WBE 1.0 0.05 10.0 2.97,1.68 3.51 3.34 11.90 11.33 II.0 bottom element -179- WBF 2.0 0.10 10.0 2.93,1.68 3.46 3.49 13.15 13.26 12.0 bottom element WCD 0.2 0.01 1.00 2.93,1.73 3.46 3.48 10.36 10.42 12.0 bottom element WGE 1.0 0.05 1.00 2.89,1.72 3.42 3.58 11.59 12.14 13.0 bottom element WCF 2.0 0.10 1.00 2.88,1.72 3.42 3.59 13.00 13.60 13.0 bottom element Note that indices, BS indicate that values correspond to H calculated by elastic solution. The analytic solution is used. Ho indicate that values correspond to H calculated numerically. The analytic solution is used, c indicates that values have been calculated completely numerically. -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 C is linear and corresponds to the loading applied to create a stress free surface.We continue loading,by applying 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=7.82^ (e=arc tan ((k /k )#cot a),which as discussed earlier in section 5.1 s n is due to the flexibility of the rock. 4 The last converging step is at (l.52,0.45).Until that point minor yielding occurs x^hich does not deviate the line from linearity.After that point partial and subsequently total yielding occurs which does 4 not allow the analysis to converge at C ’.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,WBD,WCD,WCE when run at ten times the stress level (at 1000 m depth) fail at ten times the pullout load (Ac=10), indi cating a parallel shift of the stress, path ACC1 (Figs. 5.8 and 5.14). ♦ * - 1 8 1 - * * 4- 4 ♦ o ♦ Figure 5.14 Example of stress redistribution in a symmetric wedge(WAE) -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=5^,20^,35^; E=100,10,l GPa; k =0.2,1.00, m s 2.00 are studied.The other parameters are taken to be An initial horizontal stress Cjjq =-2.7 is applied in the plane strain % elements.The initial stress applied in the joint elements to create a horizontal stress of -2.7 MPa is calculated from a=GTT •cos2a, t =Qtt *cos a»sin a, and is tabulated for the three angles a below, o Ho 63 a T a O o 5° 0.234- -2.679 o o 0.868 -2.381 ♦ 35° 1.269 -1.812 The wedge is modelled by 8 plane strain elements p^ to Pg and the jo ints by 8 joint elements jn to jQ, one side of each joint element * being fixed.In figure 5.15, the discretization of a wedge with a^=35^ 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 shown.In 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 (1—c) =(M +M0 )*2«gu /(p*g»L)=20*(M +M0 ) (5.50) gr lgr 2gr Ho K lgr 2gr -183- * 4 ♦ % * Figure 5.15 Asymmetric Wedge (WAS series) ♦ 4 4 ♦ * 4 ♦ 0 0 n 0 0 • Table 5. 5 Asymmetric wedge surrounded by rigid rock; aj.=35 ,a2=05 ; p*g=27*I0 > 20. o o H II *I03,I0’I03,I 'I03, k k /k E-IO3 Mn M0 (i-c )th s s n lgr 2gr (i-c)gr y (l-of)th u - y ( I - C f ) Comments 0.2 0.01 100 0.077 0.158 4.70 1.82 2.18 1.60 1.80 Failure starts 1.0 0.05 100 0.079 0.34-8 8.54- 2.04 3.68 1.80 3.60 in upper middle 2.0 0.10 100 0.080 0.4-63 10.86 2.26 5.17 2.00 5.00 joint elements 0.2 0.01 10. 0.077 0.158 4.70 1.82 2.18 1.80 2.40 1.0 0.05 10. 0.079 0.34-8 8.54 2.04 3.68 2.00 5.00 Failure star- 2.0 0.10 10. 0.080 0.4-63 10.86 2.26 5.17 2.00 7.00 ted in lower middle joints. -781- 0.2 0.01 1.0 0.077 0.158 4.70 1.82 2.18 - 4.60 1.0 0.05 1.0 0.079 0.34-8 8.54 2.04 3.68 7.80 _ 2.0 0.10 1.0 0.080 10.86 2.26 O.4.63 5.17 - 8.20 Note the following meaning of the indices used in this and the following tables of this chapter: gr : Value has been calculated analytically (from the graphs) th : Value has been calculated analytically from asymetric wedge solution y : Value corresponds to yield f : Value corresponds to failure. ♦ ♦ * + # + * Table 5. 6 Asymmetric wedge surrounded by rigid rock; a^=20^,a2=05^ ; p*g=27*I0 , Ac=0.25. E=I00‘I03,I0*I03,I*I03, c{)=4.00,i=00,k =20,L=I0*I03, o =2.7,k =0.2,I.0,2.0,h=22l50,W=299Q. Series WAS___. n n u s I 2 3 4 5 6 7 8 9 10 II k k /k E*I03 M (i-c)gr (1-c )th (l-cf)th (1-c ) (I-Cf) Comments s s n Mlgr 2gr y y 0.2 0.01 100 0.190 0.158 6.96 5.1 3.4 8.10 5.00 1.0 0.05 100 0.215 0.34-8 11.26 7.1 6.6 - 7.75 - 2.0 0.10 100 0.2a O.463 14.80 8.7 9.5 - 10.25 0.2 0.01 10. 0.19 0 0.158 6.96 - 3.4 - 6.75 1.0 0.05 10. 0.215 0.348 11.26 - 6.6 - 11.25 - -185- 2.0 0.10 10. 0 .2a 0.463 14.08 - 9.5 - >12.75 0.2 0.01 1.0 0.190 0.158 6.96 - 3.4 - 11.75 1.0 0.05 1.0 0.215 0.348 11.26 - 6.6 - > 12.75 - 2.0 0.10 1.0 0 .2a 0.463 14.08 - 9.5 - > 12.75 # 4 4 0 0 4 4 Table 5.7 Asymmetric wedge surrounded by rigid rock; a^=35^ ,a.2=20°; p * g=27 *10" , A c=0.25 • o M o n I03,I0*I03,I*I03, I 2 3 4 5 6 7 8 9 10 II k k /k E*I03 M (i-c)gr (i-c )th (l-cf)th (I-Cf ) Comments s s n Mlgr 2gr y (I_0y) 0.2 0.01 100 0.078 0.158 4-.72 2.40 4.60 4.00 1.0 0.05 100 0.079 0.34-8 8.54 2.50 5.10 - 4.50 - 2.0 0.10 100 0.080 0.4-63 10.86 2.60 5.50 - 5.00 - 0.2 0.01 10. 0.078 0.158 4.72 2.40 4.60 3.75 186 - 1.0 0.05 10. 0.079 0.34-8 8.54 2.50 5.10 - 5.00 - 2.0 0.10 10. 0.080 0.4.63 10.86 2.60 5.50 - 5.50 0.2 0.01 1.0 0.078 0.158 4.72 2.40 4.60 - 4.75 1.0 0.05 1.0 0.079 0.3/48 8.54 2.50 5.10 - 6.00 - 2.0 0.10 1.0 0.080 0.4.63 10.86 2.60 5.50 - 6.00 -187- Equation .5.50 has been used by Goodman,Shi and Boyle(1982).In column 7 is shown the value for (l-c) when first yield occurs calcula ted analytically from equation 5.30.In column 8 is the value for (l-c) when failure occurs as calculated analytically from equation 5.37. #- In column 9 the value for (l-c) at which first yield occurs is shown. In fact first yield occurs between (l-c y )-Ac and (l-c y ) as the load is applied in steps of magnitude Ac*W. In column 10 the value for (l-c) at which the program last converges is shown.Failure actually occurs between (l-c^) and (l-c^+Ac. As pointed out in section 5.1.2, (l-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 ■w 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/mnr 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 Unit ♦ Length mm Force N Stress MPa * The material properties assumed in the analyses are shown in Table 6.1 -189- \ Joint A Joint BJ \ I \ 11I I +1 o CD I +1.-4 ~L 20 0 V .r-- 00 0 71' '1 2 0 0 • ~453---;V It 12020 It 10360 k 13820 /]1--- 71 )J Figure 6.1 Stope and _drive georeet~y. 5 39 BEl BE2 60 3 9 13 17 23 27 Figure 6.2 Stope,drive and surroundin6 rock discretization. -190- Table 6.1 Material properties Structure Parameter Unit Value E. MPa lOOxlO3 ,10xl03 l * Rock mass V. 0 1 , 3 -4 P±xg N/mm 0.27x10 Joint MPa 20 * ’u k N/mm3 0.2,0.1 s general - 1.0,0.5 Bo V mm 1.0 * me degrees 40 , 30 Or i degrees 0 , 5 model 1 - 10 - V To * o o o m i—i Cl MPa model 2 MPa 0 s0 k N/mm3 20 % n E MPa 200xl03 s strut A mm3/mm 50,50xl0~3 s P xg N/mm3 10-5 * s 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 load steps 1 Gravitational loading - 1 2 Excavation of drive Pg to p14 and 3 jll and j18 2a Installation of struts m. to m, 1 1 6 3 First level ore excavation P19 and p2Q 3 4 Second level ore excavation P21 and p22 3 5 Third level ore excavation P23 and p24 2 ♦ 6 Fourth level ore excavation 2 p25 and p26 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 I,the wedge does not fail for any combination of $ , k , Bq , and . In subsequent analyses of the problem,Joint model 2 is used.The results are summarized below. % i. ks=0.1 , Bq=1.0 or 0.5. The wedge failed during the third step of activity 2 (excavation of drive) ii. kg=0.2 , Bq =1.0 or 0.5. The wedge failed during the first step of activity 6 (fourth level ore excavation). The case Bq=1.0 is illustrated in Figures 6.3 to 6.8. iii. Installation of struts. For very flexible initially -3 # unstressed struts (As=50xl0 , corresponding to rock bolts) 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 with s initial prestressing, the initial prestressing force being equal to 60 percent of the weight of the wedge. iv. E^=10 , Bq=0.5 or 1.0. The wedge did not fail. -193- * * ♦ EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF. INITIAL MESH LENGTHS __ = 2-0*103 UNITS ACTIVITY 0 * Figure 6.3 Initial mesh. -194- ♦ EFFECT OF SHADOHING ON A HEDGE IN A TUNNEL ROOF. DEFORMED MESH LENGTHS __ = 2.0-10* UNITS ACTIVITY 1 LOAD STEP 1 ITERATION 1 DISPLACEMENTS _ = 0.5-10° UNITS WWW //// / / / / \ / / / \ \ / / / / \ W / ✓ \ X y N ‘ 1 "/// — \ 1 / ♦ EFFECT OF SHAOOHING ON A HEDGE IN A TUNNEL ROOF. FLOH FIELD LENGTHS __ = 2-0-10* UNITS ACTIVITY 1 LOAD STEP 1 ITERATION 1 DISPLACEMENTS __ = 0.5-10° UNITS ♦ * EFFECT OF SHADOHINO ON A HEOOE IN A TUNNEL ROOF STRESS FIELD LENGTHS 2 . 0-10 s UNITS ACTIVITY 1 LOAO STEP 1 ITERATION 1 STRESSES 1.0-100 UNITS Figure 6 .4. Gravitational loading. -195- * EFFECT OF SHADOWING ON R WEDGE IN fl TUNNEL ROOF. DEFORMED MESH LENGTHS • 0 - 10 ; UNITS ACTIVITY 2 LORD STEP 3 ITERATION 4 DISPLACEMENTS 1 . 0 - 10 ' UNITS \ \ \ \ \ \ \ \ l III/// //// \ I * \ \ / / / / \ \ I / / / s / \ ♦ EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF. FLOW FIELD LENGTHS _ = 2.0-10’ UNITS ACTIVITY 2 LOAD STEP 3 ITERATION 4 DISPLACEMENTS _ = 1.0-10° UNITS X X -- + + * EFFECT OF SHADOWING ON A WEDGE IN A TUNNEL ROOF. STRESS FIELD LENGTHS _ = 2-0-103 UNITS ACTIVITY 2 LOAD STEP 3 ITERATION 4 STRESSES _ = 1.0-10° UNITS Figure 6.5 Excavation of the drive -196- e f f e c t OF SHADOWING ON A HEDGE IN A TUNNEL ROOF. DEFORMED MESH LENGTHS _ = 2.0*10s UNITS ACTIVITY 3 LOAD STEP 3 ITERATION 1 DISPLACEMENTS __= 1.0*10° UNITS \\\\\\lll III ////// \ I ♦ / / / / / / / / / / / * S / S' / ' / y " / / / / EFFECT OF SHADOWING ON A HEDGE IN A TUNNEL ROOF. FLOW FIELD LENGTHS _ = 2.0*10° UNITS ACTIVITY 3 LOAD STEP 3 ITERATION 1 DISPLACEMENTS __= 1.0*10° UNITS ^ + _ ^ ~+~ * EFFECT OF SHADOWING ON A WEOGE IN A TUNNEL ROOF. STRESS FIELD LENGTHS _ = 2.0*10° UNITS ACTIVITY 3 LOAD STEP 3 ITERATION 1 STRESSES _ = 1.0*10° UNITS Figure 6.6 First level ore excavation -197- ♦ EFFECT OF SHADOHING ON R HEDGE IN R TUNNEL ROOF. DEFORMED MESH LENGTHS . 0-10 UNITS ACTIVITY 4 LORD STEP 3 ITERATION 1 DISPLACEMENTS 1 . 0-10 UNITS /////// * / / / / / / / / / / / " / / / '/ + EFFECT OF SHRDOHING ON R HEDGE IN fl TUNNEL ROOF. FLOH FIELD LENGTHS ___ = 2.0-10* UNITS ACTIVITY 4 LORD STEP 3 ITERATION 1 DISPLACEMENTS _ = 1.0-10° UNITS X * ♦ e f f e c t o f s h r d o h i n g o n r h e d o e in r t u n n e l r o o f . STRESS FIELD LENGTHS _ = 2.0-10* UNITS ACTIVITY 4 LORD STEP 3 ITERATION I STRESSES = 1.0-10° UNITS Figure 6.7 Second level ore excavation. -198- ef fe ct OF SHROOHING ON fl HEDGE IN fl TUNNEL ROOF. DEFORMED MESH LENGTHS __ = 2.0-10* UNITS ACTIVITY 5 LORO STEP 2 ITERATION 4 DISPLACEMENTS _ = 1.0-10° UNITS \\\\\\lll I///// / /// \ V / ' / / ' 1 \ v 1 ' / / / / \ £ * / N - A s ' M l * / * J v \ \ s ' - i ? ‘ * »V ' * - « \ . « ^ / M III i i % \ V V S N EFFECT OF SHAOOHING ON A HEDGE IN A TUNNEL ROOF. FLOH FIELD LENGTHS _ = 2.0-10* UNITS ACTIVITY 5 LORO STEP 2 ITERATION 4 DISPLACEMENTS _ = 1.0-10° UNITS * * EFFECT OF SHAOOHING ON A HEDGE IN A TUNNEL ROOF. STRESS FIELD LENGTHS _ = 2.0-10* UNITS ACTIVITY 5 LOAD STEP 2 ITERATION 4 STRESSES _ = 1.0-10° UNITS Figure 6.8 Third level ore excavation -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 remains ** 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, B q 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. 1* * 4 ♦ -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, the following additional features are incorporated. (a) Quadratic joint element * (b) Exterior and interior regions modelled with boundary elements The analytical 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 encountered,when 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, even though the effect of dilation in the diagonal stiffness terms is added. For small problems with dilation,usually 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.lt 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 4 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 4 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 0 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 m 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.lt is suggested that the behaviour of the wedge be examined by using a model in which both stiffnesses vary. 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. -206- APPENDIX 1 DESCRIPTION OF INPUT FOR 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 * Restart code: 0 : initial problem 1 : problem restarted 2 : problem restarted and displacement reset to ♦ zero Save code 0 : if saving of the results not required 1 : control cards for restart will be saved in disc Second card(lOI5) ♦ 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 W Displacement printing code : 0 : incremental and total displacements are printed 1 : only total displacements are printed 2 : no displacements are printed Equation data printing code : 0 : print equation numbers and storage requirements 1 : suppress printing * Graphical output code : 0 : no graphical output 1 : plots every load case ♦ 2 : plots every 'n' steps 5 : plots every ’ n’ iterations at every step Note 'n' is the frequency given in col.41 - 45- Mesh drawing codet 0 : no mesh drawn 1 : initial mesh only 2 : initial and deformed meshes * Field drawing code- 0 : no field drawn 1 : stress field only 2 : stress and flow field -208- Frequency of graphical output- 0 : every step/iteration til n : every n step/iteration % Scaling code: 0 : scales adjusted every plot 1 ; 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 ' t 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 ♦ 26 - 35 upper limit on unbalance (divergence criterion) g (default is 10 x convergence criterion) w 6. Load information A sequence of cards headed by an activity 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 ' 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 W 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 - 20 H coordinate of temporary origin 21 - 30 V coordinate of temporary origin * For keyword 'FIX Columns 6-10 number of first node of the series 11 - 1 5 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 V POL ',’ORIG ' 'ASSOC' 'ENDND' The keyword may be left blank.In this case the previously defined « keyword is applicable. -212- A 1 .3 B o u n d a r y el em en t region Element card(A5.6l5) column 1 - 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 1 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 1 - 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 t 21 - 30 Poisson's ratio For orthotropic material* Columns 11 - 20 Young’s modulus in direction 1 21 - 30 Young's modulus in direction 2 31 - 40 Young's modulus in direction 3 41 - 50 Shear modulus 12 - 213 - * ♦ (a) Element numbering ♦ A C s 4 + material outward outward normal normal * material (-) ( + ) « (b) Orientation of element (c) Elasticity and global axes Figure Al.l 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(A5,5F1 O-O) Column 1 - 5 'JOL ' (keyword) 6 - 15 angle 'cl' 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 1 -• 5 'JOI ' (keyword) 6 - 10 direction of the normal to the joint plane(l.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 1 - 5 ’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 .1c) Elements with initial conditions(A5.1515) * Column 1 - 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 1 - 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 -2 1 6 - 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 1 - 5 ’PRT ’ (keyword) 11 - 20 height to the free surface (+) 21 - 30 ratio of horizontal to vertical stress (K ri ) 51 - 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 ' (keyword) -217- A 1 .4 Finite element region A1.4-1 Membrane elements Control card(A5,3I5) ♦ Columns 1 - 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(l5,3F10.0) Columns 1 - 5 material identification number ♦ 6 - 1 5 Young's modulus 16-25 cross sectional area 26 - 35 unit weight Member data cards(615.F10.0) <* Columns 1 - 5 member number 6-10 nodal point 1 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- A 1 .4-2 Plane strain elements Control card(A5-2I5) Columns 1 - 5 ’PLANE’ (keyword) 6-10 number of elements 11 - 15 number of different materials Material property cards(2I5,7F10.0) * Columns 1 - 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 - 30 Young's modulus in s' direction 31 - 40 Young's modulus in n' direction 41 - 50 Shear modulus in :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(1115.2A5) Columns 1 - 5 element number 6 - 10 node 1 (Fig-A1-2a) - 219- t P3 * * (a) Element [ P1 * * Figure A1.2 Plane strain element convention. * 2 6 (a) Node numbering 4 O P T nl sn ______S . s Tsn lav n (b) Positive displacements and stresses Figure A1.3 Joint element convention. -220- 11 - 15 node 2 16 - 20 node 3 21 - 25 node4 26 - 30 node5 * 31 - 35 node6 36 • 40 node 7 41 - 45 node8 ♦ 46 - 50 material identification number(default is 1) 51 - 55 element data generation K 56 - 60 keyword 1 61 - 65 keyword 2 keyword 1 may be* 'SAMEB'iall node numbers are automatically incremented by K 'N0D48':all node numbers are automatically incremented by K, * except for nodes 4 and 8 which are incremented by K/2- 'N0D26':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 1 -10 pressure on face 1 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 a yo 61 - 70 residual stress t xyo A1 .4 »3 Joint elements * Control cards(A5.415) Columns 1 - 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(l5,8F10-0) * Columns 1 5 material identification number 6 15 q^.the unconfined compressive strength of the wall rock 16 - 25 q^/T^ f°r m°del 1,or the shear strength intercept of the joint for model 2 26 - 35 shear stiffness k s 36 - 45 Blythe 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 2 66 - 75 the friction angle for a smooth joint 76 - 80 the dilatancy angle at zero(model l).or -222- low(model 2) normal pressure Element cards(915.2F10-0) Columns 1 - 5 Element number (must start from 1) * 6 - 10 node 1 (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) ■'d-0 1 45 element data generation K. as described for membrane element. 46 - 55 initial shear stress t ° (Fig-A1-3b) sn 56 - 65 initial normal stress o° n % * ♦ * -223- A 1.5 Activities A1 -3-1 Activity * GRAY .Gravity.residual stresses and pressure loading(8F10.0) ♦ Columns 1 - 10 percent of total loading for the first step n 71 - 80 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 m remaining percentage of the total load.This activity is usually applied to consolidate the field. A1 -5-2 Activity 'NOD '.Nodal point loading(l5.2F10.0) Columns 1 - 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 1EXC ’-Excavation. Nodal point cards(A5-515) Columns 1 -■ 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 (I5) Column 1 - 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(l6l5). Columns 1 - 5.6 - 10.etc.List here all elements excavated that do not have a nodal point on the excavation surface of this *k 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(l6l5)- 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.All 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 1 CON .Construction . Nodal point cards -225- First card(l5) 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 'EXCsection A1-5-5>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 A 1 -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 1EQ ’.Earthquake loading. % Acceleration cards(4F10.0.A5) Columns 1-10 minus acceleration in H direction(in g units) 11 - 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 . * 2975. 3000. 6000. 6000. 1000. 2975. HOOO. 11H00. 10350. 10350. 10350. 22206.7 22206.7 22206.7 0. 750. 3550. -750. 19800. 2H163. 26000. 17912. -3800. H816.7 2916.7 -2916.7 -2916.7 1 182 185 173 -350. 175 1 1 1 1 1 1 1 168 1 1 163 1 1 153 l l 117 1 1 Iff 1 1 1H8 1 1 106 1 1 68 1 1 61 l l 87 l l 3 -17200. 0.0 H 2 IN A TUTtCL ROOF. TUTtCL A IN -227- 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 2000. 2000. HOOO. 6000. 6000. 6000. 6000. 18000. 18000. 12H00. 11M00. IM2H0. IH2H0. 1H2H0. 1H2H0. 3 3 5 23800. 23800. 23800. 23800. 23800. 23800. 23800. 2 2 2 23800. 1 0. 0. 0. 0. 0. 0. 350. 1900. 1500. 3000. 3000. 3500. 8300. 3000. 5180. 5180. WOO. 16550. 10300. 19000. 10H00. -3900. 13000. 21600. -1500. -1900. HOOO. 31500. -3000. -3000. -3000. -8200. -3500. -5180. -5180. -5180. -3000. 22800. 29 29 2 -17200. -17200. -12200. 1 2 9 2 2 5 7 11 17 19 15 87 92 39 81 27 35 53 21 23 68 73 61 13 H7 Iff 187 182 Iff 172 171 3900. 171 3000. in 163 153 166 168 151 111 100 1H8 122 Iff 117 3600. 138 106 130 0 187 ENDND ELE C ELE BELEM CAR EFFECT OF SHADOHING ON A HEDGE A ON SHADOHING OF EFFECT W U T DATA FOR PERUSAL BY BY PERUSAL WJECHO UFOR T ROUTINE DATA PAGE 1 iiiii SfiSia 8ii!8 88888 88888 88888 88688 88868 88888 88888 -228- D W r DATA FOR PERUSAL BY ROUTINE JECHO PAGE 2 *** NOT t 100.E+03 0.0 **s CUE -I m t *** PRT 100.E+03 1. 27.E-06 «** ENDS *** i n n *** .fig i 3 2 3 5 1 *•* *B x *1 9 101 105 ss* ELI 5 12 -2 101 102 103 m Q £ 13 88 17 87 n £ 11 19 2 17 19 18 *** EtC 20 59 3 60 *** wvf ! 100.E+03 0.0 MMM CUE i *** L ttt PRT 100.£403 t. 27.E-06 ttS ENDB *** BELEM ELE 1 3 2 17 19 18 ELE 1 23 62 61 *** ELE 5 6 2 62 61 63 *** ELE 7 66 39 67 *3t* ELE 8 9 2 39 It 10 m ELE 10 13 69 68 m ELE II 18 2 69 71 70 ttt ELE 19 85 17 86 *** MAT 1 100.E+03 0.0 s*t CUE 1 ttt L ttt PRT 100.E+03 l. 27.E-06 ttt ENDB ttt BELEM ttt ELE 1 2 2 13 15 11 *** ELE 3 17 107 106 ttt ELE 1 7 2 107 109 108 ttt ELE 8 115 13 116 ttt NAT 1 100.E403 0.0 ttt CUE 1 ttt L *** PRT 100.E403 l. 27.E-06 ttt o n ttt FELEH ttt PUfC 26 1 ttt 1 1 27.E-06 100.E+03 0.0 **s 1 130 131 132 172 128 129 130 130 ttt 2 153 171 128 172 132 173 153 153 ttJt 3 132 133 131 118 153 173 132 132 ttt 1 128 17H 153 163 126 127 128 128 ttt 5 153 118 131 135 136 119 155 151 ttt 6 153 151 155 161 121 125 126 163 Bill! BBBBB BBBBB BBBBB BBfiifi BBBBB BBBBB BBBBI BBBBB IBBBB 3 PAGE ROUTINEFOR JECHOPERUSAL MW f BY DW D M A i a . r l l U EXC ENDND EXC OK*/ MEL BMP 100. joint 20 22 21 23 29 6190 26 25 20 10 12 11 17 9161 19 19 15 13 t I 15 15 IS 9130 19 7129 17 13 19 6126 16 11 7 9 5 92 7 6 V A 157 9 155 8 St 2 2 9 97 1 i 155 169 12S 157 159 159 192 120 161 190 176 113 122 115 109 17S 162 t i l 120 167 152 139 93 62 69 66 23 25 20 10 18 3 3 1 20. 20. 195 162 168 170 191 183 0 89 107 192 196 130 151 150 160 156 5 190 151 193 199 5 196 152 1S2 197 189 3116 93 75 73 97 77 26 71 69 79 29 83 81 85 11 5 5 2 5 1 1 1 1 112 129 121 2 8129 78 125 161 136 169 176 178 127 1 70 119 2 80 123 105 159 138 157 192 1 86 117 119 180 25 29 27 31199 131 93 87 31 33 15 13 12 9 3 1 1 1 8121 18 1 00. 00. 0 20.E+06 20. 20. 20. 6 169 167 166 165 137 175 171 9 11 197 199 108 9 9 193 139 195 7 178 177 7 180 179 106 181 6126 76 72 79 8115 68 28 30 2120 82 89 233 32 39 16 13 10 19 2 1 1 1 128 122 111 113 122 158 120 190 3 150 138 176 118 196 166 196 151 118 29 1186 31 90 88 536 35 37 20.E*06 20.E+06 15 11 17 17 19 -229- 0.2 0.2 168 121 2 129 123 170 119 117 132 151 152 197 182 195 185 189 109 8 69 183 0 2 107 8 180 187 32 73 12 71 51 75 922 2 69 79 71 77 31 83 11 81 38 85 12 17 9 2 166 120 122 159 196 169 157 120 118 190 176 7 177 178 161 62 66 7181 37 39 3162 13 2 2 1 1 1 1 1 167 165 169 156 2 1 120 171 190 158 6 1 160 199 175 179 1. 0 . 1 0 . 1 11 61 63 65 67 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 1. -2.93 -2.93 -2.93 -2.93 -2.93 -2.93 -2.93 -2.93 -2.93 -2.93 -2.93 -2.93 -2.93 -2.93 20. 20. IMVT BATA FOR PERUSAL BY ROUT DC JECHO 8 5 S o o S S Q o o S S S o o S S S o o 5 S II 8181 2 ! ! «| 88888 p«a "p* p " a « mp Ui * 1 IS 88888 w| UI $ i SSP3 m ^ • V4p4 Pl Q Pull 88888 " R IA a UI s • H H £ 3 S 8S p4 88888 1 N 8888 MO U> Ui R 8 f • 0 0 * 1 • • I -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 LAJRpCK’ ^ START ^ Zeroise counters and flags. Read general information. Read coordinate and boundary condition data. * Read type of region. Boundary element Finite element — ^Which typej Read information for boundary Read information for finite z o element region. elements. o c f Construct stiffness matrix and IT Construct the stiffness matrix n % load vector for it. and the load vectors. *< (Overlay 7 ) T3 (Overlay 2 ,2.1 ,2.3 ,2.4 ) © % Draw initial meoh(Overlny 5) Zeroise displacements. Calculate parameters for the solver. (Overlay 6 ) Read type of activity. ♦ ^ STOP ^ Step = 0 Step = Step + 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) * Figure A2.1 Flow chart of program AJROCK . -233- APPENDIX 3 ADDITIONAL INFORMATION.RELEVANT TO CHAPTER 3 A3.1 Orthotropic kernels The generalized Hooke's law for orthotropy may be written as J 1 ___ ----- O H l/E1 0 0 0 H \ l “V21^E2 "V31^E3 0 0 0 £22 - V E i 1/E2 ”V32^E3 °22 % -V 13/E1 0 £33 "V23^E3 1/E3 0 0 °33 • e12 0 0 0 1 / (2 .• G12) 0 0 °12 0 e23 0 0 0 1/(2«G13) 0 °23 £31 0 0 0 0 0 1/(2*G31) % _ — °31 or £ = F*0 (A3.1) Care must be taken in the notation of v.If, 3 £..= £ F . .-a.. , i=l,2,3 (no sum double index convention) 1 1 j =2_ 1J J J then Lekhnitski (1963) defines v . . from V ‘(V Ei)=-(V V 'W3 ’ whereas Tomlin and Butterfield (1974-) and Gerrard (1982b) define v . . from F.ij .=-(v../E.)=-(v../E.) ij i ji j ■ , i t Jj . In the following the latter convention is used. Due to the symmetry of matrix F > V31^ E3~ V13^E1 * V21^E2 V12^E1 ’ V32^E3 V23^E2 or v../E.=v../E. , i^j (A3.2a) ij i Ji J -234- The following inequalities must also be satisfied: E1’E2’E3,G12,G13’G31 > ° d-v^'V^) , (l-v 2 3 «v32) , > 0 (A3 .2 b) 1 “V12,V21"V2 3 ^ 1 " V21-V13"2 -V2i ’V32-V13 > ° For plane strain = e2^ = E ^ = 0 , hence °33 - ^ 1 3 ^ 3 3 ) '0u ‘ ^F23^F33^*°22 (A3.3) F „ are the compliances of equation A3.1. Thus *F F 0 * 11 12 '°ii _ z 22 21F 22 ^ 0 • o 22 (A3.A) ______P T " D GJ cv1 1 i—i — 0 0 F 1 1 i r 33m where F = F - F «F /F i,j=l»2 (A3.5a) ±3 ij i3 j3 / 33 Substituting for F . . we get 13 %■ 11 1/E1 - v 31/E3 > 0 (A3.5b) 12 _V2 1 ^ E 2 ■ V3l‘V32(/E3 22 1/e2- v 32/E3 > 0 X/(2*G12) > 0 ■ # 33 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 x x a = ((F^'F ^ ) 2 " E1 2 “ F33^ (real or imaginary) (A3 .6a) b = ((F11,F22 )i + F12+ F33)4 (real) (A3.6b) n = (b2- F ^ J ' a 2 - (a2 + F J-F^ (A3.6o ) -235- A = arbitrary = 1 (A3.6d) F = F + F A3•6e 12c *12 *33 ( ) ♦ Parameters dependent on position m = FU -*J + 2-F12c- z2. Z2 + F22-z* = X X = ((Fu )*.z* + (F22)2* ^ ) 2- 2-a2.z2-z2 (A3.7a) (A3.7b) * li = b' Zl 1 (F22'z2 + (F12 + F33 ),Z1 } * (A3.7c) 5'12=jl21=,/2,Zl -Z2 ^ ^ Fll^2'Zi + ^F2 2 ^ ’Z2 ^ ■ *22= b.z| / (F^.z* + (F12 + F33 ).z» ) (A3.7d) where * z. = x. - y. , i=l,2 (A3.8) l l •'i b is always real. This may be proven by substituting in eauation A3.6b for F.. from A3.5b. Then due to the last inequality of iJ * A3.2b we have that b2 - F^ > O’ and hence b is always real. m is always positive. This is proven as follows: If z^O equation A3.7a becomes, * m/z2 = Fl i (zl /z 2),,+ 2‘F12o‘(z1/ z2)2+ F22 (A3.9) The discriminant of the right hand side of the equation is, Di = F 2 - F-, •F^„ (A3.10) 12c 11 22 If Di < 0 , then m > 0 If Di £ 0 , then for m > 0 it is sufficient (-F + (F2 F *F )2 )/F,, (A3.11) (Zl/Z2 )2> v 12c v 12c 11 2 2 J 11 from b2> 0 we get (F..*F00)2 > -F 11 22 ' 12c from Di> 0 we get F *F < F 11 22 12c From the last two inequalities we conclude that F-j^ > 0 Hence the right hand side of inequality A3.11 is always negative, and the inequality is always satisfied. We may be proceed similarly for z^ = 0 , ^ 0. m becomes zero for z^ = z^ = 0 -236- The kernels U are the displacements due to a unit force and are given directly by Tomlin and Butterfield(1974.). The functions are different according to whether a is zero,real but not zero,or imagi- # nary. They are given by the following formulae: a zero 011 = U'+ (F*y£1;L)/U ’/2-iT*/F22) * U12 = F?3-J,12/U’Tr-b) = F?3*S,21/U*TT*b) = U21 (A3.12) U2 2 = U'+ (F|3 * Jt22) / U *^2 *7r*’A’11) a real ♦ = U 1 - 7i/(4.*/2*77*a*/F22) • arctan U12 = (a2,t)2 + :F^)/(8«Tr«a«b)«log((l + a«il12)/(l - a#&12))= U21 (A3.13) U22 = U* - ri/(4-#/2»7r«a*i/^11)-arctan (a*&22) Note that arctan is to be' specified in the interval [0,tt [ a imaginary * U11 = U* - n/(8-/2«iT*(i-a)*/F22)*log((l + i«a*il11)/(l - i-a-il^)) U12 = ^a2#t>2+ F| )/(4-ir-(l-a).b).arctan(i.a.Jl12) = U21 (A3.14-) U22 = U» - ri/(8*/2«TT* (x«a)*/F11)«log((l + i*a»&22)/(l - i*a*^22)) 4 Note that arctan is to be specified in the interval J-77/2 ,77/2 ] IP = {(b2- F33)-F33 + (a2+ F33)-b2} /(8«/2-77-b*/Fi;L)•log(F11'Att /m) ♦ (A3.15) l = /(-!) -237- The kernels T are derived from the stress field equations given by Tomlin and Butterfield (1974-). {(b2+ F33)*(F1]L*F22)^-2*(a2+ F33)*b2}*z2 + F ^ -(b2-F33)-z2 ♦ "112 ^ 2»/2*7T«m*b*i/F^ "222 z2*(^ll* (b2_ F33^*zi + l/^22*^t>2+ F33^*zP /(2*/2-ir-m«b) ♦ =122 Zl//z2-"222 (A3.16) NI—I {(b2+ F33)-(Fn *F22)^-2-(a2+ F33)-b2}-z| + F ^ ^ b 2- F ^ - z ”221 z i * ------:------2*/2*7T,m,b*/F 2 2 * =1 1 1 : Zl ’^ 2 2 * ^ 2“F 33^#Z2 + F 33)*z 2} /(2-/2»7T*m-b) =121 Z2 /Z 1 ,E 1 1 1 where =... is the stress a.. due to a unit force in the direction * k . The relation between = and T becomes t ■ (y) = T_(x,y)*ei (x) = a.k (y) *nk(y) = =j-ki(x*y) •e^.Cx)-nk (y) or ♦ Tij(x»y)=Ejki(x,y)*nk(y) (A3.17) e^(x) is a load at x in the i direction. The full expression of kernels T then becomes ^ _ {/F„*(b2- 22______F,,)-z233 2_____ + /F11 ______• (b2+ F„)-z2}*(z33 1 s s-n ) ^ 2 «/2 *7T,m*b T1 2 =-(l/(2«/2*7r-m*b))*{((b2+ F ^ ’/ F . ^ “ 2-^a2+ *3 3 )* ^ ^ 22 ) #Z2#Zl’n2+ + (Fl l / / F 2 2 )*(b 2- F 3 3 )«z1 -n 2 - /F 2 2 *(b2- F ^ ^ n ^ z * ♦ - / F n * ( b 2+ F 33)*z2*z2 «n1 } -238- T21=-(l/(2-/2*TT«m-b))*{((b2+ F^ ) m^22 “ 2*^a2+ F^)-oV/F^)-z2^ ^ + + F33^-z2-nl “ ^ n ^ b2- F33^*zl’n2 ” - /F22-(b2+ F33)-Z|.Zl.n2 } T {/Fu ’(b2- F33),Z1 + ^ 2 2 ,(b2t F33}*Z2}* (zs*ns} ^ 2#/2*TT#m*b where s takes the values 1,2. (A3.18) 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 z^ and z2 are not simultaneously zero. If a is real then, 4f F = F + F < (F *F ) 12c *12 *33 v 11 2 2 J Prove that (l+a#5,2^)/(l-a*Jl2^) is greater than zero and finite. It suffices to prove that * > 0 This is eauivalent to (a-&2 i ) 2 < 1 l+a«&2^ > 0 <^> {a-i/2»z1«z2/(/F11*z2 + /F22*z| )}2 < 1 2 -a2 *z!* zi < w ? n - zi + ^ 2 2 , z ! ) 2 which holds because m > 0. 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(a»£^) , arctan(a*&22). Let us first examine the arguments a*^2.1,8L*^22* a«b*z2 a«b«z2 a -^ ll a # £ 2 2 F • z 2 + F F • z 2 + F 22 2 12c’Zl 11 1 12c*Z2 then I f F 1 2 c > 0 a ’£l l ^ ° a ’ * 2 2 * 0 and 0 ^ arctan(a*£^) < 7r/2 , 0 ^ arctan(a*&22) < 77 /2 If f 1 2 o .<0 , then let us define the angle w from, a»£ = ±oo z2/zl = ±{(-F12c )/F22}* = ±ten 0^ n (A3.19) zl/z2 = ±{(-F12 c )/F 11}4 = ±tan “2 a ^ 2 2 = 160 For the function arctan to be continuous,we define the angle to lie within the interval [0, tt £ ,and z^/z^ = ±tan o)^ ->• arctan(a»£.^) = 7t/2 z^/z^ = ±tan 0)^ -»■ arctan(a«il22) = 7t/2 In figure A3.1 the lines defined by z ^ /z ^ = ±tan u)^ , z2/z^=±tan oj2 are shown.The sign of the functions a , achanges as point (z^,z2) passes these lines respectively,their sign being shown in the figure within parenthesis. From b2 > 0 and F^2c ^ we Se"^> /{(Fn -F22)} >-F12c -»■ (Fn /(-Fi2 c ^ > ^ “F1 2 c^ F2 2 ^ ’ or tan (tt/2 - co2) > tan oj^ or o>l + 0)2 < tt/ 2 This proves that the line defined by 0)^ is always lower than the line defined by a)2 in the first quarter of the coordinate axes. -24-0- -24-1- If a is imaginary then, F1 2 c > /{F 1 1 *F 2 2 }> 0 Prove that (l + i*a*&22)/(l - l*a*&22) is always positive and fi- * nite,so that the function log (1 + l*a*il 22)/(l - i#a«&22) has mea- ning. It suffices to prove that, 1 + i*a*S,22 > 0 and i - i-a-iZ2 > 0 this is equivalent to - a 2 ’ * 2 2 < 1 Substituting for " &22 we get —a^^b^^z**a o z2 m /(F* i ;l • z x^ + Fi?12c • z^ z2 ; ^<1 or FX l'Zl + F22*Z2 + 2*F12 c ‘ Z1 * Z2 * ° The last inequality holds , as. m > 0. m Similarly we may prove that (l + i*a*&^)/(l i#a*Jl^) is positive and finite. + Define the function arctan(1*a*£2^) to be continuous and single valued. We proceed as follows: l*a#&2^ never becomes infinite,and is continuous in the zjl’z2 * Hence arctan (i*a*Jl2^) must be defined to lie within the interval ]-7r/2,7r/2 £ ,to be continuous. In table A3.1 the function arctan for a real or imaginary is * defined Table A3.1 Determination of function arctan argument a*&„ 0 positive +oo negative arctan(a.£ .),i=j 0 ] < W 2 [ tt/ 2 J 7t/2,7t [ arctan(i*a«&..),i£j 0 ]0,ir/2[ - 3 - tt/ 2 , 0 £ -24.2- Relation between parameters of orthotropy and isotropy The compliances when the orthotropy degenerates to isotropy become, ♦ F 1 1 = F 2 2 = ( 1 - V2)/ E F12 = F21 =-v ’(1 + y )/E (A3.20) F33 = 1/(2*G) = (1 + V)/E * Substituting into A3.6 from A3.20 we get a =0 b = {2*(1 - v2)/E )4 (A3.21) ♦ n = - f| 3 = - d + v )2/ e2 Substituting from A3.20 and A3.21 into A3.7 we get, m = (1 - v 2)/Et i* * A 12 = ^21 = z1 *z2/r2*^2#E/^1 " v 2 )a (A3.22) ^ 2 2 = C^ /1*)2*^2*2^ 1 ” v2)} Ail = (z1/r)2-/{2*E/(l - v2)} % By substituting these values to A3.12,A3.15,A3.18 we get the formulae 3.15 of chapter 3 for isotropy. * ♦ -24.3- A3.2 Integration of kernel-shape function products over an element containing the first argument Kernel U 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, which 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 E, (Fig. A3.2a). Hence the shape functions are second order polynomials of r. For isotropy the logarithmic term is = A^*log(l/r), where A^ is a constant and the kernel U o - shape function product integral is a sum of terms A • / U •(r/a.,)n *dr . Let us evaluate these terms. 1 g 1 2-1 / U •(r/a..)n «dr = A, • / log(l/r) • (r/a, )n «dr = 0 * 1 1 0 1 = (-a^/(n+l)•{ log a1 - l/(n+l) }«A1 = In (A3.23) In Table A3.2 the values of I for n= 0,1,2 are shown. n Table A3 .2 Integral I for isotropy Integral value -a^« (log a. - 1)-A1 -(an/2)»(log a1 - 1/2)•A1 -(a1/3)*(log a^ - 1/3 )* £ 1 X2 -2 U - (a) /"• A * xa at I (b) I 3 2 / A 1 a i A V^ r * xa at 2 (c) I 3 2 / ai A r r a x at 3 (d) 1 3 2 ai ------ai /I — ai —/ Figure A3-2 Analytical integration of a log-polynomial product 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 : 0 = A2*log(Fu /m) , vrtiere is a constant. The kernel ,shape function product in * tegral is the sum of terms / U •(r/a^)n*dr .Let us evaluate these g , element terms• a^ a^ In = / (r/a1)n*dr = A ^ f logCF^/m) • (r/a1)n*dr = 0 g ^ “ 0 = - a1«A2/(n+l)-{4*(log a1 - l/(n+l))+ Ca} (A3.24) where Ca = log{(Zl/r)V 2.(F12o/Fll).z^z|/rV ( F ^ ) - ^ } In table A3.3 the values of I for n=0,l,2 are shown. #• n Table A3.3 Integral I for orthotropy Integral Value * -a1«A2*{4*(log a1 - 1) + Ca) xo -a1«A2/2»(4*(log a1 - 1/2) + Ca} h -a1 *A2/3-{4*(log a1 - 1/3) + Ca> *2 4 Let £ = r/a^ . In table A3.4 the analytically evaluated integrals I U (xa ,y)*Ne(y)»dr are evaluated. In column 1 the internal no- g de number of the first argument of the kernel a and the internal 4 node number e of the shape function are shown.In column 2 the relation between £ • and £ is shown. In column 3 the shape func tions are shown as functions of £ . 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 a c l is shown. In column 6 the appropriate figure for the tangent ele ment is given. -24.6- VJhen a=3 the integration is performed over two straight line ele ments and the total integral is the sum of the two integrals. As U is symmetric with respect to a , it 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 = J g N-U -dS - JaN*U .-dS. (A3-25) es S r r S. gt t r t & where subscript r denotes the real element and t the tangent one. Expressing the functions in terms of intrinsic coordinates £ , dS = J*d£ , dS, = J «k *d£ r t a a R becomes, es ^ R = / N*(U -J - H -k -J )-d£ (A3.26) es •• n r gt a a This last integration may be evaluated numerically. Table A3.4- Analytically evaluated integrals of kernels 1 2 3 k 5 75 a , e 5(?) N*(C) U rNe(c)*dS. f gt t V al/JaFig. 1 1 - 3*C + 2-S2 1 q -3 Ix +2 I2 1 2 1 - 2-C -C + 2*£2 2 A3.2b -I1 + 2 ‘I2 3 - 4 'S 2 1 -£ + 2-£2 2 2 2*£ - 1 1 - 3-C + 2*c2 2 A3.2c I0-3 *Il+2-I2 3 1 s 2 *2 3 2 1 A3.2d e c 2 X2 3 2 - 2*?2 2#I0 " 2 *I2 -247- Kernel T The principal value of the integral j T..(x ,y(£))*Ne(£)•J(£)*d£ 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 principal values of the integrals. In the program we have extended a method developed by Watson (1981,1.982) for potential flow problems,to elasticity. It relies on the rigid body translation method. Fach 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. If d(b,e) = a, and a is an intermediate node (Fig. A3.3b),then C..(xa) + / T. .(xa ,y(C))*j(5)*dC = ij bb , = - T. .(xa,y(5))*(l-;r)-J(5)>d5 - / T..(xa,y(5))*ds (A3.27) Sb ij where rg is the spiral contour. If d(b,e) = a, and a is an extreme node (Fig. A3.3a), then c . + J T (xa ,y(5))*Ue*j(5)-d5 = - / T,,(xa,y(5))-(l-Ne)-j(£)-d£ - f T,,(xa,y(?))-ds - ij ij br sr - / T (xa,y(S)).(l-Ke)*J(€)-d5 - / T..(xa,y(5))-ds (A3.28) s r n 1J bl si where now S, is the sum of S, , and S, . b ol br The integrals on the right hand side of the equations may be evaluated numerically using the G.L.Q.F. -2^8- A3.3 Particular integral The following stress field satisfies the equations of equili brium, hence it is a particular solution.• 4 0„ K. H A = 1 (p-g-xv - P'g-h^ + pQ) (A3.29) ^HV = °v 0„,T 0 HV The boundary conditions at infinity are assumed to be also satisfied by this equation. p is the vertical stress at level h , o is the ratio of horizontal to vertical stress, 4 H,V are the horizontal and vertical axes of the coordinate system. If 1,2 are the principal axes for orthotropy,then (a)12 = M-(a)HV (A3.30) _M is the stress vector transformation matrix. ♦ The constitutive law that relates stresses to strains is ^ 1 2 ^£11,E:22,y 12^ — --*^HV (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 U1 = V*l + 61*X2 + V xl*x2 + 5l'Xl + el*X2 + 5 * (A3.32) u2 = a2 Xl + e2‘X2 + W X2 + V X1 + G2*X2 + 5 Then the strains are given by, \T T = Anx-.x-.l) (A3.33) (en - E2 2 ,Y12 + where 2*0^ 61 A = Y 2 2 *B2 e2 “2 +Y]/2 b i + V2 (62 -249- By equating e from equations A3.31,A3.33 we get F*M«(a)HV = A-(x 1 ,x 2 »1)T (A3.34) Also from A3.29, ♦ (a)HV=p*g*K*« (sin 0,cos 0,-hQ+ PQ/(p*g) M x ^ x ^ l ) 1 where K* = (Ka ,1,0)T % Substituting a in A3.34 we get p'g*F*M-K*«(sin 0,cos 0,-hQ+po/(p*g) )•(x1 »x2 ,l)T=A*(x1 ,x2 »l)T (A3.35) * where ( F ^ * c o s 2 0 tF^2 «sin2 0)*Kfl-frF-|1 *sin2 0 +F^2 *c o s 2 0 p-g«F*M*K*= (F^2 »c o s 20 +F22- sin20)»KA+F^2*sin20 +F2 2 «c o s 2 0 *P*g= C2 ♦ F^«sin 0«cos 0*(l-K^) — . _ C3 (A3.36) For A3.35 to hold for any pair of (x-^,x2) ,all nine coefficients T multiplying the vector (x^,x2,l) must be identical on both sides of equation A3.35, that is 2*a^=C-^»sin 0 Y^=C^«cos 0 6i=ci*(-ho+po/(p*g)) Y2=C2*sin 0 2*32=C2* cos 9 * e2 =C2 ‘('h 0+po/ ( p ‘g)) ^ + 7 ^7 2 =0 ^* sin 0 (69+e1.)/2=G0« +p / (p*g)) 31+Y2/2=G3* c o s 9 ti .1 3 (-h 0 0 (A3.37) Three parameters may be chosen arbitrarily,to allow for a rigid body * motion. We define £^=£2=0 which imposes zero translation at the o rigin of the coordinate axes. -250- We arbitrarily also specify du^/dx^=0 at the origin which gives the following relation, * ^1 + ^2^an ^an + e2 (A3.38) Thus the parameters a to £ become a^=l/2*C^»sin 0 a^C^'sin 0- C^/2*cos 0 * 3-^=C^#cos 0- C2/2«sin 0 32=l/2*C2*cos 0 Y^=C^«cos 0 Y2=C2 *sin 0 Oo =(2«C<2«sin2 0-(C1-C )*sin 0«cos 0)* 6l=Cl‘^"h0+Po ^ p,g^ 2 3 1 ?(-h +p /(p-g)) * el=2*C3# (-ho+P0/(p*g) )”"^2 £ 2 =C2*^“h 0 + P 0^ P # g ^ (A3.39) If the displacements in the HV system are given by * W X1 + V X2 + W X2 + 6h 'X 1 + V X 2 V V X1 + V X2 + V Xl‘X2 + V X1 + V x2 (A3.4 0 ) the parameters cl. to £„ are related to the parameters a.. to £0 n V JL /d % as folloi^s : cos 0 -sin 0 “ h SH y h 6 h £H •ai Y i 61 E1 ca > sin 0 cos 0 av Y v 6V e v_ “ 2 ^2 Y 2 62 e2 * (A3.a ) The tractions are related to the stresses as follows : trr = a -(cos 0*nn - sin 0*n_) il il X t^ = Oy*(sin 0*n^ + cos 0*n2) (A3.4-2) * -251- Special cases Directions 1,2 and HtV coincide, that is 0=0, * Then °1 = P-g,(Fll-KA + F12} C2 = P'g’^F12’KA + F22^ ♦ Isotropy the angle 0 may be taken zero,hence ♦ G1 = p-g-(l+v)/E*((l-v)-XA - v) = p*g* (l+v)/E» (-vK^ + 1 - v) c3 = 0 ♦ Isotropy and K^=v/(l-v) (corresponds to lateral constraint) “h = ° Uy = p*g*(l+v)/E • (1-2*v )/(1-v )*x2*(0.5*x2-hQ+po/(p*g)) PR = v/(l-v)*p-g«(xv - hQ + PQ/(p*g)) * ♦ -252- APPENDIX A Estimate of error due to the assumption of continuous tractions * at nodes. The simultaneous equations approximating the integral equation may be written in matrix form as U * t , = T. *u (A4.1) 4 If t is discontinuous at nodes,then the equation may be rewritten as —U r *tr /sy + _ U 1 ^ 1 = — T*u /V/ (A4.2) * U r + U 1 = U (A4..3) r 1 where JJ , U_ are the matrices that have components equal to the integrals of the kernels U times the shape functions to the right r 1 or left of the node respectively, and t ,t^ are the tractions to the ♦ right and left of the nodes. Let' us say now that tr =t» + a » A t , t1 = t* + |S• A t {kU»U) 4 where A t = t^ - t1 (A4-.5) and t ’ is arbitrary.Then % a -6 = I (M.6) I is the unit matrix. Equation A4-.2 then becomes ^r+U1)-t‘ + (Ur*a + U1#3)‘At = ^u (A4.7) or ♦ U.t« + (Ur*a + = T*u^ (A4-.8) or c.t« + C.U-1. (ur«a + U1*3)*At = (A4.9) where the non-symmetrized stiffness matrix has been used for the sake - of clarity. -253- Let us define t' from the following relation: -P=C«t1 = /pN't-ds = Ol-(Nr«tr + N^t^-ds = Cr*tr + C ^ t 1 '-(A4-10) t are ,the actually applied tractions on the boundary T. r l Substituting for t^ ,1^ from A4-.4- in equation A4-.10 we get Cr *a + C 1 * ! =0 (A4-.ll) Also Cr + c1 = C (A4-.12) From equations A4-. 6, A4-.ll, and A4-.12 we get a = C - ^ C 1 , §_ = -C_-1‘Cr (A4-.13) From equation A4-.9 it can be seen that an additional term AP is needed for the correct answer to be obtained.Substituting for a and 3 from A4-.13 into A4-.9 we get - U - 1 • (Ur • c -1 • C 1 -!!1 . C-1. Cr). At= (C^C-U^.D1) ^ = =_(Cr-G«U“1*Ur )*At= = (1 / 2 ) • { (C1 -Cr )-C • U -1 • (U 1 -Ur )} • At (A4-.14-) If the elements are all of equal length then Cr =C1 =0.5‘C (A4-.15) - AP=(l/2) •C,U ~1# (U1*-U1 ) *At (A4-.16) The error in displacements is found from K^«Au + AP =0 (A4-.17) APPENDIX 5 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 /kn , and dilation angles 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. t ££Z- ¥ I ¥ * -256- * « * t 4 4 4 - l £ Z - 0 4 4 -258- i' 0 4 . ♦ * * * 6£2- ♦ ■* • * # * * -260- * # * * ♦ ♦ -192- -Z9Z- -263- * * ♦ * t * -4 M'O (1*0 Ot'O IL'O »9*0 99*0 9»*0 0»*0 ZCO *Z*0 91*0 90*0 OO'IT 90*0- UVS O VROS H-AU O PI N LA AXIS ALFA ON PHI OF PHT-VALUE VARIOUS FOR CURVES 0.08 eP.'OO 0..08 0.. I8 0.24 0 . J2 0.40 0.48 O.M 0.84 0.72 0.80 0.88 O.M — LF—► FA— AL ►— f UVSFRVROSP-Ab FPI ONAXISOFALFAPHI VARIOUSPW-VALbE CURVES’FOR * + —ALFA-*- •— * -V92- o.oa J.00 0.08 O.I» 0.24 0.32 0.40 0.41 O.M 0.84 0.72 0.80 0.08 O.M 4 # * 4 * -265- 0.08 _0.OO 0.08 0.)S 0.24 0.J2 0.40 0.48 0.M 0.14 0.72 0.80 * + * + * + + + 266 - - • .. .. ~ • • • CURVES FCft V~ftlOUS ~H)~VRLU! e'~Hr eN ~L'~ '~XJS EUl¥IS Jot vaRlaus 'HI-VQlUE OF PH) QN 8LfR AX)S ~ ~ OIK 5 / K N= 0 D 000 9 lOT R= 5 olK 5 / KN = 0 . 0 0 1 , lOT R= 5 ~ ":• o o .to: •flo a o . to: ": o ·a flo flo ": ": a a a ": a ~ a t t z: z: I l\) 1 1 0" -...J I ~ a ~ ·a ~ a ~ , !! it' ) ~t.oo .1. u ., u -f u -, - ~" • .. • .. .. •• • CURVES FOR VRRIOUS PHI-VRLUE~'·O·F·pHTOH'Ali·A·AXIS CU~VES fe~ VA~leus PHI-YALUE OF PHI ON ALFA AXIS ~ ~ .olK S / KN = 0 . 002 , lOT R= 5 .olK S / KN = 0 . 0 0 4 , lOT R= 5 ": ": o o ":...... o o ...... ": o o ...... ": ": o o i t ~ ~ I l\) 1 1 0' 00 I •~ o .. ~ o • • • .. .. • , CURVES FOR VRRIOUS PHI-VRLUE OF PHI ON RLFR RXIS C~U~VES' ,'o'''vAiil'ous -PHr:"YA'L,Uf'Of "HI ON "L'" AXIS ~ ~ arK S / K N= 0 D 0 0 5 , lOT R= 5 olK 5 / K N= 0 • 006 , lOT R= 5 ": • ·a to: a to: ~ ·a a N .to: ·a a N N ": to: a a t t ~ ~ I l\J 1 1 0" '"I ~• a ~ a ~ ·a .. • • ~ • • , CURVES FOR VRRIOUS PHI-VRLUE OF PHI ON RLFR AXIS CURVES fOR VARIOUS PHI-VALUE OF PHI ON ALFA AXIS o "! ~ O'K S / KN = 0 · 0 0 8 , lOT R= 5 DIK S / K N= 0 · 009 , lOT R= 5 ":• ~ o o •": =; o D ": .": ·D o N N ": ": o o t I t l\) z: z: ....,J o 1 1 I ~ • .. • .. fI • CURVES FOR VARIOUS PHI-VALUE OF PHI ON ALFR AXIS CURVES fOR YAR J OUS PH J -VRLUE-OF",. "'AX'IS' .:, ~"f...:.::t'.",:.: .. l ...... I1... ~~..... · ~ it ~"K 5 / KN = 0 · 0 1 0 ,.. lOT R= 5 ~.lK 5 / KN = 0 · 020 , lOT R= 5 C'O• "!• o o "! ·o "! ·o C'O c~ o t t J: J: I l\) -J 1 1 f-J I ♦ 4 ^ ♦ 4 -272- UVS O VROS H-AU O PI N LA AXIS ALFA ON PHI OF PHI-VALUE VARIOUS FOR CURVES 0.04 0..00 0..I2 0.16 0..20 0.24 0.26 0..J2 O.M 0.40 0.44 0.46 0.62 0.66 0.60 KS/KN = 0.0 5 0 , I 0T f l = 5 CURVES FOR VARIOUS PHI-VALUE OF ONPHIALFAAXISPHI-VALUEVARIOUS CURVESFOR * ♦ ♦ -273- * + * * * ♦ 4 -Uz- * * < 4 * * -275- 4 4 4 * I 4 -9^2- CURVES FOR VARIOUS PM1-VAIUE OF PHI ON ALFA AXIS CURVES FAR VRRfBUS'FHl-VRLUeBF PHI ON ALFA AXIS -277- “lb.oo i4.oo ib. oo ib. oo ib.oo ib.oo jb.oo ib.oo 4bm .oo .oo ib.oo 4b.oo ib.oo *Tb7oo ib.oom .oo iToo sb.oo4b.oo 4V.00 -278- REFERENCES BANERJEE. P,K. & BUTTERFIELD, R. (1981). Boundary element methods in engineering sience,McGraw Hill. BANDIS, S., LUMSDEN, A.C.,& BARTON, N. (1983). Fundamentals of rock joint deformation. I.J.R.M.Min.Sci.,Vol.20,No6,pp.249-268 BARTON, N,R. 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