block theory and its application to rock engineering

Block Theory andIts Application toRock Engineering

Prunice.Hall Intenutlorul Sedcs in Civil Ergfunuiry and Engfuuertry Meclu.nics

Williarn J. Hall, Editor

1

Block Thoor>l and

Rock Enginoering

Richard E. Goodman

Gen-hua ShiBoth of University of California, Berkeley

Its Application to

PRENTICE-HALL, INC., Englewood Cliffs, New Jersey 07632

Library of Congress Cataloging in Publication Data

Goodman, Richard E. (date)Block theory and its application to rock engineering,

Bibtiography: p.Includes index.l. Rock mechanics. 2. Block theory (Rock mechanics)

3. Rock excavation. I. Shi, Gen-hua, 1939-II. Title.TA706.c64 1985 624.1',st32rsBN 0-t3-078189-4

84-3348

Editorial/production supervision and interior design :

Sylvia H. SchmokelCover design: Edsal Enterprises

Manufacturing buyer: Anthony Caruso

G)1985 by Richard E. Goodman

All rights reserved. No part of this book may bereproduced, in any form or by any means,

without permission in writing from the publisher.

Printed in the United States of America

10987654321

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Prentice-Hall of Southeast Asia Pte. Ltd., SingaporeWhitehall Books Limited, Wellington, New Zealand

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Contents

PREFACE

NOTATION AND ABBREVIATIONS

1 INTRODUCTION

Excavations 3

Modes of Failure 4

Assumptions of Block TheorY 9

Comparison of Block Theory with Other AnalyticalApproaches 11

The Key Block System 19

2 DESCRIPTION OF BLOGK GEOMETRYAND STABILITY USING VECTOR METHODS

Equations of Lines and Planes 25

Description of a Block 3lAngles in Space 40

The Block Pyramid 4lEquations for Forces 43

Computation of the Sliding Direction 45

ExampleCalculations 48

xill

24

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GRAPHICAL METHODS: STEREOGRAPHIC PROJECTION

Types of Projections 57

Stereographic Projection of Lines and Planes 64Stereographic Projection of a Joint Pyramid 75

Additional Constructions for Stereographic Projection 78

Projection of Sliding Direction 83

Examples 85

Appendix: Important Properties of the StereographicProjection 92

THE REMOVABILITY OF BTOCKS

Types of Blocks 98

Theorem of Finiteness l0lTheorem on the Removability of a Finite, Convex Block 108

Symmetry of Block Types 112

Proofs of Theorems and Further Discussion ll2Shi's Theorem for the Removability of

Nonconvex Blocks l2lJOINT BLOCKS

Joint Blocks in Two Dimensions 130

Joint Blocks in Three Dimensions 134

Stereographic Projection Solution for Joint Blocks 138

Computation of Emptiness of Joint Pyramids UsingVectors 147

Applications of Block Theory: An Example 149

6 BLOCK THEORY FOR SUFFICIAL EXCAVATIONS

Basic Concepts 157

Conditions for Removability of Blocks IntersectingSurfaceExcavations 166

Identification of Key Blocks Using StereographicProjection 169

Evaluation of Finiteness and Removabilitv of BlocksUsing Vector Methods 179

The Numbers of Blocks of Different Types in a SurfaceExcavation 183

Procedures for Designing Rock Slopes 188

Removable Blocks in an Excavated Face 198

Appendix: Solution of SimultaneousEquations (6.19) to (6.22) 201

Contents

125

157

98

txContents

7 BLOCK THEORY FOR UNDERGROUND CHAMBERS

Key Blocks in the Roof, Floor, and Walls 204

Blocks That Are Removable in Two Planes

SimultaneouslY: Concave Edges 206

Blocks That Are Removable in Three Planes

Simultaneously: Concave Corners 2lIExample: Key Block Analysis for an Underground

Chamber 214

choice of Direction for an Underground chamber 225

Intersections of Underground Chambers 232

Pillars between Underground Chambers 236

Comparison of Vector Analysis and Stereographic

Projection Methods 238

8 BLOGK THEORY FOR TUNNELS ATTD SHAFTS

Geometric Properties of Tunnels 241

Blocks with Curved Faces 244

Tunnel Axis Theorem 249

Types of Blocks in Tunnels 249

The Maximum KeY Block 251

Computation of the Maximum Removable Area Using

Vector Analysis 253

Computation of the Maximum Key Block Using

Stereographic Projection Methods 271

Removable Blocks of the Portals of Tunnels 279

Appendix: Proofs of Theorems and Derivations

of Equations 287

9 THE KINEMATICS AND STABILITYOF REMOVABTE BLOCKS

Modes of Sliding 296

The Sliding Force 301

Kinematic Conditions for Lifting and Sliding 304

Vector Solution for the JP Corresponding to a Given

Sliding Direction 306

Stereographic Projection for the JP Corresponding to aGiven Sliding Direction 307

Comparison of Removability and Mode Analyses 310

Finding the Sliding Direction for a Given JP 310

Mode and Stability Analysis with Varying Directionfor the Active Resultant Force 313

Appendix: Proofs of Propositions 325

2&

296

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Prefaco

As the authors began their collaboration, which happened by the lur'cky accident

of a well-timed inlernational visit, it rapidly became obvious that the unfolding

story would require more than the limited space and scope of journal articles'

Although the succession of development and applications could be stated in a

long seiies of articles, it would be the rare reader who could make his or her way

through all in the proper order. Therefore, the authors decided to write a book'

This work was intended, right from the beginning, to point toward the practical

applications of interest to tunnelers, miners, and foundation engineers' Yet,

because the threads of the developments are new' it seemed important to show

the theoretical foundation for important steps, lest there be disbelievers' In the

end, the methods are so easily applied, and the conclusions so simple in their

statement, that one could entertain doubts about the universality, completeness,

and rigor of the underpinnings if the complete proofs of theorems and proposi-

tions were not included. It is not necessaryto follow all the steps of all the proofs

in order to use the methods of block theory, but they are there, in appendices at

the end of most chapters, and in some cases' within the chapters themselves

when the material is especially fundamental'

The applications of btock theory in planning and design of surface and

underground .^ruuations and in support of foundations are illustrated in alatge

number of examples. Computer programs (available from the authors*) can be

used by the reader to duplicate these illustrations. Alternatively, the examples

*c/o Department of Civil Engineering, University of California, Berkeley, C-alifornia

94720.

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can be worked using manual stereographic projectiqns. It should be understoodthat the theory is independent of the methods of its application. This book is notabout stereographic projection or computation as such, but about the geometricfacts of intersecting discontinuities penetrating a three-dimensional solid. Theadvantage of stereographic projection is its ease and economy. The distinctadvantage of computational solution is automatic operation, which allows itsincorporation in larger enterprises.

The authors have developed a number of computer programs for use withthis material, but limitations of space forbad their inclusion here. Interestedreaders are encouraged to write for further information. (See footnote on pre-vious page.)

The notation employed here may be somewhat new to many readers, sincetopology and set theory are not standard components of courses in engineeringmathematics. However, the level of mathematical skills actually required tounderstand the material is not advanced, and the book presents all the back-ground required to understand all the developments. Readers should be familiarwith matrix notation and vector operations. The mathematicat basis of stereo-graphic projection is presented more completely here than in any other bookknown to the authors. However, some familiarity with simple steps in applyingthe stereographic projection, as presented in some of the references citid,

-will

certainly speed comprehension.On first learning of the ideas of block theory, after hearing an introductory

lecture, a valued colleague suggested that we were then "at the tip of an iceberg.;'We have since exposed considerably more material but have hardly begun toexhaust the possibilities. We hope that some of you who read this materialwill discover yet new directions and possibilities in your particular specialties.

Many colleagues have assisted in the development of this book. We areespecially grateful to Dr. Bernard Amadei, Dr. Daniel Salcedo, William Boyle,and LapYan Chan. Partial support was provided by the California Institute forMining and Mineral Resources Research, Douglas Fuerstenau, Director, andby the National Science Foundation, Civil and Environmental Technology pro-gram, Charles Babendreier, Program Officer. A grant from Horst Eublacher andAssociates was also appreciated. The authors wish to thank Ms. Marcia Golnerfor devoted typing of a laborious manuscript.

Richard E. Goodman

Gen-hua Shi

Notation andAbbreviations

AaA^B

Av BAeBorB=AA+BorBfAAcB

A+B-B

u(n)

L(n)

A vectorA unit vectorThe intersection of ,4 andBThe union of A and .BI is an element of set B

I is not an element of B

,,{ is a subset of .B; ,4 iscontained in BI is not a subset of .B

^B is a subset of A; Acontains -B

B is not a subset of IThe complement of .B-the set of elements notincluded in .B

The upper half-space ofril, the boundary of whichpasses through the originThe lower half-space ofrC', the boundary of whichpasses through the origin

U(ft, Q) The upper half-sPace

whose boundarY is

normal to 2 and Passesthrough QThe empty set

Joint pyramid, a closed

set

Excavation pyramid, aclosed set

Block pyramidSpace pyramid (:- EP), an open set

,4 is not equal to ,B

,{ is parallel to ,B

A is parallel and in thesame direction as ,B

Dip and dip direction of

u)

JP

BPSP

A+ BA=B A+B

AII BAIIIB

arfa plane

(A)' The matrix transPose of(A)

(x, y, z) Coordinates of the tiP ofa vector whose tail is at(0' 0' 0)

xill

xtv

Radius of the referencesphereRadius of a circle on thestereographic projectionResultant applied forceThe number of joint setsThe unit normal vectorpointed into a rock blockThe unit normal vectorpointed into spaceUpward unit normal toplane iCross product of A andB

Notation and Abbreviations

Scalar product of A andB (dot product)Wall iThe interior edge formedby the intersection of W,and W,The interior cornerformed by the intersec-tion of W1, W1, and, WoAn exterior edge at theintersection of two cham-bersAn exterior corner at theintersection of two cham-bers

R

fti

AxB

A.B

wiEti

Crto

Ett

cr*

rnU

w

chapter I

Introductlon

Rock is one of the oldest yet one of the least well understood construction

materials. Although ancient civilizations demonstrated remarkable ability incutting and erecting stone monuments, excavating tunnels, erecting fOrtifica-

tions, and creating sculptures in rocks, it is doubtful that these architects and

builders ever created analytical procedures to govern their engineering activities.

In an age oftechnological advance, with precise and powerful analytical tools atour fingertips, it is reasonable to hope for a more fundamental basis for rock

engineering. Yet we must cope with hard facts that rhake our task complex.

First, rock as an engineering material is variable and all-encompassing since we

find, within the earth, rock materials possessing all classes of mechanical behav-

ior. Second, rocks differ significantly from most other materials with which we

build in possessing numerous flaws and weaknesses that together tend to inter-rupt the continuity of material and divide it into domains of difrerent types.

Common rock displays so many planes of weakness as to be essentially a collec-

tion of separate blocks tightly fitted in a three-dimensional mosaic. We call such

material "discontinuous rock."An example of how the network of discontinuities in rock affects engi-

neering performance of excavations is shown in Fig. 1.1, a photograph of the

spillway for old Don Pedro Dam, California. The concrete training wall of the

spillway can be seen in the upper right-hand corner of the figure. This structure

was designed to conduct water along a single bedding plane down'the'limb of a

roughly cylindrical fold in quartzite. In this ingenious way the water would flow

along a natural surface all the way to the river. Unfortunately, crossing joints

carried water down to a parallel surface forming the bottom of the folded layer.

Introduction Chap. 1

Flgur€ 1.1 Rock block eroded from spillway of old Don pedro Dam, california.

The system ofcrossjoints and bedding surfaces above and below the uppermostquartzite bed delimited a physical rock block, hundreds of cubic meters involume, which was lifted and removed from its position by the running water.consequently, instead of a smooth water course, the picture shows a cliff belodthe edge of the training wall. Further spills are now able to quarry the rock evenmore effectively, and if unchecked, could destroy the training wall and workbackward to release the entire contents of the reservoir. This will not happenbecause the dam has been replaced by a larger structure downstream and theplace is now submerged. But it serves to illustrate how the movement of a kevjoint block can create a worsening situation.

In discontinuous rock, we assume that the motions of points withinindividual blocks are derived mainly from rigid body motions of the blocksystem. This is an acceptable assumption in hard rocks such as granite, quartzite,gneiss, Iimestonen slate, and other rock types, where the rock material is oftenconsiderably stronger than good concrete. There are softer rocks, however,in which soil-like deformations of the blocks are at least as important as inter-block translations and rotations. This is true, for example, in Tertiary age sand-stones, claystones, tuffs, in chalk or very young limestone, and in decomposed

Excavations

or altered rocks of any kind, in which relatively large rock deformation accom-

panied by new crack growth cannot be ignored even if the rock mass is highly

fractured. As in all systems that have ideal end members, the usual rock mass

encountered in excavation work lies somewhere betweenthe extremes considered.

To those who live and work in the real world, we say only that the methods

described in this book apply to an idealized mode, whose validity will always

have to be tested in practice. On the other hand, we would not burden you with

reading this unless our experiences demonstrated a wide applicability for this

theory.The types of discontinuities that chop up the rock mass into blocks depend

on the scale of interest. The theory of blocks can be applied to explain fabrics

observed in rock specimens, in which case the controlling planes of discontinuity

arefissures and microfauh.r. In detailed rock excavation, to shape the interior ofan underground gallery, for example, individual ioints parallel to the bedding

planes, or from sets in other spatial attitudes, may control the outcome of the

work. In constructing a large underground room in a mine or for a hydroelectric

power project, important through-going joints, sheared zones, disturbed zones

ulotrg contacts of different rock types, andfaults will bound the individual blocks

of the rock mass. Finally, the structural geologist addresses systems of blocks

kilometers on edge formed by systems of major and minor faults and major

formational contacts. Most of the examples in this book and the main experience

of the authors concern excavations in which it is the joint system that decides

the rock blocks. This impties that the most immediate applications of this

theory will be in rock engineering for underground and surface space, trans-

portation routes, hydroelectric power, water supply, and mining.

EXCAVATIONS

An excavation is a new space created by removing earth or soil from its natural

place. The objective may be to use the excavated material for fill or in manufac-

lor., or to gain space at or below the surface. Occasionally, by advanced plan-

ning, both objectives are combined in one project: for example, when a quarry

is intended to be converted into a park or development after the product has

been removed to a predefined contour. In any event, the space created byremoval

of rock has to behave according to the design, or the project or excavation willfounder in delay and excessive cost. In civil engineering work, the design invari-

ably requires that the excavated space remain stable during the process ofexcavation, and usually for the lifetime of the engineering work, which may be

on the order of a hundred years. Examples of permanent civil engineering

excavations are underground space complexes for industrial use or storage;

chambers for hydroelectric turbines and transformers; tunnels for roads, water

supply, and water power; shafts and tunnels for subterranean sewer systems;

open cuts for highways, railroads, pipelines, and other transportation routes;

open cuts for spillways of reservoirs; and open cuts to site surface structures.

lntroduction Chap. 1

Temporary excavations that will eventually be backfilled with earth or concrete,or allowed to collapse, provide surface cuts for foundations of buildings anddams and for quarries attached to specific construction projects. In all theseexcavations the loosening of rock blocks along the contour of the excavation isto be avoided to achieve safe, sound conditions for the users of the space.

In the mining field it is sometimes preferred to create intentionally unstableexcavations so that the costs of rock breakage will be minimized. The idea willbe to initiate self-destruction of the rock through a properly executed initialexcavation that triggers the caving process. There are also many uses for per-manent underground mine openings: for example, as shafts to gain access to along-life mine, for hoist machinery, and for underground crushers. Moreover,even though mines that exhaust the ore in a few tens of years can one day beabandoned, the requirements for safety usually dictate that they be excavatedas if they were to be stable, permanent openings.

The theory developed in this book applies to all the types of excavationsmentioned above. In the case of cuts and underground openings thatareintendedas permanent features, block theory will permit the design of excavation shape,size, and orientation, and a system of internal supports (if necessary) thatprovides stability at minimum cost. In the case of mining openings that areintended to cave, block theory will help choose initial undercuts to maximizethe chance of natural rock caving. This theory relates to the movement of jointblocks that are liberated by the artificial surfaces to be excavated. This is thecorrect approach wherever the mode of failure of the excavation involves themovement of rock blocks. To appreciate the place of this approach in the totalscheme of things, it is useful to examine briefly all the important failure modesthat can actually occur in an excavation.

MODES OF FAITURE

Failures of underground excavations are uncommon, fortunately, but all under-ground excavations undergo localized "overbreak" as the desired dimensionsof the opening readjust themselves to geological realities. Surface excavationsare less stable than underground openings and experience failures more fre-quently. But rock falls in the confined space of an underground chamber ortunnel are apt to be more troublesome than even large rock movements intosurface cuts. It is the flow of stresses tangentially around the undergroundopening that makes it relatively safer than its surface counterpart, which lacks acompletely encircling contour to confine rock movements driven by gravity.In very deep mines, or excavations into weak rock, these stresses may themselvesinitiate new cracking, or even violent rock bursts. More often, however, thetengential stress around the opening tends to hold the potentially moving rockblocks in place, thereby acting as a stabilizing factor. In fact, it is usually theabsence of continuous compression around the opening, rather than its presencein elevated magnitudes, that permits serious overbreak and failure. For example,

Modes of Failure

Figure 1.2 Rock fall from a Norwegiantunnel. (From Brekke, 1972.)

consider the slide of some 200 m3 of rock, including blocks several cubic meters

in volume, that fell from the roof of a Norwegian tunnel in 1960 (Fig. 1.2).

The slide initiated from a tabular mass of altered metamorphic rocks between

two seams of swelling clay whose squeeze into the tunnel left the rock free ofnormal stress and therefore unsupported along its sides.

One form of "failure" of both surface and underground excavations that is

not addressed by this book is destruction of the surface by erosion from water,

or gravity alone. This is likely to be critical when running water is allowed to

travel along the surface of altered or weathered rocks, or poorly cemented

sediments. Figure 1.3(a) depicts the formation of deep gullies by the action ofrainwater charging down a face in friable, tuffaceous volcanic rocks. Flatterslopes are more seriously affected than very steep slopes. Loosening and ravelling

of shales in underground galleries, with damage to support systems, is frequently

experienced in these weaker rocks, or in harder rocks laced by seams or affected

by hydrothermal alteration or weathering. As was seen in Fig. 1.2, such behavior

can eventually undermine contiguous rock blocks to permit larger cave-ins.

Another rock failure mechanism not examined in this work is rock"squeeze" depicted in Fig. 1.3(b). The inward movement of the walls, roof, and

floor, due to creep, slow crack growth, or slowly increasing load on the rock, can

crush the support and lining structures under the worst conditions. Similar

results accompany the swelling of active clay minerals in certain claystones,

shales, and other rocks with disseminated clays (usually smectites). Occasionally,

problems of active weathering, for example of sulfide-bearing rocks, can cause

related problems.In layered rocks, in relatively weak rocks at moderate oreven shallow depth

lntroduction Chap, 1

(d)

Figure 1.3 Modes of failure of rock excavations: (a) erosion, (b) squeezing,(c) slabbing, (d) buckling, (e) toppling, (f) wedge sliding, (g) tension cracking andsliding after loss of toe support.

(c)

Modes of Failure

(f) (s)

Figure 1.3 (Continued)

and in hard rocks at very great depth, growth of cracks can undermine portions

of the excavation and permit local rock failures. The interaction of new cracks

with preexisting fracture surfaces presents an important, but intractable analyt-

ical probtem. This mode of failure is therefore not discussed here. In the absence

of governing geological structure, new cracks tend to develop parallel to the

excavated surface, as shown in Fig. 1.3(c). Therefore, new crack growth will lead

to major failures underground only if these cracks intersect preexisting discon-

tinuities. When that happens, major cave-ins can materialize, as shown in Fig.

I.4. It would be possible to enlarge the scope of block theory to analyze a

situation like that of Fig. 1.4 if the orientation of the new cracks could be known

in advance.Excavations in regularly layered rocks present failure modes due to flexure

of the layers, as shown in Fig. 1.3(d). This occurs in those regions of the excava-

tion surface that make small angles with the rock layers. A maximum inclination

of the tangential stress with the normal to the layers is determined by the angle offriction between the layers (see Goodman, 1980, Sec. 7.4). Flexural movements


Figure 1.4 Rock fall in diversion tunnel for Castaic Dam, Calif. (From Arnoldet al., 1972.)

of a cantilever type occur in surface excavations, leading to a style of failureknown as "toppling," shown in Fig. 1.3(e), while bending of both symmetricalbeams and cantilevers occur in undetground openings (see Goodman, 1980,Chaps. 7 and 8).

All the modes of failure discussed above are ignored in this book. What lsdiscussed is the movement of rock blocks from the surfaces of the excavation.The movement of a first block creates a space into which previously restrainedblocks may then advance. Thus a serious failure can occur retrogressively,sometimes very quickly. An important example is afforded by the Kemanotunnel in British Columbia (Cook et al., 1962),where an unlined tunnel originally7.5 m in diameter suffered a rock fall 20,000 m3 in volume. The cave extended42m above the original roof of the tunnel. Such a large mass of rock could caveinto the small space of the tunnel section only by retrogressive action.

The main thrust of this book is that modes of failure like that of Kemanotunnel cannot occur if the initiating block is held in place. Thus we need notanalyze a complex, potentially caving mass but only the first block to go. Casesarise where it is the movement of the first block alone that causes the totalfailure. For example, the movement of a large wedge in the left abutment ofMalpasset arch dam caused that tragic failure (Bernaix, 1966). Figure 1.3(f)illustrates the movement of a single rock block from a surface excavation.

Assumptions of Block Theory

Figure 1.3(g) suggests a more complex mechanism in which the release of the

block is possible only with local rock crushing (at the toe of the lower block) and

were release of the lower block triggers tensile cracking and a further block fallfrom above. Corresponding examples of single block movements with or withoutnew crack growth can be found in underground excavations.

Many failures of rock excavations develop from the movement of a single

rock block previously defined by joint intersections. On the other hand, even

relatively pure block sliding modes begin to look more complicated by the time

the failure has developed because the movement of each block gives birth to

new block movement potentials, new rock cracking loci, shifting loads, and

changing strength values. Some excavation failure modes are truly hybrid inthat the initiation of failure arises from the combination of two or more mecha'

nisms acting simultaneously. For example, it was new cracking and tensile

failures of the rock that permitted the tunnel support failure sketched in Fig. 1.4.

The rock was interbedded friable sandstone and shale, dipping steeply into the

tunnel and striking parallel to the tunnel axis. Overbreak occurred in the roofwhen the top heading was driven and undermined the beds. When the bottomheading was removed, heightening the excavation, tensile stresses were created

due to the sliding tendency of a rock wedge along the bedding. The rock then

cracked, releasing the wedge, whose movement into the excavation destroyed

the steel supports and triggered a complete collapse of a long section of the

tunnel.The examples above demonstrate that excavation engineering demands

attention to the rock structure. The engineer can deal with most rock conditions

if the geological data are properly incorporated in the design process. Althoughmany interrelated factors affect the design of an excavation, sufficient flexibilities

will usually exist to accommodate geological requirements. For example, in the

case of block movements we will be able to show in this book that rational

procedures can determine the best orientation of the excavation in order toobtain stability with minimum support cost. Turning a tunnel, an open cut, oran underground chamber may be allowable even in the construction stage, and

almost surely in the design stage. The procedures to be presented here demon-

strate further how to modify the shape of a gallery, or the contours of a surface

cut, to enhance stability. Conversely, in a mining scheme involving intentional

rock caving, the procedures determine the most efficient orientation and shape

of the initial cuts to assure caving.

ASSUMPTIONS OF BLOCK THEORY

The thrust of this book is to produce techniques to specify the critical jointblocks intersecting an excavation. It applies to rock engineering for excavations

in hard rock where the movement of predefined blocks precipitates failure.

Even though other modes of failure are ignored, the possible applications are

many and the problems addressed are significant in mining, civil engineering,

and other fields of application.

Introduction Chap, 1

The problem posed is limited in scope-to find the critical blocks createdby intersections of discontinuities in a rock mass excavated along defined sur-faces. Yet this problem is sufficiently difficult that a series of simplifying assump-tions must be adopted in order to gain workable solutions. The principalassumptions follow.

1. All the joint surfaces are assumed to be perfectly planar. This approachesreality for most joints and faults, but not for all and can be quite wrong forbedding surfaces, for example, on the limbs of folds. We assume perfectplanarity in order to describe block morphology by linear vector equations.On the other hand, it would be a straightforward extension of the theory topermit certain planes to be curved.

2. Joint surfaces will be assumed to extend entirely through the volume ofinterest; that is, no discontinuities wilt terminate within the region of akey block. The implications are that all blocks are completely defined bypreexisting joint surfaces so that no new cracking is entailed in the analysisof block movements. In view of the preceding discussion, this limits theapplications to a specific type of failure mode, excluding failures with newcracking, as in Fig. 1.4. However, the latter could be studied using themethods to be developed if the locus of new cracking were defined initially,that locus then being input as an additional discontinuity plane. Inpractice, it will be necessary to consider the finiteness ofjointsinattemptingto apply the block theory to successively larger excavations.

3. Blocks defined by the system of joint faces are assumed to be rigid. Thismeans that block deformation and distortion will not be introduced.The key-block problem is formulated entirely through geometry andtopology. Subsequent examination of the stability of the key blocks, whichare found through block theory, will then introduce strength properties forthe discontinuities. Since the development of frictional resistance along thefaces of the key blocks actually entails deformation along the surfaces ofthe blocks, it therefore implies accumulation of strain and stress withinthe blocks. Deformation of the blocks under limiting surface forces couldbe examined by coupling block theory to a numerical analysis embracingmechanics, in which case constitutive relations for the rocks would haveto be defined. This will not be examined in this book.

4. The discontinuities and the excavation surfaces are assumed to be deter-mined as input parameters. If joint set orientations are actually dispersedabout some central tendency, some one direction will have to be taken asrepresentative of the set. Through Monte Carlo simulation techniques, itshould be possible to examine the influence of variations in these anglesand to relate the output results statistically in terms of probabilities. Butthis will not be pursued here. Since the result of the block theory is a listof key-block types, none of which include by themselves more than twojoints of any set, it is entirely reasonable to treat the discontinuity orienta-tions as if they were precisely determined quantities.

Comparison of Block Theory with Other Analytical Approaches

In summary, block theory will be developed on the basis of geometricinformation derived from structural geology and equilibrium calculationsusing simple statics. It is assumed that continuum mechanics is second in impor-tance to the calculation and description of key blocks. Only block movementmodes are to be considered; we will not attempt to assess other modes of failure,such as erosion, rock cracking, and flexure of layers.

GOMPARISON OF BTOCK THEORYWITH OTHER ANATYTICAL APPROAGHES

A number of analytical tools are available for engineering calculations involvingexcavations. These include numerical methods (finite element analysis, finitedifference analysis, and discrete element analysis), physical model techniques,

and timiting equilibrium analysis. Most engineering decisions involving rockexcavation are conditioned as well by intuitive assessment based on experience

or informed judgment. Block theory is new. fn some respects it is more immedi-ately applicable and potent than any of the older approaches. It will be instruc-tive then to compare block theory with the various alternatives.

Block Theory and Finite Element Analysis

By means of block theory, we.will be able to analyze the system of jointsand other rock discontinuities to find the critical blocks of the rock mass when

excavated along defined surfaces. The analysis is three-dimensional. With the

determination of the key-block types, the theory then provides a description ofthe locations around the excavation where the key block is a potential hazard.An example of the output of a block analysis for a tunnel is given in Fig. 1.5.

Figure 1.5 Intersection of a key block and a tunnel; typical output of analysis

using block theory.

11

12 Inttoduction Chap. I

The largest key block, defined by the input sets ofjoints, the tunnel section, andthe tunneling direction, is drawn in relation to the tunnel. The next step willbe either to provide timely support to prevent the movement of this block, or toanalyze it further in hopes that available friction on the faces will hold it safelyin place. Alternatively, the theory could be redirected to a new set of input, forexample, by changing the tunnel shape or direction.

An example of finite element analysis is shown in Fig. 1.6, where a two-layer rock beam spanning a symmetrical excavation is modeled. The inputinformation required begins with a computing 'omesh" establishing the size andshape of the domain to be studied, which in this symmetrical example defineshalf of the roof. Also to be input is the set of mechanical properties for eachelement of the mesh. In this particular program the input properties to besupplied are the deformability and unit weight properties of the rectangular andtriangular elements, and the stiffness and strength properties of the lines ofjointelements between the two layers. Figure 1.6(b) shows one part of the output-thedeformed mesh-obtained from the displacements of each "nodal point," eachcorner of an element in this case. Note the shear deformation of the joint ele-ments over the abutments of the beam and the opening of a gap between thelayers in the center of the beam. Figure 1.6 shows the state of stress in the centerof each element, as revealed by vector crosses aligned and proportioned to thedirections and magnitudes of the principal stresses.

We have seen in these examples of block theory and finite element analysisthat there are fundamental differences.

1. Finite element analysis determines strains and displacements throughoutthe model, whereas block theory does not determine strains or displace-ments anywhere. The latter determines, rather, a list of dangerous orpotentially dangerous blocks behind the surface of the excavation.

2. Finite element analysis determines the stresses and, with difficulty, thesecan be manipulated to find regions of potential danger. Block theoryimmed,iately locates danger points and provides an estimate of the supportforces needed to avert failure. Block theory does not find stresses in orbetween elements.

3. Finite element analysis can be used parametrically, once a model has beenset up, to find a suitable shape for an excavation. But it cannot providemuch help in charting the wisest direction for the excavation. Block theorycan handle both tasks very well.

4. Finite element analysis must always compute from a specific mesh, withpredefined directions and spacings of joints. Generic studies can be madeonly if numerous meshes are generated. [n contrast, block theory analyzesthe essential, generic aspects of the problem posed by a given joint systemwithout calling for a specific joint map. Then, as an optional secondstage, the theory can be applied specifically to actual joint locations (ifthe data are available).


tt+f

+-y+

i+

t*+

ii-+

I

+

I

+ | +++rl|l

(cr

Figure 1.6 Results of a finite element anatysis of an excavation in jointed rock:(a) initial mesh; (b) deformed mesh; (c) stress field. (From Hittinger, 1978.)

13

(al

(b)

14 Introduction Chap. 1

5. In general, finite element analysis is a larger computation than blocktheory and will always require a computer. In contrast, block theory canbe applied entirely manually with graphical methods, for example stereo-graphic projection. The graphical techniques required to apply almost allaspects of the theory will be demonstrated in this book. On the other hand,speciflc applications can be freed of tedium by making use of a computer.In contrast to finite element analysis, these programs are suited to micro-computers of the type now available to most engineers. Since theseprograms are relatively small, block theory computations are far lesscostly than finite element runs.

Block Theory and Distinct Element Analysis

The distinct (or discrete) element method is a numerical model approachwith reduced degrees of freedom compared with finite element analysis. Byremoving deformational modes from blocks outlined by joint elements, onlyrigid-body modes remain. A finite difference or finite element analysis of thesystem can then trace the rotations and displacements of the block system asconditioned by the load/deformation relations adopted for .the joints. In thedistinct element programs pioneered by Cundall (1971), and Voegele (1978),relatively large two-dimensional block systems are calculated by integratingfinite difference approximation of the equations of motion for each block, withchanging boundary forces calculated at each time step from the changing blockinteractions. The analysis can be performed with a microcomputer and displayedinteractively. Figure 1.7(a) shows the initial positions of the blocks. This infor-mation had to be input as well as the friction properties of each joint and the unitweight of all blocks. As the computation begins, the blocks displace and rotateunder gravity and the deformed mesh can be followed through large deforma-tions. An early stage of output is shown in Fig. 1.7(b). The model achievedstability through arching as the program continued to run. In contrast, a secondmodel with joint AB rotated clockwise to a more nearly vertical positionproduced instability and collapse. (The reason for this can be demonstratedusing block theory.)

Distinct element analysis is an instructive tool for excavation engineeringin that it permits analysis of large block movements in geologically complexsections having many joint blocks. It is restricted to two dimensions, however,unless very large computers are used. As with flnite element analysis, it is stillnecessary to compute from a predetermined mesh, incorporating precise loca-tions of all joints. As noted earlier, block theory does not require premapping ofthe joints and it is fully three-dimensional. On the other hand, block theory doesnot offer an analysis with large deformations. Block theory is better equippedto help choose the direction and shape of an excavation, both because it isindependent of the precise joint map in the section of interest, and because it isthree-dimensional.


6 = 26.5"

{bl

Figure 1.7 Distinct element analysis of a tunnel in blocky rock by Voegele

(1978): (a) input information; (b) output after an early stage of deformation.

Block Theory versus Engineering Judgment

Engineers have been siting excavations in jointed rock for a very long time

-far longer than the period in which numerical tools have been available.

Presumably intuition, experience, and judgment were called and, hopefully,

some specific information about the directions and properties of the major jointsets. It is noteworthy that relatively few excavation engineering experiences haYe

been well documented in the literature so that a newcomer cannot easily acquire

such experience from study alone. Further, it is not clear what procedures, ifany, have been used to relate excavation directions with joint set orientations,

except for the general rule that major rock walls should not be cut parallel to

the strike of a major joint set.

The excavation siting problem is truly three-dimensional. Block theory,

which is tailored precisely to the three-dimensionality of the problem, can

address the siting of excavations in better focus than can intuition. Experience

offers no alternative to rational procedure when designing an excavation ofunprecedented shape, size, or function.

Introduction Chap, 1

Block Theory and Limit Equilibrium Analysis

In the field of soil mechanics, the value of analyzing the limiting equi-librium condition is well appreciated. Methods of limiting analysis for slopesand foundations are available (e.g., Bishop's modified method of slices, Morgen-stern and Price's method, etc.) to determine the critical condition when the soilis just about to pass from a state of stability into one of instability. Referring to

Figure l.E Londe's analysis of stability of a dam abutment. (From Londe andTardieu, 1977.)

(";

(__:

fr\loo€I

tl

2

/60'7O U3 Wl U1 l%l


the limiting state saves the trouble and uncertainty associated with trying todefine the actual state of stress for a specific set of conditions. Methods forassessing limiting equilibrium also enjoy use in rock mechanics, and can be

quite sophisticated. For example, Fig. 1.8 shows the results of Londe's equi-

librium analysis for a tetrahedral rock wedge within the abutment of an arch

dam. The wedge has three joint faces initially in contact with rock. Each face

plane l, 2, or 3 is potentially subjected to a water force equal to a given propor-tion (U) of the head of the reservoir acting over the entire plane. Entering

appropriate values for the three water force coefficients (Ut , (Jr, and Ur) in the

diagram produces a point R (or R') whose position determines two strength

parameters, ,1, and p, that determine the degree of safety of the wedge. It is

beyond our present purpose to describe the solution or these parameters at thisjuncture. We wish to point out, however, that such an analysis can be run only

after a particular tetrahedral block has been singled out. Block theory is not a

substitute for the limiting equilibrium analysis but, rather, a necessary prereq-

uisite since it will allow you to determine which block to analyze. It will also

evaluate the blocks' dimensions, face areas, volumes, and other factors that are

required for further analysis.Limit equilibrium analyses for rock wedges are discussed in detail in

Chapter 9. We will extend the previously available analytical methods to permit

stability analysis of blocks with any number of faces and any shape.

Block Theory and Physical Models

Three-dimensional physical models of rock excavations containing joints,

and embracing full similitude in all important physical quantities, provide the

only three-dimensional alternative to block theory with respect to practical rockproblems. Such models permit structural engineers to assess and design for rockweaknesses due to discontinuities. Figure 1.9 gives examples of some impressive

physical models tested in the laboratory ISMES in Bergamo, Italy. In Fig. 1.9(a)

the rupture of an arch dam model bearing on an abutment in layered rock isshown to follow from interlayer sliding in the rock. This model was made withcarefully fitted prismatic blocks. Figure 1.9(b) shows an even more complex,

three-dimensional block model used to design a reinforcing structure to stiffen

the abutments of a high arch dam abutting against limestone with inclined jointsets.

Although such models are useful in practice, they are too cumbersome and

costly for everyday problems. Furthermore, they are not convenient for under-ground excavations, in which the region to be observed is hidden from view.

No three-dimensional models have yet incorporated block shapes more complex

than rectangular prisms. Thus the inherent anisotropy of the rock mass is notaccurately represented except in those cases where the jointing is actuallyprismatic. Nevertheless, because three-dimensional physical models have great

visual impact, they are here to stay.

17

(a)

(b)

Figure 1.9 Physical models of blocky rock foundations after testing to failure:(a) a dam on gently inclined layers; (b) a dam on steeply inclined layers. (FromFumagalli, 1978.)

The Key Block System

THE KEV BTOCK SYSTEM

The objective of block theory, as stated earlier, is to find and describe the mostcritical rock blocks around an excayation. The intersections of numerous jointsets create blocks of irregular shape and size in the body of the rock mass ; then,when the excavation is made, many new blocks are formed with the added

surfaces. Some of these will not be able to move into the free space of the

excavation, either by virtue of their shape, size, or orientation, or because they

are prevented from moving by others. A few blocks are immediately in a positionto move, and as soon as they have done so, other blocks that were previouslyrestrained will be liberated.

Figure 1.10 shows two foundation arches of a Roman aqueduct in Spain

that stands and supports load without bolts, or pins (see Barton, 1977).In this

Figure 1.10 Sketch of part of a Roman aqueduct in Spain. (From Barton, 1977.)

masonry arch every block is key because the loss of any one of them wouldcause the entire structure to collapse. Another type of masonry arch is sketched

in Fig. 1.11(a). One last block, with a shape different than the rest, is restrained

by bolts. As long as it is kept in place, and a moderate force may be sufficient toachieve that, the arch will function. This model is more appropriate for excava-

tions than the Roman Voissoir arch of Fig. 1.10 because the joint blocks aroundan excavation are not perfectly similar in shape. Figure 1.11(b) shows key

Introduction Chap. I

Figure 1.11 Key blocks in: (a) an arch; (b) an underground chamber; (c) asurface cut; (d) and (e) dam foundations.

blocks around an underground opening. Loss of the shaded blocks (1) wouldpermit movement of blocks (2), then (3), and so on, destroying the chamber.Rock slopes of surface excavations show, similarly, dependence on a fewcritically located blocks, as in Fig. 1.11(c). It may not be certain, without dynamicnumerical modeling, exactly what will be the sequence of failure afterthe removalof block (1), but it is definite that survival of block (1) wilt curtail movement ofany higher-numbered block.


Figure 1.11(d) shows key blocks in a dam foundation. Plane p below the

dam would appear to be a possible sliding surface to be calculated for the

equilibrium of the dam. But the rock above p cannot move as long as block (1)

remains in place. Even afterward, the large foundation mass above plane pcould not move without lifting the dam, but it could be destroyed byretrogressiveaction, with first (1), then (2), then (3), and so on, translating or rotating indi-vidually.

All these examples try to show in two dimensions what is usually under-

standable only in terms of three. They are therefore greatly simplified. Figure

1.11(e) is a more realistic example, showing the key block of a foundation and its

relation to the design of anchors to hold the structure down. The block shown

is to be restrained from uplift of water and earthquake forces by the downward

forces of cables that are anchored below the key block. The block drawn is the

largest of its type that can fit into the space of the excavation or natural valley

in which the structure is located.Another example where the three-dimensional character of the key blocks

is essential is shown in Fig. 1.L2. Arectangular tunnel6 m wide by 3.5 m high is

intended to induce caving of the overlying rock to permit its mining. The jointdirections and spacings of the rock mass are given in Fig. 1.12(a). If the tunnel

is oriented in the direction of azimuth 105", the rock will not cave, except for

(a)

Figure 1.12 Influence of the direction of a tunnel on the extent and nature ofrock falls. The dips and dip directions of the joints in the rock are as follows: set

(1) 60'and 285'; set (2) 48" and 105'; set (3) 90" and 15o. In (a), the tunnel

azimuth is 105"; in (b) the tunnel azimuth is 15'.

21


(b)


the single shaded block above the roof. If the tunnel is turned to azimuth l5',it caves as shown in Fig. 1.12(b). The portals of tunnels are another environmentthat requires three-dimensional theory for solution. Figure 1.13 shows theintersection of a key block with the excavation created at a portal. This block

,=" -

Figute 1.13 A wedge that is daylighted when a portal is created at the toe of aslope.

-a- -

-f-

__L


cannot slide into the tunnel, nor into the free space above the valley side, but itcan slide where one meets the other.

Movements of blocks like those sketched in these examples have produced

numerous important failures, or near failures, and are responsible for routine

overbreak and rock falls accounting for great cost. The methods of block theory

can be applied to prevent these types of problems if the requisite informationabout the joint sets can be obtained. The particular applications to be discussed

in later chapters are: foundations and surface and underground excavations,

including complex three-dimensional openings at tunnel intersections, portals,

enlargements, bifurcations, and so on. The first chapters present the methods

we will use to treat these problems, including vector analysis and stereo-

graphic projection.

23

Fchapter z

Description ofBlock Goornotryand StalcilityLJsing VectorMethodsBefore proceeding with the fundamentals of block theory, some mathematicalpreliminaries will be helpful. In this chapter we develop vector equations thatpermit computational solution for the basic problems of block theory. The nextchapter develops an alternative graphical solution procedure. In practice, weoften combine both methods, using the vector equations to produce and displaygraphical solutions on the screen of a microcomputer.

The methods of vector analysis provide relatively simple formulations forall the quantities relating to block morphology, including the volume of a jointblock, the area of each of its faces, the positions of its vertexes, and the positionsand attitudes of its faces and edges. The use of vectors also permits kine-matic and static equilibrium analysis of key blocks under self-weight, appliedforces, and friction, inertia, and support reactions.

Solution of a set of vector equations with a computer is facilitated byentering them in coordinate form, according to a Cartesian basis. More compactvector formulations, from which the coordinate equation are generated, arewritten here as well.

The fundamental information required by block theory is the descriptionof each joint plane's orientation. The joints are collected in a modest numberof joint "sets," whose average orientations are described by two parameters:the dip and the dip direction. Figure 2.L explains these terms and their relationto the geological quantities known as "strike" and "dip." An inclined plane,ruled on the figure, intersects the horizontal xy plane along the "strike [ine" andplunges most steeply in the "dip direction," which is perpendicular to the strike.The dip direction is defined by horizontal angle p from y toward x. Throughout

24

Equations of Lines and Planes

Figure 2.1 Terms describing the attitude of an inclined plane: a, dip angle from

horizontal; f, dip direction (clockwise from north)'

this book, we adopt the convention that y is north and x is east, with z up. In

this manner, the angle B is the same as the compass bearing of the dip direction

as conventionally given in geological reports. The amount of dip, or simply "the

dip," is measured by vertical angle (a) between the dip direction and the trace

of the joint in a horizontal plane. If lines in the upper hemisphere are preferred,

the opposite to the dip vector is defined, rising at a in direction f + tgO'.

An alternative to expressing the orientation of a plane by its dip vector is

to state angles s and p tor its normal, as shown in Fig. 2.2 (with a measured

from vertical). Let om be the direction of dip of a plane passing through o and

dipping a with horizontal. Then the upward-directed normal to the plane is the

tiii on,with direction p from / (towards x) and inclination c from z (vertical).

Either an upper hemisphere normal or its opposite in the lower hemisphere

could Ue required in a particular analysis. UsuallY, w€ will select the upward

normal, corresponding to the choice of a > 0.

EOUATIONS OF LINES AND PLANES

The general case with a plane or line not passing through the origin will be

treated first. Later we examine the important special case with all lines and

planes containing the origin. Coordinates are represented by values of the

intercepts X, Y, and Z along the basis directions x, y, and z'

[-

and

Description of Block Geometry and Stability Using Vector Methods Chap. 2

x east

Figure 2.2 Coordinate system and direction cosines of a normal: z, normal ofjoint; z, projection of n to plane OXY; a, dip angle; f, dip direction (clockwisefrom north).

Equation of a line. Let x, be a "radius vector" from the origin topoint (X,, Y,, Z). A line in direction x, through point (Xo, Yo, Z) is defined bythe set of points along the tips of a family of radius vectors x such that

X:Xo*txr Q.T)

where xo is the radius vector from the origin to point (xo, Yo, z) (Fis. 2.3).The parameter / takes any negative or positive real value. The vector equation(2.1) can be transformed to coordinate form by replacing each radius vectorby the coordinates of its tip. Substituting

x:(X,Y,Z)xo : (Xo, Yo, Zo)

Xr : (X1, Yr, Z)

(2.2)

in vector equation (2.1) generates three parametric equations that are its coor-dinate equivalent:

X:Xo*tX,Y:Yo*tY,Z:Zo*tZr

(2.3)


Figure 2.3 Equation of a straight line.

The equation of a plane. Let fio be the unit vector directed normal

to plane P and x be a radius vector from the origin to any point in plane P.

The plane P is defined as the set of tips of radius vectors x such that

x. fio- D (2.4)

where D is constant. As shown in Fig. 2,4, D is the length of a perpendicular

from the origin to the plane.*Equation (2.4) can be converted to coordinate form by the following

substitutions:

and

to yield

x: (x, Y,z)

fio : (A, B, C)

AX+BY+CZ:DAs shown in Fig. 2.2,the values of the normal's coordinates are

I : sin usin p

B:sinacospC: COS fl

*The symbol ^ over a lowercase letter will always signify that the letter represents a

unit vector (a direction),

(2.s)

(2.6)

(2.7)

x0

["

and

Description of Block Geometry and stability using Vector Methods chap. 2

The intersection of a plane and a line. A point like c of Fig.2.4,where a line pierces a plane, can be described by solving equations Q.3) and(2.6)simultaneously. Let(Xo, Yo, Zo) be a point on the line that has the direction of aradius vector to point (xr, Yr, zr). Substituting the values for x, y, andz ftom(2.3) in the equation of the plane (2.6) and solving determines I as

t:t-- D-(AXo+BYotCZo)5

- to

- (2.8)

Figure 2.4 Equation of a plane.

and the radius vector from the origin to the point of intersection of the line andthe plane has its tip at point (X, Y, Z) gpven by

X: Xo * tox,Y:Yo*toY,Z:Zo*toZr

(2.e)

The line of intersection of two planes. The intersection of twojoint planes creates an edge of a joint block. Consider planes p, and p" (Fig.2.5)with line of intersection f ,r. Let fi, and firbe unit normals to planes p, and pr.Since the line of intersection is contained in each plane, and each plane containsonly the lines perpendicular to its normal, then f,, is perpendicular to bothfi, and frr. A line that is perpendicular to two other lines is generated by thevector ctoss product. Therefore, the line of intersection of P, and Pris parallel to

Ire : fit x fiz (2.10)*

*If the planes are not perpendicular to each other, then I12 has magnitude l(f r x fiz) l,which is different from unity. If only the direction of the line of intersection is needed, it willbe noted by Ir".


Figure 2.5 Line of intersection of two planes.

To convert this to coordinate form, let fi, : (X r, Y r, Zr) and frz : (Xz,Yr, Zr)

and let ft, j, and 2 be unit vectors parallel to the coordinate axes. Then since

fi, x fir:xxrx2

it 2 |

IYr ZrlIYz Zrl

z'l* - lx' Y'll,zrl' - lx" Yrl"

(2.11)

Irz: Yt Zrl^ lX,Y2 z'l* - l*,

In coordinate form, then,

lrr: l(YrZ, - YrZr),(XrZ, - XrZr),(XrY, - X"Yt)l (2'12)

A corner of a btock. Figure 2.6 shows a polyhedral block. The coor-

dinates of its corners (vertexes) are each simultaneous solutions for the equations

of three intersecting planes. For example, vertex A, defined by the intersection

of planes Pr, Pz, anilPr, is determined by point (X,Y,Z),which solves the set

ArX*BrY+C&- Dl

ArX*BrY+CzZ:DzArX*BrY+CiZ:Dz

(2.t3)

30 Description of Block Geometry and stability using vector Methods chap. 2

D Figwe 2.6 Corners of a block.

The description of a half-space. consider plane p of Fig. 2.7. Apoint like C2 is in its upper half-space, that is, it is above plane p; point C1 is inthe lower half-space of P. Determining whether a point is situaied above orbelow a plane is a cornerstone of block theory. Let the equation of plane p be

AX+BY+CZ:Dwhere (A, B, C) defines the coordinates of the tip of the radius vector fio perpendicular to planeP. If normal fi, is upward (i.e.,co> 0), a point x: 1x, i,z1will be said to belong to the lower half-space of p ii

or, in coordinate form,

similarly, a point x : (x, y, z) will be said to belong to the upper half-spaceofPif

fi-. x> D

AX+BY+CZ>D

fi-.x<D

AX+BY+CZ<D

Q.t4)

(2.rs)

Q.r6)

(2.17)

or, in coordinate form,

\{f\\'\Dz\ \ \\\',\\ \

31Description of a Block

Figure 2.7 Half-spaces determined by a plane'

DESCRIPTION OF A BLOGK

We are now in a position to determine all the relevant features of a block-thenumbers, locations, and areas of its faces, the locations of its corners, and its

volume.

The Volume of a Tetrahedral Block

A four-sided block can be thought of as one part of the division of aparalletepiped into six equal volumes, as shown in Fig. 2.8. Consider the paral-

lelepiped drawn in the upper half of Fig. 2.8(a), with corners 0t, Q2,031o4s o5e

or,h,and ar. First, *. * divide it into two equivolume triangular prisms by

."tting along plane a2qjasa6. Each of these, in turn, can be divided into three

equivolume tetrahedra, as shown in Fig. 2.8(b). This yields tetrahedra with

corners ara2a3q4)azasa4as,andarana.ao. Since the volume of the parallelepiped

is the area (S) of its base [e.g. , area ata2atai of Fig. 2.8(a)] multiplied by its

I92 Description of Block Geometry and stability using vector Methods chap. 2

la1 a2asaTaaa5 a6 ag)

Volume = hS

a4

a6

-A.'/ \.,tt / tr

(a1 a2a3aaasa5)

Volume = * hS

a5

(araa a, auau ar)

Votume = jtrs

(a)

Figure 2.8 Subdivision of a parallelepiped into six equal tetrahedral volumes:(a) subdivision into two triangular prisms; (b) division of each prism into threetetrahedra.

Description of a Block 33

votume = * sh

(b)


(a1 a2 a3aaa5a6l

(a,, ara3 aa)

Volume = t trS

Volume = 5

I 34 Description of Block Geometry and Stability Using Vector Methods Chap.2

height (h), it follows that each tetrahedron must have volume tSU in vectorform,

Zrorr.o"n.oo:*a.Oxc) (2.18)

where, as shown in Fig. 2.9, t, b, and c, are the three edge vectors radiating fromany vertex of the tetrahedron. Letting the four corners of the tetrahedron be

?4 lx4, ya, za)

bXc

a1 (x1, Y1, z1)

a3 {x3, y3, 23}

Flgure 2.9 Vector names for edges of a tetrahedron.

ar, ct2r 43' and a4' and taking ar as the vertex from which radiate vectors a, b,and c,

a : (X, - Xr,Yz - YrrZ, - Zr)

b: (X, - Xr,Ys * Yr,Zs - Zr) (2.1e)

and c : (Xn - Xr, Y4 - yr, Zn - Zr)

Substituting (2.I9) into (2.18), the tetrahedral volume is expressed in coordinateform by

Ix"-x, Yr-Y, Zr-2,v-*lxr-x, Yr-Y, Zr-2,

lxn-xr Yn-Y, Zn-2,(2.20)

Alternatively,

l1 Xt Yt 'rl

v : +lr x' Y2 zrl"ll xs Y3 zrll1 x4 Y4 znl

a2 (x2, y2, z2l

Q.2r)

Description of a Block

The Volume, Edges, and Gorners of a Polyhedral Blockwith tt Faces

The intersection of joint planes creates blocks with various shapes, most

of which, in general, will have more than four faces. Our procedure for calcu-

lating the volume of any such block is to subdivide it into tetrahedra and then

make use of (2.20).

Consider a three-dimensional block with n faces formed by portions of nplanes. Each plane (i) divides the whole space into an upper half-space, denoted

[f,, and a lower half-space denoted L,. The intersection of one or the other ofthese half-spaces of each plane (i: I to n) determines the dimensions and

morphology of the block. For example, a block may be created by Lr, Ur, Lr,Ln, (Jr, and Lu. In later chapters we show precisely how to choose which of the

many combinations of Z's and U's will define the critical blocks. For the

present, we assume this to be given.

1. For each plane i, i: 1 to n, determine the constants A,,8,, C,, and Dr;

the coefficients A,, B, and C, are calculated from the dip and dip direction

of plane f using (2.7).The dip angle o is always between 0 and 90o, and C,

is always positive, meaning that of the two possible directions for the

normal the upward one is selected.* The coefficient D, must be input. An

example calculation of A,, Br, Cr, and D, using sample field data will be

presented later.

2. Calculate the coordinates of all possible block corners. A corner Cqp is

calculated as the point of intersection of the three planes i, i, and k, as

described by (2.13).

We must now determine which of the corners actually belongs to the

block (i.e., which are real). The number of corners calculated in step 2 equals

the number of combinations of n objects taken 3 at a time (Cj), which equals

nr.l(n - 3)! 3!). For a parallelepiped (i.e., n - 6) there are then 20 possible

corners; only eight will be real. The procedure for making this selection, amount-

ing to simultaneous solution of n inequalities, is presented in steps 3 and 4.

3. Consider face m. Examine every possible corner C,ro in turn and retain itas a candidate-real-corner if its cordinates Xro, Y,to, and Z,,o satisfy

A^X,to * B^Yur, + C^Zt*) D* (2.22a)

if the block is defined with U^, or

A^X1o * B^Yrtr * C*Z,t* I D^

if the block is defined with Z-.

(2.22b)

*In the case of a vertical plane, the normal is determined as positive in one direction

or the other automatically according to the value input for the dip direction angle (f).

Description of Block Geometry and stability Using vector Methods chap. 2

4. Repeat step 3 for every face in turn (i.e., m : ! to n). The real cornersare those candidates that survive step 3 for every single face.

At this point, a two-dimensional example will be helpful. Figure 2.10 showsthe polygons created by five lines, p, through pr. There are

c?:r==#16r: ro

intersections of these lines, labeled Crr, Crr, and so on (the order of the indexeshas no significance). Now consider one specific block i ur, Lz, Ls, [Jn, (Jr.(In this example, Umeans the half-plane above a line andLmeans the half-planebelow a line.) Considering, first, half-plane (J1,the real vertexes (Cri) must allsatisfy

ArX,t * BtY,t) D,This step eliminates Gc. Considering half-plane L, next eliminates Crr, and soon. The real corners of block (J,,,Lz,Lr,(Jo,andU, are thus identified as Ca1,cnr, crr, crr, and crr.

x

Figure 2.10 Real corners of a given polygon.

ul L2 L3 U4 U5

Description of a Block 37

We have now succeeded in identifying the coordinates of all the corners

C,,o of apolyhedral block. Next, we will find all the real faces of the polyhedron.

Considering again the two-dimensional example of Fig. 2.10, the intersections

of the five lines produced polygons of three, four, and five faces. In three

dimensions, blocks created by intersections of n planes may have from fourto n faces.

5. Determine which faces belong to the given block. A real face is defined by

any subset of three or more real vertexes (block corners) that have a com-

mon index. For example, face m (of plane m) is the triangular region

between corners Cr-r, Cro*, and C-r5.

6. Determine all edges of the block. A real edge is a line between a pair ofreal vertexes Clyp that have two common indexes. For example, one edge

is the line connecting corners C,,, and D ut. This line is parallel to the line

of intersection (I'.') of planes i and i.

The next steps will divide the polyhedron first into polygonal pyramids,

and then into tetrahedra by subdivision of the polygonal bases into triangles.

7. Choose one real corner C11p as an apex (the summit of a polygonal

pyramid). The choice of corner is arbitrary and only one corner is to be

selected. C,,oisthe point of intersection of face planes i,i,andft. Excluding

these three, subdivide each of the other (n - 3) faces of the block intotriangles as follows. Each face, m, is in general a polygon with t corners.

The corners of face m are the subset of the t real corners of the polyhedron

that have m as one of its indexes. Now subdivide face m into triangles by

selecting one corner and connecting it in turn with the end points of each

edge of face m. (The edges of polygon m are the subset of all edges, found

in step 6, that have m as one of the common indexes.)

Figure 2.11 illustrates the procedure described above. By choosing corner

al as the vertex of all triangles, polygon (arararanssae) is divided into triangles

(ararar), (ararao), (aranar), and (ararau).

8. Finally, connect the corners of every triangle for all (n - 3) faces (excluding

faces i, j, and k) with the apex, cotnet C,,1,. This creates the set oftetrahedra, the sum of volumes of which is the volume of the polyhedron.

Figure 2.12 gives an example of the procedures described in steps 7 and 8.

We will subdivide, into tetrahedra, the five-sided block shown in Fig. 2.t2(a).

First, we arbitrarily select corner C1r, as the apex. This excludes faces P1,P3,

and P, from decomposition in triangles. Faces P" and Pn remain. The former is

shown in Fig. 2.12(b). It divides into two triangles : I (Cs35, Crro, Cr"n) and

II(Cnr, Crrr, Crrn).Face Pn is already a triangle-(Ct ,n, Cr2a, Czto). The five-

38 Description of Block Geometry and stability using Vector Methods chap. 2

Areaa'a6auai=1bXcl

Figure 2.11 Subdivision of a polygon into triangles,

sided block is split into three tetrahedra by connecting these three triangles withapex Crrr.These tetrahedral volumes are shown in Fig. 2.IZ(c), (d), and (e).

The Faces of a Polyhedral Block

The corners, edges, and areas of each face of a general n-faced block canbe computed using the procedures described in the preceding section. A poly-gonal face is defined as the planar region between all corners (C,rr) that shareany one index. Each polygon is then divided into triangles by the procedure ofFig. 2.11. Consider a triangle with corners (ar, az,a3) and sides L: atazand

a1 a2 a3 a4 a5 a6

a1 a2 a3

a1 a3 a4

a1 A1 85

a1 O5 E6

Face 2

ilage lz

Cr ss

Figure 2.12 Subdivision of polyhedron into tetrahedra.

b : erar. The area of the triangle is

U: +la x bl (2.23)

In vector form with a, : (Xr, Y r, Z r), a2 : (Xr, Yr, Zr), and os : (Xt, Yt, Zt):

a : (X, - Xt, Y, - Yr, Z, - Zr)

b : (X, - Xr,Y, - Yr,Zs - Zt)

nq\ts

cnq

(b)

Czss

and

giving(lv

A: lrll" -- IIY' -Yl 22 - trl' *Yt Z, - Zrl

z, - z, x, - xrl'zr-2, xr-xrl

,lxr-x| Y2--lxr-xL Y3- ';,''']["'

/r/rr

Edge 24

(2.24)

40 Desctiption of Block Geometry and Stability Using Vector Methods Chap. 2

(The formula can be simplified by choosing new coordinate axes x', !', and z'uirth z' perpendicular to the plane of the face.)

AilGLES IN SPACE

The angle between lines, between planes, or between a line and a plane will berequired routinely in computing the sliding resistance of blocks.

The angle between lines. Consider two intersecting vectors n, andn, in space.

Ilr : (Xb Yb Z)and [2 : (Xz, yz, Zz)

The angle (a) between nl and n, is given by

Q.2s)cos d - -=!=;+lnrllnzl

\ /n'\1\6 /\+L/\/

(b)

Flgure 2.13 Angles between lines and planes: (a) the orthographic projectionof a line on a plane; (b) the angle between two planes.

The Block Pyramid

or' in coordinattj.

" : xg2 + Y\Yz +JA. :-cosd:ffiffi

If n, : fi, and a2 : fi', (unit vectors),

COS d : fir.fi.2: XrX, * ItrY" -f ZrZ,

X:/X1

X: tXr

Y: tYr

Z: tZt

The angle between a line and a plane. The angle between a line

and a plane is defined in terms of the angle between the line and its orthographic

projection in the plane [Fig. 2.13(a)]. Let n, be a vector inclined with respect to

plane P2, whose normal is nr. The angle y between n, and its line of projection

in P" is the complement of the angle d between n, and nr.

T:90-6 (2.28)

The angle between two planes. As shown in Fig. 2.13(b)'the angle

d between two planes P, and P2 is the angle between their normals n, and nr.

d:{(n,,nr) (2.2e)

THE BTOCK PYRAMID

Consider a real block formed with one each of n different faces, that is, a block

bounded by z nonparallel surfaces. Recall that a particular block is created by

the interesection of upper or lower half-spaces corresponding to each of itsfaces. For example, a block might be given by UrUr(JrLoLr. Now let each half-

space be shifted so that its surface passes through the origin. The set of shifted

half-spaces UIUEU\L\LZ will create a pyramid-the "block pyramid"-withapex at the origin, as shown in Fig. 2.I4. The superscript (0) signifies that the

plane in question has been shifted to pass through (0,0, 0).

The importance of this construction will become apparent in Chapter 4.

But it is appropriate at this point to describe the block pyramid using the

formulas established earlier in this chapter.

Lines through the origin. All the edges of the block pyramid are

lines passing through the origin [i.e., xo : (Xo,Yo,Zr): (0,0,0)]. Therefore,

the equations of the edges, obtained from (2.1) and (2.3), are

(2.26)

(2.27)

(2.30)

Q.31)

Faces through the origin. Any plane (i) of the block pyramid willinclude the origin. Hence Dt :0, since D, is the perpendicular distance from

the origin to the plane. The equations of a plane, (2.4) and (2.6), simplify to

Description of Block Geometry and Stability Using Vector Methods Ghap. 2

(o, o, o)

Figure 2.14 Block pyramid: U?UgUgLlLg.

and

X.trf:0A,X t B,Y * C1Z: O

trr'X)0A,X*B,Y*C&>0

(2.32)

(2.33)

Q.36)(2.37)

where tt : (An 8,, C) is normal to joint plane f.

Description of Ll and Ul. When plane (i) cuts through the origin,its lower half-space (^Ll) is determined by (2.14) as

n,.x<0 (2.34)

or A,X*B[+C&<o (2.35)

Similarly, the upper half-space (U,0) of plane f is determined by (2.16) as

Edges of the block pyramid. The normal to plane P, (i.: 1 to n) is

at : (Au Br, Ct) (2.38)

I-et Fo'F| ... F9 be the block pyramid corresponding to a particular set ofupper and lower half-spaces (U,0 orz,0). Every pair of indexes of F0 (i and j * i)defines a potential edge vector Irr

/\II P. /I^

| '05 ,,,ul

\/ es4.//

ItJ: n, X nJ Q.3e)

ltr: ry X nr - -lttWith n block pyramid faces (n planes), the total number of possible edges equals

2C3:n2-n.In fact, a specific block pyramid has fewer edges, determined as follows,

To be the edge of block pyramid F?, Fg, . . . , Fl,frr : (Xr1, Y,1, Zrt)

must satisfy, for each pyramid face (m) (*: 1to n),

A*X,t * B^Y,t + C-Ztt>0 when FI: Ul"

A^X,t * B^Yu + C-Z.J <0 when F?" : L2'

The intersection vectors f,, that satisfy all n simultaneous equations of (2.42) arcreal edges of the block pyramid. There are no more than n such solutions. Thesequence of these edges around the pyramid are determined by the numericalsequence of indexes (r,i) since each pyramid face lies between two edges thatshare a common index. For example, in Fig.2.l4 the edges in order areIrr,Irr,Iur, Irn, and I23.

EOUATIONS FOR FORCES

Vector analysis facilitates the analysis of block stability under self-weight, waterpressure, support forces, inertia forces, friction, and cohesion.

Equations for Forces

According to the rule for the cross product,

lF l : (X' + Y2 + 22)t/2

and the direction of F is given by

43

(2.40)

(2.4t)

Q.a2a)(2.42b)

Representation of amagnitude and direction of a

coordinate values,

The magnitude of F is

force by a vector. We represent both theforce F by the symbol F. Its components are its

F: (X, y,Z) (2.43)

(2.M)

(2.4s)+ Yz * Zzyrz

The resultant of two or more forces. A series of intersectingforces F,., Fz, . . . , Fn can be replaced by a resultant R:

R:iF,:(,.i*,fy,,Et,)Figure 2.I5(a) shows a graphical solution for two-dimensional summations.(It is always possible to use the two-dimensional solution for two intersectingforces by transforming the basis directions to coincide with the plane commonto the directions of the two forces. We make use of this in the next chapter.)

z

M Description of Block Geometry and stability using vector Methods chap. 2

Figure 2.15 Forces as vectors: (a) theresultant of forces; ft) equilibrium underforces.

The equilibrium of forces. If a system of n forces F,, Fr, . . . , F,are in equilibrium, their resultant R has zero magnitude [see Fig. 2.15(b).Therefore,

n\-F:0L/ \'t -j=1

n

E (X,, Y,, Z,) : 0i=l

Friction forces. Friction provides a resisting force that opposes thedirection of motion or of incipient motion. We will call the latter the "slidingdirection" (^f). Then the direction of all friction forces is -.i. Let B signify apotentially sliding rock block and suppose that {,., i : l, . . . , fl, are the magni-tudes of the normal reaction forces from each sliding face P,, i : !, . . . , fl,of .8. Then the resultant friction force is

Q.46)

(2.47)

(2.48)Rr : -E (t, tan f,)^i

where f, is the friction angle for sliding in direction ^f on face i.

Gravity and other body forces. Gravity acts remotely and its forceis proportional to the mass. Its direction is vertically downward (-2).Inertiaforces act in a direction opposite to an applied acceleration and are also pro-portional to mass. If the weight of a block is W, the inertia force of the blockthat is accelerated by (kg)d is

F.r : -(kW)A Q.4e)

Computation of the Sliding Direction

where g is the magnitude of the acceleration of gravity. If the direction ofacceleration (r?) is uncertain, the summation of F, with other known forcesproduces a circular cone in space that contains all possible resultants.

Water forces and cohesive forces. The integration of waterpressure (Ft-tl acting over the face of a block produces a water force in thedirection of the inward normal to the block. Cohesion (Ft-'lproduces additionalresistance to motion. If cohesion is constant over a face, the total force iscomputed with the known area of the face. The procedure for calculating thearea of any face of a polyhedral block was given earlier in the chapter.

Water pressures in rock produced by hydraulic structures tend to varywith time. Suppose that P, are the faces of a polyhedral block, each with areaA, and inward unit normal. fi,r. Then the resultant r,, of all water forces is

n

I* : E S,ft,j=l

(2.50)

where ,S, is the integral of water pressure acting over face i. In many cases it issufficiently accurate to substitute

S; : P;Ai

where P, is the water pressure acting at the centroid of face i.

GOMPUTATION OF THE STIDING DIRECTION

(2.51)

The direction of incipient motion of a block is determined by the mode of failure.Ltft@ or fallout occurs when a block loses initial rockirock contact on all facesto advance toward free space. Sliding may occur on any face individually or ontwo nonparallel faces simultaneously along their line of intersection.

Lifting. The action of water pressures, structural pull, or inertia forcemay pop up a block as shown in Fig. 2.16.If the block exists in the roof, it mayfall out under gravity alone. In both cases, the direction of initial block motioncoincides with the direction (l) of the resultant force (R) acting on the block

.i:f Q.s2)

Figure 2.16 Lifting.

Description of Block Geometry and Stability Using Vector Methods Chap. 2

Single-face sliding. A block may tend to slide along a single one ofits faces, as shown in Fig. 2.I7.In this case the initial sliding direction is parallelto the direction of the orthographic projection of the resultant force (r) on thesliding plane P,. Denote the normal to the sliding plane by fi,. The projectionof r is along the line of intersection of plane i and a plane common to r and fi,.

Figure 2.17 Single-face sliding.

Therefore, the sliding direction ^i is

slll (fi,xr)xt, (2.s3)

where the symbol lll means "is in the same direction as." The double use of thecross product is justified in Fig. 2.t8.

Ftgure 2.18 Sliding direction under single-face sliding.

Sliding in two planes simultaneously. If a block slides on twononparallel faces simultaneously (Fig. 2.I9), the direction of sliding is parallelto their line of intersection. Let n, and n, be vectors normal to each of the

\' ', \['l \\l

hXn,//l s

h=fiixR

Computation of the Sliding Direction

Figure 2.19 Sliding on two planes.

sliding planes P, and Pr. The direction of sliding (s) is either the same as (nl

x nz) or its opposite (-n, X nz). The actual sliding direction is the one thatmakes the smaller angle (i.e., less than 90") with r (Fie. 2,20). Let sien (/)signify *1 if/is positive, - I iflis negative, and 0 if/is zero. Then the direction(s) of sliding along the intersection of planes i and 7 is

s lll sign [(n, x nr) . r](n, x or)

n2

(2.s4)

Figure 2.20 Sliding direction for dou-ble-face sliding.

The sliding direction under gravity alone. The analyses of blockstability under the action of self-weight only will be examined as a special case.

Without other forces, the resultant force on the block is

s /l/ nt Xn"

48 Description of Block Geometry and Stability Using Vector Methods Chap. 2

r : (0, 0, -W) (2.55)

where W is the weight of the block (W > O).

For a block fall, the sliding direction must then be

s lll (0, 0, -w) (2.s6)

For sliding on face P, alone, we must substitute r from (2.55) together with

fi, : (Ar, Bt, Cr)in (2.53).

ft,

At

0

fi1 xt:

and (ft,xi)xfi,:

, 2lBt C, I W(-8,, A,,0)0 -wl

* , 2l

-B,W A,W 0 |At Br C,l

s lll w(Arc,, B,cb -U? + B?)) (2.s7)

For sliding simultaneously on planes and j, we substitute in Q.5$:

fi,1 xfi1 :*t2At B, Ct

AJ BI CJ

glvlng

S lll [sign (Afi, - A,B)j(B'C. - BtC,,AtC' - AlCy A,Bt - Afi,) (2.58)

EXAMPTE GATCULATIONS

ln the following examples, x is east, y is north, and z is up.

Example 2.1. The Equation of a Plane

Consider plane P with dip angle d : 30" and dip direction F :320'clockwise from north. The plane passes through point (I,2,1). The equation ofthe plane is

AX+BY+CZ:DFrom (2.7), A: -0.32139, ^B

: 0.38302,and C: 0.86607. Since (1,2, L) is on

this plane, it satisfies its equation, so I + 28 * C : D, giving D : 1.3106.

Then the equation of plane P is

-0.32139x + 0.38302Y + 0.866072 : 1.3106

Example Calculations 49

Example 2.2. The Intersection of a Plane and a Line

Consider plane P, whose equation is

2x+3Y+z:4and a straight line passing through (1, 1, -2) in direction (2, -2,3). Equation(2.3) gives

X:l*2tY:L-2tZ:-2+3t

The piercing point of the line on the plane is found by making the followingentries in (2.9); A:2,8:3,C: l, D:4,X0: l,Yo:L,Zo: -2,Xt: 2, Yr - -2, and Zt: 3, givinEto : 1. Then equations (2.10) determine the

radius vector (X, Y, Z), from the origin to the point of intersection, by

X:L!2:3Y:t-2:-I

and Z: -2 + 3 : 1

Example 2.3. The Intersection Vector of Two Planes

Assume thatP, has ilt:20" and Ft:280oP, has &z: 60" and Fz: 150"

From (2.7) the unit normal vectors ft, and fi, ate

fir : (-0.33682,0,059391, 0.93969)

and ftz : (0.43301, -0.75000, 0.50000)

vector ltz: ftrx h, is parallel to the line of intersection of Pr and Pr. From(2.12),

I r z : (0.7 6494, 0.59918, 0.23631)

Example 2.4. A Tetrahedron Greated by Planes Pr, Pz, P3, and Pa

Assume that a tetrahedron is the region common to the intersections of 21,

L2, Ls, and (J 4, where Lr, L", and L, are the half-spaces below planes 1,2, and 3

and Un is the half-space above plane 4. These planes are defined by values of dand f as follows:

Plane o(deg) f(deg)

145902 4s 330

3 45 2lo4090


The half-spaces are then

Lt. 0.70710X + 0r + OJ07t02 < 0.70710 (a)

L,zi -0.35355X + 0.61237Y + 0.707102 < 0.70710 (b)

Lsi -0.353ssx - 0.61237Y + 0.707102 < 0.70710 (c)

U+i 0x+ 0r+ Z>O (d)

Now we must compute the coordinates of every corner C,1r of the block.C1r, is the intersection point of planes I,2, and3. It is found by the simultaneoussolution of

0.70710x + 0r + oJo7Ioz: 0.70710

_0.35355X + 0.6L237y + 0.707t02: 0.70710

and -0.35355X - 0.61237Y + 0.707102 :0.70710

The solution is Ctzs : (0, 0, 1).Similarly, Clrn is the intersection point of planes 1,2, and 4. Simultaneous

solution of the corresponding three equations gives

Ctzt : (1, L.7320,0)

Using the method for the other corners gives

Ctb* : (1, -1.7320, 0)

and Czst : (-2, 0, 0)

In this case, with only four planes, the block has exactly four corners, so aselection test using (a), (b), (c), and (d) is not necessary.

The volume of the block can now be computed by entering the coordinateof Ct zzt Crz+t Crrn, and Czz+ in Q.Zl), giving

lt 0 0 1ltl.ll I r.7320 0lV:*l t:1.7320"ll I -r.7320 0ltlll -2 0 0l

Example 2.5. The Area of Every Face

Consider further the tetrahedron of Example 2.4. The area A of the facein plane Pt is the area of the triangle with corners Crro all sharing the commonindex 1. These ate Cp3,Cr"o, and C, ,n, Let r: CtzzCrrn andb: CpsClsa[see Fig. 2.21(a)]. According to the discussion following Q.23),

x: Crzr - Crrr: (1, t.732, -I)b : Cr s4 - Crzt : (1, - 1.732, -L)

51

Face 1

Figure 2.21 Area of a triangular face.

By (2.23),

Ar: Ila x bl

: +ll r'7320 -t l' *[| -r.zrzo -11

:2.4494

Similarly, the area A, of the face in plane 2 is the triangular area between

Cr"r, Crrn, and Crrn.

Az : + lC, ,rCrrn X C, ,rCrr+l : 2.4494

[see Fig. 2.21(b)].The area I, of the face in plane 3 is the trangular area between Cr Lgt Ctgqt

and C2.s a.

As: tlC,rrCrrn X C,rrCrr+l:2.4494

and finally, A+ is the triangular area between C1 ,n,C134, and Crrn.

A+: *lCrrnCtrn X C,rnCrr+l:5.1961

Example 2.6. The Angle between Two Vectors

Given two vectors fi,t and fi2,

fir : (Xt, Yr, Zt) : (9,8,7)

ftz : (Xr, Yr, Zr) : (1,2,l)

Crzs a Crzc

2

(a)

c'"o

1

{b}

L 1.7320l'zltrz

1 -1.73201 )

-1-1

1r ,

1l -r


the angle (d) between fr,t and frz is calculated from (2.26): cos 6 :0.93793,giving 6 : 20.292".

Example 2.7. The Angle between Two Planes

Given Pr with dr : 30", f , :320", and PL with d2 : 50o, Fr:160o.First compute the normal vectors fi, and fi, of P, and Pr, respectively.

fir : e032139, 0.38302, 0.86602)

fiz : (0.26200, -0.7t984,0.&278)Then use (2.26), giving cos d : O.9675 and d : 78.653.

Example 2.8. The Angle between a Plane P anda Vector v

Given planeP with d : 30o, F : 320", vector v : (1, 2, l). First use (2.7)to compute the normal fi of plane P,

i : (-0.32139, 0.38302, 0.86602)

The angle between fi and y is calculated using (2.26): cos d : 0.84480 and d: 32.349". This is the complement of the desired angle, which is between v andplane P.It is then 57.651".

Example 2.9. Find the Block Pyramid with Four Planes

Given four planes:

Dip Angle(dee)

Dip Direction Angle(dee)

PtPzPsPq,

dt:30dz :4Odr :50d+:10

Fr: 90

Fz :320fs : l9Of+: 80

We will compute the block pyramid created by the intersection of half-spacesu r, Lr, Lr, and u n.

Using (2.7), we compute the upward unit vectors of these planes:

fir : (0.5, 0, 0.86602)

fiz : (-0.4131 7, 0.49240, 0.7 6604)

fit : (-0.13302, -0.75440,0.64278)ft+ : (0.17101, 0.030153, 0.98480)

The equations of planes Pr, Pr, Pr, and Pn in the block pyramid are

Pri 0.5X + 0f + 0.866O22 : 0

Pz: -0.41317 X + 0.49240Y + 0.766042 : 0

Psi -0.I3302X - 0.754r';0Y + 0.642182 : O

P+: 0.17101X + 0.030153Y + 0.984802 : 0

Example Calculations 53

The terms right of the equal signs are zero because the planes Pr, P2, Pr, and Pn

pass through the origin (0, 0, 0). The equations of half-spaces Ur , Lr, Lr, and U n

areU rt 0.5X + 0r+ 0.866022>0 (a)

(b)

(c)

(d)

Lz: -O.413I7 X + 0A924frY + 0.766042 < O

Lti -0.13302X - 0.75440Y + 0.642182 <0(J n: 0.17101X + 0.030153 r + 0.984802 > 0

Compute all of the intersection vectors:

ItJ : fi4 x fi1, irj : 1,2,3,4, i+iIrz : e0.42643, -0.74084,0.24620)Irr : (0.65333, -0.43659, -0.37720)Ir c : e0.126113, -0.34430, 0.015076)

Izs : (0.89442,0.16368, 0.37720)

lz+ : (0.46182, 0.53790, -0.096664)Ir + : e0.7 6232, 0.24092, 0. 12500)

Finally, we test all of the vectors -flrz, *Irs, *Ir+, tlrr, t[z+, and flrnone by one, substituting the coordinates of fI,, : (X, Y, Zj in equations (a),

(b), (c), and (d) simultaneously. The intersection vectors from :[Ir/ that satisfy

simultaneously all the inequalities (a), (b), (c), and (d) are Irr,lrn, and -Irn.So the joint pyramid cut by (Jr, Lr, Lr, and (/n has only three edges.

Izr : (0.89442,0.16368, 0.37720)

Izt : (0.46182, 0.53790, -0.096664)

-frn : (0.76232, -0.24fr92, -0.12500)These three edge vectors completely define the corresponding joint pyramid.

Example 2.1O. Determine That a Block Pyramid ls "Empty"

In the next chapter we will learn that an empty block pyramid (i.e., one

without any edges in space) means that the intersection of these half-spaces willalways create a finite block. The method of this example will be used frequentlyto judge the finiteness of rock blocks.

Given four planes Pr, P", Pr, and Pn, which have the same dip angles anddip direction as in Example 2.9, we compute the block pyramid created byLr, Lr, Lr, and Un. Referring to Example 2.9, the equations of half-spaces Lr,Lr, Lr, and U n are

0.s0000x+ 0.00000r+0.866022<0 (a)

-0.413r7X+ 0.49240y +0.766042<0 (b)

-0.13302X - 0.75440Y + 0.642782 <0 (c)

0.17101X+0.030153r+0.984802>0 (d)

Lr:Lz:Ls:

Ui


Compute all of the intersection vectors:

lrl:fi,1 xfiy i,j:Ir2,3,4, i+j(this is exactly the same as in Example 2.9). Finally, we test all of the vectors*lrr, *Irr, -flr+, *lzt, *lzo, and +I34 one by one, substituting the coor-dinates of f Ir7 : (X, Y, Z) into equations (a), (b), (c), and (d).

None of the vectors fI,, satisfies simultaneously equations (a), (b), (c),and (d). Therefore, the blockpyramid cut by Lr,Lr,Lr,and Un has no edge;such a pyramid is said to be empty.

Example 2.11. Gompute the Resultant of Forces

Suppose that there are three forces acting on a rock block, arising

from weightt w : (0,0, -5)from water: p : (4, 1,0)

and from inertia i € : (2,2, l)Then the resultant r of forces w, p, and e is

J:lr*p*egiving

r : (6, 3, -4)Example 2.12. Gompute the Sliding Direction for Uplift

Here we assume it is known that the resultant force tends to cause uplift.Suppose that the resultant is

r : (0, 3,4)

Then the sliding direction is

Ji , ' r:f:Til

from formula (z.sz). : (o' 0'6' o'8)

Example 2.13. Gompute the Sliding Direction for the Gaseof Sliding on a Single Face

Suppose that the resultant is

t : (I,2, l)rhe sriding nt*il

i;tJl - 50o, dip directi on p : ze,^o

The unit vector of plane P is

fi : (-0 .7 tgg4, 0.26200, 0.64279)

Example Calculations

From equation (2.53) we know that the sliding direction unit vector J is

,flll (,axi)xfifr x i: (-0.0235,1.3626, -1.7017)@ x f) x fi: (1.3217,1.8829,0.71'270)

.f is the unit vector of (fi x l) x fi.; then

i : (0.54880,0.78181, 0.29593)

Example 2.14. Gompute the Sliding Direction in a Case

Where Sliding ls Simultaneously on Two Faces

Suppose that the sliding faces are planes P, and Pr. The orientations ofthese faces are given by

Pra a"r : 20",

Pz: &z: 60",

The resultant force is given as

r: (0, _1,1)

The unit vectors of planes Pt and P, arc

fir : (-0.33682,0.059391, 0.93969)

fiz : (0.43301, -0.75000, 0.50000)

From equation (2.54) we know that the sliding direction unit vector ^f is

s lll sign [(fi' x ft,r)-rl(fi' x ftr)i is computed by the following steps:

fit x fiz: (0.73446,0.57531, 0.22690)

(fi, x fir) . r: -0.34841sign [(fit x fir). rl : -1

sign [(fi, x fir). r](ft, x fir): (-0.73446, -0.57531, -0.22690)i is the unit vector of sign (fit x fi"). rl(fi, x fir);then

^i : (-0.76494, -0.59918, -0.23631)

ft :280"

fz: 150"

chapter 3

GraphicalMethods:StereographicProjection

The use of vector analysis to describe and examine blocks permits speedysolution of real problems by means of computers. An alternative solutionmethod incorporating the stereographic projection is presented in this chapter.The techniques discussed are complete in themselves and can serve as a fullsubstitute for the previous theory. On the other hand, they can be adopted inpart as a complement to vector analysis to provide a semigraphical solutiontechnique placing the entire computational process inside a glass chamber forobservation. The use of graphics to examine the geometric relations in projec-tion at any stage of the computation offers a clearer perception of the geometricand physical relations being manipulated inside a computer.

Stereographic projection as a graphical device for solving geological,crystallographic, and other spatial problems has been discussed in a number ofbooks and articles, some of which are listed with the references. However, whilesome of the essential material already available in this literature will be repeatedhere, most of the content of this chapter is novel. The thread that connects thesections of this chapter is that stereographic projection procedures can berefined and enhanced by supporting them with initial calculations. The use ofa stereonet is therefore not necessary, although it may be preferred by customor previous bias. Our approach is to make use of the power of a modest cal-culator-the common, faithful companion of almost anybody who reads thisbook-to establish the points and lines to be plotted. The tools required forthe creation of a plot, besides the calculator, are paper, pencil, compass, pro-tractor, and scale. Alternatively, these plots can be drawn with computer gra-phics on a video screen or plotter. Before plunging into this procedure, it will

56

Types of ProjectionsTypes of Projections 57

be instructive to place sterographic projection into context with other graphicaltechniques, including orthographic and oblique projections.

TYPES OF PROJECTIONS

Projections can be organized into two groups, parallel and perspective. The well-known orthographic projection belongs to the first type, in which the construc-

tion lines transferring the points of the subject to the surface of projection are

all parallel rays. The second class of projective techniques, ofwhichstereographicprojection is an example, gathers the construction lines together at one or more

foci behind the surface of projection.

Projection of Distances

Orthographic. An example of orthographic projection of a three-

dimensional object is given in Fig. 3.1. This technique is probably the most

usual basis for the drawings prepared by engineers. The lines and planes definingan object are transferred to the drawing by means of rays drawn perpendicular

to the projection plane. These lines and planes are fixed uniquely by multiple,nonparallel views. In Fig. 3.1, a polyhedral block is determined in size, shape,

and position by three orthogonal projection planes (views).

SURFACE "A" SURFACE "8"

7 8,9 8,7 IFHONT RIGHT SIDE

POINT ANALYSIS

SURFACE "C" SURFACE "TY'

Figure 3.1 Orthographic projection of a three-dimensional object using mul-tiple views. (Warren J. Luzadder, FUNDAMENTALS OF ENGINEERINGDRAWING: For Technical Students and Professional Draftsmen, 3rd Ed., @1952, renewed 1980. Reprinted by permission Prentice-Hall, Inc., EnglewoodCliffs, N.J.)

1,9

TOP

Graphical Methods: Stereographic Projection Chap. 3

Oblique. Figure 3.2 shows a related parallel projection technique inwhich the construction rays are oblique to the plane of projection. This methodproves useful in conjunction with orthographic projection to provide inclinedviews of objects.

Figure 3.2 Orthographic and oblique projections. (From Luzadder, 1952, withpermission.)

Perspective. Nonparallel construction rays are used to create three-dimensional views hpving the appearance of true shape. Figure 3.3 shows sucha projection of a complex polyhedral block. Perspective projection of this typeis quite useful to convey shapes of objects but is not appropriate when it isintended to convey measurements of distances or angles.

Otthographic Projection of Angular Relations

The orthographic projection can be used to show the relations betweenlines and planes in space and to measure the angles between them. Imagine alldirections (unit vectors) radiating from a central point. This bundle of unit

PENCILS SHOW THE DIRECTIONOF PROJECTORS

;,hfiQe

Types of Projections

Figure 3.3 Perspective projection. (From Luzadder, 1952, with permission')

vectors produces a sphere and the tip of any one of them (e.g., I in Fig. 3.4) is

a specific point on the surface of the sphere. Orthographic projection produces

u piun of a diametral plane of the sphere by means of construction rays directed

perpendicularly to the diametral plane. Thus unit vector OA is projected to

point Ao. A series of planes inclined with a common line of intersection through

the center of the sphere creates a family of great circles on the reference sphere.

These project to the diametral projection plane as a family of elliptical curves,

as shown in Fig. 3.4(b). They are analogous to the lines of longitude of the

globe. Small circles of the reference sphere that are generated from a family

of cones about the axis of the great circles will project to the diametral plane as

straight lines (lines of latitude) as shown in Fig. 3.4(b).

Orthographic projection of the sphere is used commonly for cartography.

It has the disadvantage, however, that equal solid angles can produce greatly

unequal areas on the projection. Furthermore, because of the crowding of the

lines of longitude near the edges of the projection, measurements of interplane

angles can be inaccurate. An even stronger reason to reject orthographic projec-

tion of the sphere for the development of block-theory graphics is its failure to

59

PIERCING POINTS

PICTURE PLANE

GROUND LINE

60 Graphical Methods: Stereographic projection Chap. 3

ffia \ \\\\\v//// / | \ \ \\\\// I I \ \ \\\\

4 '// / I \ \\\ \r/t I I I

tt I

tl \

\

t\I

IIN\

\t I //k\ \\\ \ / lt

\\\\ \ \ I I / ///Z\\\\ \ \ | / / ,/.//z

(bl

Figure 3.4 Orthographic projection of a reference sphere: (a) basis for theprojection; (b) a projection net of lines of longitude and latitude.

distinguish symmetrical points in the upper and lower hemispheres. If line OAis inclined o with either the upward vertical or the downward vertical, distanceOAo from the center of the projection to its representation, point ,,{0, will havethe same value, R sin d, where R is the radius of the reference sphere. Aperspective projection technique will be required to differentiate symmetricalpoints in the upper and lower hemispheres.

Equal-Area Projection

Figure 3.5(a) shows a homogeneous projection of the sphere called equal-area ptojection. Line OA, piercing the reference sphere at point l, is projectedfirst to a tangent plane at the top of the sphere along a circular arc whose centeris at O' at the top of the sphere. This projection on the tangent plane of thesphere is then produced on a parallel equatorial plane by construction rays toa perspective point (F) at the bottom of the sphere; the representation of lineOA on the equatorial plane of projection is then point Ao. If line OA is inclinedat a with the vertical, distance OAo will be equal to RJT sinul}.

The advantage of equal-area projection is its homogeneity, meaning that asolid angle produces a unique projection area (but not a unique shape); thatis, the area of its projection is the same anywhere on the sphere. This propertyfacilitates statistical operations with lines. However, great circles and smallcircles of the sphere [Fig. 3.5(b)] project as nonconic loci that are difficult todraw. In contrast, the stereographic projection produces loci that are the ulti-mate in simplicity.

Rsina As

OAo=Rsina

(a)


o'A=AB=znsinfr

OrB = OtA, = ZrrD.A ti"t

oAo = .'fr.a"i"$

(a) (b)

Figure 3.5 Equal-area projection of a reference sphere: (a) basis for the projec-

tion; (b) a projection net of lines of longitude and latitude.

StereograPhic Proiection

Figure 3.6(a) presents the basis for stereographic projection, which we

will examine closely in this chapter. Consider line OA inclined at angle a with

the vertical. The piercing point A of this line on the sphere is projected to a

horizontal equatorial plane along perspective line AF, with the perspective

point (r) located at the bottom of the sphere. The point ,,{o representing the-.t.r.ogruphic

projection of line oA is located at distance R tan ulL ftom the

center of the projection Plane.Figure j.O(U) shows the families of great circles and small circles corre-

sponding to lines of longitude and latitude, as discussed in the previous figures.

All these lines are portions of true circles in the projection plane. The proposi-

tion that any circte on the reference sphere projects to a circle in the projection

plane is proved in the appendix to this chapter. This important property makes

,trr.og.uphic projection far easier to draw and formulate than either ortho-

graphic or equal-area projections.Note from Fig. 3.6(b) that a given solid angle will project with different

areas in different regions of the sphere; that is, the projection is not homogene-

ous. In this ,.rp..t, its quality is intermediate between orthographic and

equal-area projections.Compared with orthographic projection, both stereographic and equal-

area projection methods will produce a unique point corresponding to every

61

(ttN

(a)

(b)

Figure 3.6 Stereographic projection of a reference sphere: (a) basis for the projection;(b) a projection net of Iines of longitude and latitude.


unique direction radiating from the center of the sphere. Thus there is no con-

fusion between symmetrical points in the upper and lower hemispheres.

Figure 3.7 contrasts a stereographic projection embracing a focal point

at the top of the reference sphere with a stereographic projection having the

focus at the bottom of the sphere. Regardless of which focal point is chosen,

the equatorial plane projects as a circle, which is denoted the reference circle.

All points on this circle represent horizontal lines. If the focal point is at the

bottom of the reference sphere, as shown in Fig. 3.7(a), a line Iike OA that

pierces the sphere in its upper half will project to a point that is inside the

ieference circle [Fig. 3.7(b)]. If the top of the sphere is taken as the focal point,

a point like OB of fig. 3.7(c) that pierces the sphere in its lower half will lie

wittrin the reference circle, as shown in Fig. 3.7(d). Because of these properties

of the stereographic projection, the upper and lower foci are said to generate,

o /oo

Referencecircle

(a)

(c)

Figure 3.2 Upper- and lower-focal-point stereographic projections of a line

und uplane: (a) section of thereference sphere; (b) projection plane; (c) section

of the reference sphere; (d) projection plane. (a) and (b) give lower-focal-point

projections of a line; (c) and (d) give upper-focal-point projections of a line'

f tlontt

t trontr

o

64 Graphical Methods: Stereographic Projection Chap. 3

respectively, lower and upper hemisphere projections. This can be misleading,however, as either choice of focus permits projection of lines in both hemi-spheres. In the case of the lower focal point, lines contained in the upper hemi-sphere will project inside the reference circle and lines in the lower hemispherewill fall outside the reference circle. tn this work we adopt the lower-focabpointprojection, tmless otherwise indicat ed.

STEREOGRAPHIC PROJECTIOIU OF LINES AND PTANES

We have seen in Fig. 3.7 that stereographic projection of a sphere permits theconstruction of points and loci representing lines and planes in space. Rememberthat all such lines and planes are assumed to pass through the center of thereference sphere. A line will then project as a point somewhere in the projectionplane (except for a line from the center of the sphere to the focus of the projec-tion)' A plane passing through the center of the reference sphere generates agreat circle on the surface of the reference sphere and projects to the plane ofprojection as a circle. Several procedures for constructing these projections aredescribed in the following sections.

Stereographic Projection of a Vector

Consider the vector v in Fig. 3.8(a), with coordinates

v: (X, Y,Z) (3.1)

Orthographic projection of v on a horizontal plane (xy) gives vector u' in directionf with y; v makes an angle a with the vertical (z). The stereographic projectionof v is constructed by means of a vertical section through the reference spherealong the direction of r.r'. This section is shown in Fig. 3.8(b). We arbitiarilyassign the reference sphere a radius R. Then the distance OV to the projectionpoint of v is given by

(3.2)

where, as noted, a is the angle between v and f. Point Zrepresentingvector vis plotted in the projection plane [Fig. 3.8(c)] by its polar coordinates (o V, p),where B is measured clockwise from north (j).

Alternatively, V can be plotted from its Cartesian coordinates (X", Y").In a vertical section of the reference sphere along the x axis [Fig. 3.8(d)] we haveshown OV",the orthographic projection of v in this plane. From similar triangles

R+Z - R_T:n

Similarly, a section of the reference sphere alongyz will give

R+ zT

OV: Rtan$

RYo

(3.3a)

(3.3b)

Stereographic Projection of Lines and Planes

F

(b)

x

)

z

(c

F

(d)

ligure 3.8 Stereographic projection of a vector.

o

*r{ \/

/

y north

TI

I

z

Y

OV": orthographicprojection of vin the xz Plane

o

R

xo


Given a point (x, Y, z) on the surface of the reference sphere, R2 : xz + yzI 22, we can determine the stereographic projection of the point from itsCartesian coordinates Xo, Yo in the projection plane, with

xo: RXR+ z

and y- Rr(s:FvLet X : xlR, f : YlR, and Z: z1n rhen the coordinates xo, yo of thestereographic projection of a unit vector (F, t,Z) are given by

rl RIz16:

-:

- l+zvRfl al

- --..:

- r+zIf the vector v establishes a point (X, Y, Z) not on the surface of the sphere(i.e., R2 + xz + Y2 + z'), the length of v is proportioned so that it does. Thenthe representation of v in the stereographic projection will be given by

- RY

' JXr+Y2+22+Z

-g" JXr+Y2+22+ZAs an alternative to (3.5), when vector v is given in coordiate form (3.1),

we can plot its stereographic projection in polar coordinates (oV, a) with

and

and

and

Therefore,

(3.aa)

(3.4b)

(3.5a)

(3.5b)

(3.6a)

(3.6b)a: tan-t #Stereographic Projection of the Opposite of a Vector

A vector v : (X, Y, Z) has an opposite -v : (- X, -y, -Z). In Fig.3.9, the stereographic projection of v is point V, and the stereographic projec-tion of -v is point V'. From Fig. 3.9(a),

OV: Rtanl

ov':Rtan(no-#)(ov)(ov'): pz

Given any vector v, through the center of the reference sphere, we can calculateOV from (3.2) and then determine OV' from (3.7).

(3.2)

(3.7)


V'

{b)

Figure 3.9 Stereographic projection of the opposite to a vector.

Stereographic Proiection of a Plane

A plane that passes through the center of the reference sphere projects as

a true circle in the stereographic projection. To construct this circle, we must

find its center and radius.Consider an inclined plane dipping a below horizontal in direction p

measured horizontally from the x axis. Figure 3.10(a) is a vertical section

through the reference sphere along the direction of dip, that is, along the direc-

tion OC in the projection plane [Fig. 3.10(b)]. Line POP'in Fig.3.10(a) is an

edge of the inclined plane, and OP'is its dip vector. The stereographic projec-

tion of OP is point p, and the stereographic projection of OP' is p'. Therefore,

line pp' in Fig. 3.10(b) is a diameter of the required circle and its bisector,

point C is the center. The angular relationships in Fig. 3.10(a) offer simple

formulas for the center and radius of the circle.

First note that

4.POp:{P'Op':q

< OFP: 45o - +Also, the distance p'C: distance Cp: distance CF because triangle p'Fp isa right triangle and line FC bisects its hypotenuse, making triangles p'CF and

pCF isosceles. It follows, then, that { P'FC also equals 45" - dlz, so that

< CFO: c. The stereographic projection of the inclined plane is a circle whose

center, at C, and whose radius, r, ate accordingly determined by

OC: Rtana (3.8)

(3.e)

67

Then

R|.--cos d

o)

/

x

and


OF=r

o =45" -tCF=p'C=Cp=r

(a)

Referencecircle

Stereographicprojectionof plane

{b) (c)

Figure 3.10 Stereographic projection of a plane.

The coordinates of the center of the circle representing a plane with dip aand dip direction p arc

IN

il>l+lx

5

|+5

C*:RtanasinpC":Rtand,cosp

(3.10a)

(3.10b)

These apply to a lower-focus (upper-hemisphere) stereographic projection ofradius R. If the upper focus is adopted, the coordinates will be (-C*, -Cr).

Let fi: (N, f ,2) be an upward-directed unit vector, normal to the


inclined plane that dips a in direction B. Then

and Z: CoSd,

Since Z>O and X, * f, a 2z : l, (3.11c) determines the

relations according to Fig. 3.10(c) as

, - -. JV, +V _A -Z;tan d - -----..-.-----=_-:

-

Z

sin a - JFTTfrThen (3.11a) and (3.11b) give

cos d : ---:L-' ,\/Xzt Yz

sinB:---L:. JX'+Y"

69

.f : sin usin B

f : sin ucos p

(3.11a)

(3.1lb)

(3.1lc)

trigonometric

(3.rZa)

(3.12b)

(3.12c)

(3.12d)and

Substituting (3.12) into (3.8), (3.9), and (3.10) gives the following coordinates

for the center (C)and radius (r) of the stereographic projection of the plane:

OC: RJr-;

(l.l3a)

(3.13b)

(3.13c)

(3.13d)

Figure 3.11 summarizes the various ways to construct the stereographic

projection of a plane, given its dip (a) and dip direction (f) and reference circle

radius R.

1. Calculate fi with (3.9) or (3.13a). Then find the strike line's projection

points A and.B [Fig. 3.11(a)] as either end of the diameter of the reference

circle in directi on p + 90". Find center c with AC : BC : R. Finally'

draw a circle through C with radius r.

2. Calculate OC with (3.S) or (3.13b) and plot the center Cin direction p

from y [Fig. 3.11(b)]. Calculate r using (3.9) or (3.13a) and draw the circle.

3. Calculate the coordinates of center C with (3.10) or (3.13c) and (3.13d)

and calculate r from (3.9) or (3.13a); then draw the circle from C with

radius r [Fig. 3.11(c).

Ra

---: Z

/1 _RX-Z

c,: R!'z

Graphical Methods: Stereographic Projection Chap, 3

(cl

Figure 3.11 Alternative methods to construct a great circle.

The Line of Intersection of Two Planes

Since all planes considered here pass through the center of the referencesphere, any two planes will have a common line. An example is shown in Fig.3.12(a), where plane 1 (P,) has dip angle (a) equal to 60' and dip direction (p)equal to 100" and plane 2 (Pr) has a : 50o and p: 260o. The stereographicprojection of these planes [Fig. 3.12(b)] yields two circles, whose intersectionpoints I and.f' represent the two directions along the line of intersection of P1and Pr..Iis inside the reference circle (the dotted circle) and is therefore directed

AC=BC=r=R/cose


(b)

Figure 3.12 Line of intersection of two planes.

into the upper hemisphere. I',lyrng outside the reference circle, is directed intothe lower hemisphere, and is opposite to L

A Small Circle

The locus of lines (through a common origin) making an equal angle witha given direction through the origin is a cone. This cone pierces a sphere about

the origin along a circle; it is denoted a small circle because it can be generated

also by the intersection of the reference sphere with a plane that does notcontain the origin. By the fundamental property of stereographic projection(Section 2 of the appendix to this chapter), any small circle on the reference

sphere becomes a circle in the projection plane.

Consider a unit vector fi from the origin with coordinates fi : (X, Y, Z).We will construct the stereographic projection of a small circle representing the

locus of lines making an angle f with n [Fig. 3.13(a)].

Figure 3.13(b) shows a section of the reference sphere along a verticalplane through fi. The upper and lower limits of the conical locus are lines Oa

and. Ob, with stereographic projection points A and B. AB is the diameter ofthe small circle in stereographic projection and C, the bisector of ,,{.8, is itscenter. Then the radius of the small circle is

Graphical Methods: Stereographic projection Chap. 3

(b)

Figure 3.13 Stereographic projection of a cone.

r:AC:BC

<OFA:n;Q

<OFB:ryOA:Rtan+

OB:Rtanryr: t(oB _ oA)

Substituting (3.14) in (3.15) and simplifying gives

;_ Rsindcos@ cos d

The distance from the origin to the center of the projection is

oC: *@a + OA)

Substituting (3.14) in (3.17) and simplifyrng gives

oc: 4sinacos I -F cos c

Since

and

and

Then

(3.laa)

(3.14b)

(3.15)

(3.16)

(3.17)

(3.18)


The coordinates of the center of the small circle are

r _ Rsinasin/vx-cosf*cosa

and

Rsindt

- ___________J_=

cosS * Zt-'=-------'----'-4X2+Y'

Since fi is the unit normal to plane P with dip o and dip direction B

[Fig. 3.13(a)], (3.11), (3.I2), and (3.13) all apply here. substituting these in

equations (3.16), (3.18), and (3.19) gives the following simplecoordinateformulasfor the radius r and center C of the small circle of lines making an angle t' about

fi.:(X,f,21:

(3.19a)

(3.1eb)

(3.20)

(3.2r)

(3.22a)

(3.22b)

OC:

c*:cosS I Z

1/rl-.-.-----..-----

cnsS * Z

r_ YuY - cos6 + Z

ln summary, we can project the locus of lines equidistant from a normal

fi to plane P as follows. Knowing the dip (a) and dip direction (B) of plane P,

calculate OC from (3.18) and lay it off along a line making angle a with x

[Fig. 3.1a(a)]. Then use (3.16) to calculate r and draw the circle with C as

center. If you are using a computer, the coordinate formulas for the center (3.22)

and the radius (3.20) are preferable [Fig. 3.14(b)].

(a) {b}

Figure 3.14 Alternative methods to construct a small circle.


Example: Gonstructing a Stereonet

A stereonet is a projection of the longitude lines of one-half of the referencesphere. The net is used to obtain approximate readings of angles between linesand planes by tracing great circles and rotating the tracing about the center ofthe projection. Procedures for doing this are described by Phillips (1971), Hoekand Bray (1977), and Goodman (1980).

We will discuss the construction of a stereonet as an example of methods

(b)

Figure 3.15 A stereonet, showing the lines of longitude and latitude of a refer-ence sphere: (a) for one hemisphere only; (b) for the entire sphere except a regionabout the position of the focus.

Stereographic Projection of a Joint Pyramid

explored in the preceding two sections. In the stereonet, one finds two types ofcircles: the great circles, representing the projection of a family of planes witha common intersection; and the small circles, representing the projection of afamily of cones about the line of intersection of great circles.

Let the angle between each successive great circle be equal to d. Then the

planes have dip 6 : akd with kd : 0, d,2d,. . . , 90o. Substituting these

values, together with p : 90o,270' into (3.9) and (3.10) yields

C* : *Rtan kd

C":0

(3.23)

The small circles arc a series of cones about fi : (0, R, 0) or fi : (0, -R, 0).

The angular increment is kd, so that Q: d,Zd, "',90o; substituting these

values into (3.20) and (3.22) gives

r : Rtan kd

cr: o (3.24),1 +RuY: cdkd

Figure 3.15(a) is a stereographic projection net constructed with d: 10o,

using the formulas above to locate only those points within the reference circle.

Figure 3.15(b) is a more complete coverage of the sphere, with the same value

of R as in the hemisphere net.

STEREOGRAPHIC PBOJEGTIOil OF A JOINT PYRAMID

Stereographic Proiection of a Half'Space

In Chapter 2 we defined a joint pyramid as the set of points common to allthe half-spaces bounded by the plane of each face of a rock block when these

planes are shifted to pass through a common origin. All the joint planes plotted

in stereographic projection satisfy the requirement that they contain the center

of the reference sphere. Therefore, the stereographic projection can serve torepresent joint pyramids.

Assume that P, is a joint plane that passes through point (0, 0,0). Anyvector from (0, 0, 0) that is not contained in plane P, is on one side or the other

of P,; that is, it lies in either of two half-spaces created by P,. If h dips a(0 < d < 90") in direction B (0 < B < 360'), the unit normal vector to plane

P, isfi, : (sin w sin p, sin d cos p, cos c) (3'25)

We will adhere to a rule that cos d > 0. Therefore, fi, is always upward or

Ry-_'- coskd

and


horizontal. Since P, contains point (0,0,0), its equation is ff, . x : 0 [as in(2.32)1.

We will define the upper half-space (U,) of P, as the set of all vectors xobeying

fi,.x)0 (3.26)

The lower half-space (L,) of P,, is similarly, the set of all vector x obeying

fir.x(0 (3,27)

When P, is a vertical plane, the terms "upper" and "lower" may seem arbitrary.But they are still adequately defined by (3.26) and (3.27) when fi, is determinedby one of the two horizontal lines normal to P,.

The stereographic projection of a plane, Pr, is a great circle. [n the lowerfocus projection that we have adopted, the half-space U, of all vectors from(0,0, 0) to points above P, is the region inside the circle of plane P, (Fig. 3.16).Similarly, the half-space below plane P, is all the region outside the great circleof plane Pr.

Figure 3,16 Lower-focal-point stereo-graphic projection of half-spaces ofplane P1.

The Intersection of Half-Spaces to Form Joint Pyramids

Suppose that there are n nonparallel sets of joints, each determined inorientation by a plane, P,, passing through the origin (0, 0,0). The system ofplanes P,, i: L to n, cuts the whole sphere into a number of pyramids allhaving their apex at (0, 0, 0). Each of these planes is represented by a great circleon the reference sphere and therefore by a great circle in the stereographicprojection. The intersections of all these circles, as shown in Fig. 3.17a, generates

a series of regions on the projection plane. On the figure, these regions arenumbered arbitrarily. The reference circle, the dashed circle on the figure, has

no part in the bounding of the regions, but is shown for clarity. Each numberedregion in the stereographic projection can be thought of as a set of radius vectorsinside a particular joint pyramid. The corner points of a region are then the

Stereographic Projection of a Joint Pyramid 77

t/Lt4 /

P,P,NN

P, 3Pt

Ir')'.,"

a- -- a'1"',

---i---'\---jto \ t4 '\. 11 ,'

S7(c)

Figure 3.17 Stereographic projection of joint pyramids: (a) lower-focal-pointpro!rction of the whole sphere; (b) projection of UrUzUtUc (region 1); (c) the

upper hemisphere, using a lower-focal-point projection; (d) the lower hemisphere,

using an upper-focal-point projection.


stereographic projections of the edges of the corresponding joint pyramid andthe circular arc boundaries of a region are the projections of the faces of thepyramid. For example; region 1 on Fig. 3.17(a) is the joint pyramid formed bythe intersection of half-spaces (/, , (f z, (f s, and, (In. fts corners are the projectionsof the four lines of intersection, Irs, Izs, I2a&rrd,Inr, and these form the edgesof the pyramid as shown in Fig. 3.17(b).

We have found it easiest to plot all the regions of the sphere from a singlefocal point at the bottom of the reference rphrrc. That is, fig. 3.17(a) is anupper-hemisphere stereographic projection of the joint pyramids. An alter-native representation can be made by separating the upper and lower hemi-spheres in two separate figures. Figure 3.17(c) is the upper-hemisphere portion,projected from the lower focal point, and Fig. 3.17(d) is the lower-hemisphereportion, projected from the upper focal point of the reference sphere. In Fig.3.17(d) the region inside the circle of P, belongs to the half-spaceZ, whereasin Fig. 3.17(c) the region inside circle P, belongs to half-space t/,. This cancause unnecessary confusion; thus we prefer to project with a single focalpoint.

ADDITIONAT CONSTRUCTIONS FOR STEREOGRAPHICPROJ ECTION

Several additional procedures will make it possible to carry out all of the stepseventually required in the graphical representation of block theory.

The normal to a given plane. Given plane p, we wish to find theprojection of its normal fi. In Fig. 3.18(a), a vertical section through the refer-ence sphere along the dip direction of P, the plane is seen as diameter PP',

<oFA =qs" -Z

(a)

Figure 3.18 Normal to a plane.

Additional Constructions for Stereographic Projection

inclined d with horizontal. Because < OFN : ul2,

79

(3.2e)

Suppose that we have a great circle in the projection and want to plot the

normal [Fig. 3.18(b)]. First draw the diameter of the great circle that passes

through O, at the center of the reference circle. This diameter is AB, andtheradius of the great circle r : ABl2. The normal is at point i[, distant d fromO along OB.

The plane normal to a given line. The procedure described above

can be reversed to draw the projection of a plane given its normal. Suppose thatwe are given point N in Fig. 3.18(b), so we know that d: ON. Solving (3.29)

for r in terms of d yields

:*_!4p (3.30)- R2 - dz'-

From (3.8),

OC:Rta -. n 2tan@12)tr'--n7-1nP616

and from (3.28), tan(ulT) : dlR. Combining these two relations gives

oc:#ryFThus knowledge of d permits calculating the radius, r, and the distance to the

center OC, of the great circle.

The vector represented by a point (Xolro) in the stereographicprojection. Suppose that we are given a point V: (Xo, Yo) in the stereo-

graphic projection. Z represents a unit vector 0 : (X, f , Z) and therefore

From (3.9), cos d : Rlr.Inserting this in (3.28) gives

Xz+yz+Zz:lAlso, equations (3.4) apply. With (3.32), these can be inverted to

oN:d:Rtand' ^ ll-cosd'T:n1l 1a.ouo

l, - RON : d: Rnlvr-l-R

i zRXo^:I{2+XT+yXfr zRYot : pzlj{yEo+ R2-X3-Ytz': I{2+XT+y4o

(3.28)

(3.31)

and

(3.32)

(3.33a)

(3.33b)

(3.33c)

Multiplying equations (3.33) by (R' + XA + Y&)|QR), we obtain


X: Xo

Y:Yo

Z- Rz-X'"-Y',

(3.3aa)

(3.34b)

(3.3ac)

(3.35a)

(3.35b)

(3.37a)

(3.37b)

(3.38a)

(3.38b)

and

and

2R

Vector v : (X, Y, Z) and its unit vector (t, f , Z) both have the projectionpoint V:(Xs,Yr),

The center of a great circle through two points. It is oftennecessary to find the great circle through two points in the stereographic projec-tion. This corresponds to finding the plane containing two intersecting, non-parallel vectors.

Assume that we are given two points V, : (Xr, Ir) and Vz: (X", Yr),anywhere in the stereographic projection. Equations (3.3a) permit us to cal-culate their corresponding vectors vl : (Xr, Yr, Zr) and v, : (Xz, Yr, Zr) with

7 -R2-X?-Y',pr----T

7 _R2-X?-Yi22--____zF-

The normal ft to the plane common to y1 and v, has coordinates given by

ll : Yr X Yz : (YtZ" - YrZr, XrZ, - XrZr, XrY, - XrYr) (3.36)

Equations (3.13c) and (3.13d) give the coordinates (C* and Cr) of the center ofa great circle representing a plane with unit normal N, ?,2. fhe ratio of twocoordinates of a unit vector equals the ratio of the corresponding two coordi-nates of any vector parallel to and in the same direction as the unit vector.Therefore, we can rewrite (3.13c) and (3.13d) as follows:

/1 RXW*: -T

and Cr: VSubstituting the cooldinates of fi from (3.36) into (3.37) gives

/1 -XrZt-XtZ,vy-Wn

In summary, to find the center of a great circle through points Vr :(Xr, Yr) and, V, : (Xr, Yr) in the stereographic projection, compute Z, and Z,from (3.35) and calculate the coordinates of the center from (3.38).

An alternative graphical procedure to construct the great circle throughtwo points is to find the opposite to one of them. In Fig, 3.19,V|is the oppositeto point 2,. Then the circle is constructed through I/1 , Vr, and V\.

Additional Constructions for Stereographic Projection

Figure 3.19 Great circle through two points.

The orthographic projection of a vector on a plane. Given a

plane P and a line v, we will find the trace of the line projected into the plane byparallel, orthographic projection. This trace is the line of intersection of plane

P with the plane that contains both v and the normal to P. In Fig. 3.20, pointZrepresents line v and the circle for plane P is shown. To find the orthographic

eferencecircle

Figure 3.20 Stereographic projectionof the orthographic projection of a lineon a plane.

81

l

I

I

I

.t "'tQ...\LV\\

IN\ l'\R"tfc

)'_v'-

ane


projection of v on plane P:

1. Plot the normal (N) to plane P.

2. Draw a great circle through V and N.

This great circle intersects the circle for plane P at points e and e'. Theorthographic projection of v on P is point Q, and Q' is its opposite.

Measurement of angle between two vectors. Given the projec-tion pointsvrand v, of two vectors v, and vr, to find the angle (d) betweenthem: (1) measure the coordinates of Z, and Vr: Vt:(X0,, yor) and Vz:(X_or,Yor); (2) then using equations (3.33), calculate 0r: (Fr,f r,Zr)andAr:(Xr,fr,Zr); and (3) calculate d from

cosd:Irtr*7rfri2rZ, (3.3e)

The angle between two planes, or betw€€n a line and a plane.In Fig. 3.21(a), Z is the projection of a vector v and P is the projection of aplane. Construct the normal N to plane P. Then determine the angle between.l[and Zusing (3.33) and (3.39). The required angle is the complement of < NV.

In Fig. 3.21(b), P, and P, arc two planes. The angle between them is theangle between their normals, .l/, and.tr/r. There is another, simpler way to mea-

(b)

Figure 3.21 Angles between planes and lines.

(a)

Projection of Sliding Direction

sure this interplane angle. The stereographic projection has the property thatthe angle between two planes is exactly equal to the angle between two tangents

to the great circle projections of the planes, constructed at their intersection.(This property is established formally in the appendix to this chapter.) In Fig.3.21(b) the two planes are represented by the great circles P, and P". The angle

between them can be read with a compass between the trangents to Pt and P,at either point of intersection, as shown.

PROJECTION OF STIDING DIRECTION

The direction of sliding of a block under a given set of forces was discussed,

using vector analysis, in Chapter 2. This direction can be determined by stereo-

graphic projection procedures as well.

Lifting. Suppose that the direction I of the resultant force applied to

the block is oriented such as to lift the block from every joint plane. The sliding

direction .i is then identical with the direction of l.

Single-face stiding. If the resultant force r acts in a direction such

that a block tends to slide along one of its faces, the sliding direction is parallei

to the orthographic projection of r on the plane of that face. In Fig. 3.22(a) iis the direction of the resultant force and sliding is along plane P. Construct

N

Figure 3.22 Sliding directions for single-face sliding.


.l/ normal to P. Then construct the plane common to N and f , intersecting P atsr and sr. Now construct plane P, perpendicular to f The sliding direction, ^f,is the choice of st and s, that is contained in the same half-space of P, as is thepoint f. The sliding direction is then s,.

In the frequent special case where the direction of the resultant force isthat of gravity (0,0, -1), f cannot be plotted. However, we know that thehalf-space of the P, circle that contains i is the region outside the reference circle.Also, the normal to P lies along a diameter of the reference circle extendedthrough C the center of the circle for plane P [Fig. 3.22(b)]. The plane commonto i and Nis vertical and is therefore a straight line in direction OC. The slidingdirection, s, is the intersection of the extension of OC and circle P.

Sliding in two planes simultaneously. In Fig. 3.23(a), f is theprojection of the resultant force direction. Sliding is on planes P, and Pr, whoseintersection lines are the two crossing points (/,2 and -.I,r) of circles P, andPr.Construct the plane P, whose normal is at f . The sliding direction is the choiceof Ir2 and -.I1, that lies in the half-space of P,, containing f. Thus s is f,r, asshown in the figure.

Under only its own weight, f is downward and cannot be plotted. But P,

(a) (b)

Figure 3.43 Sliding directions in double-face sliding.

P2

Let

Examples

is the reference circle and the half-space of P, that contains I is the regionoutside the reference circle, that is, the lower hemisphere. Therefore, s is the

line of intersection of planes 1 and 2 that plots outside the reference circle

[Fig. 3.23(b)].

EXAMPTES

Example 3.1. Stereographic Proiection of a Unit Vector

Given a unit vector

6 : (0.38683, 0.51335, 0.7 6ffi4)

R:1Using formulas (3.a) with

.f : 0.38683, f :0.51335, Z:0.76604We calculate the coordinates of the projection point V of unit vector 8,

y: (0.21903,0.29068)

so we can plot point Z in the projection plane. Another method to plot pointZ is to use formulas (3.6), by which

OV :0.8391

and d,:53"

Erample 3.2. Stereographic Proiection of a Vector

Given a vectory : (1,2, 1)

and the radius R of the reference circle

R:1Using formulas (3.5) with

X:1, Y:2, Z:1, R:1the coordinates of the projection point Z of the vector Y are

y: (0.29999,0.57975)

Then we can plot the projection point V of v in the projection plane.

Example 3.3. Stereographic Proiection of a Plane

from its Dip and Dip Direction

GivenaplanePwithdip angle il.: 42o

and dip direction f : L44", let the radius of the reference circle be unity:

R:l


From equation (3.9) we compute the radius of the projection circle of plane p:

r :1.3456From equation (3.8) the distance from the origin (O) to the center (C) of theprojection circle of plane P is

OC :0.90040

Alternatively, using equations (3.10) we can compute the coordinates of thecenter of the projection circle of plane p:

C*: 0j2924

C' : -0'72844We can draw the projection circle by three methods:

1. Using F : lM" and r : t.3456 [Fig. 3.11(a)J

2. Using F : 744', r : t.3456, and OC: 0.90040 tFie. 3.11(b)l3. Using C,:0.52924, Cr: -0.728{/,, and r : I.3456 [Fig. 3.11(c)

Example 3.4. Stereographic projestion of a planefrom its Normal

Given the coordinates of the unit normal vector fi of plane p,

i : (0.6L2j7,0.35355, 0.20710) and .R : Iwe compute the data of the projection circle of plane P as follows. Usingequation (3.13) with

and

R:1X : O.61237

Y :0.35355

Z : 0.70710

the radius of the projection circle of plane p isr : 1.4t42

The distance from origin o to the center C of the projection circle is

OC:lThe coordinates of the center C of the projection circle of plane p are

C* :0'86602

C' :0'50000

We can now draw the projection circle by any of the three methods of Example3.3.

Examples

Example 3.5. Draw a Small Gircle Representingthe Stereographic Projection of a Cone about the Normalto a Plane

Given fi is the normal to a plane with u: 1-16" and P: 50", and thecone makes an angle 6 : 20" with ,. For R : 1.0, (3.16) determines the radiusof the small circle &s r : 0.68223, and (3.18) gives the distance from the center(O) of the reference sphere to the center of the small circle (C) as OC : 1.7928.Alternatively, the coordinates of C arc determined by equations (3.22) as C" :L.3734 and C" :1.1524.

Example 3.6. Draw a Small Circle Representinga Gone about a Vector v

Given v = (0, 1, -1), R : 1, and S : 20". The unit vector correspondingto v is 0 : (0, 0.70710, -0.70710). Entering R : 1, X : 0, Y : 0.70710, andZ : -0.70710 in equations (3.22) gives C* :0 and C":3.0401 and from(3.20), r : I.4705.

Example 3.7. Constructing a Stereographic Projection Net(Equal-Angle Projection Net)

In this example we compute the centers and radii of all the circles of thestereonet as shown in Fig. 3.15. Assume that the radius of the reference circleR : 1, and the degree step is 10'. Using (3.23) with

d: lO", k: I,2r3,4,516,7r9

we obtain the radius r and coordinates C* and, Cy of the great circles.

cyc*

I23

45

67

8

00000000

1.0154t.064r1.15471.3054l 55572.00002.92385.7587

+o.r7632+o.36397ltro.s7735+0.83910tL.l9l7+r.7320+2.7474+s.6712

Similarly, from (3.24), with d : 1.0o, the values of the radius r and coordinatesC, and C, of the centers of the small circles are:


cycx

I23

45

67

8

o.176320.363970.577350.839101.19t71.73202.74745.67t2

+1.0154*1.0641tr.rs47*1.3054+1.5557+2.00m+2.9238+5.7587

00000000

Example 3.8. Stereographic Projestion of a Joint Pyramid

Given four sets of joints:

Dip, a(dee)

PrP2PtP+

Assume that the radius of the reference circle is Rwe compute the radius r and the coordinates C,circle of planes Pr, Pr, Pr, and Pn.

: 1. Using (3.9) and (3.10),and Cy for each projection

Dip Direction, p(dee)

70zffi

0150

@508020

Plane cyC,

PtPzPsP+

21.55575.75871.0641

o.59239

-o.2Q6945.6712

-0.31520

r.6n6-1.t736

00.18198

Then construct all projection circles as shown in Fig. 3.17(a). Each region isthe projection of a joint pyramid.

For example:

. Region I is the joint pyramid U r, Uz, Ur, (J n.

. Region 2 is the joint pyramid Ur, Ur, Lr, (Jn.

. Region 3 is the joint pyramid Ur, Ur, Us, Ln.

. Region 10 is the joint pyramid Lr, U", Lr, Ln.

. Region 14 is the joint pyramid Lr, Lr, Lr, Ln.

Examples 89

where Z, means the lower half-space of plane P, and (/, means the upper half-space of plane Pr.

Example 3.9. A Plane Gommon to Two Lines

Given two points in the stereographic projection with R : 1:

A: (_0.99,2.3)

B : (_2.2,0.64)

as shown in Fig. 3.18. Our task is to draw a great circle through points A and B.

Let Xo: -0.88 and Yo : 2.3; formulas (3.34) determine a vector N -(X, Y,Z) corresponding to point l:

a : (-0.88,2.3, -2.5322)

Similarly, for point .B input Xo : -2.2 and Yo : 0.64 in (3.3a) defining vectorb corresponding to point B:

b : (_2.2,0.64, _2.1249)

Then equations (3.38) give the coordinates of the center (C) of the projectioncircle for the plane common to a and b:

c: (_0.72639,0.82303)

Example 3.10. Finding a Normal of a Given Plane Pin the Projection Plane

Given a projection circle of plane P as shown in Fig. 3.18(b). Measure the

distance AB; line AB is the diameter of the circle of P along the line throughthe origin (O). The radius is then

t.tl44

Assume that R : 1. Using (3.29), compute d: ON :0.23260; the positionof the projection N of the normal to plane P is now known.

Example 3.11. Finding the Plane (P) Normal to a Given Vectorin the Stereographic Proiection

Given a poiqt i/ in the stereographic projection plane as shown in Fig.3.18(b) Suppose that R : 1. We can measure d: ON :0.23260. ConsideringN as the projection of a vector n, equations (3.30) and (3.31) give r : l.LI43and OC: 0.4918. Now plot C and draw plane P as the circle of radius rabout C.

AB-T:


Example 3.12. Stereographic Projection of the OrthographicProjection of a Vector v on a Plane

Given plane P with normal fin and vector v. The problem is to find thestereographic projection of v in plane P. The construction is shown in Fig. 3.20.Let R : 1 and let the stereographic projection of fio be N : (-0.24, -0.15).The vector v is represented by V: (-0.42,0.44). Then with (Xr,Yr):(-0.24, -0.15) and (Xr,Yr): (-0.42,0.44), use equations (3.35) to cal-culate Z, and Zr. Finally, substituting Xr, Yr, Zr, Xz, Yr, and Z, in (3.38)calculate the corrdinates (C") and (Cr) of the center of the great circle throughV and, N. (C,, Cr) : (1.4805, 0.6973b). Constructing this circle, we find inter-section points Q and Q', which are the orthographic projections of v on plane P.

Example 3.13. Measuring the Angle between Two Vectors

Given two pointsVt : (1, 1)

and Vz: (-0.5, 0)

in the stereographic projection plane, which are the projections of vectors v,and vr. Using formulas (3.34), the coordinates of vectors V, and V, are deter-mined as follows (with R : 1):

Vl : (1, 1, -0.5)yz : (-0.5, 0, 0.375)

Convert v, and v, to unit vectors 0, and Srby dividing each component by thelength. Then from (3.39) the angle::Xn vectors V, and V"is

Example 3.14. Measuring the Angle between a Vector vand a Plane P

See Fig. 3.21(a). Let R : 1. Given a projection circle of plane P, measurethe radius r and the dip direction f from this circle: Suppose that

and

From (3,29), compute

The coordinates of N are

r : 1.15

F :205"

d: ON : 0.26413

nf - (d sin p, d cos f) : (-0.11 162, -0.23938)Given the projection I/ of vector v,

v : (_0.29,0.37)

Examples 91

Using (3.33), compute unit Yectors fi and D, which have projections 'n/ and %respectively.

fi: (-0.20868, -0.M753, 0.86957)

0 : (-0. 47 502, 0.60606, 0.63800)

From (3.39), comPute< (fr'D) :67'500

The angle between P and V is

90' - 67.5" :22.5"

Example 3.15. Measuring the Angle between Two Planes

Let R: 1. Given the projection circles of planes P, and P"as shown in

Fig. 3.21(b). Measure the radius and dip direction from these two circles:

Prt rr : l'!5, Fr: 205o

Pzi rz:1.55, fz:0"From (3.29), comPute

dr: ONr:0'264t3dz: ON2: 0'46442

where l/, and N, are the projections of normal vectors fi, and fi, of P, and Pr,

respectively.

l[r : (dr sin Fr, d, cos B,) : (-0'11162, -0'23938)Nz : (d, sin Fr, d, cos Pr) : (0, 0.46442)

From (3.33), compute

fir : (-0.20868, -0.44753' 0'86957)

fiz : (0, 0.76404,0.64516)

Using (3.39), we have

4 (fit' fi") : 77 '345"

which is the d angle between P, and Pr. This could have been read directly as

shown.

Example 3.16. Finding the Direction of Sliding

on a Single Face

See Fig. 3.22(a). Given a projection circle of plane P and projection f of

resultant force R. Let R : 1. Measure the radius and dip direction of P. The

radius r : 1.55; the dip direction F :210". From (3.29), compute

d: ON :0.46442

N : (dsin p, dcos P): (-0.23221, -0'40220)


Measure the coordinates of f in the stereographic projection

f : (_0.24,0.13)

Using (3.35) and (3.38), compute the center C of the circle through points Nand f:

g : (I.9711, _0.10525)

Then draw this circle, which intersects circle P at points ,Sr and 52. Calculate

d:Or:ffi:O.21294Then using (3.30) and (3.31), compute the radius ro and center Co of the planenormal to f :

/o : 1.1609

and OCo :0.58983

Draw circle Po with radius ro and center Co. Note that points ^9, and r are on

the same side of circle Po ; therefore, ,S1 is the projection of the sliding direction.

Example 3.17. Finding the Direction of $llding on TwoFaaes Simultaneously

See Fig. 3.23(a). Let R : 1. Given the projection circles of planes P, andPr, respectively, and given the projection r of resultant force r. Measure

Or:0.25Using (3.30) and (3.31) with

d: Or : O,25

compute the radius ro and distance OCo of the projection circle of the plane(Po) normal to vector f .

ro : 1.1333

OCo :0'5333

Draw circle Po. The intersection point I* and i lie on the same side of circle Po,so .f12 is the projection of the sliding vector.

APPETUDIX

IMPORTAIUT PROPERTIES OF THE STEREOGRAPHIC PROJECTIOil

t. THE STEREOGRAPHIC PROJECTIOIU OF A PLAIUEIS A CIRCLE

Given plane P, with dip a and dip direction p (with sign conventions as inChapter 2), passing through (0, 0, 0). From (2.6) and Q.7) the equation of P is

sin a sin pX * sin d cos pY + cos uZ : O (1)

Suppose that (Xo; fo,0) is a point in the stereographic projection of plane P

tmportant Properties of the Stereographic Projection

Figure 3.24 Stereographic projection.

(Fig. 3.24). The focus of the projection is at (0, 0, -R) and a straight line from

the focus to (Xo, %,0) is given bY

X: Xot

Y:Yot Q)

Z:R(t-1)The piercing point, I, of this line on the reference sphere is defined by a value

of t : to, obtained by equating the line and the reference sphere

93

Xz+yz+Zz:RzSubstitutin1 (2), with lo in place of /, into (3) gives

ts: zRzYTW#T

So the coordinates of the piercing point, I, of the line on the sphere are, from

Q), ,:$qffi-p.'n#le'ffi) (s)

Since 1is also on planeP, it satisfies (1); combining (5) and (1) yields

(sinasinp 2R2)Xo*(sindcos P2R2)Yo*cosdR(Rz - XA - fA):0 (6)

which simplifies to

(X, - Rtan asin f)' +(% - Rtandcos f)' :(#J' (7)

Equation (7) defines a circle with center at (Xr' %): (R tanu sin B,

R tan d cos B) and with radius -R/cos c.

2. THE STEREOGRAPHTC PROJECTION OF A CONE

ls A clRctE

Given unit vector d:(X*Yo,Zo), which is the axis of a cone with vertex

(0,0,0) on d: the angle between d and any straight line on the cone is f (Fig.

3.25). The equation of the cone is

(3)

(4)

(0,0, -R)


(X, Y,zl

0 a

(0,0,0)

Figure 3.25 A cone about line ri.

JXry-We cos d : XoX + Y.Y + Z.Z (B)

x?+Y:+Z?:l (9)

As in the preceding section, let (xo, yo,O) be a projection point of plane pand let the focus be located at (0,0, -R) so that a straight line from the focusthrough the projection point intersects the reference sphere at I (Fig. 3.24).The coordinates of I are given by (5). If point .Iis also required to be on a conethrough 0, (5) can be equated to (8). This gives

2R2X"X. 2R2Y"Y.Rcosd:7ffip+ffffi*ffiYt +:w (10)

(l 1)

which simplifies to

(x. - ffi.)' * (r' - #.)' : (4_d-.)'Equation (11) defines a circle with center at

(xo,Y):(#-t,#r")and , : _4 gio d:_

cos@ - Zo

r-et a be the normal vector to plane p, with dip a and dip direction p. Then,from (2.7),

Xo: sin a sin BYo: sin a cos p

and Zo: cos d

Substituting (12) in (11), the radius of the circle is

'- 4sinr0cos d+ cos d

and the coordinates of the center of the circle are

(x,,r,):(#F+#,##)

(r2)

(l 3)

(14)

App. lmportant Properties of the Stereographic Projection

3. THE STEREOGRAPHIC PROJECTIOTU PRESERVES

THE ANGTES BETWEEN PTANES

This property can be stated: *The stereographic projection is an

projection."Suppose that P, and P2 are two planes defined by normals

respectively:forP,i trr.x:0forPrr trz.X:0

Let

and

where

and

equal angle

n, and nr,

and

The planes are represented, in the stereographic projection, by great circles Pt

and pz Fig. 3.26). If the proposition is true, the angle d, between planes Pt

(a) (bl

Figure 3.26 Stereographic projection of the angle between planes.

and. P, [Fig. 3.26(a)] is exactly equal to the angle d2 between the tangent vectors

to great circles P, and P" lEie.3.26(b)] at their points of intersection.

Ir: (Xr,Yt,Z)D2 : (X2rY2, Z2)

and suppose that Zt ) 0 and Zr2 0. Let

(xr,Yr, zt): (#r'f.r'?)R,:6?+Y?+Z!)trzi: 1,2


Then x?+y?+z?:l, z,)0 (1S)

suppose that the radius of the reference circle is unity (R : 1.) Let (A,, B,) bethe coordinates of the center of great circle P, and let its radius be rr. Using(3.13),

A, : -xt-'21

B,: + (16)

and-l ,r: zthe equation of circle p, is

(x - A,), * (Y - B,), : r? (L7)Solving equation (ll) for Y,

y:+m*n, (18)

Differentiating with respect to X, the tangent vector (t ) to the great circle forplane P, is

t':(l T(X-A') \'\'ffi) (re)

t, is parallel to(m,+(x-A)) (20)

From equation (18),

JFW:*(r_8,)so from (20),

tr : [*(F - B,), +(X - A,)lor t, : +[(f - B,), -(X - A,)l el)From (17), lt,l : r, and accordingly,

f t : *lft" - B,), -(x - A,)l (22)' - tt'Let (x, Y) : (xo, ro) be the point of intersection of circles p, and p, so that(&, %) satisfies (Z2)withboth i: ! and i :2. At(xo, %), the angle betweenf, and f, is given by

cos d, - f r.f z: I#n*, - Ar)(Xo - Ar)+ (% - s,Xro - Br)l

I+;K&- A,)'*(Yo-8,)'

+ (& - Ar)(A, - Ar)+ (% - Br)(8, - Br)l (23)

: + *rn,+ (& - A,)(A, - A;)

+ (% - 8,)(^B, - Br)l

lmportant Properties of the Stereographic Projection

This yields

cosd2: f[x,xr*yry, I zrzz* Xs(xiz- xzzr) * Yo(ytzz-!zzr)l Q4)

The line of intersection of Pr and P, is found from

fir x fiz: (Itr) : (Xrr,Yrr, Zrr)Irz: (yrr, - !zzr, xzzr - xrzz, xr!z - xz!)

(2s)

Q6)

(27)

The stereographic projection of I,, is (Xo, Io); using (3.5)'

But

(xo,ro):(+,+)

where

Substituting for

k- l\zli- xtYr- xrYr

X and Y from (26),

Consider the term

Xo(xr 22 - xzzt) * Ys(ytzz - fzzt)which is found in Q$. Substituting for Xo and Yo from (28), this term becomes

P r*)(x,r, - xzz t) * W)(y,r, - !zz)

which is equal to zero. Therefore, (24) simplifies to

cos d, : f [xrx, * yry, * zrzzl

lxrxr*yJz* ztzzf :frt,fir: cosdl

(28)

Qe)

So d, - d1 and the angle between the vectors fi, and fir,hence between Pt and

Pr, is exactly equal to dr. Q.E.D.

chapter 4

The Removabilityof Blocks

The central idea of this book is that stability analysis of excavations can skipover many of the conceivable combinations of joints and proceed directly toconsider certain critical blocks, denoted as keyblocks. This efficient approachis workable by virtue of a theorem on the finiteness of blocks. tn this chapterwe establish and demonstrate this theorem and some associated propositions.

TYPES OF BTOCKS

You will recall from Chapter 2 that a block is determined by the intersection ofa particular set of n half-spaces. Considering orientations alone, there are 2,unique half-space intersections. Not all of these define potentially critical blocks.We will now establish criteria for the relative importance of blocks. A keyblockis potentially critical to the stability of an excavation because by definition, it isfinite, removable, and potentially unstable. Table 4.1 uses these terms to recognizefive types of blocks. An infinite block (type v), as depicted in Fig. 4.1(a), pronid.tno hazard to an excavation as long as it is incapable of internal cracking; wehave ruled out consideration of internal cracking of blocks in the basic ur*tttp-tions introduced in Chapter 1. As we shall see, most of the 2' half-space intei-sections produce infinite blocks. Finite blocks are divisible into nonremovableand removable types. An infinite block that does not crack obviously cannotbe removed from the rock mass. However, a finite block may also be non-removable because of its tapered shape. We shall prove later than any finiteblock intersected by an excavation surface so as to increase the totat number of

98

Types of Blocks

TABLE 4.1

aPEffi 8rc95!

v\lnfinite F inite

Nonremovable

Stable aren withoutfriction

IV -/ l\ra'ered

rr/ I \,

Removable

Stable withzufficientfriction

Potential

Key block

Unstabld withoutsupport

Key block

faces acquires a tapered shape and cannot be removed from the rock mass.

An example of a tapered block (denoted type IV) is presented in Fig. a.l(b).

All tapered blocks are nonremovable unless they are undermined by movement

of an adjacent block.Nontapered, finite blocks are removable but they are not all critical to

the survival of a given excavation under a given set of loading conditions. We

can distinguish three classes of removable blocks. A type III block has a favor-

able orientation with respect to the resultant force, so that it tends to remain

stable even without mobilizing friction on its faces. Figure a.1(c) presents an

example. Although it would be possible to lift this block from its home, the

block is of no concern under purely gravity loading since its virtual movement

is away from the excavated space. Figure 4.1(d) shows a type II block; it is

defined as a block that is potentially unstable (since the tendency for movement

is toward the free space) but unlikely to become unstable unless the frictional

resistance on the potentially sliding face is extremely small, or there are loads

in addition to the block's self-weight driving the displacement. A block like this

is a potential key block. A true key block (type I) like that of Fig' 4.1(e) is not

only removable but oriented in an unsafe manner so that it is likely to move

unless restraint is provided. In the case shown in Fig. 4.1(e), the restraint would

have to be constructed before the excavation has completely isolated the block.

The actual shape of a rock block is governed not only by the numbers and

relative orientations of the planes forming its faces but by the numbers of these

planes that are paired. A block may be formed with one each of n discontinuities.

br it may have one or more sets of discontinuities represented twice as opposing

100 The Removability of Blocks Chap, 4

(e)

Figure 4.1 Types of blocks: (a) infinite; (b) tapered; (c) stabre; (d) potentialkeyblock ; (e) keyblock.

faces. Figure 4.2 illustrates this with a two-dimensional example. The block inFig.4.2(a) is determined by the intersection of three hatf-planes, defined by threenonparallel joints. Repeating one of these joints creates a block like that of Fig.4.2(b). Repeating all the joints outlines a block like that of Fig. 4.2(c), whichbegins to resemble a crystal. Blocks with parallel faces tend to be more stablethan those lacking parallel faces because the range of directions of kinematicallypossible movement become diminished.

Two-dimensional tapered blocks are shown in Fig. 4.3. In Fig. 4.3(a) atriangular block is to be cut by the excavation surface shown. The resultingfour-sided block will subsequently have a shape such that it cannot move into

(b)(a)

(d)

Theorem of Finiteness

Figure 4.2 The influence of the number of parallel sides on block shape: (a) noparallel sides; (b) one set of parallel sides; (c) all sides parallel.

the excavated space; that is, any displacement of the block toward a point inside

the excavation would generate a larger width so that the block cannot fit throughthe available space. This is also true of the block produced by four nonparalleljoints in Fig. 4.3(b). When the block is excavated as shown, the resulting flve-sided block becomes tapered. If the excavation surface eliminates more thanone corner of the originat block, so that the resulting number of faces is notincreased, the resulting block may not be tapered. This is depicted in Fig. 4.3(c),

where two corners of the original block have been removed, and in Fig. 4.3(e),

where three corners have been excavated. Neither of the resulting blocks are

tapered. These examples are two dimensional but they apply equally to three-

dimensional blocks.

THEOBEM OF FINITENESS

An important theorem will allow us to establish whether or not a given block is

finite. Let a block be defined by the intersections of half-spaces defined byplanes L,2, . . . , n. Using the terminology introduced in Chapter 2, we defrne

an associ ated block pyramid by moving each plane to pass through the origin.

A convex block is finite if its block pyramid is empty. Conversely, a convex block

is infinite if its block pyramid is not empty.

It will be recalled that a block pyramid is generated from the system ofinequalities for a block by setting all terms D, to zero, as illustrated in Example

2.9. An'oempty" pyramid is one that has no edges. In Example 2.9 we saw

how to determine whether or not a block pyramid is empty through the use

of vector analysis. It is also possible to judge the emptiness of a block pyramid

using the stereographic projection. Before proceeding to demonstrate how to do

this, it will be instructive to pursue some two-dimensional examples.

101

(b)(a)

102 The Removability of Blocks Chap. 4

(c)

Figure 4.3 Two-dimensional tapered blocks.

Theorem of Finiteness 103

Figure a.a@) shows a free surface (3) and two joint planes. Consider the

block determined by the intersection of half-spaces Ur, Lr, and L3. (The symbols

L and (J are used, as in Chapter 2, to identify the half-spaces respectively belowand above a given plane.) The block is obviously infinite. To apply the theorem,

we determine the block pyramid corresponding to UtLzL3 by moving the half-spaces, without rotation, as required to pass the bounding planes through a

common point. This has been achieved in Fig. 4.4(b). The block pyramid cor-responds to the region common to (Jl, Ll, andZ! on this shifted diagram. Since

there is a region common to these three half-spaces, block UtLzLiis determined

to be infinite.

Region common tou?, L8,and L!

u!

(al

t2l

(b)

finitenessFigure 4.4 Application of thedimensional example.

theorem to an infinite block; two-


The converse is shown in Fig. 4.5. Block (/1(JzL3 in Fig. a.5(a) is finite byinspection. To determine this formally using the theorem, planes I,2, and 3are shifted without rotation to pass through a common origin. It is then seenthat the region common to U! and, Uorhas no points overlappingregionZ! exceptfor the origin itself. There is accordingly no edge to the block pyramid UoJl\L?,which is therefore "empty." By the theorem, block (Jr(JzL3 is finite.

In these two-dimensional examples, the finiteness of the blocks consideredwas obvious by inspection. [n real, three-dimensional cases, inspection is not sosimple. However, the formal application of the theorem does permit a directdetermination of finiteness. The method is particularly simple when the stereo-graphic projection is used, as will be shown.

Free surface

An "irclated block";the common part of

Ul, U2, Ls

(b)

Figure 4.5 Application of the finiteness theorem to a finite block; two-dimen-sional example.

QI

(al

The only point common toU?, Ug and L!

Theorem of Finiteness 105

In each example, the block was defined partly by joint-plane half-spaces

and partly by free-surface half-spaces. Each block pyramid was therefore

formed of planes parallel to both joints and free surfaces. The joint-plane subset

of the half-spaces determining a block pyramid will be designated as a iointpyramid (JP). The set of shifted excavation half-spaces will be designated as the

excavation pyramid (EP). The block pyramid (BP) is then the intersection (n)of the joint pyramid and the excavation pyramid for a particular block:

BP:JPNEPFor a block to be finite, the block pyramid must be empty.

A block is finite if and only ifJPnEP:@

An alternative statement is possible if we define a space pyramid,

set of directions that is complementary to EP, that is,

SP: -EPThen equation (a.2) is equivalent to stating that a block is finite if and only ifits joint pyramid is entirely contained in the space pyramid, that is, if and only if

JPcSPLet us now reexamine the previous examples in Fig. a.a(b): U\LZ defines

the joint pyramid JP: Lt is the excavation pyramid EP; and U! is the space

pyramid SP. Since JP is not included within SP, the block is infinite. In Fig.4.5(b), JP is Uor(lL and SP is U3. Since JP is entirely included in SP, the block is

finite.

The Finiteness Theorem on the Stereographic Proiection

Given a series of joint planes and free surfaces, it is possible to produce

a stereographic projection of the series of planes by placing each in position topass through the center of a reference sphere of radius R. Each shifted plane

projects as a circle crossing the reference circle at opposite ends of a diameter.

A typical projection of the great circles for a series of planes was demonstrated

in Example 3.8 [Fig. 3.17(a)].As another example, we will construct the stereographic projection for

blocks defined by the three joint planes and single excavation surface listed inTable 4.2. Figpre 4.6(a) shows the projection of the four planes using a lower-

TAB,LE 4.2

(4.1)

(4.2)

SP, as the

(4.3)

Dip, a(dee)

Dip Direction, p(deg)

Joint plane IJoint plane 2

Joint plane 3

Free plane (excavation)

9050

130

90

3065

65

15


Referencecircle

\111;

:

It0I\\

(a)

(b)

Figure 4.6 Application of the finiteness theorem in three dimensions.

Theorem of Finiteness

4: Free plane

{d}


focal-point stereographic projection. The reference circle (horizontal plane)is the dotted circle. The free plane is represented by the dashed circle and thethree joint planes are represented by the three circles with solid lines. For now,consider only the joint planes. The regions of intersections of these three circlesdefine joint pyramids. Each is identified by a string of three binary digits. Thenumber 0 corresponds to the symbol U and defines the half-space above a plane.The number L corresponds to the symbol Z and identifies the half-space belowa plane. The digits are arranged in order. Thus 100 signifies the joint pyramidLorU\.U!, which is simultaneously below plane 1, above plane 2, and aboveplane 3. In this lower-focal-point (upper-hemisphere) projection, the regionabove a plane is the area inside its great circle.

108 The Removability of Blocks Ghap. 4

Consider now the excavation of an underground opening having theindicated free plane as its roof. Blocks in the roof are formed of joint-plane half-spaces together with the half-space above the free plane. In Fig. 4.6(b), thefree surface has been redrawn as a solid circle and portions of the joint greatcircles have been removed, leaving only the two joint pyramids, 011 and 100,that lie entirely inside or outside the free-plane circle. The space pyramid of theexcavation with the free plane as its roof is the region below the free plane.Therefore, SP is the region outside the free-plane circle. Since joint pyramid 100is entirely outside the excavation circle, it is entirely contained in SP. Therefore,by the finiteness theorem, joint pyramid 100 corresponds to a finite block-LrU"Us. All the other regions identified in Fig. a.6@) are at least partly insidethe excavation circle (i.e., partly inside EP) and are therefore not entirely con-tained in SP; they are, accordingly, infinite.

With the same line of argument, we will see that region 011, which liesentirely inside the free-plane circle, determines a block that is finite below anunderground opening that has the free plane as its floor. When tie blocks areformed below the free plang, the space pyramid becomes the region above thefree plane and therefore inside the free-plane circle. We can determine at aglapce that 011 is the only joint pyramid entirely contained in the SP abovethe free plane.

Figure a.6@) shows a section through the roof looking south. Block 100is shown both in this section and in Fig. a.6(d), an isometric view. Such a blockis quite clearly a potentialhazard for the excavation. It is a type I block-finite,nontapered, and probably unstable without support unless it has a considerablefriction angle on the bounding planes. The drawings of the block are helpful inplanning rock reinforcement, but it is not necessary to draw a block to determinethat it is removable. That property is determined by a second theorem.

THEOREM OtU THE REMOVABILITV OF A FINITE,cotuvEx BLocK

A finite, convex block is removable or not removable according to its shaperelative to the excavation. We have previously termed a nonremovable finiteblock as tapered. Necessary and sufficient conditions for the removability ornonremovability of a finite block are established by the following theorem.

A convex block is removable if its block pyramid is empty and its joint pyramidis not empty. A convex block is not removable (tapered) if its block pyramid isempty and its joint pyramid is also empty.

By the finiteness theorem, the block considered in both parts of theremovability theorem must be finite. The new theorem states that a finite blockdetermined by a series of joint planes and free surfaces will be tapered if the

Theorem on the Removability of a Finite, Convex Block 109

joint-plane half-spaces themselves determine a finite block; that is, if the blockcorresponding to the joint pyramid is finite, the block corresponding to thejoint pyramid plus one or more free surfaces will be tapered. We saw this pre-viously in Fig. a3@) and (b).

A two-dimensional example will help to demonstrate the theorem. Figure4.7 shows a series of blocks defined by joint planes and a free surface (plane 5).

Consider first block l, determined by UrUrUnLr. JP for this block is U? UZU|.As shown in Fig. 4.7(b), there is a region common to these shifted half-spaces,

and therefore JP, is not empty. On the otherhand,theblockpyramid UiUgU|L?ls empty. By the first part of the theorem, block I is removable because JP,is not empty and BP, is empty. L

{!ock B in Fig. a.7(a) is determined by half-spaces LrUr/rZr. JP" is

L?U\F,?.The only point common to these shifted half-spaces is the origin, as

showir in Fig. a.7@). Therefore,IP" is empty. BP, is also empty. Therefore, by

r-"*ffi

u+

4

,,J

A(U1, U2, Ua) not taperedB(L1, U2, Us) tapered

Region common to 1

uf, u!, u!

(bl (c)

Figure 4.7 Application of the removability theorem in two dimensions.

L2

u4The only point common to

L?, ug, and L!


the second part of the removability theorem, block -B is nonremovable. Since itis finite, this means that it is tapered.

The case of a nonconvex block is treated at the end of this chapter.

Application of the Removability Theorem in Three DimensionsUsing Stereographic Projection

Recall that the joint pyramids belonging to a given block pyramid ploton the stereographic projection as a series of regions enclosed within portionsof great circles. Given n nonparallel joint planes, there are T possible blockscreated by their intersections. However, when n is greater than 3, not all thesepossible regions appear in the stereographic projection. It will be proved inChapter 5 that the number of regions actually appearing on the stereographicprojection (lf.) is given by

l[r:n(n-l)+Z (4.4)

The stereographic projection can represent lines and planes in space. But itcannot project points in space, except those that just happen to lie on the surfaceof the reference sphere. If a block is finite, its block pyramid is empty. Whenthat block is defined entirely by joint half-spaces, its joint pyramid is empty.An empty joint pyramid is one that has no edge. If it lacks an edge, it cannotbe represented on the stereographic projection. In other words, the only pointin common to the half-spaces defining an empty joint pyramid is the originitself and the origin is absent from every stereographic projection except one ofzero tadius-a trivial case. The regions that are absent from the stereographicprojection are therefore the joint pyramids corresponding to finite blocks. Thenumber (lft) of these blocks defined by n nonparallel joints is therefore

i'rfr:2"-[n(n-1)+2] (4.s)

Consider the system of four joint planes listed in Table 4.3. The stereo-graphic projection of these joints is shown in Fig. 4.8 (a lower focal pointprojection). Since n equals 4, N^ : L4 and N" : 2. Each of the joint pyramidsis identified by a four-digit binary number. Checking them off in turn, it will

TABLE 4.3

Joint PlaneDip, a(dee)


7060

Q20

1

2

3

4

l0il0230330

Theorem on the Removability of a Finite, Convex Block

Figure 4.8 Lower-focal-point stereographic projection of the joints given inTable 4.3.

be confirmed that only 14 regions appear in the projection plane. The two thatare missing are 0001 and 1110. The two finite blocks determined by this systemof joint planes are therefore (J.(J2(J3L4 and LtLzLr(Jo. When either of theseblocks is cut by a free surface to define a new block whose block pyramid is theappropriate joint pyramid (0001 or 1110) plus the free surface, the resultingblock will be tapered and nonremovable from the rock mass.

The significance of finite blocks defined only by joint planes is exploredin Chapter 5. Tapered blocks are important to excavations because they cannot be key blocks. An example is shown in Fig. 4.9, where a circular tunnel hasa finite block in the roof. Since this block is tapered, it cannot fall into thetunnel.

1r2 The Removability of Blocks Chap. 4

Figure 4.9 Tapered block above a tunnel.

SYMMETRY OF BLOCK TYPES

In the preceding examples, you have no doubt noticed a certain symmetry inthe solutions to the removability and finiteness criteria. For example, we deter-mined with respect to Table 4.3 that two blocks were tapered: 0001 and 1110.In the example shown in Fig. 4.6, furthermore, block 100 was finite in the roofabove a given free plane, while block 011 was finite in the floor below the samefree plane. There is, in fact, a general proposition concerning these observations:

Consider the convex block formed of n joint planes and k free surfaces. Theblock whose joint pyramid is tttttg ... tf and whose excavation pyramid isil*i}*z . . . i2** is finite if and only if the block whose joint pyramid is jli|jg . . .

j9 and whose excavation pyramid rb j9* r j|+z .. . il*r is finite. The symbol i iseither I or 0. The symbol j is I if i is 0 or 0 if i is I.

Blocks that are related as in the conditions for the proposition will betermed cousins. Since any tapered block i, iz . . - in+k has an empty joint pyramid,the symmetry proposition establishes that the corresponding symmetricalblock j, j, . - . ja+r has an empty joint pyramid. Thus the cousin of any taperedblock is also tapered.

PROOFS OF THEOREMS AIID FURTHER DISCUSSIOil

Finiteness lheorem. Let there be m planes, including a mixture of jointsand free surfaces. The equations of these planes are

ArX * BrY + Cp: Dr

A".X * Bzr.* "r7:

Dz

::::A^X + B^Y + C^Z: D^

(4.6)

Proofs of Theorems and Further Discussion

Consider the following system of half-spaces:

ArX * BtY + C&) D,

u:* * Bz.Y + C|Z2 Dz(4.1)

A-X + B*Y + C^z> D-

There is no loss in generality in adopting > for all inequalities, since the con-

stants A,, Br C,, and D, can be positive or negative.

A convex block is defined as a region within the intersection of a number

of half-spaces such that a line between any two points in the region is contained

completely within the block. Figure a.10(a) shows a convex block; Fig. 4.10(b)

shows a block that is not convex ("concave'n). In a convex block, all interiorangles are less than or equal to 180'.

FigUre 4.10 Convexity of a block: (a) aconvex block; (b) aconcave (nonconvex)

block.

A half-space is an unbounded convex block. This can be seen intuitively.However, it can be established formally as follows.

Consider two points, 1: (X,,Y,, Zt) and J : (X1,Y1, Z1), bothwithin the half-space:

ArX * BtY + CLZ) D, (4.8)

Any point on a straight line between 1 and J has coordinates given bylinear interpolation as

v,x, -l (1 - x)xi, v,Y, * (1 - l)Y i,llz, * (1 - I)zi (4.9)

where I is a constant varying from 0 to I in moving from I to L If allvalues of I between 0 and 1 satisfy the inequality (4.8), then by definition

the region between .I and J is convex. The entire half-space is convex if.Iand J are permitted to be anywhere in the half-space. Substituting (a.9)

1.

The Removability of Blocks Chap. 4

in the left-hand side of (4.8) gives

LM+(1 -r)r/tuf-Arx,*BrY,*CrZ,N:ArXt*BtYt+CAt

Since by design both I and J are in the half-space, they each satisfy (4.8),SO

MlD, and NlD,,Therefore, if 1is between 0 and 1,

IM + (1 - 1)N> Dl (4.10)

and the block is convex.

The intersection of any two convex blocks is also e convex block. The"intersection" means the part of each block that is common to both.Since any two points in the intersection are in both blocks and each blockis convex, the line between these points is within each block and thereforeis in the intersection. Thus the intersection is convex.

The intersection of any number (m) of half-spaces defines a convex block.Since a half-space is a convex block, by (1), the intersection of any twohalf-spaces is a convex block by (2). Its intersection with a third half-space is then convex, and so on, until all m half-spaces have been inter-sected. According to this lemma, the intersection of all rows of (4.7) is aconvex block.

If the convex block

ArX*BtY+CAlD,Ar.X*Bz.Y+CrZ2D.z

(4.1la)

A^X + B*Y + C^Z> D^

is finite, then the convex block

ArX * BrY * CrZ> O

A"X * BrY + C2Z> O

(4.11b)

A*X + B^Y + C^Z> 0

represents a series of z inequalities in three unknowns with only one root-X: Y:Z:0.Proof of (4). Suppose that there is a point (Xr, Yr, Zr) satisfying (4.lLa)

and another point (Xo, Yo, Z) not equal to (0, 0, 0) that satisfies (4.11b). Then(tXo, tYo, tzo), with t ) 0, also satisfies (4.1lb) because

where

and

2.

3.

4.


tArXo

tAr.Xo

tA-Xo +Since, for i: 7r2,. . . ,m,

^ ^:,;, ;:'l:,?,"; :;,i'l:; :X' :,',1i,' * c,z.l)D,*0

then (X, * tXJ, ( yr + t fo), (Zt * tZ) satisfies (4.11a) for any positive valueof r. But then (4.11a) is not finite. Therefore, (Xo, Yo,Zo)

=a 0 can not satisfy

(4.11b).

5. If the convex block given by the m simultaneous inequalities of equation(4.11b) has only one root, (0, 0,0), then the convex block given by the msimultaneous inequalities of (4.1la) is finite.

Proof of (5). Suppose that the block determined by the simultaneoussolution of the inequalities (4.11a) is infinite; then it has at least one infiniteedge, meaning that

X: Et * Frt, Y: Ez* Fzt, Z: Es* Ftt

with (tr'r , Fr, Fr) ?€ (0, 0, 0)

satisfies (a.11a) for any t> 0. Therefore,for i:1,2,...,ffi,A,(8, * Frt)* B,(E, * Frt) * C,(Er + Frt)) D,

and t(AFt * B,F, + Cf)) D, - (A,E, + Bfiz + CEs)

foranyt>0Then, for all i:1,2,. . . ,m, A,F, * B,Fz + C[3 cannot be ( 0; therefore,(Fr, Fr, Fr) satisfies (4.11b). But we supposed that only (0, 0, 0) satisfies (4.11b).

Therefore, there is a contradiction and (4.11a) is finite.

Blocks, Block Pyramids, and Associated Regions

Assume that there is a block .B formed by n joint planes and k free planes.

The inequalities of the block are

A,X * BrY + C&) D,

u',*

A,X + B,Y + C,Z> D,

*tBr%+tCrZo)0* tBz% + tC2Zo > O

(4.r2a)


An*tX * B,*rY * Co**Z) Dn+t

The system of inequalities (4.12a) are for the half-spaces cut by n joint planes;they define what we will designate as a joint block. The system of inequalities(4.12b) are for the half-spaces cut by excavation planes. Figure 4.11 shows atypical block, in two dimensions, with n :2 and k :2.

Free planei=3

Joint planei=l

Joint planei=2

Figure 4.ll Block created by theFree plane intersection of two joint half-spaces and

i = 4 two free-plane half-spaces.

The system of inequalities determining the block pyramid (BP) of A is

ArX*B,Y+C&>Ou:.**u'..'*':.t=0.

A"X + B"Y + C"Z> O

An*rX I B,*rY I Cn+tZ> O

(a.13a)

:::: (4'13b)

An**X * Bn*Y * C,*rZ> 0

The system of inequalities (4.13a) determines the joint pyramid (JP) of B, whilethe system (4.13b) determinesthe excavation pyramid (EP) of B. The intersectionof the entire system (4.L3) defines the block pyramid (BP).

1. The rock mass is determined by the system of inequalities (4.12b) as shownin Fig. 4.12. The rock mass is then the intersection of the last (ft) half-spaces of (4.12), that is, the common points of these half-spaces.


Free planei=3

\ ,,'\//./\

Rock

Figure 4.12 Rock mass created by theintersection of two free-plane half-spaces.

Free planei=4

2. The free space consists of all points lying outside the rock mass. The free

space is therefore in the union of the half-spaces opposite to those ofinequalities (4.12b) as shown in Fig. 4.13.

Figure 4.13 Nonconvex rock mass

created by the union of two free-planehalf-spaces.

The union of ft half-spaces is the set of points in any one of them.

The free space is the union of all half-spaces i : n + t through n + kof inequalities (4.I2b) with ) replaced by <; that is,

A,X*B,Y*CtZ<Dti:n*l,n+2,...,n+k

(4.r4)

3. The removable spece, Fig. 4.14, is the union of a block (B) and the free

space. It is therefore the union of (a) the intersection of inequalities (4.12),

and (b) the union of the (ft) inequalities of QJa).


Figure 4.14 "Removable space" of a

block.

4. The intersection of all joint half-spaces of a block (the joint block) belongsto its removable space (Fig. 4.15).

1.

2.

Space defined byinequalities

14.12al

Figure 4.15 Joint block belongs to theremovable space of a block.

Removability Theorem

A finite block cut by joint planes and free planes is removable if it can bemoved along a direction without colliding with the adjacent rock mass

[Figure a.16(a)]. A finite block cut by joint planes and free planes is non-removqble (tapered) if it cannot be moved in any direction without collid-ing with the adjacent rock mass [Figure 4.16(b)].

The necessary and sufficient condition for a block (8) to be removableis that BP is empty and JP (a.13a) is not empty; an equivalent statement isthat the block formed by joint planes and free surfaces together (4.12)is finite and the block formed by joint planes alone @.12a) is infinite.


/l

I

i

I

I

I

i

i

I

-l\ xo: direction ofmovement

(a) (bl

Figure 4.16 Test for removability of a finite block under sliding.

Proof of (2)

The necessary condition: If a block is removable, there is a direction(Xo, Yo, Zo) : xo * (0, 0, 0) such that when the block moves along x0 everyjoint i : l, . . ., nwill not tend to close. Let n1 : (At, 8,, C,), i : 1,2, . . .,ft,be the normal to joint r pointing to the interior of the block (Fig. 4.17), that is,

nr. xo > 0 fot i:1,2,...,n (4.1s)

xo: direction ofmovement

Figure 4.17 Removability of a finiteblock

-necessary condition.

119

Adjacent rock mass


ArXo * BrYo * CtZo> O

ArXo * Bzyo + C2Zo> 0

AoX, + B,Yr + CZo) O

(4.16)

Since (XoYoZ) * (O,0,0) is a solution to (4.16), and (4.16) is the joint pyramidof the block, the joint pyramid is not empty.

The sfficient condition' If the JP is not empty, there is a vector Xo :(Xo, Yo,Z) * (0,0,0) such that (Xo, Yo,Zo) is a solution of (4.16). For anypoint (Xr, Yr, 21) of the block defined by (4.L2),

A,Xr * B,Y, * CtZt> Dt,

After moving in direction xo, the coordinates of point (Xr, Yr, Zr) become(X, + Xot, Y, * Yot,Z, * Zot)with t > 0. Inserting this in the left-hand side

of (a.17) gives

^ ^! i, X :'l:,7,'"; : ;:'.:,lf ,' i :,1, * c,z.)) D, for i:7,2,...,n

So the moved point (& + Xot, Y, * Yot, Z, * Zot), t > 0, satisfies inequal-ities (4.12a). From paragraph 4 of the preceding section, the intersection of alljoint half-spaces belongs to the removable space. Thus the moved point does

not lie within the rock mass outside the block. Therefore, the block is removable.

Symmetry Theorem. If the convex block formed by the intersection of

A,X * B,Y * CtZ> D,, i:1,2,...,n * k (4.18)

is finite, the convex block formed by the intersection of

A,X*B,Y*CtZ<Db i: lr2,... rtr * k, (4.19)

is finite.

Proof. Assume fhat block (4.18) is finite; then the intersection of the sys-

tem of inequalitieq_

A,X * B,Y * CIZ> O, i:1,2, " ',n + k, (4'20)

has only the solution (0, 0,0). Therefore, (0, 0,0) is the only solution to the

systemA,X*B,Y*CA<0, i:1,2,...,fl*k, (4.21)

since if (Xo, Yo, Zo) satisfies (4.20), then (- Xo, - Yo, -Zo) must satisfy (4.21).

Therefore, the convex block (4.19) is finite.

Shi's Theorem for the Removability of Nonconvex Blocks

SHI'S THEOREM FOR THE REMOVABILITYoF tlomcoilvEx BLocKs

United Blocks

A united block is a union of convex blocks; the number of convex blocksis flnite. In general, a united block is nonconvex. Figure 4.18 shows a two-dimensional example of a nonconvex eight-sided block.

Figure 4.18 Decomposition of a united block.

The boundary of a united block consists of faces of joint planes or freeplanes.

Assume that. B is a united block.. 0,,i: 1,. ..,fl, is the normal ofjoint face F, of ,Bwhich points toward

the inside of .8.- 8,,i: n* 1,... ,n * k, is thenormal of free face F, of Bwhich points

inside.B.. UB@) is a half-space; UB(t') is bounded by face F, and includes f'.

Let8,: (A,,8,,C,), i: tr...,n+ k

The equation of ^F, is

| ^:,!\ mB

@.

4X + B'Y * C1Z: D' i: lr. , . rn I k


quation of UB(8,) is

A,X * B,Y * C,Z> Dt

Suppose that A is a point of B; we define a convex block B(A) as follows:

The equation of UB(O,) is

B(A): /J uB@) (4.22)

where D: the set of all j, j:1,...,n+k, such that I e UB(0). ^B(,a) isa convex block. Denote by JP(,a) and EP(l) the joint pyramid and excavationpyramid of block B(A), respectively. Then we have

JP(/) : nu@,)where .E : the set of all, j, j : 1, . . .,fl,such that rd e UB(0) and

EP(/): n U@,)

where F:the set of all j, j:n* 1,... ,n* k, such that A e UB(0).

Theorem on Removability of United Blocks. Assume that B is a unitedBlock; and A1, Ar, . . . , A, are the points of B.

Are B, i:Ir...rhsuch that

(a6n;:nThe necessary condition for B to be a finite removable block is

(4.23)

(4.24)

(4.2s)

(4.26)

(4.27)

(4.28)

(4.2e)

(4.30)

(4.31)

and

where

(l) IP+ s(2) EPnJP: @

h

(3) JP:nJP(IJ,::

(4) EP:UEP(r,)

Proof. One can prove an important fact that

B(A) c. B

The proof depends on elemental theory of general topology, and is given byshi (1e82).

The fact that B(A) c .B is apparent from the two-dimensional case dia-grammed in Fig. 4.18. In this example, an eight-sided united block (B) has been

decomposed into three convex subblocks B(At), B(Ar), B(At), each of whichis entirely within .8.

Because .B is finite and B(At) c B, then

B(At), i:t,...,h

Shi's Theorem for the Removability of Nonconvex Blocks

is finite (see Fig. 4.18). Because B(A,), i : l, . . . , h, is finite, then fromtheorem on finiteness,

JP(A;) n EP(,4,): @, i:1,...,h

123

the

(see Fig. 4.I9).

Since

Figure 4.19 JP and EP for each component of the united block.

JP n EP : JP n t0 gp(,1,)lt=l

h: [J EP(,4r) n JPJ=r

h

JP : n JP(lj)

(4.32)

(4.33)

l=L

JPcJP(lr), i:1,...,hthen, from 3,

IEP(,4,) n JPI c [EP(11) n JP(,{ )l : a

From (4.33) and (4.34) we know that

(4.34)

JPnEP:@(see Fig. 4.20).

The above is shown in Fig. 4.20. This figure also shows that the set of alldirections (.f) along which B can move without invading the rock outside of .B


_* \

"/A

%

Figure 4.20 Application of shi's theorem for a united block.

is the set of all vectors of JP. It can be seen that if

i+JP:AJp(A)then there is an l, such that

s + JP(,4r)

and there is a joint face Fo of B(Ar) such that

s # u(60)

So when .B is moving along J, ,B will invade the rock outside of ,B at face Fo.For further detail, see the compatibility theorem described in Shi (1981):

I

II!tr

chaptet 5

Joint Blocks

Joint blocks are rock blocks determined entirely by joint planes, that is, withoutfree surfaces. In other words, joint blocks exist inside a rock mass, behind the

exposed face. A type of rock block that is nonremovable from an excavation

surface because of a tapered shape was discussed in Chapter 4. Delete the

free surface from such a block and the remaining joint planes will determine

a joint block. In this chapter we explore the geometrical properties and numbers

of joint-block types produced by given joint systems.

Joint blocks are the building blocks of a rock mass and must therefore be

linked fundamentally to rock behavior. However, to our knowledge the subject

of blocks within a rock mass has not received any significant, general discussion

in rock mechanics literature, and there are no theories of rock behavior that rely

on an accurate description ofjoint-block shapes. When, in physical or numericalstudies, rock blocks have been depicted as building up a model rock mass, onlycubical or prismatic blocks were stacked. Such blocks do exist in the orthogon-ally jointed, uniformly dipping siltstones of western New York. But more gen-

erally some variation in extent and spacing, and nonorthogonality of jointsproduce variable and nonprismatic block shapes, as in Figs. 5.1 and 5.2(a).

When blasting in a rock mass like that of Fig. 5.2(a), the joint blocks atleast partly determine the sizes and shapes of the debris as idealized in Fig.5.2(b). The design of a blast to achieve a particular shape of excavation mustthen be influence by the shapes, sizes, and arrangements of the in situ jointblocks. The properties of the rubble will depend at least partly on these same

factors. For example, a single rock type may yield tabular blocks [Fig. 5.3(a)]

t. * *'*'*

.S";'{ffitun,ri*-"i

Figure 5.1 Joint blocks: (a) prismatic joint blocks exposed in theexcavation for Upper Stillwater Dam (USBR), IJtah, in Precambrian(b) a section through joint blocks in the Upper Stillwater Damexcavation.

foundationsandstone;foundation

Chap. 5 Joint Blocks

Figure 5.2 Effect of blasting in jointed rock.

(a)

Figure 5.3 Block shapes produced by excavationshapes: (a) a tabular block; (b) a cubical block.

(b)

depend mainly on joint block

I

128 Joint Blocks Chao. 5

or roughly cubical blocks [Fig. 5.3(b)], depending on relative joint spacings andorientations in the undisturbed rock mass.

All rock mass properties that are affected by jointing are probably influ-enced by the shapes of the natural rock blocks in situ. Not only comminutionand excavation properties, but probably wave propagation, permeability, grouttake, and other system properties should depend at least partly on block shape.Although a mathematical discussion of block shapes may be ahead of engineer-ing practice, we expect that applications will emerge as the subject becomesestablished.

The geological data that are required to determine the system of jointblocks are the spacing, orientation, and extent of each joint set. The averagespacing of joints in a set will dictate the average dimensions of the block per-pendicular to these joints. The average extent of joints will dictate the probablesizes of the largest blocks. These quantities are all nondeterministic, and statis-tical distributions are required to describe them. For simplicity, we witl assume,for the time being, that particular values of spacing, extent, and orientation areassignable.

There is an important difference between the joint blocks discussed in thischapter and the key blocks forming the main interest of all the other chapters.Key blocks occur on the surface of an excavation and one or more of their facesare created by the excavation, as shown by blocks KB in Fig. 5.4. The joint

Figure 5.4 Keyblocks (KB) and jointblocks (JB).

blocks (JB) are rock blocks that do not contact the excavation surface. Thisdistinction is more than surficial. It is rare that a key block will be formed withparallel faces, because blocks with parallel faces have greatly restricted move-ment directions and are generally stable until they are undermined; any rough-ness on the parallel surfaces creates rock bridges that must be broken before theblock can move along them. On the other hand, joint blocks will usually haveparallel faces. In fact, joint blocks without parallel faces are the rarity.

Intuition suggests certain principles on the relative occurrences of different

Chap. 5 Joint Blocks 129

sorts of joint blocks. First, blocks with fewer joint sets are more likely to occur

than those involvin g alarger number of joint sets. The latter requires the inter-section of a greater number of planes, and the probability thatn planes intersect

each other varies inversely with r. Figure 5.5(a) depicts a block formed of fournonparallel joints. This block is the intersection of LlLrUr(In and, using the

digital notation introduced in Chapter 4, it is labeled block 1100. We now intro-

duce the digit 2 to signify omission of a particular joint set, so that block 1120

[Fig. 5.5(b)] signifies a block formed of joints 7,2, and 4 without any face fromjoint 3. Similarly, block 1200 [Fig. 5.5(c)] is formed of joints 1,3, and 4 only.Intuition says that blocks 1120, and 1200 will be more numerous than block

1100.

Figure 5.5 Use of digit 2 to describe omission of a particular joint: (a) block

1100; (b) block 1120; (c) block 1200.

Another intuitively derived principle says that a block with opposing faces

formed by two joints of the same set is more likely to occur than a block formed

without parallel joints of the same set. In the case of n nonparallel joints, thejoint pyramid is a spherical polygon. With n - 1 nonparallel joints, and one

joint set repeated to produce a block with one set of parallel faces, the JP is an

arc of a great circle. To create a finite block in an excavated space, a JP must be

contained in that space's SP. In general, it is easier to accommodateo within an

SP, an arc of a great circle than a spherical polygon. Therefore, the block withparallel faces will tend to be more numerous in an excavation. (However, such

blocks are more stable than those lacking parallel faces because the directions

(c)(a)

130 Joint Blocks Chap. 5

of motion are greatly restricted.) If blocks with parallel faces are more numerousin an excavation, they are probably more numerous inside the rock mass as well.

We will use the digit 3 to indicate that a block is formed with both the upperand lower half-space of a given joint sef. Thus block 1120 of Fig. 5.6(a) can betransformed into block 3322by forming two faces from each of joints 1 and 2,and omitting joints 3 and a [Fig. 5.6(b)]. Similarly, block 3323 involves two eachofjoints 1,2, and 4 and has no face ofjoint 3. According to our second intuitiveprinciple, blocks 3323 and 3322 are more likely than block 1120. Combiningthis with the previous principle, we would predict that 3322 is more likely than3323, which is more likely than 1120. If this be so, the joint-block system formedby a determined system ofjoints is relatively regular; however, the block shapeswill not be prismatic unless the joints are mutually orthogonal.

(a) (b) (c)

Figure 5.6 Use of digit 3 to describe doubling of a particular joint: (a) block1120; (b) block 3322; (c) block 3323.

JOIIIT BLOCKS IN TWO DIMETUSIONS

Two-dimensional blocks build planar models of rock mass behavior. Someexamples for discrete element analysis and finite element analysis appear inChapter 1. All real problems are three-dimensional, yet for some a two-dimen-sional approximation proves sufficient. In any case, a two-dimensional discus-sion ofjoint blocks will serve as an informative introduction to the mathematicalrelationships required for three dimensions. We will begin with an analysis ofthe number of block types generated by z nonparallel joints, that is, with nojoints repeated in any block.

Joint Blocke When No Joints Are Repeated

For analysis in two dimensions, a joint is a straight line in a plane P.To examine blocks bounded by a number of such joints we will appty the theoryof Chapter 4 and judge the emptiness of the corresponding joint pyramids.

Joint Blocks in Two Dimensions

Given n joints, we must move each to pass through the origin (0, 0) in plane P.

The n joints thus shifted will then create 2n angles subtended at the origin.Figure 5.7, for example, shows three joint sets passing through (0, 0) that gen-

erate six angles. We may regard these angles as two-dimensional analogs to

Figure 5.7 Nonempty joint pyramids.

joint pyramids. So, in two dimensions, we can draw the joint pyramids directly,without the stereographic projection. According to the theorem of finiteness,the finite blocks correspond to the empty joint pyramids. There are 2" half-plane intersections with n joints. Since there are 2n joint pyramids, there mustbe 2" - 2n empty joint pyramids. By the fi.niteness theorem, then n joints define2n - 2n joint blocks.

In the example of Fig. 5.7, the three joints define six nonempty jointpyramids. Continuing the ordered digital notation of Chapter 4, with 0 and Irepresenting the half-planes above and below a joint, respectively, the nonemptypyramids, and therefore the infinite joint blocks are the three pairs of cousins:000, 111; 100,011; and 001, 110. There are23 -2(3):2 flnite joint blocks.By the process of elimination, the two finite blocks are the cousins 010 and 101.

Representatives of these block types are shown in Fig. 5.8.

Joint Blocks When One Set of Joints ls Repeated

Now consider the possible joint blocks having n + I faces with only njoint sets, that is, with one joint set repeated. Such a block is determined by theintersection of half-planes corresponding to all faces; so it includes both theupper and lower half-planes of the repeated joint. When both of the latter areshifted to pass through the origin, they determine a pah of joint pyramids alongthe line of the repeated joint set as shown in Fig. 5.9. In this example, whichuses the same three joint-set orientations as Fig. 5.7, the repeated joint is set 1.

When the number ofjoint sets forming a block is greater than 1, the num-

131

Joint Blocks Chap. 5

Figure 5.8 Finite cousins: (a) block 010; (b) block 101.

al

--2

Figure 5.9 Joint pyramids with re-1 peated joints.

ber of nonempty intersections is not increased beyond the two rays identifiedby the repeated set because the condition that the repeated half-planes be inter-sected already limits the nonempty regions to these radii and they cannot befurther subdivided by new radii from the origin. So the number of nonemptyjoint pyramids, when n 2 2, is 2.

Suppose that one set out of n is repeated. The remaining n - l sets define2o-1 unique half-space intersections. Any one of these defines a nonempty jointpyramid if and only if it contains one or the other of the rays corresponding tothe intersection of half-planes of the repeated joint set. So there are only twononempty joint pyramids. The number of empty joint pyramids is therefore2n-t - 2. Again using the digit 3 to represent a repeated joint set, let us examinethe example of Fig. 5.9. Considering the intersection of all four half-planes,there are only two nonempty joint pyramids, the cousins 300 and 311. Thenumber of empty joint pyramids is 22 - 2 : 2, which are the cousins 310 and301. Typical blocks corresponding to the latter are shown in Fig. 5.10. It shouldbe appreciated that the shapes of blocks with a repeated joint set now dependon the spacing of the repeated joint set. When no set is repeated, as in Fig. 5.8,three joint orientations uniquely determine the shapes of the finite blocks.

(a)

Joint Blocks in Two Dimensions 133

Figure 5.10 Typical blocks belonging to the finite cousins: (a) 310; (b) 301.

Consider now the different types of joint blocks that could be fashionedfrom n joint sets when only one joint set is repeated. Given that the first joint set

is repeated yields a corresponding 2"-t - 2 joint blocks. Proceeding to makethe second joint set the only repeated one yields another 2"-t - 2 joint blocks.Continuing in this fashion through all possibilities, we must conclude that thereare nQ"-t - 2) individual joint-block types.

Joint Blocks When Two or More Sets of Joints Are Repeated

Let there be n sets of joints of which m are repeated, with m ) 2. Toconstruct joint pyramids, as before, we pass all these joints through (0,0).If set 1 is repeated, all of the nonempty joint pyramids must lie in its line. Butif the second set is repeated, all of the nonempty joint pyramids must lie alongjoint 2 as well, and therefore in the intersection point of the two repeated joints

[Fig. 5.11(a)], which is always point (0,0). Therefore, all the possible jointpyramids are empty. This is true for any additional repeated joint sets. Thenumber of half-space intersections produced by the nonrepeated joints is 2o-*.So the number of finite joint blocks is 2"-*. This many joint blocks are createdby assigning the first z joints as repeated and an equal number is created byselecting any other combinations of z joints as repeated. Therefore, the numberof unique joint-block types when m joints out of n are repeated is 2"-- times thenumber of combinations of n joints taken m at a time (Cf). The total number ofjoint-block types is then (2'-^)nUK" - m)t. m!). For example, suppose thatn : 3 and m: 2, as shown in Fig. 5.11(a). There are then (23-2)3UQl t!) : 6

types ofjoint blocks corresponding to 033, 133, 303, 313, 330, and 331. If n:3and m:3, there are correspondingly 2o(3UQ! 3!)): l, namely block 333,

which is shown in Fig. 5.11(b).

1U Joint Blocks Chap. 5

(a) (b)

Figure 5.11 Multiple, repeated joint sets.

Formulas for Joint Blocks in Two Dimensions

Expressions for the numbers of joint-block types in two dimensions havenow been developed for all possible cases of repeated joints. These are sum-marized in Table 5.1.

It will be noted that all the entries of Table 5.1 are even numbers. Thisis because all block types come in pairs since the theorem of symmetry assuresthat every block has a cousin of the same type formed by changing its 0's into1's, and vice versa. If a joint block is finite, its cousin is finite. The interchangeof 0 and 1 to form cousins does not alter 3's and 2's. For example, block 3L20has cousin 3021.

JOI]UT BLOCKS IN THREE DIMETIISIONS

The types of joint blocks in the three-dimensional world will now be discussedformally. First we consider blocks having no repeated joints as faces and thenextend these results to cases with one or more repeated joint sets. An incrementalanalysis is used in which we determine the number of half-spaces added to theexisting number when an additional joint is allowed to cut through the origin.In this wily, the results of the preceding section can be incorporated.

Joint Blocks When No Joints Are Repeated

In three dimensions, the joint pyramid is a solid angle made by a series ofplanes passing through the origin (0, 0,0). We will accumulate the number ofnonempty joint pyramids formed by a series of joints as they are added one byone to the list.

TABLE 5.1 Numbers of Block Types in Two Dimensions

Number of EmptyNumber of All Combinations Number of Non- Joint Pyramids

of Half-Planes (All Joint empty Joint (Number of FiniteNumber of Repeated Sets Pyramids) Pyramids Joint Blocks) Condition

0 repeated sets

I definite repeated setAny I repeated set

(n choices)rn definite repeated sets

(m> 2)Any m repeated sets

(m2D Qf choices)

2"2,'- |

nTo- |

2n-^

f m.7n-m

2n22n n(z-r - 2) n) 2

2 -2nzn-r - 2

20--

cm.)n-m-n

n)ln)2

nlm)Z

nlm)2

Note:C!: rrnl ,,:@.n ml(n-m)! 1.2.-.m

(.)ol


one joint: A single joint divides the whole space into two joint pyramids.Two joints; Addition of a second joint to the first subdivides each ofthe previous joint pyramids. Thus there are two additional nonemptyjoint pyramids [Figure 5.12(a)], and the total number of joint pyramidsis 4.

Figure 5.12 Nonempty joint pyramids with up to three planes.

. Three joints: Addition of a third joinf to the first two subdivides thepyramids as shown in Fig. 5.12(b). The previous joints cut plane 3 intofour angles so that the addition of plane 3 must provide a total of fouradditional joint pyramids. This may be written as 2(3 - 1). There arenow 8, or 2 * 2(2 - l) + 2(3 - 1), nonempty joint pyramids (Fig.5.13).

Plane 2

Figure 5.13 Nonempty joint pyramids with addition of a fourth prane.

. n joints (n > l).. The nth joint plane is cut into 2(n - 1) angles by theprevious (n - 1) joints. Therefore, the addition of the nth joint to thefirst (n - 1) joints yields 2(n - 1) additional joint pyramids. The totalnumber of nonempty joint pyramids is then

(b)(a)

Plane 1

I

I

I

L__ __

t/\ltl

-_J_7_:_---*---)

Joint Blocks in Three Dimensions

n-l2(i-r):2[r + Er]i=1

:z(1 t n(n- l)\:r, -n-Lz-\- 2 / ','

The number of combinations of half-spaces is 2". Since there are nz - n

f 2 nonempty joint pyramids, the number of empty joint pyramids is

2n - (n' - n t 2).This is then the number of joint block types withn joint sets and n joint-block faces.

Joint Blocks with n Sets of Joints and One or MoreRepeated Joint Sets

One repeated joint set. Assume that the first joint set is repeated.Then any nonempty joint pyramid is simultaneously in the upper and lowerhalf-space of joint plane 1. Each nonempty joint pyramid is therefore an angle

subtended at the origin and measured in plane I between two limiting jointplanes of the other sets (Fig. 5.14).

Figure 5.14 Nonempty joint pyramids with one repeated joint set.

If there are njoint sets, with n ) 2, the n - 1 planes 2 through n will cutplane 1 into 2(n - 1) angles (as noted in the preceding section). Therefore, thenumber of nonempty joint pyramids is 2(n - l). The total combinations of the(n - 1) joint planes of sets 2to n is 2(n-1). Therefore, the total number of emptyjoint pyramids is 2h-') - 2(n - 1).

Considering each set in turn as the repeated joint set, there will be n timesthis many types of empty joint pyramids. The total number of types of finitejoint blocks is therefore equal to nl2r'-t) - 2(n - l)1.

Two repeated joint sets. If there are more than two joint sets

(n ) 3) and both sets 1 and 2 are repeated, any nonempty joint pyramid must

lie simultaneously in the planes of joint set 1 and joint set 2 and therefore alongtheir line of intersection. Plane 3, which passes through the origin, divides thisintersection line into two rays from the origin (Fig. 5.15). Therefore, there are

two nonempty joint pyramids. Planes 1 and 2 are repeated but planes 3 to n

n?_L\-| .L/


Line of intersectionofl and2

Figure 5.15 Nonempty joint pyramidswith two repeated joint sets.

are not. Since we can choose either half-space of the latter (n - 2) planes, thetotal number of combinations of half-spaces in this case is 2b-D. Accordingly,the number of empty joint pyramids, and therefore the number of finite jointblocks, is 2h-2) - 2. This was the result upon choosing joints 1 and 2 as therepeated sets. Considering all possible combinations of n joints of which 2 arerepeated gives as the total number of empty joint pyramids

c!1}r"-zt - 21 :n(n: l)r2h-2, - 2l

2

or finally, n(n - 1)[2<'-st - 1].

More than two repeated joint sets. If the number of repeatedsets is m, with m) 3, each nonempty joint pyramid must contain the line ofintersections of the repeated joint sets. This means that only the origin can becontained in such joint pyramids. Therefore, all joint pyramids are empty inthis case. The number of combinations of the upper and lower half-spaces ofthe (m + 1) to n nonrepeated joints is )tn-n). This is the number of empty jointpyramids, and therefore the number of joint blocks when the first m joint setsare repeated. Considering all possible choices of m repeated sets from n jointsets yields the total number of joint blocks as Ci.)("-m).

Summary of Formulas for Joint Blocks in Three Dimensions

Table 5.2 summarizes the numbers of joint pyramids for the qases con-sidered. In the following section we use this table to identify all the joint pyramidson the stereographic projection, corresponding to a selected example.

STEREOGRAPHIC PROJECTIOIU SOTUTION FOR JOITIT BLOCKS

Using the formulas in Table 5.2, we are now in a position to explore a three-dimensional example. These are based on the sy$tem of four joint sets (n - 4)shown in Table 5.3. We will discuss all the joint pyramids in every case ofTable 5.2.

TABLE 5.2

Number of Repeated Sets

Number of All Combinations Number of Non- Number of Jointof Half-Spaces (All Joint empty Joint Pyramids (Joint

Pyramids) Pyramids Blocks) Condition

0 repeated setI selected repeated setAny I repeated set2 selected repeated setsAny 2 repeated sets

m selected repeated sets(rn > 3)

Any m (m> 3) repeatedsets

2n

2r-rn2-r2-zn(n - l)2-t2n-^

?m)n-m

n2-n*2 2-(rf -n*2) n>l2(n - l\ zo-t - 2@ - t) n)ZZn(n - l) nf2n-1 - 2@ - l)l n) Z7

n(n - l)0

0

2n-^

C!2-^

2-z -2 n)jn(n - lX2"-3 - 1) n) 3

n) mi_3

n) m ;"-3

gt(D

Joint Blocks Chap, 5

TABLE 5.3

Dip, a(des)

Dip Direction, p(deg)

Finite Joint Blocks with No Repeated Joint Sets

1. First compute the radius and center of projection of the circles of each

of the four sets. They are shown in Fig. 5.16(a).

2. Number each region delimited by circles. For example, region 1010 is thespherical surface outside of circle 1, inside circle 2, outside circle 3, andinside circle 4. Since we have used a lower-focal-point projection, region1010 corresponds to the intersection of the lower half-space of set 1, theupper half-space of set 2,the lower half-space of set 3, and the upper half-space of set 4.

3. Identify in order all the combinations of upper and lower half-spaces ofthe four joint planes and indicate by the letters JP the fact that a particularjoint pyramid has been located on the stereographic projection. This yields

the following binary list:

0000 JP 0110 JP 1100 JP0001 JP 0111 JP 1101 -0010 - 1000 JP 1110 JP0011 JP 1001 JP 1111 JP0100 JP 1010 JP0101 JP 1011 JP

The number of combinations of half-spaces for the four joints is 2a : 16. The

number of joint pyramids (JP) is given by Table 5.2 as nz - n + 2: 14 andthe number of empty joint pyramids is 16 - L4: 2. Considering the binarylist developed above, the empty joint pyramids are 0010 and 1101. Blockscorresponding to these are shown for equal spacing between joints of each set

in Fig. 5.16(b) and (c).

Finite Joint Blocks with One Repeated Joint Set

1. Compute the radius and center of each projection circle and draw thefour circles for Table 5.3.

2. Label each segment of each circle. The digit 3 in the place ofjoint i means

that the segment belongs to circle i. For example, 1030 identifies the cir-cular arc segment along circle 3 that is outside circle 1, inside circle 2,

and inside circle 4. This has been done in Fig. 5.17(a).

Set

I23

4

8033030

270

7565

4010

Referencecircle

Figure 5.16 Analysis of joint blocks with no repeated sets: (a) joint pyramidswith no repeated joint set; (b) block corresponding to empty joint pyramid 0010;(c) block corresponding to empty JP ll0l.


(b)

Figure 5.17 Analysis of joint blocks with one repeated set: (a) JPs with one

repeated joint set; (b) block corresponding to empty JP 3010; (c) block corre-

sponding to empty JP 3101.

-:.14 --. \ \

Stereographic Projection Solution for Joint Blocks 143

3. Identify in order all the combinations of upper and lower half-spaceswith a given repeated joint set and indicate by the letters JP the fact thata particular segment has been located on the stereographic projection.For example, assume that set I is repeated. The number of combinationsof half-spaces stated in Table 5.2 is ){n-1) :23 :8. The list of theseeight combinations is as follows:

3000 JP3001 JP3010 -3011 JP

3100 JP3101 -3110 JP3111 JP

According to Table 5.2, the number of nonempty joint pyramids with set 1

selected as repeated is 2(n - 1) : 6. Therefore, the number of empty jointpyramids is 8 - 6: 2. The empty joint pyramids can be identified from thelist as 3010 and 3101. Blocks corresponding to these have been drawn for thecondition of equal joint spacing in each set in Fig. 5.17(b) and (c).

The preceding analysis was based on the repeated joint being from set L.

Now assume that set 2 is repeated. Following the same procedure, inspection ofFig. 5.16(a) yields the list of half-space combinations shown in the third columnof Table 5.4. From this table, the empty joint pyramids are identified as 0310

and 1301. Pursuing a similar analysis with joint set 3 as the only repeated set

produces the fourth column of Table 5.4, determining the empty joint pyramidsas 0030 and 1131. Finally, a choice of joint 4 as the repeated set produces thelist of joint pyramids in the fifth column of Table 5.4 and identifies the emptyjoint pyramids as 0013 and 1103. All of these empty joint pyramids representfinite joint blocks.

Finite Blocks with Two Repeated Joint Sets

1. Again prepare a stereographic projection for the four joint sets of Table5.3.

2. Now identify the joint pyramid number corresponding to each intersectionpoint. For example, 3031 identifies the intersection of circles 1 and 3; thisdirection is inside circle 2 and outside circle 4. Similarly, 0331 means theintersection point of circle 2 and circle 3, which is inside circle L andoutside circle 4.

3. Identify in order all combinations of half-spaces with two planes repeatedand note which ones appear as intersection points on the stereographicprojection. These are nonempty joint pyramids. All the others are emptyjoint pyramids, and therefore finite joint blocks. A tally of this procedureis given in Table 5.5. There are two empty joint pyramids for each specificchoice of repeated joint sets and six such cases (C? : 6); so there arc 12joint block types with two faces repeated. The two finite blocks 3301 and3310 for the case of joint sets 1 and 2 repeated are shown in Fig. 5.18(b)and (c). These are drawn assuming all joints have equal spacing.

D5

TABLE 5.4 Empty and Nonempty Joint Pyramids with No or One Repeated Joint Sets for the Joint System of Table 5.3r

No Joints Repeated Joint Set I Repeated Joint Set 2 Repeated Joint Set 3 Repeated Joint Set 4 Repeated

000000010010 empty001101000101011001ll10001001

10101011

il001101 empty11101111

300030013010 empty301131003101 empty31103111

030003010310 empty03r113001301 emptyl3l0l3l I

0030 empty0031

0130013110301031

ll30l13l empty

00030013 empty010301131003t0t31103 empty1113

rAIl the empty joint pyramids are labeled "empty." The other joint pyramids can be found on the stereographic projection.

Stereographic Projection Solution for Joint Blocks

Figure 5.18 Analysis of joint blocks with two repeated sets: (a) JPs with tworepeated joint sets; (b) block corresponding to empty JP 3301; (c) block corre-sponding to empty JP 3310.

145

tt., 3 1

I

TABLE 5.5

Joint Blocks Ghap. 5

Empty and Nonempty Joint Pyramids with Two RepeatedJoint Sets for the Joint System of Table 5.3

Repeated Joint Sets

land2 land3 land4 2and3 2and4 3and4

33003301 empty33103311 empty

3030 empty30313130 empty3l 31

3003 0330 empty3013 empty 03313103 empty 13303113 1331 empty

0303 0033 empty0313 empty 01331303 empty 1033

1313 1133 empty

Finito Joint Blocks with Three or More Repeated Joint Sets

All combinations of half-spaces with three or more joint sets repeatedform empty joint pyramids. Table 5.6 lists all such combinations. Figure 5.19shows the finite blocks 3330 and 331.

TABLE 5.6 Empty Joint Pyramids with Three RepeatedJoint Sets for the Joint System of Table 5.3

Repeated Joint Sets Empty Joint Pyramids

2,3,41,3,41,2,41,2,3

0333,13333033,31333303,33133330,3331

Figure 5.19 Finite blocks corresponding to: (a) 3330; (b) 3331.

If all four joint sets are repeated (* :4), the only half-space intersectionis 3333. This combination forms an empty joint pyramid. Figure 5.20 shows thefinite block 3333.

tr, --- 1\ --

\. --4\. '-_\_ -_\\\.l--

Computation of Emptiness of Joint Pyramids Using Vectors

Figure 5.20 Finite block 3333.

Symmetry of Joint Pyramids

In all cases all the combinations of half-space intersections can be dividedinto symmetric pairs-for example: 0100 and 1011; 0030 and 1131; and 0133

and 1033. If a combination of digits defines a nonempty joint pyramid, changing

the number of the joint pyramid by interchanging 0's and 1's produces anothernonempty joint pyramid. For example, 0000 and 111L are nonempty; so are

3110 and 3001, and 0303 and 1313. Since all combinations of joint half-spaces

can be divided into symmetric pairs, the empty joint pyramids must also possess

the same symmetry. For example, 0010 and 1101 are both empty joint pyramids.

So are 0310 and 1301;0313 and 1303; and 3303 and 3313. Thejoint blockscorresponding to these joint pyramids are cousins, related to each other by acenter of symmetry. [Compare: Figs. 5.16(b) and (c); 5.17(b) and (c); 5.18(b)

and (c); and 5.19(a) and (b).1

COMPUTATION OF EMPTINESS OF JOINT PYFAMIDSUSIIIG VEGTORS

Consider the four sets of joints of Table 5.3. The corresponding upward normalvectors are shown in Table 5.7.

TABLE 5.7

Joint Set

I2

3

4

0.95r2

-0.45310.3213

-0.1736

o.t6770.78480.5566

0

0.2588o.4226o.76ffi0.9848


Computation of Joint Pyramids with No Repeated Joint Sets

Consider joint pyramid 0100. The coordinate inequalities of 0100 are

0.9st2x + 0.1677 v + 0.25882 >00.4s3rx - 0.7848 Y - 0.42262 > 0

0.3213X + 0.5566 Y + 0.76602 > 0

-0.1736X + 0r+ 0.98482> 0

(5.1)

We will compute the edges of joint pyramid 0100.First all the intersection lines of these four planes are calculated and listed

in Table 5.8. For each line of Table 5.8, (X, Y, Z) and (- X, - Y, -Z) are two

TABLE 5.8 Lines of Intersection of the Joint Planes

Intersected Planes

1,2113lr42r3214314

-o.13470.01940.1658

-o.46410.8895

-0.7661

-0.52890.8049

-0.e857--0.6125

0.42910.6282

0.8378

-0.59300.02920.63980.1568

-0.1350

opposite vectors in the corresponding intersection line. Substitute each (X, Y,Z)and (- X, - Y, -Z) into inequalities (5.1). Those vectors that satisfy inequalities(5.1) are the edges of the joint pyramid 0100. The edges of joint pyramid 0100

are

Lines of Intersection Sign of (X, Y, Z)

1,21,32143,4

i+

If there is no vector satisfying (5.1), the joint pyramid is empty. Consider jointpyramid 0010; its inequalities are

0.95t2X + 0.1677Y + 0.2s882>0

-0.4531X + 0.7848 Y + 0.42262 > 0

-0.3213 X - 0.s566 Y - 0J6602 > 0

-0.r736X + or+ 4.98482> o

(s.2)

For each line of Table 1, no vector (X, Y, Z) or (- X, - Y, -Z) satisfies all ofinequalities (5.2); therefore, 0100 is an empty joint pyramid.


Gomputation of Joint Pyramids with One Repeated Joint Set

Consider the joint pyramid 1310. The coordinate equations of 1310 are

-0.9512X - 0.L677 r - 0.25882 > 0

-0.4531X + 0.7848 Y + 0.42262 : O

-0.32r3X- 0.5566 y _ 0.76602>O (5'3)

-0.r736X + 0r+ 0.98482> 0

Joint pyramid 1310 is a sector of plane 2. Therefore, the edges of 1301 are

intersection vectors of plane 2 with other planes, and in Table 5.7 we need toconsider only the intersection lines 12,23, and 24. Substitute (X, Y, Z) and(- X, - Y, -Z) of intersection lines 12,23, and 24 into equations (5.3). Vectors(X, Y,Z) of 23 and (-X,-Y, -Z) of 24 satisfy equations (5.3). These twovectors are therefore the edges of joint pyramid 1310.

Using the same method, we will find that joint pyramid 3010 is empty.

Gomputation of a Joint Pyramid with Two Repeated Joint Sets

Consider the joint pyramid 3031. The coordinate inequalities of 3031 are

O.95l2X + 0.1677 r + 0.25882 : O

-0.4s3rx + 0.7848 Y + 0.42262 > 0

o.32r3x+0.5566 Y+0-76602:o (5'4)

0.r736X + OY - 0.98482> 0

Since joint pyramid 3031 is a ray of intersection line 13, we only need to test

the two opposite vectors of intersection line 13 of Table 5.1:

(X, Y, Z) : (0.0194, 0.8049, -0.5930)(-X, -Y, -Z): (-0.0194, -0.8049,0.5930)

The second vector satisfies (5.1), so joint pyramid 3031 is the ray determined byvector

(-0.0194, -0.8049, 0.5930)

Using the same method, we will find that joint pyramid 3030 is empty.

We do not need to compute joint pyramids with three or more repeated jointsets since all such joint pyramids are empty.

APPLICATIONS OF BTOCK THEOBY: AN EXAMPLE

As noted in the introduction to this chapter, joint-block theory is new and thereare no examples of its actual application. In this section we demonstrate the

steps in a hypothetical calculation and suggest potential applications relevantto energy requirements for rock comminution.


For the example problem we will reuse the data of Table 5.3 with fourjoint sets (n - 4). If each joint set has close spacing and long continuation(extent), there will be blocks in the rock mass corresponding to all the emptyjoint pyramids. More often, however, one or more joint sets has limited extentand all the sets are spaced differently. Often one set, related to bedding orfoliation, has close spacing and long extent, while at least one of the othersets has wide spacing and limited extent. In such cases, some of the empty jointpyramids will lack corresponding joint blocks.

Joint Spacing

A statistical evaluation procedure for joint spacings was presented inpapers by Priest and Hudson (1976, 1981) and Hudson and Priest (1979) forwhich data were obtained by "scan-line surveys." For a scan-line survey, areference line is laid across an enlarged photograph of an outcrop; one thenmeasures the frequency at which joints cross the reference line. Priest andHudson concluded that most joint-frequency data are fit by the negative expo-nential distribution. Using this distribution as a model, and obtaining data fromat least n nonparallel scan lines or drill holes, it is possible to make a formalanalysis of a system of n joint sets to characterize the spacing distributionparameters of every set.

Whether or not a scan-line survey is made, joint spacing values can bedetermined directly from outcrops or excavations. It is possible to calculatethe true joint spacing from the apparent spacing measured on a rock face. Onehas to know the dip and dip direction of both the joint set and the plane ofobservation. The true joint spacing is measured in the direction of the normalto the joint set. The apparent spacing is the separation of parallel joint tracesin a plane that does not contain the normal to the joint planes. The procedurefor calculating the true spacing from the apparent spacing and planar orienta-tions is based on Fig. 5.2I, a section perpendicular to the line of intersection ofa joint plane (1) and a rock face (2). The true joint spacing is ft and the apparentspacing of plane 1 measured in plane 2 is d. The normal vectors of planes 1 and2 are

Dr : (sin a, sin pr, sin a1 cos p,, cos d1)

[2 : (sin a, sin fr, sin c, cos p2, cos ar)

where & are B the dip and dip direction of the respective planes.The angle between n, and n, is 6, calculated as follows:

cosd: nl . n2: sina, sinarcos(8, - f") * cosdl cosd2

The apparent spacing d is measured normal to I,2, the line of intersectionof the two planes. The true spacing can therefore be seen by rotating f ,, aboutn, (Fig. 5.21), from which

d:J-=:L-- sin d Jl - cos2 d

Therefore,

Joint, plane I

Applications of Block Theory: An Example

Figure 5.21 Section perpendicular to the intersection of a joint plane and arock face.

d: n =* ^'=, (5.5)4 | -Itloai sind2cos(Br - fr) * cos il"rcosa,zlz

or h:d (5.6)

Equation (5.6) can be used to calculate the true spacing from joint traces

seen in a rock face of known orientation. Equation (5.5) can be used to calculate

the separation of joint traces in a given section through a rock mass when thejoint spacing is already known. We will illustrate both procedures in a hypo-

thetical example.Table 5.9 lists the dips and dip directions of the four sets of joints previ-

TABLE 5.9

151

Plane

Dip,u

(dee)

DipDirection,

fl(dee)

ApparentSpacing (d)in Plane 5

(m)

CalculatedSpacing,

h(m)

TraceLength inPlane 5

(m)

I)3

45

75

65

Q10

60

80

33030

27050

2.8l05.53.5

t.469.382.33

3.24

Very long6.59.57.8

162 Joint Blocks Chap, 5

ously analyzed, together with the dip and dip direction of a section plane,denoted as plane 5. Plane 5 may represent a rock outcrop, the wall, or an ex-ploratory excavation, or the roof or wall of an actual underground gallery.On plane 5 we proceed to measure the spacings and trace lengths of the jointsof each set.

The true spacing values (ft), corresponding to the apparent spacings andorientation data, were calculated from equation (5.6) usinE crz : 60o and f z :50". From these spacings, we can now draw the various joint blocks generatedwith different numbers of repeated joint faces. Knowing the true spacing valuesof the different joint sets, we can also calculate the apparent spacing values inany other section plane.

Gonstructing a Trace Map in Any Section Plane

The first step in constructing a trace ffi&p, for a generally inclined sectionplane, is to plot the stereographic projection. This has been done in Fig. 5.22.

Otll ,/

Referencecircle

t\\\

II

I/

,b

\\.1000

a\\-d'

4

i.1

1 1ro tor rdl

Figure 5.2| Lower-focal-point stereographic projection of data in Table 5.9.


The section plane (5) being inclined, it will not be obvious which side of a jointtrace corresponds to the upper half-plane (U) and which to the lower (I) when

seen in the section. However, this is readily established by the following proce-

dure. On Fig. 5.22, points a and b mark the intersections of the section, plane

(5), and the reference circle. They therefore represent the two opposite horizontalvectors in the plane of the section. In Fig. 5.22 d is the projection of vector 4the dip vector of plane 5; we determine in which half-space of each joint plane

the point d lies. The results of this step are given in Table 5.10.

TABLE 5.10

Plane(Great Circle)

Location of Point dwith Reference to the

Great Circle

Half-SpaceCorresponding

tod

Half-Space Corre-sponding to the Lower

Side of the JointTrace in the Section

I23

4

InsideOutsideOutsideOutside

ULLL

ULLL

In the section plane, the direction corresponding to d is now identified as

either L or U (Table 5.10, column 3). Next we must determine the apparent dipof each plane in the section and draw the trace orientations in the plane of the

section (Fig. 5.23). The direction of / was determined to be in the upper half-

space of plane 1. Therefore, the lower side of trace 1 corresponds to the upper

half-space of plane 1. In like manner we find that the lower side of joint traces

2,3, and 4 all correspond to the lower half-spaces of these planes.*

Knowing the apparent spacings and the half-spaces of all the joint traces,

we can now draw a section in plane 5 and locate all the finite joint blocks. Asection has been drawn in Fig. 5.23, making use of the added information onjoint extent hypothesized in Table 5.9. Plane L corresponds to bedding and

therefore set 1 is assumed to have almost infinite extent. The other joints have

finite extent. Obtaining such data is not as simple as obtaining joint-spacing data,

and there is a statistical problem with bias and truncation, as discussed by

Priest and Hudson (1981), Baecher and Lanney (1978), and Warburton (1980).

Since there is freedom in placing the joints of finite extent within the sec-

tion, Fig. 5.23 is only one possible solution from an. infinite population. It may

be considered a generic trace map.In order to discuss the joint blocks cut by the section, we have assigned

a number to every closed polygon of Fig. 5.23. In Table 5.11 the joint pyramids

corresponding to each of these polygons are named. To determined the name ofa joint pyramid, we made use of the labeling of half-spaces carried out previ-

*An alternative method of identifying (/ and.L in the trace map is presented in the last

section of Chapter 6.

(n$

72.1"

),,

\68i1" r\

Figure 5.23 Section in plane 5, looking toward the southwest.

Applications of Block Theory: An Example

TABLE 5.f l Joint Pyramids and Joint Blocks fntersected

bY the Section (Fig. 5.23)

155

Polygon Numberon Fig. 5.23

JointPyramid

Probable JointBlock

Finite orInfinite?

I23

45

6

7

8

9l0lll2l3l4l5l6t7l8l9202l22232425262728293031

3233

343536

37

0201I l0llt233Z2l1t0l00100023000132123223320132lO3223tt23322332233223oo2332013210I I0l0003rzlo02013210ll0l0001t 1l00010320132233ZO3

121032013210fi230023

32013t0l310332233r0130r 3

30033001321 1

3r2330013013302331233203312332133013300132t33103

30033l 13

30013210310330033113301332013223300331103ZO3

32133l l3301 3

InfiniteFiniteFiniteInfiniteFiniteFiniteInfiniteInfiniteInfiniteInfiniteInfiniteFiniteInfinitelnfiniteInfiniteInfiniteInfiniteFiniteInfiniteInfiniteFiniteInfiniteInfiniteInfiniteInfiniteFiniteInfiniteInfiniteFiniteInfiniteInfiniteInfiniteInfiniteInfiniteInfiniteInfiniteFinite

ously. Thus polygon 4, involving only pairs of traces of planes 1 and 4, islabeied as 3223; polygon 20, created by a pair of planes of set 1, the lower side

of trace 3 and the upper side of trace 4, has been labeled 3210.

Up to now we have ignored the possibility that a joint whose trace is out-

side the polygon might truncate the underside of a joint block, behind the

section. This is certainly expectable in the case of bedding (joint set 1). Therefore,

every joint pyramid should have its first digit replaced by 3, meaning that every


joint block lies between two bedding planes. Close inspection of every polygonwith respect to proximate, nonintersecting traces will suggest other modificationsto recognize that some other joint will slice that block behind the section. Forthis analysis, it is necessary to apply judgment, assisted by drawings of thedifferent joint blocks as seen looking at the section. Polygon 7 was labeledinitially as 0023. Recognition of the likelihood that bedding will qeatetwo facesof this block requires changing the label to 3023. Digit 2 means that joint plane3 does not bound polygon 7 in the section. However, the trace map shows thata joint plane of set 3 lies just below polygon 7. Thus the proper joint block isnot 3023 but 3003. In this w&y, by methodical study of all the polygons, thejoint blocks corresponding to the joint pyramids are found to be those listed inthe third column of Table 5.11. Comparison of the joint-block numbers withprevious results of this chapter (Tables 5.4 and 5.5) determines whether theblock is finite or infinite.

The Ratio of Finite to Infinite Blocks as a Rock Mass Index

From Table 5.11 the section parallel to plane 5 intersection 37 completepolygons of which 11 were finite and26 infinite. This rock mass thenhas rela-tively few finite blocks and the rock mass is semicontinuous in nature. Progres-sive collapse of tunnels or slopes following loss of a key block is less likely thanin a fully discontinuous rock mass, with completely finite joint blocks insidethe rock. From the point of rock communution, the powder factor (i.e., theconsumption of explosives per unit volume of rock) should be rather large.

chapter 6

Block Thooryfor SurficialExcavations

BASIG CONCEPTS

In this chapter we explore the application of block theory to engineering for

surficial excavations. Cuts into rock, for various purposes, range from small rock

faces to those whose value rivals that of impressive concrete structures' Rock

excavations provide space for buildings, factories, and powerhouses [Fig' 6'1(a)]

and for routes of pipeiines, canals, railroads, and roads tFig' 6'l(b)l' Excavations

on the sides of hills are made for foundations of bridges [Fig. 6.1(c)J' for spill-

ways, and for abutments of dams [Fig. 6.1(d)]. In the latter works, the engi-

neering designer must address not only the stability of rock masses adjacent to

a sloping free surface, but the influences of added forces and the effects imposed

by the u-.tion of the structure and the percolation of water through fractures in

the rock.Rock excavations are also made to gain access to undergtound openings,

as in Fig. 6.1(e). Stability problems are common in portals to the underground

because the intersections of underground and surface excavations create addi-

tional freedoms for movement of iock blocks. The portal design problem will

be discussed in Chapter 8 in connection with tunnels. Here we concentrate on

movement of blockr into excavations that are entirely at the ground surface'

Failure Modes

Rock slopes present an infinity of failure modes' But on close analysis'

these derive from mixtures of a few fundamental modes : sliding of a block along

one face ; sliding of a block along two faces, parallel to their line of intersection;

167

158 Block Theory for Surficial Excavations Chap. 6

(e)

Figure6.l Rock slopes: (a) for a powerhouse; (b) for a highway; (c) as abridge foundation; (d) as a dam abutment; (e) as the portal of a tunnel.

rotation ofa block about an edge; and fracture ofthe rock because ofshear orbending stresses. Figure 6.2(a) shows a typical wedge failure in which a singleblock slides along two joint planes simultaneously, advancing in the directionof the line of intersection of the two surfaces. The direction of movement mustbe parallel to the line of intersection of the sliding surfaces, because this is theonly direction in space that is shared by both surfaces. In many rock masses,the surfaces of sliding create multiple, parallel lines of intersection, formingedges of a series of similar wedges. These may slip in progression, or simultane-ously in a complexly shaped body. Figure 6.2(b) shows a rock slope after theremoval of a series of geometrically similar wedges.

Tunnel

Basic Concepts

Figure 6.2 Wedge sliding: (a) a single

united wedge.

(b) a slope after removal of a

A second fundamental mode of failure is sliding along a single rock face,

either planar, as in Fig. 6.3(a), or curved as in Fig. 6.3(b) and (c). Although only

one orientation of sliding surface is involved, the actual block in motion may

glide along different planes with that orientation, as in Fig. 6.3(d).

Another fundamental failure mode for rock slopes is block rotation. An

\ \\\

(c)

Figure 6.3 Single-face sliding: (a) on a planar face; (b) on a convex face; (c) ona con@ve fa@.

160

Basic Concepts

Figure 6.3 (Continued) (d) A united block. ld)

example of a slope failure involving mainly block rotation is "toppling failure,"

depicted in Fig. 6.4. Failure of slender columns of rock occurs because each

column receives an overturning moment about its base. In Fig. 6.4, this arises

naturally under gravity alone because the layers incline into the hillside. How-ever, columns with other orientations can overturn in response to ice or water

161

Figure 6.4 Sliding and toppling of multiple blocks.

162 Block Theory for Surficial Excavations Chap. G

forces, or forces transmitted from adjacent rock masses. For example, topplingof steeply dipping layers is often induced by downslope sliding of soil layersacross the bedrock surface.

Excavations at the surface also create mechanisms for new fractures togrow in rock masses. This may occur by virtue of the stress concentrations atthe boundaries of the excavation, or in response to bending stresses accom-panying movements of rock blocks. In Fig. 6.4, the bending of the topplingcolumns was facilitated by opening of previously existing cross joints. [n columnslacking cross joints, the bending leads to flexural cracks. Figure 6.5 shows ablock that has cracked in bending after initial sliding along an edge.

Figure 6.5 Beam bending.

Actual rock slope failures often encompass combinations of the funda-mental modes and may involve some new fracturing as well. For example, theslope in Fig. 6.6 has failed by coupled movements of four blocks, each withdistinct mechanisms. In this case it is quite likely that all movement would havebeen curtailed had block I been restrained. Complex failure modes like this tendto develop progressively, and can be prevented by retention of a key block. Inthis work, we assume that there is a set of key blocks and that more complexfailure modes that may progress from their movement need not be considered aslong as the key block is kept in place. It will be the reader's responsibility tojudge the validity of this assumption in any practical case. It cannot be deniedthat some failures could not have been prevented by a simple key-block approach.The list of exceptions is shortened, however, when modes nessitatfig newfracturing or shearing through intact material are invalidated as candidates forkey-block analysis.

Key-block Analysis

Potential key blocks of an excavation differ in important respects fromfinite joint blocks within the body of a rock mass. Key blocks, such as (l) ofFig. 6.7, possess at least one face belonging to the excavation surface (i.e., a

Basic Concepts

6

163

Figure 6.6 Progressive slope failure.

Figure 6.7 Types of blocks in a slope:(l) key block; (2) removable block withparallel faces; (3) joint block.

164 Block Theory for Surficial Excavations Chap, 6

"free surface"), whereas joint blocks, such as (3) of Fig. 6.7, do not. If a jointblock is cut through by an excavation surface, it tends to produce either a taperedblock that is not removable, or a removable block having one or more pairs ofparallel faces, such as (2) of Fig. 6.7. A block with parallel sides offers fewerallowable sliding directions than a block lacking parallel sides and thereforeproves more stable against sliding. Consequently, joint blocks do not yield keyblocks of excavations. For this reason, the methods to be developed in thischapter are different from those of Chapter 5.

The analysis for locating key blocks of a surface excavation is initiallygeometric, the first input being the dips and dip directions of each set of jointsand of each planar segment of the excavation surface. The first stage output isthen a list of half-space codes (e.g., 0110). As defined in Chapter 4, each blockis identified by a code (Dr) comprised of a string of numbers 0, l, 2, or 3, descrip-tive of each joint set in turn. The ith digit in D, is 0 if the block is formed by anintersection with U, (the upper half-space of plane t); it is 1 if the block isformed by an intersection with I,. The ;th digit of D" is 2 if neither (1, nor L,is intersected with the other half-spaces (i.e., ifjoint i is not a face of the block);and the ith digit is 3 if both lJ, and L, are intersected with the other half-spacesof the block (i.e., if the block has parallel faces of set i).

After identifying the block codes D, for potential key blocks, subsequentanalysis can identify the most critical key blocks. The spacing of eachjointsetis input ind the volumes, and shapes are computed for all convex blocks cor-responding to the potential key-block codes. If the actual locations of the tracesof joints on the excavation surface are known, the key-block analysis can thendetermine the actuallocations of key blocks on the surface. At this step ofoutput,the key blocks will include some with nonconvex shapes formed by the union ofconvex key blocks.

In this procedure, the input can be specified deterministically, but in somecases statistical input would be more natural. At the initial stage, for example,the joint-set orientations might be specified in terms of parameters of a spatialdistribution of planes about a mean joint-set orientation. For deterministicanalysis, only the mean orientation, or an extreme value selected from thedistribution, would be entered. In subsequent study, the spacings might bedescribed by an appropriate distribution (e.g., log-normal or negative exponen-tial) in which case the volumes and shapes of key blocks would be stated prob-abilistically. A distribution of joint extent could also be introduced to calculatemore accurately the probability of any key-block type having a volume largerthan a given size. Finally, simulation procedures can be used to generatehypothetical trace maps on the excavation surface. Analysis of key blocks in thegenerated trace maps, together with Monte Carlo repetition, can then generateprobabilistic support procedures for use in design. A program for statisticalsimulation of trace maps and subsequent key-block analysis was reported byChan and Goodman, (1983).

Computer programs such as those available from the authors (see thefootnote in the Preface), can be used to calculate the dimensions and support

Basic Concepts

requirements for the largest key block of any type, even though the input data

are selected from a statistically defined data set. The recognition that there can

be a maximum key-block size is very important and will be stressed in this and

later chapters.

Design

The results of key-block analysis, and subsequent stability analysis forthe determined key blocks, permit economical design of rock reinforcement(Fig. 6.8). For a particular key-block code, there is a specific set of "danger

,IFigure 6.8 Rock reinforcement: (l) L

support of key blocks; (2) general

reinforcement.

165

,{

zones" in the excavation. Although general reinforcement of the entire excava-

tion, on a regular pattern, may well be justified, it will also be feasible to treat

the danger zones with additional special support. Calculation of the support

force and its optimum direction are discussed in Chapter 9.

The extent to which support is actually required for a surface excavation

depends greatly on the direction and inclination of the excavation. Figure 6,9

shows a railway cut in which the same rock mass requires different slope angles

to be self-supporting on the two sides of the rail line. The difference is connected

with the kinematic control of failure by the system of blocks. If the designer has

freedom to choose the direction and/or inclination of a surface excavation,

analysis of the key blocks will permit an optimum choice to be made. This isprobably the most significant application of block theory. The freedom to change

directions need not be extreme; sometimes only a 10 to 20" shift in the strike ofanexcavated slope will greatly reduce the number and severity of the key blocks.

As the direction or inclination of the slope is changed, large, sudden changes inthe degree of stability can be realized as the key-block types and the danger


oo

Figure 6.9 Asymmetric excavation.

zones shift. In this respect, rock slopes are greatly different from soil slopes.In the latter the factor of safety varies continuously as the slope inclination ischanged. In blocky rock slopes, the factor of safety moves discontinuously fromone function to another as the key blocks shift from one block code to another.Occasionally, a steep slope in rock proves more stable than a flatter slope.

GOTTIDITIOIUS FOR REMOVABITITY OF BTOCKSIilTERSECTIITIG SURFAGE EXCAVATIOilS

In this chapter we consider only those blocks that intersect an excavationsurface. To start, let us assume that the excavation is made by a single plane, Pr,with normal 0, pointed into the rock mass, as shown in Fig. 6.10. By P,(Or) wemean "the half-space of plane P, that contains vector 0r". Assigningthe subscript

i+i

Figure 6.10 Tapered block,

Conditions for Removability of Blocks lntersecting Surface Excavations

i to the free surface, the excavation pyramid, EP, and space pyramid, SP, are

related by

167

and

Alternatively,

and

Since

where

EP: U,otLtSP:L,orU,

EP : P,@t)

SP : P,(-0,)

Most of the blocks intersecting a free surface are infinite or tapered.

Infinite btocks. By the theorem of finiteness, Chaptet 4, an infinite

block must have a nonempty block pyramid (BP). Denoting the empty set by

the symbol @,

(6.1a)

(6.1b)

(6.2)

(6.3)

BP* @

BP:JPNEPn means "intersected with," the criterion for a block to be infinite is

JPNEP*@ (6.4)

As shown in Figure 6.!1, a planar diagram (not necessarily a stereographic

projection), the excavation pyramid, EP, plus the space pyramid, SP, accounts

Figure 6.11 Diagrammatic representa-tion of requirements for an infinite,convex block.

for all space. When (6.a) is true, JP lies only partly in SP, meaning that it is not

contained in SP:JP+SP

Finite btocks. To be finite, a convex block with one face in the free

surface must have an empty block pyramid, by the finiteness theorem of Chapter

4; that is,

(6.s)

BP: @ (6.6)

Introducing (6.3),

and

JPnEP:@JP: fr

Block Theory for Surficial Excavations Ghap. 6

JPnEP:@ (6.7)

Equation (6.7) is true if and only if JP lies entirely outside EP, meaning that itis completely contained in SP (Fig. 6.12), or

JPcSP (6.8)

Equations (6.7) and (6.8) are equivalent, as are (6.4) and (6.5). Equations (6.7)and (6.4) prove more convenient for vector solution, whereas (6.5) and (6.8)are better suited to stereographic projection solution.

Figure 6.12 Diagrammatic representa-tion of conditions for a finite, convexblock.

Tapered blocks. The condition for a finite convex block to be non-removable \ilas established in Chapter 4. Since it is finite, its block pyramid isempty, while the condition of nonremovability dicates that its joint pyramid isalso empty. Thus tapered blocks satisfy both

(6.e)

Equation (6.9) is also the condition for the finiteness of a joint block. A taperedblock is diagrammed in Fig. 6.10. Note that the block would be finite with thejoint planes alone (i.e., without the excavation surface). This demonstrates thevalidity of equation (6.9).

Removable blocks. To be a potential key block, a convex block mustbe finite and removable (Fig. 6.13). According to the previous discussion, afinite block satisfies (6.7) and (6.8). Moreover, since it must not be tapered,a potential key block must not satisfy (6.9). The conditions for removability ofa block are, therefore,

JP+ @

JPnEP:@JPcSP

and

or

(6.10)

ldentification of Key Blocks Using Stereographic Projection

Figure 6.13 Potential key block.

IDENTIFICATION OF KEY BTOCKS USINGSTEREOGBAPH IG PROJECTION

Stereographic projection offers a direct, graphical solution of the equations

above. Thus it is possible to use stereographic projection to determine the

infinite, tapered, and removable block codes corresponding toaparticular system

of discontinuity and excavation planes.

It will be recalled that both joint sets and free planes are projected as

great circles. In the lower-focal-point projection, (J,is the area inside circle i;its code is 0. Z, is the area outside circle r; its code is 1. Both U, and L, arc half-

spaces. JP (if not empty) is an intersection of joint half-spaces, and therefore

occupies a region between arcs of gteat circles in the projection.

A series of examples will apply the criteria for infinite, tapered, and

removable blocks. For these examples we again use the joint data of Chapter 5.

The system of joints and free surfaces is listed in Table 6.1.

TABLE 6.1 Joint and Slope Orientations for Example Problems

Joint Set orSlope Plane

(Free Surface)Dip, a(dee)

Dip Direction, B(dee)

I (joint set)2 (joint set)3 (joint set)4 fioint set)5 (free surface)6 (free surface)

80330

30270

090

75

654010

6080

Block Theory for Surficial Excavations Chap. 6

A Slope Formed by a Single Plane

In Fig. 6.14, a lower-focal-point projection, the fourjoint sets are projectedas great circles together with the gteat circle for a free surface corresponding toplane 5. The rock mass is in the lower half-space of plane 5. Therefore, EP : Lsand SP : Us. In other words, EP corresponds to the area outside circle 5,while SP corresponds to the area inside circle 5. In Fig. 6.14, the four joint-planecircles intersect to create regions each of which corresponds to a JP. Each ofthese is simultaneously inside or outside every one of the joint-plane circles, andtherefore each region can be assigned a code. On the figure, the regions ofintersection of joint-plane circles have been labeled with appropriate half-spacecodes.

According to equation (6.5), infinite blocks satisfy the criterion that JP

+ SP. This is true for any region that is not entirely contained inside the circle

face (5)

Figure 6.14 Stereographic projetion of data of Table 6.1, with free surface ofplane 5 only.

q"

T


for plane 5. By inspection, the infinite blocks correspond to joint pyramids withcodes 1111, 0111, 1110, 1011, 0110, 1010, 1100, 0101, 0100, 0000, and 1000.

The tapered blocks satisfy the criterion JP - @. This means that the jointpyramids of tapered blocks do not appear in the stereographic projection(because an empty pyramid exists only at the origin, and the origin is excluded

from the stereographic projection). There are four joint planes, each of which

creates two half-spaces, so the number of possible joint-plane intersections is 16.

Inspection of the stereographic projection of the four joint planes shows thatthere are only 14 regions of intersection of the four joint-plane circles. Thus two

half-space intersections are not represented on the projection. By the process ofelimination, these are found to be those corresponding to codes 1101 and 0010.

Thus the tapered blocks are those formed with joint pyramids corresponding to1101 and 0010.

The removable blocks satisfy the criteria JP * @ and JP c SP. Every

JP with a code shown on the stereographic projection is nonempty and therefore

satisfies the first criterion. The second requires that a removable block have a

JP that falls entirely inside the circle corresponding to plane 5. An inspection ofFig. 6.14 establishes the codes meeting this criterion as 0011, 1001, and 0001.

Potential key blocks are therefore those whose joint pyramids correspond toone of these three half-space codes.

Gonvex $lopes

The stereographic projection also offers a solution for the infinite and

removable blocks of slopes formed by more than a single excavation plane.

Suppose that a surface excavation is formed by planes 5 and 6 of Table 6.1. We

will consider first the case where the rock mass is the intersection of the lower

side of plane 5 and the upper side of plane 6. As shown in Fig. 6.15, this intersec-

Plane 5

Figure 6.15 Convex slope of planes 5

and 6, viewed along the intersection ofPs and Po.

171

I


tion creates a convex rock surface. These planes are projected as great circles inFig. 6.16, together with the four joint planes. The regions of intersection ofjointplanes have been labeled as in the previous example. The addition of excavationplanes does not change the JP codes since they are defined exclusively by thegreat circles ofjoint planes. Since the rock mass is the intersection of l., and (J5,

EP: Lt)UuSP:UreLu

where A v B indicates "the union of A and B." The limits of SP within Fig. 6.16are shown by the ruling, which lies inside SP. The infinite blocks correspond to

and*

Bounds ofSP=UuULu

N

\IIII

I

\\

Figure 6.16 Stereographic projection of data of Table 6.1 with both free surfaces5 and 6.

*Since SP : - EP where " - " means "the complement of" and is interpreted here as

"the space not included in" (i.e., "the other part of"). In general, if @a1is the space notincluded in l, and (-A) is the space not included in -8, then -(l n 8): (-A) U (-B)and - (A v B) : (* A) n (-B).

ldentification of Key Blocks Using Stereographic Projection 173

JP regions that do not lie completely inside SP. By inspection of Fig. 6.16 they

are found to be 0110,0111,0100,0101, 1111, 1110,0000, 1100, and 1000.

The tapered blocks are those having JP abS€nt from the stereographicprojection. Since the projection of the JP regions is independent of the

choice of excavation surfaces, the tapered blOcks are the same as in the

previous example: 1101 and 0010.

The removable blocks are formed with JP corresponding to regions

entirely included in SP. They are, therefore, 1011, 1001, 0011, 0001, and

1010.

Concave Slopes

Now consider a rock block in a rock slope formed by planes 5 and 6 but

with the rock mass in the shaded region of Fig. 6.t7. This concave region is

determined by the union of the lower half-space of plane 5 and the upper half-space of plane 6. We can use the theorem of nonconvex blocks presented inChapter 4 to find EP. First choose two points, a and b, as shown in Fig. 6.17.

Figure 6.17 Concave slope of planes 5

and 6, viewed along the intersection ofPs and Pe .

Point a is in convex block .B(a) with free surface Lr. Point b is in convex block8(b) with free surface (Js. Then EP(a) : Ls and EP(6) : Uu. By Shi's theorem,

equation (4'30)' Ep : Ep(a) u Ep(b)

Therefore' Ep:Lsuua

SP is convex. It is determined by*

SP:UrAL,Figure 6.18 shows the space pyramid corresponding to this case. The

limits of SP are ruled, with the rulings inside SP. Using the same criteria as inthe previous examples,

*See the preceding footnote.

SP=Uuft La0011

011 1

1111

\\)/\

Plane 6 \

Figure 6.18 JPs when there are no repeated joint sets and the SP for a concaveslope of planes 5 and 6.

. The infinite blocks are 0000,0001,0011,0100,0101,0110,0111, 1000,1010, 1011, 1100, 1110, and 1111.

. The tapered blocks are the same as before (i.e., 1101 and 0010).

. The removable blocks are reduced in number to one, namely 1001.

That is, only the single block 1001 is entirely included in SP.

Removable Blocks with One Repeated Joint Set

Removable blocks sliding between parallel joints of the same set areinherently more stable than those lacking a repeated set, as noted previouslSbecause the restriction of sliding directions mobilizes an increasing normalstress as a consequence of a small initial motion. Sometimes, however, suchblocks are critically located and need to be analyzed.

Figure 6.19 shows a convex block between parallel joints. Suppose that therepeated set is joint plane 1. The possible JP codes are then 3000, 3001, 3010,and so on. If any one of these JP codes identifies a region on the projection, the174

\\\\sI

,

III

/

\\\\

ldentification of Key Blocks Using Stereographic Projection 176

Figure 6.19 Convex block between par-allel joint planes.

region must lie along a segment of the gxeat circle for joint plane 1. Let plane 5

be the free plane, with EP: Lr.In this case, SP is the region inside the circle

of plane 5 (Fig. 6.20).

Plane 6(free plane)

Figure 6.20 JPs with one repeated joint set and the SPs for convex and con-

cave slopes.

3)8t

NBounds of

SP=UuUL.Bounds of

SP=Urr^tLu

Plane 5(free plane)

Ns-2.s----3100 otgo


The joint pyramids corresponding to repetition ofjoint 1 are nonempty ifthey appear as segments of circle 1. A particular segment corresponds to a finiteblock if and only if it lies entirely inside SP. Codes that do not appear as asegment correspond to tapered blocks and those that are entirely or partlyoutside of SP correspond to infinite blocks. In Fig. 6.20, all the segments ofjoint plane t have been labeled with the appropriate JP code. The infinite,tapered, and removable joint blocks for EP : Ls were then determined byinspection of the figure. The infinite blocks are 3111, 3110, 3100, and 30@. Thetapered blocks are 3010, and 3101; and the removable blocks are 3011 and 3001.

The same analysis can be repeated for joint plane 2 serving as the singlerepeated set, then for joint plane 3, and finally for joint plane 4. The results arepresented in Table 6.2.

TABLE 6.2 Identification of Block Types for Any One RepeatedJoint Set and Excavation along Plane 5 (EP : Ls)

Repeated JointSet Infinite Tapered Removable

31 ll31 l031003000

30103101

30113001

0300130013101311

03101301

03010311

0t3l0r3011301030

0030I 131

00311031

0103011311 13

l0t3

00131103

00031003

Now we consider a block, having a single repeated joint set, sliding intoan excavation produced by two free surfaces. First a convex slope will be exam-ined, with EP formed by the intersection of L, and, (Ju, as in Fig. 6.15. ThenSP : Us U 2.. The JP codes are not affected by the choice of SP, so the list oftapered blocks is unchanged. But the division of the blocks into infinite andremovable types is changed. Table 6.3 lists the infinite and removable blocks asdetermined from Fig. 6.20 with SP : (Js U Le.


TABLE 6,3 Identification of Block Types for Any OneRepeated Joint Set and a Convex Rock Slope Formed

by Planes 5 and 6 (EP : Ls o Uc)

BlocksRepeatedJoint Set Infinite Removable

3l l13l r03100

301 I30013000

1310131 1

1300031103010300

1130

01300131

003110301031

1113

01130103

000310031013

In the case of a concave slope, with EP - Ls U Uu, as in Fig. 6.17, SP is

reduced to the area of intersection (Js I Lu, as noted previously. The list offinite, remoyable blocks is then shortened to four: 3011, 3001, 1031, and 1003'

When joint set 2 is repeated, there are no removable blocks in the concave slope.

Removable Blocks with Two Repeated Joint Sets

If a block with a single pair of parallel faces tends to be more resistant

to sliding than a block lacking parallel faces, a block with two sets of parallelfaces must be even more resistant to sliding. Although unlikely to slide, we willdiscuss such blocks because they are potential key blocks under special con-

ditions. We have seen that a nonempty joint pyramid of a block with one set ofrepeated surfaces is an arc of a great circle, as shown in Fig. 6.20. In the case oftwo sets of repeated joint surfaces, a nonempty joint pyramid will be represented

by the point of intersection of the arcs of great circles corresponding to the tworepeated joint sets. The code for such a JP contains the digit 3 in two positions.

Figure 6.21 shows all the JP's corresponding to blocks of this type (for the jointsystem of Table 6.1).

Let plane 5 be the only free plane, with EP equal to Lr, and SP - Usi

examination of Fig. 6.21then establishes which of the blocks corresponding tothese JPs are infinite, tapered, and removable. The results are given in Table 6.4.

fl'''1

l

I

17E Block Theory for Surficial Excavations Chap. G

3113

\.-A

Plane 6 \

\\

Figure 6.21 JPs with two repeated joint sets.

TABLE 6.4 Identification of Block Types for Blocks with Any TwoRepeated Joint Sets for an Ercavation Corresponding to

EP-LsandSP:Us

/III

I

\\\

\\I

I

I

If*-- Plane 5

I I

l

I

331 |

0303 /

i-gts 1330

BlocksRepeatedJoint Sets Infinite Tapered Removable

1,21,31,42,32,43,4

3300313031 13

1330

l3l30133

3301,33103030,31313013,31030330, 1331

0313, 13030033,1133

331 I30313003

033103031033

Evaluation of Finiteness and Removability of Blocks Using Vector Methods 179

Now consider a conyex slope with EP : Ls A Uu and SP - Us \J Lu.Since the JP representations'are unaffected by changing the nature of SP, thelist of tapered blocks is the same as in Table 6.5. The list of infinite blocks isshortened by two and the list of removable blocks lengthened by two becauseblocks 1330 and 1313 become removable. In the case of a concave slope withEP : L, \) Uu and SP : U, ) Lu, the list of removable blocks is reduced toa total ofthree: 3031, 3003, and 1033.

TABLE 6.5 Direction Cosines of Normals to Joint Planesand Free Surfaces Listed in Table 6.1; fi : (Xr Yu Zi

I23

45

6

o.9512

-0.45310.3213

-o.17360

0.9848

0.16770.78480.s566

00.8660

0

0.2588o.42260.76600.98480.5000o.r736

EVATUATION OF FIIUITENESS AND REMOVABITITYOF BLOCKS USING VEGTOR METHODS

In the preceding section, methods were developed to test for key blocks usingthe stereographic projection. For a computation of finiteness and removability,formal analysis using vector equations can be used. Mathematical procedures

establishing the emptiness of a joint pyramid were presented in Chapter 2. lnthis section we establish an efficient mathematical shortcut that simplifies codingand speeds the solution.

Table of Normal Vector Signs

Again let us consider the joint sets and free planes with orientations listedin Table 6.1. The basis directions are as in the examples of Chapter 2: x is east;y is north; and z is up. The dip angle (a) is measured from horizontal and the dipdirection (/) is measured clockwise from north. Using equations (2.7) we firstcompute the direction cosines for the normals to all the planes. The results are

reported in Table 6.5. It will prove useful to introduce a direction-orderingindex /f; defined by

I'l: sign[(n, x n). no]

where sign (F) : (1, 0, -1) when -F is (> 0, : 0, < 0).The values of ft computed from Table 6.5 are shown in Table 6.6. Note

that the values of Il depend on the order in which the joints are written down,which is arbitrary. However, all possible unique combinations of f and j are

(6.11)

L GI

, C6 = ,lrJ\* '

(7 = n(n-l)-n ---T--Block Theory for Surficial Excavations

examined in Table 6.6 for all choices of ft. For n planes, there arecombinations (90 values of It in the present case).

TABLE 6.6 Values oI I'/ : Sigr [(ni X n;) . n*] for the Planesof Table 6.1

\/n -

nl-'.(n-")l L I

7zoZ,I'Z

* lLtl

Chap. 6l"'ti

nC| such

j

1.

We will be able to use Table 6.6 to determine which half-space combina-tions determine finite blocks; we will also be able to describe the edges of blockpyramids.

Finiteness of a Block

Consider a particular block defined by the intersection of half-spaces asgiven by the block code (D"), for example, (1 0 0 1 1 2). Using Table6.6, the finiteness of any such block will be judged by the following steps.

1. Choose a block code (Dr):

D": (a, a2 q3 a,) (6.12)

2. For the selected D", determine a signed block code (D") obtained bytransferring each element a, of (Dr) to a value I(a), defined by

I22223

3

3

445

20304050603l4l5 -l614l5 -l615 -l6161

0

-1-l

I-1

o000

-1

0II

-l0I

-l0

-l00

l I -101-l

-10-l110-1 -1 -l0 -1 1

l0l-l -t 0

-l -1 |00-10100-1 r

-1 0 0l0l

-l-10-t

if a,: g

if a,: 1

if a,:2ifa,:3

I(a,):{:j (6.13)

so thatD, : (I(ar), I(ar), I(or), . . ., I(a,)) (6.14)

Evaluation of Finiteness and Removability of Blocks Using Vector Methods 181

3. Using the terms It/ defined by equation (6.11), form a testing matrix (Tii)corresponding to block Dt for each unique combination of i and i : 1,2,3, . . , , n, ard if*i. (7/i) is a row of n numbers defined by the term-by-term multiplication of Il and l(ar):

(Tti) : (It/ - I(ar), l.]. I(ar), . . ., ry. I@")) (6.15)

Forexample,forD":(I 0 0 I 1 2),withi:Iand7:2,Table6.6gives II' : (0 0 1 1 -1 1); then D", corresponding to D" of (1 0 0

I 1 2), is given by application of (6.13) as (-1 1 1 -1 -1 0). Nowmultiply the corresponding elements of lt*z and D" to determine (Tt2):

(Trr):(0 0 1 -1 I 0)

Similarly, computing the row matrixTti for every unique combination of i andidetermines the complete testing matrix (T) as given in Table 6.7. g) is the(10 x 6) matrix of elements for each row i, j and each column k of Table 6.7.

Rule for Testing Finiteness: If every row of (T) includes both positive and negative

terms, the block Ds corresponding to (T) is finite.

According to this rule, block (1 0 0 1 I 2) isfinite since every row of (?),given in Table 6.?, contains a mixture of *1 and -1 terms. A contrary example

is offered by block 100122. Following the same steps as in the previous example

will yield the same row elements (Z'r) as in Table 6.7 except that eYery term inthe fifth column becomes zero. This results in a matrix (T) whose nonzero terms

are those corresponding to ft from I to 4 in Table 6.7. The second, third, and

eighth rows of this testing matrix lack positive terms and therefore block (L 0

0 I 2 2) must be infinite. [We knew this to be true by the flniteness theorem

of Chapter 4 since (1 0 0 | 2 2) is the joint pyramid for block (1 0 0

1 1 2) and the latter was previously found to be finite.l

TABLE 6.7 Testing Matrix (?) for Block Da: (l 0 0 1 | 2)

j

0000000000

0

-l-t

0I00

2001-l30-10-14 0 -1 -l 05 0 I I -13 -1 0 0 I4-1 0105 I 0 -1 I4-1-r005110-l5 I I -1 0

2223

3

4


Rule for Finding the Edges of a Block Pyramid: If a block is infinite, its blockpyramid is not empty, meaning that it has at least one edge. The edges of ablock pyramid correspond to the lines of intersection of planes i and j for which(Ttt) lacks positive or lacks negative terms. If (TtJ) contains only negativeterms,one edge of the block pyramid corresponding to (T) is -ni X rJ : -111. Con-versely, if (ftt1 contains only positive terms, one edge of the block pyramidcorresponding to (T) is n; X n7 : *fu

Using the rule above, the edges of the block pyramid of Du: (100122)are vectors parallel to

-f":-rl1 Xtr3

-f,o - -nr X tr+

and -frn - -n3 X tr+

(In the expressions above, Iry indicates the line of intersection of planes i and j.The normal vectors n, are positive upward; in the case of a vertical plane, thenormal is directed parallel to the direction input as the "dip direction.")

Finiteness of a Block with Repeated Joint Sets

The method outlined above also applies to analysis of the finiteness of ablock with one or more repeated joint sets. In the block code, D", a repeatedjoint set is indicated by the symbol 3. Equation (6.13) gives (+t) for the .I codecorresponding to an element a, : 3. Consider block I2O3I2, with the joint planesof Table 6.1. Using (6.13) the signed block code, D", corresponding to D": (12 0 3 I 2) is D": (-1,0, 1, *1,-I,0). Table 6.6, determining valuesof It/, still applies to this block since the same joint planes are being investigated.The testing matrix, (7), is produced as in the previous example: For each rowof Table 6.6, multiply column k by the /cth term of D". (Since planes 2 and 6are not involved in the formation of the block being investigated, the pertinentvalues for i and j are only 1, 3, 4, and 5.) The testing matrix (T) determined bythis multiplication is shown in Table 6.8. Since every row (fii) of matrix (7)has both positive and negative terms, block Da: (l 2 0 3 L 2) is finite.

TABLE 6.8 Testing Matrix (") for Block Ds : (l 2 0 3 | 2)

3

3

4

3000+l+10400-10+10500+l+1004 -l 0 0 0 +l 05+100*1005+r0-1000

The Numbers of Blocks of Different Types in a Surface Excavation 183

Now consider block 300321. Using (6.13), D, : (+1, 1, 1, *1, 0, -1).For each row (Z'i) of Table 6.6, we must multiply column fr by the ftth term ofD". Because plane 5 is not involved in the formation of the block, the applicablerows correspondto i,j: I,2,3,4, and 6. The complete testing matrix (fl isgiven in Table 6.9.

TABLE 6.9 Testing Matrix (7) for Block Dn: (3 0 0 3 2 1)

All the rows of T now have both positive and negative elements: except

row 1, 4, which is entirely negative. Block 300321 is therefore infinite. Its blockpyramid has one edge, parallel to -Ir4: -llr X tr+.

THE NUMBERS OF BTOCKS OF DIFFERENT TYPES

IN A SURFACE EXCAVATION

In Chapter 5 we discussed the numbers of blocks of different types occurringwithin the rock mass. Now we examine the numbers of different types of blocksin the presence of one free plane, corresponding to a surface excayation. Blocks

fall into the following categories: (1) infinite, (2) tapered, and (3) removable.

Blocks with No Repeated Joint Sets

Given that there are n sets of joints and one free plane separating the rockmass from free space, we will consider blocks formed from one half-space ofeach of the n + 1 planes. The codes for such blocks will be expressed, generally,

AS

D" : (qt az a3 an an+r)

where a, is 0 or 1 and i:1,2,...,n and a,*1 describes the rock half-space

determined by the free plane; that is, an*1 describes EP. Each of these blockshasn*lfaces.

2001+10-l30-10+10-14 0 -l -l 0 0 -160-1-1+1003 +r 0 0 +1 0 -l4+10100-16+10-1+1004 +1 -1 0 0 0 -16*ll0+1006+1 11000


1. The number of all half-space combinations of the n joint planes is 2"since each a, has two possible values. This is the number of possible jointpyramids.

2. The number of infinite blocks is the number of nonempty block pyramids.It was established in Chapter 5 that the numberofnonemptyjoint pyramidsdetermined by n joint sets, none of which is repeated, is n2 - n + 2.Since the number of JPs is unaffected by the presence or absence of freeplanes, this result is valid here as well. The question to be determined ishow many of these JPs remain when a free surface cuts through.

K+1

K+1

Angle infree plane

(?)

Figure 6.22 z joints cut a free plane into 2n angles.

(a)


3. Suppose that there is a free plane cutting through n sets of joints. Asshown in Fig. 6.22(a), the joint planes cut the free surface into 2n angles.

Every angle in the free plane is the intersection of the free plane and some

JP, as shown in Fig. 6.22(b). Therefore, among the JPs created by the njoint planes there are 2n nonempty JPs that intersect the free plane.

4. The nonempty JPs that do not intersect the free plane can be counted by

subtracting the result in (3) from the result in (2). This yields nz - 3n * 2

joint pyramids that are nonempty and do not intersect the free surface.

Half of these are on one side of the free plane and half are on the other;or in other words, half are entirely included in SP and half are entirelyinclude in EP (Fig. 6.23).

Removableblocks

Free space

F ree plane

Rock mass

Figure 6.23 Diagram representing conditions for infinite and removable blocks.

The removable, finite blocks are those whose JPs are entirely included inSP. From (4), the number of JPs entirely contained in SP is (nz - 3n

+ 2)12 (Fis. 6.23).

The infinite blocks are those whose joint pyramids are not entirely includedin SP. Subtracting the result in (5) from the total number of nonemptyJPs given in (2), the number of infinite blocks is equal to (nz 4- n * 2)12

(Fis. 6.23).

The number of tapered blocks is the total number of half-space combina-tions of the four joint planes less the number of nonempty joint pyramids.Subtracting the result in (2) from that of (1) yields the number of taperedblocks as equal to 2o - (n" - n * 2). This is the same as for joint blocksas deduced in Chapter 5. The results (1) to (7) are summaized in the firstrow of Table 6.10.

5.

6.

7.

lnfinite blocks

I ntersecting free plane

I@o

TABLE 6.10

Numb€r of AllCombinatiols Numbec of Number of Number of

Numb€r of R€peat€d of Haf-Spac€s llf[ite Tapercd RenovableJoint Sets (All JPs) Blocks Blocks Blocks Condition

Orepeatedset 2n nzllt*Z )n- (nz-z-.r-?\ n2-3n*2-z- z"-(42-n+2) -f

n>lI selected repeated set 2o-r n zo-r - 2@ - l) n - z tt>zAny 1 repeated set n2o-r n2 nlln-t - Zfu - l)l n(n - 2) n)22 selected repeated sets 2n-2 | Z,-2 - Z I n>3Any 2 repeated sets n(n --l)2-z

I n(2n - 1X2z-3 - 1) n(n; l'l2 n23

n(n - l\2

z selected repeated sets 2o-^ 0(m>2)

Any m repeated sets Cf .2-^ 0(m>2)

Cf;.2"-- 0

2r-^ n2m) 3

nlm)3


Blocks with One RePeated Joint Set

Now assume that the first joint set is repeated. The possible blocks all have

the code Dr: (3 o2 a3 an dn*r). Each of these blocks has n * 2

faces, comprising 1 face of the free plane, 2 faces of the first joint set, and I face

each of planes 2 to n.

The number of possible joint pyramids is T-' .

Since the first joint set is repeated, all nonempty joint pyramids must

intersect this joint plane and each nonempty JP is therefore an angle

subtended at the origin in plane 1. The n - Ljoint planes 2 through n willcut plane 1 into 2(n - 1) angles, each of which can be considered to be a

nonempty JP.

The free plane intersects plane 1 along a line that lies in two opposite JPs.

Therefore, the total number of JPs that intersect the free surface is 2.

The number of JPs that are nonempty and that do not intersect the free

plane is the number of nonempty JPs given in (2) less the number that

intersect the free surface given in (3). Half of these are entirely contained

in SP and therefore belong to the finite, removable blocks. The number

of removable blocks is therefore equal to n - 2.

5. The number of infinite blocks is the number of JPs not entirely included in

Sp. This is the total number of nonempty JPs as given in (2) less the

number of JPs included in SP as given in (a). This yields the number ofinfinite blocks as equal to n.

6. The number of tapered blocks is the total number of JPs, given in (1) less

the number of nonempty JPs, given in (2). Therefore, there ate 2"-r

- 2(n - 1) taPered blocks.

Results (l) to (6) for the blocks with one assigned repeated joint sets are

summarized in the second row of Table 6.10. If any joint plane can be repeated'

there will be z times as many blocks of each type, as summaizedin the third row

of the same table.

Blocks with Two Repeated Joint Sets

Consider that the first two joint sets are repeated, so that blocks all have

codes D" : (3 3 a3 a4 an a,+r). These blocks have n + 3 faces,

including the free plane, each of the joint sets 3 through n, and the first two joint

sets taken twice.

1. The number of all combinations of joint half-spaces is now 2"-2.

2. All JPs must be in the intersection of both half-spaces of joint set 1 and

therefore must lie along joint set l. However, this is also true for joint set

3.

4.


2, so the nonempty joint pyramids must lie along the line of intersectionof sets I and 2. The number of nonempty joint pyramids is therefore equalto 2.

3. Since only one of these is within the rock, the number of infinite blocks is 1.

4. The number of tapered blocks is the total number of half-space combina-tions of the joints less the number of nonempty JPs and is therefore equalto 2"-2 - 2.

5. The number of removable blocks is the number of nonemptyjointpyramidsthat are entirely contained in the space pyramid. It follows from (2) and(3) that there is only one JP contained in SP and therefore there is only oneremovable block.

When any two joints can serve as the repeated set, there are C| cases likethe one considered and therefore all the number characterized for two assignedrepeated sets must be multiplied by (n)(n - I)12.

These results are summafized in the fourth and fifth rows of Table 6.1.The sixth and seventh (bottom) rows generalize the results for three or morerepeated sets.

PROCEDURES FOB DESIGNING ROCK SLOPES

The design of a rock slope entails choosing geometric properties for the excava-tion and temporary or permanent support measures in the face of real constraintsof time, precedent, and land use. This section considers only the application ofblock theory to the geometric design of the rock slope, all other practical con-siderations laid aside. The question of slope support, involving computation offorce equilibrium, can be addressed more efficiently in chapter 9.

The first step in a practical design problem with a new rock slope is toidentify the critical key-block types. This will allow you to examine the con-sequences of a change in dip directions and dip angles of the slope planes. Asnoted previously, abrupt changes in the degree of rock slope stability accompanyshifts in either the rock slope's dip or dip direction.

Most Gritieal Key-Block Types

In the previous sections it was demonstrated that a certain number ofblock pyramid codes dictate finite, removable blocks. These are all potentialkey blocks of the rock excavation. But not all are equally critical. Two measuresof relative importance of key blocks are their size, asmeasured by block volume,and their net shear force, as measured by the difference between sliding andresisting forces per unit volume. Given a particular key-block code, the volumeof the largest probable block is dependent on the aerial extent of the free planes

Procedures for Designing Rock Slopes

and the joint planes forming the faces of the block. A joint plane's probable

maximum area cannot be measured, but it can be assigned a relative "extent";joint sets that tend to have long traces on exposed surfaces are assumed to have

a large aerial extent, and vice versa. The net shear force per unit volume depends

on the friction angles of the sliding faces and on the orientation of the sliding

direction. The manner in which these factors can be weighed in ranking key

blocks is illustrated by means of an example.

Consider again the four joint planes and the first free surface (plane 5)

of Table 6.1. For convenience, the dips and dip directions of the planes are

listed in Table 6.11. The last column of this table states an estimate of each

plane's relative extent.

TABLE 6.11 Dips and Dip Directions of Planes consiilered in Example

189

PlaneDip, a(dee)

Dip Direction(dee) Relative Extent

I23

4

5 (freeplane)

75

65

4010

60

80330

30270

0

LargeLargeLargeSmallLarge

Figure 6.24 is a lower-focal-point stereographic projection of the fourjoint sets and the free plane (plane 5). Assuming that the rock mass isbelowplane

S, Sp is the area inside the great circle for plane 5. The only removable blocks are

therefore those corresponding to joint pyramids 0011, 1001, and 0001 (i.e., block

pyramids 00111, t0Ott, and 00011). Assume that the resultant forces are due

io gravity only. Then the key blocks are also these three, because each of these

JPs contains vectors in the lower hemisphere, therefore making an acute angle

with the direction of the resultant force (0, 0, -1). This will be discussed further

in Chapter 9. (Also, see the analysis of the sliding direction in Chapters 2 and 3.)

The question is: Which of the three key blocks is most critical? JP 0011

is a spherical triangle meaning that it has only three joint faces, whereas JPs

1001 and 0001 are spherical rectangles having four joint faces. In other words, a

block of type 0011 is comprised of the free plane and one face each of joint

planes 1,2,-and3. Each of these planes has a relatively large extent, so the block

formed by them can run to large size. The other two key blocks involve plane 4

also and lhis plane has limited extent. Therefore, blocks of class 00111 are the

largest expectable key blocks of the three. Moreover, the steepest vector in 0011

plunges steeply and therefore has a relatively large net sliding force per unit

volume. The other key blocks have no vector as steep. For these two reasons,

blocks of code 00111 are more critical than either of the other'

\


\n\\ Free plane

\r'r011

Figure 6.24 Lower-focal-point stereographic projection of data in Table 6.11.

stereographic Projection construction for the Limiting slopeAngles When the Slope Direction ls Given

We have now established which blocks are potentially critical for a particu-lar rock slope. Suppose that the slope is fixed in direction but there is freedom toadjust the slope angle. This is frequently the case in engineering for transporta-tion routes. For every potential key block, represented by a JP, and the assignedstrike for the rock cut, it is possible to construct two great circles serving asenvelopes to extreme vectors of the joint pyramid. These envelope circles bracketthe range of slope angles for which blocks corresponding to the enveloped JPare removable. To perform this exercise it is necessary to construct a great circlepassing through both: a given point anywhere on the projection plane;and agiven point on the reference circle (the strike vector for the cut).

To construct a great circle with given dip direction (given strike) throughany assigned point (see Fig. 6.25):

ol 11

/


Rt.=

-

' sin {6}

,/:

..' Reference!4- circle

Figure 6.25 Great circle with assigned

strike that passes througlr a given point'

1. Plot points ,4 and -B representing opposite strike vectors for the cut. (/4

and B are the intersection points of the reference circle and a diameter of

the reference circle oriented perpendicular to the dip direction of the cut

slope.)2. Loiats the point P through which the great circle is required. P is an

extreme point of a JP.

3. Let R be the radius of the reference circle and let < APB: d (Fig. 6'25)'

Then if C is the center of the required great circle: 4 ACB : 2 6; and

< ACO: { BCO : d. Calculate the radius r of the required great circle

from

If a is the dip angle of the required great circle,

R'-cosd

Equating (6.16) and (3.9) the dip angle of the required great circle is

a:g0-6 (6.17)

4. Construct the required great circle at center C and with radius r deter-

mined bY AC : r-

Returning to the example of Fig. 6.24, it has been established that JPs of

interest for a cut with dip direction corresponding to plane 5 (dip direction equal

0) are 0011, 1001, and obot. In Fig. 6.26, the method described above was used

to construct greai circles enveloping extreme points of these three joint pyramids.

Rt

- -------Tsln o

(6.16)

(3.e)

\


0111

1111

Figure 6.26 Great circles of assigned strike that just contain three Jps.

These gireat circles dip 57.7",46.3", and 36.4'. If a cut slope is steeper than57 .7", all three JPs are potential key blocks. But if the cut is sloped at an angleless than 57.7o,0011 no longer determines a removable block. f ihe cut is flatterthan 46.3",0001 ceases to define a removable block. And if the cut is flattenedto less than 36.4", 1001 ceases to determine a removable block as well. Table6.12 summarizes these conclusions. This example shows that rock slope safety,measured in terms of a ratio of resisting to driving forces, must be a discontinu-ous function of slope angle. As stated earlier, this behavior makes rock slope

TABLE 6.12 Key Blocks of Each Dip-Anete Interval"

l,t tt l/t

NumberDip, a(dee) Key Blocks

RemovableBlocks

Dip Direction, fl(dee)

0000

I2

3

4

0-36.436.446.346.3-57.757.7-90

1001

1001,00011001,0001,0011

0000,10000000

tSee Figure 6.26.

/


stability analysis distinct from that of soil slopes. With respect to the example,

Table 6.L2 implies that there is no change in the safety of the rock slope with the

assigned strike when the dip angle of the free surface is steepened from 60o to90". Such a conclusion is particularly valuable when choosing a construction

method for the rock excavation, for it infers that steep benches can be used.

Use of Vector Methods to Calculate the Limiting Slope

Angles When the Slope Direction ls Given

The problem solved in the preceding section was to find the dip of a cut

slope, given its dip direction, such that a given JP ceases to determine a remoY-

able block. A solution can also be obtained using vector methods. The procedure

will be to compute the edges of the JP and for each edge to compute the dip of a

free plane containing both the line of strike of the free plane and the line of the

edge of the JP. The edges are determined as lines of intersection of the jointplanes, as discussed previously [equation (2.11) and Example 2.3)J. The solution

will be worked out for the joint sets of Table 6.11 and free planes dipping north(i.e., striking east-west).

1. Compute the unit normal vectors fi, : (X,, Y,, Z,) for each joint set, withequation (2.7).The computed values of fi, arc given in Table 6.13.

TABLE 6.13 Unit Normal Vectors

Planei

193

Xi

I2J

4

Compute the line of intersection f,, of each pair of planes, using equation(3.7), and convert each to a unit vector, I,r.The results are listed in Table

6.14.

o.95t2

-0.4531o.3213

-o.t736

0.16770.78480.55660.0000

0.25880.42260.76600.9848

TABLE 6.14 Lines of Intersection lii : (Xij, Yii, Zii)

Plane

a

-0.5289-0.8049-0.9857

0.61250.4291

-o.6282

ZijYtiXij

2J

43

44

-o.1347-0.0194

0.1658o.464r0.8895o.t66l

0.83780.59300.0292

-0.63980. I 568

-0.1350


3. If the cut planes all have the same dip direction, the strike line, ,S, is acommon line of intersection of all the cut planes. Since the x axis is east,,S : (1, 0, 0). The plane that contains b"th

-S ;; 7,;;;; normal n,,

calculated bv

n;J:Sx1,, (6.18)

Let Pq be the plane having fi,, as its unit vector. The equation of p,, isAux I B,tY + c,fi: 0 with the parameters as listed in Table 6.15.

TABLE 6.15 Parameters of planes ps7 Containins 4,, and (1, 0, 0)

Plane

Au Bu CiJDip, a Dip Direction, p(dee) (dee)

III)23

23

43

44

000000

-0.8456- 0.5931

-0.0296o.7223

.-0.3433

-0.2102

-0.5338-0.8050-0.9995

0.6915o.9392

-0.9776

57.73

36.381.69

46.2520.0712.13

0000

1800

4. Having computed a series of planes containing each line of intersection,the next step is to choose each one of these as a fiee plane, in turn and usingthe methods of this chapter or of Chapter 2, findittr t.y blocks of eachP,v' only three of the Prr's generate key brocks i prs, prr, andp,r. The solu-tion for the key blocks of each of these free planes willgenerate Table 6.12.

stereographic proiection construction for the Limiting slopeDirections When the Slope Angle ls Fixed

In the preceding section it was assumed that the slope direction was givenand it was possible to adjust the slope angle. Here we examine the alternativecase where the slope angle is prescribed and the cut can be shifted in direction.

The problem posed corresponds to constructing a great circle of determinedradius through a given point. In Fig. 6.27,1 is the poioitnrough which the greatcircle is desired. The radius of the reference circle is n. f the dip angle of the cutslope is given as c, the radius of the great circle for the cut slope cuo t. calculatejfrom equation (3.9):

To draw the great circle, it is convenient first to locate point 4 the opposite ofpoint A. As shown in Fig. 6.27, point B is located at distance frfrom the centerof the reference circle along the extension of line AV and, from (3.7):

r: R-cos d


R

'- "oto

Figure 6.27 Great circle of assigned inclination that passes through a

given point.

The required great circle can be drawn with radius r from point C, where

CA: 6: r.With the construction above it is possible to draw a great circle through

the extreme points of joint pyramids corresponding to potential key blocks. As

the direction of the cut is shifted through these limiting orientations, the key-

block type changes from removable to nonremovable, or vice versa. Consider

JP 0011 in Fig. 6.28. The corners of this JP are lrr, I"r, and lrr, which are the

projections of i, z, tzt,and f,3, respectively. Given that the angle of the cut is 60",

with dip direction 0o as listed in Table 6.11, we can construct great circle P,

representing the plane of the cut slope. As noted previously,JP 0011 then belongs

to a potentially critical key block. Now, using the procedure outlined above,

great circle P,, is constructed to pass through.I12. With a shift in orientation ofthe cut slope to that of plane P t z, JP 001L ceases to determine a removable block.

A further shift to Pr, (Fig. 6.29), passing through lrr, drops JP 0001 fromthelist of removable blocks. The dip direction of the original cut slope design was

360". Turning it a mere 13.3" to direction346.7 (that of Prr) Fig. 6.28 deletes key

block0011. Turning it24.1o to dip direction 335.9 deletes keyblock 1001 (Fig.

6.29). We see that slight adjustments in orientation may significantly enhance

stability.

p2OB:h

r96 Block Theory for Surficial Excavations Chap. 6

,';"//

0t 11

1111

Figure 6.28 Great circle of assigned inclination that just contains Jp 0011.

Use of Vector Methods to Find the Limiting SlopeDirections When the Slope Angte ls Given

The stereographic projection solution presented above has a parallelcomputation using vector methods. It is required to find the dip direction of aplane passing through a unit vector a : (A, B, c) when the dip an gle, 6,, of theplane is given. Let the unit normal to the desired plane be ff : (i, y, z). Thesystem of equations to be solved is

/lI

I 1011

\\\\

\

tu)t,\\

\jt\ I

(6.1e)

(6.20)

(6.2r)

(6.22)

AX+BY+CZ:O(x, Y,Z),(0,0, 1) : cos d

X'+Y2+22:lA'+B'+C2:land

In the above, A, B, C,components of i6.

and a are known and it is desired to determine the


.//,/ /,

/

Figure 6.29 Great circle of assigned inclination that just contains JP 0001.

The solution, given in the appendix to this chapter, is

Il{ : T+tl,-AC cos d :E .B(sin2 e - c2)t/21

IY:i7zl-BccosdT,4(sin2 d- Cz)t/21 6'23)

and Z:cos&Using equations (6.23), the equation of plane P,, can be computed, given

unit vectol I,t : (A, B, C) and dip a. Then the dip direction of P,1can be calcu-

lated. The intersection vectors (t) are computed as in the previous example

(Table 6.14). For each intersection line formed by the system of planes, and for

a,: ffio, the components of the normal to the required great circle have been

calculated using (6.23). Then the dip direction has been calculated from the

components of the normal. The results are stated in Table 6.16.

The first plane on Table 6.16 is Pr2 of Fig. 6.28. The second is Prt of Fig.

6.29. The results for dip direction agree with those determined previously using

the stereographic projection.

\,I

I

I

/I/

lIpnI

\

/brot,':'/


TABLE 6.16 components x1, y1, 21, or Normal unit vector fu1 ofthe praneContaining fll anilHaving Dip Angle c

Plane

ZtYtXiDip, a Dip Direction, p(dee) (dee)

223

3

443

3

4444

-0.r9920.5781

-0.7746o.7924

*0.85630.85140.8566

-0.3s380.3031

-0.4462-0.6001

0.4947

o.8427o.64480.3871o.3492

-o.12920.1580

-0.1268o.7904

-0.8112o.7422

-0.62430.7108

346.69

4r.87296.55

66.21261.4r

79.4898.42

33s.88159.50328.98223.86

34.83

0.50.50.50.50.50.50.50.50.50.50.50.5

606060606060606060606060

REMOVABTE BTOCKS IN AN EXGAVATED FACE

All of the preceding analysis has been focused on determining the combinationsof joint half-spaces that create potential key blocks in an excavation of statedorientation. Whether or not a key block actually exists once an excavation ismade depends not only on the orientations of the individual joint planes but ontheir positions in the face. Suppose that a map is available showing the lines ofintersection (trace9 of actualjoint planes with the excavation face. How canthe key-block analysis be applied to identify polygons belonging to key blocks ?We will now demonstrate how to solve this problem.

To demonstrate the method of delineating the actual removable blocks,we will again use the sets of joints pr, pr, pr, and, pn and the slope face p, oiTable 6.11. The rock mass occupies the lower half-space of plane fr, so the spacepyramid, SP, is the region within the great circle for plane 5. The Jps that areentirely included within the SP are 0011, 1001, and 0001. These Jps define theremovable blocks given the four sets of joints. From Fig. 6.14 we can also findremovable blocks corresponding to any subset of three joints. When plane p1is deleted, the remainingjoint planes determine a removable block 2001. Simitartydeleting in turn the second, third, and then the fourth plane and examining Fij.6.L4 fot JPs entirely included in the SP yields JPs 1201, 0021, and 0012 as detei-mining removable blocks.

Figure 6.30 is a geological map of slope face Pr. All of the joints in Fig.6.30 belong to one or another of the sets P, to Po and the traces are coded byIine type. Using the list of critical JPs, we can analyzethetrace map to locateall the removable blocks and all combinations of removable blocks. It is impor-tant to search for all the blocks because the one missed might be the one that

Removable Blocks in an Excavated Face

Map code

JP

ru:0121:rZtZ: 0101

-t\\\\\t; 2101

F--1 : 1201

-=IT]TTTIM

llllllllll : 0112

JOrnt1

234

tD

Figure 6.30 Geological map of traces of joints as seen in plane 5 of Table 6.1 I .

initiates a progressive collapse of the entire face. The following procedure willlocate all the removable blocks of the face:

L. Establish the downward direction of the geological map (FiS. 6.30).

This direction, shown ur 6 in the margin of the ffi&p, is the dip of planePr.

In Fig. 6.14, direction 6 is r.p.esented by point D on circle 5. In Fig. 6.14 we

can determine that point D is inside circle 2 and outside circles 1,3, and 4.

Therefore, the upper half-planes of joint traces 1, 3, and 4 as seen on the ffiap,

correspond to the upper half-spaces of joint planes 1, 3, and 4, respectively. But

the upfer half-plane of joint trace 2 on the map corresponds to the lower half-

space of joint plane 2, This suggests creating a special map code (MC) corre-

sponding to each of the critical JPs as shown in Table 6.17. For example, JP

TABLE 6.17 Map Codes (MC) for Face Ps

PlaneCorresponding to Code

0, Assign Code:Corresponding to Code

l, Assign Code:Corresponding to

Code2, Assign Code:

I2

3

4

2

222

I0II

0I00

0001 generates map code (MC) 0101 for a map of plane Pr. Table 6.18 shows the

Jp codes and the MC codes for all the removable blocks in this example.


TABLE 6.18

JPJoint Pyramid

MC(Map Code)

Number of Zones inFig. 6.30

01010t l11l0tott2ot2l12012lol

2. Locate the removable block zones corresponding to four sets of joints.The map codes are those that lack any 2's. They are 0101, 0111, and 1101.

A removable block zone is a closed line-segment loop that satisfies both of thefollowing conditions:

(a) It consists of the joints of all sets P, , Pr, pr, and p4.

(b) For any point P, of the loop the loop area is on the upper side of the jointtrace if the map code of P, is 0 and on the lower side of the joint trace ifthe map code is 1.

For example, loop ABCDEFA of Fig. 6.30 is a zone corresponding tomap code 0101. The segments of this loop are as follows:

. Segment AB is along joint set 4. Corresponding to MC 1, the zone mustbe on the lower side of the segment, which it is.

. Segment BC is along joint set 1. The MC is 0 and the zone must thereforebe on the upper size of the segment in order to satisfy the condition fora key block. It is.

. segment cD is along joint set 3; MC is 0, so the zone must be on theupper side, which it is.

. Segment DE is along set 2 and MC is 1; the zone is below the segment.

. Segment EF is alongjoint set 3; the MC is 0 and the zone is above thesegment.

. Segment FA is along joint set 2, with MC 1, so the zone must be belowthe segment which it is.

Since all segments satisfy condition (b), the zone constitutes a removable block.3. Locate the removable block zones corresponding to critical JPs with

three sets of joints.The map codes for these JPs are 0112,012L, 1201, and 2101. A removable

block zorrc with three joints is a line-segment loop that satisfies the followingconditions:

000100111001

0012002112012001

3

00222

2

(a)(b)

Chap. 6 Appendix

It includes segments from any three of the four joint sets.

For any joint set P, of the loop, the zone is above or below the segment

corresponding to map codes 0 and 1, respectively.

For example, consider loop GHIJKG of Fig. 6.30 with respect to map code

0121. Examination of Fig. 6.30 establishes the following properties for the zone:

Segment GH is along set 1 and the MC is therefore 0, so the zone must

lie above the segment, which it does.

Segment HI is along set 4 and the MC is therefore 1. The zofie must be

on the lower side of the segment, which it is.

Segment /Jis along joint set 1 and the MC is 0, so the zone mustlie above

the line segment in order to satisfy the condition to be a key block.Segment "/i( is along joint set 2 so the MC is 1. The zone lies below the

segment, which it must to be a key block.Segment KG follows joint set 4 and the MC is therefore 1,. Since the zone

lies below the segment, it satisfies the condition to be a key block.

All segments of the loop satisfy condition (b).

Table 6.18 gives the number of closed loops that satisfy the key-blockconditions for each segment for all critical JPs. These numbets can be checked

against Fig. 6.30.

According to the assumptions of block theory, the entire face will be stable

if all the zones detimited in Fig. 6.30 remain in place. One way to assure this is to

establish a rock reinforcement system that, at least, will maintain each of these

blocks. However, some or all of these blocks may in fact be stable because oftheir friction properties or their attitude relative to the free space. Procedures

for kinematic and stability analysis are presented in Chapter 9.

APPENDIX

soturroN oF slMUtrANEous EOUATIONS (6.19) TO (6.22)

and

From (6.20),

AX+BY+CZ:0(x, Y, z).(0,0, 1) : cos d

X2+yr+22:lA2+82+C2:L

(6.1e)

(6.20)

(6.2r)

(6.22)

(1)lt:'"* |

2OZ Block Theory for Surficial Excavations Chap. 6

Combining (1) with (6.19) gives us

v_ I, - E(-C cos q, - AX) (Z)

Combining (1) with (6.21) yields

Xz + Yz:1-cos2d:sin2d (3)

Then substituting (2) in (3) and expanding, we have

(A'+ Bz)Xz +QACcosd)X+ C2 cos2d - Bz sin2c:0 (4)

The solution to (a) isI

X:plVtl-Accosn +B (s)

From (6.22), Az + 82:1. - C2.Insertingthis in (5) gives

r_X : fttl- AC cos d + BN/sin-' d, - C1

Substituting (6) into (2) gives us

(6)

Y:+ulrycosd+AN/wE=ef (7)

From (6.22),

Ar+Cz_l:_82Substituting this in (7) gives

Y:*reBccosd +AN/W@

Equations (1), (6), and (8) constitute equations (6.23) given as the solution.

(8)

chapter 7

Block Thooryfor LJndergroundChambers

This chapter shows how to apply block theory to engineering for undergroundchambers. Underground space is being used increasingly for storage, industrialoperations, power generation, defense, mining, and other purposes. Chambers

are mined to store water, compressed air, oil, nuclear waste, and other com-

modities. Complex arrangements of openings create space for offices, ware-

houses, sports facilities, power plants, rock crushers, ship docks, and even

aircraft hangars. Advantages of underground space include proximity to the

need; invulnerability to attack, landslides, storms, or earthquakes; constant

temperature and humidity; and high resistance to physical, chemical, or thermalloads. If the rock cooperates, underground space may also be more economical

than space for equivalent uses aboveground. This depends largely on the joints

and the system of rock blocks.One way to realize economy in development of underground space is to

choose an arrangement for the three-dimensional network of excavations thatrequires only minimal artificial support. In a self-supporting underground open-

ing, the initial state of stress will be concentrated by the opening and flowaround it in such a way as to preserve the closure and interlock of the jointplanes. Block movements, triggered by loss of key blocks, promote opening ofjoints and looseoiog, which, in turn, may create expensive stability problems.

Block theory provides optimum choices for the orientations, shapes, and

arrangements of openings to minimi ze the danger of block movements. Since itis often feasible for the designer to make adjustments to the layout, particularlyas regards orientation of long excavations, practical use can be made of theprinciples to be demonstrated here.

203

204 Block Theory for Underground Chambers Chap. 7

Figure 7.1 Layout of a rock cavern to store 20,000 cubic meters of water. (FromBroch and Odegaard, 1980, with permission.)

Figure 7.1 shows a plan through an underground water storage complexin volcanic rocks. Although every underground project is unique, most containsome of the features shown in this example, including large, essentially prismaticrooms, enlargements, branches or bifurcations, bends, pillars, entries, andintersections. The elements of these openings are planes, edges, corners, andcylinders. This chapter shows how to determine the key blocks of all types ofdetails formed by the intersection or union of planar excavation surfaces. Thefollowing chapter, devoted to tunnels and porta.ls, adds the complexity of curvedexcavation surfaces.

KEY BLOGKS IN THE ROOF. FLOOR, AND WALLS

Let t be a vector and P(a) be the plane normal to a. P(a) divides the wholespace into two half-spaces denoted U(a) and Z(a). Vector a points into U(a)and vector -a points into Z(a). Therefore, U(-a) : L(a) and U(a) : L(-a).In the following discussion ,let 2 : (0, 0, 1) be the unit vector pointed upward.

Key Blocks in the Roof, Floor, and Walls

Removable blocks in the roof. Since the rock is found on the upperside of the roof plane, EP : U(2) and SP : L(2).The criteria for removabilityof a block are stated in equations (6.10). For a block to be removable in theroof:

and

where

and @ means "empty."Figure 7.2(a) presents a lower-focal-point stereographic projection of

a horizontal plane representing a roof; SP is the region below the roof, andtherefore outside the great circle for this plane. In this and subsequent figures

of this chapter, the region corresponding to EP will be shaded while the regioncorresponding to SP will be left unshaded. Any JP that is entirely included inthe unshaded region determines a removable block and, therefore, a potentialkey block.

Figure 7.2 SP and EP for: (a) a horizontal roof; (b) a horizontal floor.

Removable blocks in the floor. Since the rock is in the lower side

of the floor plane, EP : L(2) and SP : U(2). The criteria governing removabilityof a block now dictate that its JP plot in the region above the great circle for theplane of the floor, that is, inside the great circle shown in Fig. 7.2(b). If theresultant force is given by weight only, such a block can never be a key blockbecause every vector inside such an SP is upward.

Removable blocks in the walls. Figure 7.3 shows a rectangularunderground room. Let wall i be determined by its unit normal vector r?, point-ing toward the free space, so that

EP : L(fi,) and SP : U(fr')For each of the four walls of a rectangular opening, the appropriate stereo-graphic projection is shown in Fig. 7.3. Since the walls are vertical planes, theyeach project as a straight line along a diameter of the reference circle parallelto the strike of the wall. A block is removable in wall i if and only if its JP projectsinside the unshaded region of the stereographic projection for wall i.

JP+@JPcSPSP : L(2)

(7.1)

(7.2)

Block Theory for Underground Chambers Chap. 7

Wall 1

Wall3

Figure 7.3 SP and EP for the walls of a prismatic underground chamber.

BLOCKS THAT ARE REMOVABTE IN TWO PTANESSIMULTANEOUSLY: GONGAVE EDGES

A block that is removable simultaneously in two faces of an undergroundexcavation can become very large and still fit inside the excavated space. There-fore, such a block is potentially very dangerous. However, it is possible to choosean orientation that minimizes the risk of encountering such a key block sincethe number of key blocks of this type is limited.

A block that is simultaneously removable from two adjoining planes willbe said to be "removable on the edge" of the two planes. We will consider firstconcave edges, that is, those in which the rock mass is concave. For a prismaticexcavation, there are 12 such edges, belonging to the intersections of two walls;a wall and the floor; or a wall and the roof. In the following, the roof and floorwill be called, respectively, walls W, and Wu.

Wall/wall edges. Figure 7.4 shows the four vertical edges formed bythe lines of intersection of the four walls of a prismatic gallery. Consider E1r,formed by the intersection of walls 1 and 2. Any block that is removable on this

Wall4

Blocks That Are Removable in Two Planes Simultaneously: Concave Edges

Figure 7.4 SP and EP for the wall/wall edges of a prismatic

underground chamber.

edge is concave. Figure 7.5(a) shows an example of such a block. Using Shi's

theorem (Chapter 4) it is possible to view a concave block of this type as the

union of two convex blocks, block / and block 2. Block I is contained in the

-r0, side of wall 1,, and block 2 is contained in the -r?, side of wall 2. Forblock 1,

and

For block 2,

and

EPr : L(fr t)

BPr: L(fr') n JP: g

EP2 : L(frr)

BP2: L(frr) n JP - @

(7.3)

(7.4)

(7.s)

(7.6)


Figure 7.5 Blocks intersecting edges of a chamber: (a) wall/wall; (b) roof/wall.

The criteria of removability of a block containing edge 1 are

and

From Shi's theorem for nonconvex blocks (4.30),

and

where tt -

tt

Chapter 6,

EP:EP,uEP,SP: -EP

Since EP is concave, SP is convex, so

JP+ @

EPnJP:@ or JPcSP

SPl : U(frt)SP2 : U(frr)

SP:SP,nSP,SP: U(frr)nU(tt,z)

denotes "the other part of" or "the complement of." As noted in

^,(B U C): (-B) n (-C) and -(^B n C): (-A) U (-C)

(7.7)

(7.8)

(7.e)

(7.10)

(7.11)

(7.r2)

(7.13)

(7.r4)

Then

and

htr

Blocks That Are Removable in Two Planes Simultaneously: Concave Edges 209

In general, if an edge is the line of intersection of walls i andn and the

space is convex,SPir: U(fi,)^U(fi1)

Figure 7.4 shows the stereographic projection solution for the removable

blocks of the four wall edges. A JP belongs to a removable block in edge E y

if and only if it projects entirely inside SP,1 and therefore in the unruled area

of the appropriate stereographic projection.

Wall/roof edges. A concave block that is removable in the edge

between a wall and the roof is shown in Fig. 7.5(b). Such a block must be

simultaneously removable in both the wall and the roof. Figure 7.6 shows a

Figure 7.6 SP and EP for the wall/roof edges of a prismaticunderground chamber.

(7.15)

21O Block Theory for Underground Chambers Chap. 7

plan of a rectangular opening drawn in the plane of the roof. Consider edge.Er ,, which is the line of intersection of the roof (wall 5) and wall 1 . Each remov-able block of E1r is the union of block l in the 2 side of the roof, and block 2in the -r0, side of wall 1. For block 1,

and

while for block 2,

EP, : 9121

BPr: U(2)^IP: @

EP2 : L(fr t)

BP2 : L(fr') n JP - @

(7.t6)

(7.t7)

(7.18)

(7.re)

Figure 7.7 SP and EP for the walffloor edges of a prismaticunderground chamber.

Blocks That Are Removable in Three Pfanes Simultaneousfy: Concave Corners

For the concave block, Shi's theorem (4.30) gives

EP: EP, u EP,

and the criteria of removability (7.7) and (?.8) apply with

SP:L(t).U(fir)In general, for wall i intersecting the roof,

SPi,"oor:L(2) n U(lA,)

and

where

and

For block 1,

and

Figure 7.6 shows the regions corresponding to SP for each of the wall/roofedges. Again, if a JP plots in the white area of one of these stereographic projec-tions, all blocks formed with this JP are removable in the corresponding edge.

Wall/floor edges. The same arguments apply for the removability ofconcave blocks in the edge of a wall and the floor of an underground chamber.Any block in the edge of wall i and the floor is the union of two blocks: block 1

in the -2 side of the floor plane, and block 2 in the -rD, side of wall i. UsingShi's theorem, the criteria for removability of such a block are

JP+@JP c SP,,r,oo,

SP,,r,oo. : U(2) ^

U(ri))

(7.23)

Figure 7.7 shows a plan through the floor of the rectangular opening andthe four wall/floor edges. The SP regions corresponding to each of these edges

is the unshaded region of each stereographic projection.

BTOCKS THAT ARE REMOVABLE IN THREE PTANESSIMULTANEOUSLY : CONGAVE CORNERS

Blocks that are removable simultaneously in three intersecting surfaces aresaid to be "removable in a corner." A concave rock mass is created in eachcorner where two walls intersect the roof or where two walls intersect the floorof the prismatic chamber considered previously in this chapter. There are eightsuch corners. Figure 7,8 shows the four roof/wall/wall corners and Fig. 7.9shows the four floor/walfwall corners.

First, consider corner Cr* of Fig. 7.8, formed by the intersection of theroof, wall 1, and wall 2. Each block of corner 1 is the union of three blocks:

Block 1 in the 2 side of the roofBlock 2 in the -r0, side of wall IBlock 3 in the -rC'" side of wall 2

EP, : 9721

BPl :U(2)olP:@

211

(7.20)

(7.2r)

(7.22)

(7.24)

Q.2s)


Figure 7.8 SP and EP for the wall/wall/roof corners of a prismaticunderground chamber,

For block 2,

and

while for block 3,

and

EP2 : L(fr t)

BP2 : L(fr') n JP - q

EP3 : L(frt)

BP3: L(ftr) n JP - @

The criteria for a block to be removable in cornat Cp5and (7.8), but in place of (7.9),

(7.26)

(7.27)

(7.28)

(7.2e)

are the same as (7.7)

Then, since SP: -EP,

EP: EPr u EP, u EPt (7.30)

Blocks That Are Removable in Thtee Planes Simultaneously: Concave Corners 213

Figure 7.9 SP and EP for the wall/wall/floor corn€rs of a prismatic under-ground chamber.

or, finally,

SP:-(EPl uEP2uEP3): (*EP,) o (-EPr) o (-EPr): (- U(2)) A (-L(fr')) n (-L(ftr))

SP: L(2).U(fr')^U(tnz) (7.31)

In general, if a concave corner is formed by the roof, plane f, and plane 7,

SP,,r,oor : L(2) a U(t,) n UQD j) (7.32)

The regions of SP for each of the roof corners are shown in the stereographicprojections in Fig. 7.8.

Similarly, for a floor/wall/wall corner a block is removable if and onlyif it satisfies conditions (7.7) and (7.8) together with (7.26) to (7.30) and

(EP') : L(2)

BP1 :L(2)^JP:@(7.33)

(7.34)


If the concave corner is formed by the intersection$ of the floor, wall i,and wall7,

SP,,r,r,*, : U(2) n U(fij) A U(fri) (7.35)

and a finite block is removable in the corner if and only if its JP is containedin SPr,r,r,*,. Figure7.9 shows the SP regions corresponding to the four floorcorners.

The stereographic projections shown in this chapter to test whether aJP is contained in an SP offer a simple and direct method to judge the removabil-ity ofany block in a concave corner, concave edge, or a face ofan undergroundgallery. Vector methods may also be used to judge if a particular JP is containedin a particular SP. First, all the edge vectors ej are computed for the JP. Then atest of each half-space of the SP must be made to see if it contains the e,. If SPis convex, JP belongs to a removable block if and only if each edge vector er

belongs to each half-space of SP, without touching any boundary plane.If SP is not convex, it is necessary but not sufficient that all e, be con-

tained in SP in order that JP be contained in SP. Since concave space pyramidsarise at intersections of chambers, they will be discussed later in this chapter.

EXAMPTE: KEY BLOCK ANALYSISFOR AN UNDERGROUND CHAMBER

The principles stated in the preceding section will now be illustrated by meansof an example. Consider the underground excavation shown in Fig. 7.10. Thiscombined gate and surge chamber for a hydroelectric power station is approxi-mately a flat, prismatic box 12 by 47 by 80 meters with its long dimensionoriented vertically. During construction, there was freedom to rotate the orienta-tion only around a vertical axis. The joint sets and excavation planes of thechamber are described in Table 7.1.

From Table 7.1, the critical joints of the chamber are joints I,2, and 3,

because of their steep dip. The critical excavation surfaces are the roof andwalls 1 and 3, which are very large.

TABLE 7.1 Joints and Excavation Surfaces of the Chamber Shown in Fig. 7.10

Dip, a Dip Direction, pPlane (dee) (deg)

Extent Spacing(m) (m)

Joint IJoint 2

Joint 3

Joint 4Walls Wr, WsWalls Wz, W+

Roof [ZsFloor We

7l6845t3909000

163243280

343I l8r28r00

505020l247x8012x8012x4712x47

I15

l0l0

aThe direction given is that of the normal to the wall.

Example: Key Block Analysis for an Underground Chamber 215

crlo

E46

Figure 7.10 Underground chamberdiscussed in the example.

In order to apply the relationships discussed previously, it will be helpfulto establish the directions of the inward normals to the walls. These are given

in Table 7.2.

TAB.LE 7.2 Inward Normal Vectors for the Walls

Dip ofNormal (deg)

Dip Direction of theNormal (deg)Wall

Key Blocks of the Roof, Floor, and Walls

The space pyramid of the roof is SP : L(2) (7 .l). The four sets of jointsare projected in Fig. 7.11 and the reference circle is plotted with a dotted line.L(2)is the region outside the reference circle. Therefore, the JPs that are entirelyincluded in SP for the roof are 110L and 1011; these JPs determine the removable

0000

I23

4

118

28

298208

ora6

czso

Wr --\ |

JiE.*w;iI


Figure 7.ll Projection of joint data given in Table 7,r and key brocks ofthe roof.

1111 EEJoint 3

FFigure 7.12 Key blocks of wall 3.

Example: Key Block Analysis for an Underground Chamber 217

blocks of the roof. The removable blocks of the floor are those whose JP lies

entirely inside SP : U(2) and therefore entirely inside the reference circle. These

are 0010 and 0100. Because the roof and the floor are opposite half-spaces, the

JPs of the removable blocks are cousins (see the section on symmetry of blocktypes in Chapter 4).

From (7.2), the space pyramid of wall i is SP: lJ(fr,). In Fig. 7.l2thedashed line is the projection of walls 1 and 3. The space pyramid for wall 3 is the

side of this line that contains a horizontal vector whose direction is 298", as

given in Table 7.2. Therefore, the SP for wall 3 is the region on the unshaded

side of the dashed line. The JPs that lie entirely in this region are 1001 and 1101.

Similarly, the JPs entirely in the SP for wall 1 are the regions 0110 and 0010,

which are cousins of the preceding.Figure 7.L3 shows the stereographic projection of walls 2 and 4. The space

pyramid for wall 4 is the region below the dashed line. The joint pyramids con-

Figure 7.13 Key blocks of wall 4.

tained in this SP are 0001, 0010, and 0011. For wall2, the space pyramid is on

the ruled side of the dashed line, and the JPs entirely included within it are 1110,

1101, and 1100.

Now consider removable blocks with one repeated joint set. The JP ofany such block includes the intersection of both half-spaces of the repeatedjoint set and, therefore, projects along the circumference of the great circle for

1111

/ -$t*$/


{.tt

01 13

fsotg

0031 /I

//

oc)

o(r)

Figure 7.14 JPs with one repeated joint set.

that joint set. Figure 7.14 identifies the JP codes for all segments of all the jointgreat circles. As previously, a block is considered removable if its JP plots entirelyin the appropriate SP. Applying this test to the JPs identified in Fig. 7.I4produces a list of additional removable blocks. For example, the following areentirely outside the reference circle and thus determine removable blocks in theroof: 1131, 1301, 1103, 1311, 1031, and 3011.

All the results of this section are summarized in Table 7.3.

Key Blocks of the Edges of the Underground Ghamber

First, consider the wall/wall edges. From (7.1s), the Sp for edge E* formed,by walls 2 and 3 is given by the intersection of u(frr) andu(fir). on Fig. 7.15,this is the region on the unshaded sides of the dashed lines. There is only oneremovable block in Err; it is formed with JP 1101. Similarly, JP 0010 deflnes aremovable block in edge Etn, formed by the intersection of walls 1 and 4. Thesewalls are opposite walls 2 and 3, so edge E, n is the cousin of the removable blockof edge Err. Ed,ges Erz and Ern have no removable blocks.

The key blocks of the roof/wall edges, Err, Err,Err, and Eor, are poten-tially dangerous. Not only can they be large, but they occur high in the excava-

/////

---t S,t --

Example: Key Block Analysis for an Underground Chamber

TABLE 7.3 Summary of Removable Blocks for the Example Considering Roof,Floor, Walls, Concave Edges, and Concave/Concave Corners

Removable Blocks with:

Position No Repeated Joints 1 Repeated Joint Reference Figure

219

Roof (lZs)

Floor (We)

Wall | (W)

Wall2 (Wz)

Wall3 (LTg)

Wall4 (I4r+)

EdEe ErzEdEe EztEdge Er+Edge,Er+Edge ErsEdge EzsEdge.EssEdge E+sF,dge ErcEdge EzcEdge EtoEdge,EeeCorner CzrsCorner C1a6

All other corners

1101, 1011

0010,0100

0110,0010

1101,1100,1ll0

1001,1101

0001, 0010, 001 I

None1torNone0010NoneI1011101

None0010NoneNone00101101

00r0None

7.13, 7.14

7.14, 7.157.14,7.157.14,7.157.14, 7.15

7.12,7.147.13, 7.147.13, 7.147.13,7.147.13,7.147.13,7 .147.12,7.147.13,7.147.14,7.157 .14,7.15

113I, 1301, 1103, l3ll, 7.11,7.141031,30113100, 0300, 0310, 0130, 7.1r,7.140030,00133110,0130,0310,0113, 7.12,7.140030,00131103, 1300,3100, 1130, 7.13,7.141 301 , 1 1 3 1, 3 1 10, 1 1 1 3

1301, 1003, 3001, 1031, 7.12,7.141131,11033001, 0031, 0003,0030,0310, 0013, 0311, 3011

31101131,1301,110330010013, 0030,0310NoneI 131, 1301, I 103

1131,1301, 1103, 1031

301 I0030, 0013, 0310, 01303100None0030,0013, 03101131, 1301,11030030,0310, 0013

None

tion. Using Fig. 7.6to test the data on Fig. 7.15, the removable blocks of each

of these edges have been determined. JP 1101 lies in the SP for both edges Ert(Fig. 7.I2) and ELs (Fig. 7.13) and therefore determines removable blocks ineach edge. Edges E* and Errhave no removable blocks.

The key blocks of the floor/wall edges are found similarly, using Fig. 7.7

as a guide. Since all the vectors of the space pyramids of the floor edges are

directed upward, there can be no key blocks in the floor edges if gravity supplies

the only force. The removable blocks of all edges are summarized on Table 7.3.

RemoYable Blocks of the Corners

There are eight corners in the gallery. These have been numbered as shown

in Figs. 7.10 and 7.15. The most important regions are the roof cornersl C125;

Crrr, Crnr, and, Crnr. The space pyramids for the roof corners ate shown in


1111

Joint 3

1011

Figure 7.15 Key block of edge 84.

Fig. 7.8. Only corner Crr proves to have a removable block-determined byJP 1101. Although there is only the one potential key block, it could prov;critical and will have to be examined further.

The JP diagrams for the floor corners were shown in Fig. 2.9. Oppositecorner C23, is corner Crnu. Therefore, removable blocks of C2s, have cousinsin Ctnu. Since 1101 is a removable block of C4 s, we determineiio* symmetrythat 0010 will be a removable block of Crnu. This can be verified on the stereo-graphic projection.

Using the List of Removable Blocks to Locate"Danger Zones"

The previous analysis determined the list of removable blocks presentedin Table 7.3.If a joint trace map is available, showing the edges of joints on thewalls, roof, and floor, the removable blocks can be delineated, as will be shown.Such a map can be developed during construction as the excavation surfaces areexposed. For the current example, Fig. 7.16 shows the traces of joints in each ofthe walls. We will demonstrate how to delimit zones of potential key blocks("danger zones") in the walls and in the four wall/wall edges.

/s>€/

'%1001

=

NN

Figure ?.16 Geological trace map of walls (compare with Fig. 7.10).


In the case of the geological map of Fig. 7.16, all the map planes arevertical walls, and the upper half-plane of any trace belongs to the upper half-space of the respective joint. In the case of a horizontal geological map, eachjoint's dip direction points into that half-plane of its trace belonging to its upperhalf-space. With inclined geological map planes, either half-plane of a jointtrace may correspond to the upper half-space of the joint. This more com-plicated situation was discussed in Chapter 6 and a special map code (MC) wasintroduced to facilitate solution.

The removable blocks of the four walls and the four watl/wall edges (edges1 to 4) are listed in Table 7.3. Consider wall 3; the removable blocks withoutrepeated joint sets are 1101 and 1001. The joint planes are identified by numberin the geological map. For block 1101 a search must be made to locate anypolygons that are simultaneously below joints 1, 2, and 4 andabove joint 3. Forblock 1001, the search is for polygons simultaneously above planes 2 and,3 andbelow planes 1 and 4. The polygons stipled with dots, labeled A and -8, are oftype 1001 and are therefore removable blocks.

Similarly, in wall 1, the stippled polygon c, of type 0110, is identified as aremovable block. There is no removable block to be found in wall 4. WaIl, Zpresents block D, of type 1120. Since both 1100 and 1110 are removable blocksof wall 2,ll20 must also be a removable block in wall2. Wall2 also has a blockof type 1210 (block E). Since 1110 is in wall 2, and a short trace of joint 2liesabove the stipled region, it would be prudent to consider block .E as a potentialkey block also.

Edges Errand, Ernwere determined to have no removable blocks. Block1101 was determined to be removable in edge Err. Examination of all thepolygons fails to reveal any one with this block code extending across the inter-section of walls 2 and 3. Similarly, although there is a removable block type inedge .Er 4 (0010), search of the trace map fails to turn up a polygon of this iype.Therefore, there are no potential key blocks in the vertical edges of the chamberwhen it is oriented as given.

Note that the stereographic projection line for walls 1 and 3 pass throughthe point of intersection of great circles for joints 1 and 2 (Fig. 7 .I2). This is whythe traces of joints I and 2 are parallel in the geological maps of walls 1 and 3.Because walls 1 and 3 do contain the line of intersection ofjoints 1 and 2, blocks1011, 0011, and 0111 were determined to be nonremovable. The long polygonsbetween joints 1 and 2 in these walls belong to blocks with a long Jdge thatfails to "daylight" into the excavation.

Table 7.4 lists the real and potential key blocks of the walls of the gallery.(There were none found in the wall/wall edges.) Figure 7.17 shows the shapisof these blocks and their positions relative to the chamber. If the chamber isrotated 90', the long walls take up the orientations of the present short wallsand some of the key blocks shown in the short walls become more important.Figure 7.18 shows a very large key block (0011) in wall 3 after such a rotation.

4\

\g

/\

NNg, Figure 7.17 Key bbcks actually located in the walls.

hrN5

Figure 7.18 Key block when the excavation is rotated through 90'.

Choice of Direction for an Underground Chamber

TABLE 7.4 Real Removable Blocks of the Underground Chamberof Fig. 7.16

Location Name on the Map

225

Wall 1

Wall 2Wall 2Wall 3

Wall 3

Wall 4Edges Erz, Ezt,

Et+, Et+

0110tt201110.1001

1001

NoneNone

CDEAB

NoneNone

aPotential key block if joint 2, above polygon E, cuts off the block Efrom behind.

CHOICE OF DIRECTION FOR AN UNDERGROUND CHAMBER

From the preceding it is apparent that the choice of direction for the long dimen-

sion of an underground chamber can markedly influence the types, numbers,

and sizes of potential key blocks. In the following sections we will consider how

to make a wise selection for the direction of a chamber in order to promote

safety and reduce need for supports. Similar arguments, in reverse, can be

martialed to choose an optimum orientation for instability as required formethods of mining that employ self-caving of the rock mass.

The Most Gritical KeY Blocks

The most critical key blocks tend to have the following attributes:

1. They belong to the largest free planes.

2. They involve joints of large extent.

3. Their space pyramids contain steep vectors.

The size of the free surface provides an upper bound on the dimensions ofremovable blocks and the dimensions dictate the maximum volumes. Forexample, Fig. ?.19 shows two similar blocks with edge lengths in the proportion2: l;the volume of the larger block is then eight times the volume of the formerand since both have the same sliding direction, the net sliding force of the larger

block is eight times that of the smaller. Of course, the maximum key-block

dimensions allowed by wall dimensions can only be realized if the joints have

sufficient extent.In the previous example, the largest walls are W r and W 3, with four times

the areas of W z and W 4. The most extensive joint sets are 7, 2, and3. Figure 7.20

presents the JPs of these three joint sets alone as well as the projection of walls


Figure 7.19 Relationship between relative volumes and relative edge lengths.

///

/' 0012

///

Volume = 8 V

-'/\/

/0002

Figure 7.20 JPs of joint sets I ,2, and 3 only.

Choice of Direction for an Underground Chamber 227

1 and 3. It can be seen that none of the eight JPs are removable in either wall L

or wall 3. Indeed, the direction of these walls was chosen precisely to contain

the line of intersection of joint sets 1 and2. In this way, the most critical key

blocks. of the largest walls have been avoided.

Relationships between Key Blocks of Walls, Edges' and Corners

Since geometrical relationships connect corners, edges, and planes of a

prism, there must also be relationships connecting removable blocks of corners,

edges, and planes of a prismatic underground chamber. This section explores

such relationships. We consider here only concave rack edges and corners, in

the interior of a prism. In the next section, concerning intersections of chambers,

convex edges and corners are discussed.

Notation: W, is a free plane i i : 1,4 arc walls; i :5 identifies the roof;and i : 6 identifies the floor. E, is the edge formed by the intersection of planes

i and j. C,to is the corner formed by the intersection of W,, W1, and W* Theorder of indexes has no significance.

SP (l/r) is the space pyramid of W,, for i : l, 6

SP (E i) is the space pyramid of edge E"and SP (Crir) is the space pyramid of corner Crro

In a concave chamber, the space is convex and its edges and corners are

accordingly the intersections of half-spaces as follows:

SP(E r) : SP(P|") n SP(fj)SP(Gio) : SP(I4IJ n SP(I4lj) n SP(14/e)

As a consequence of (7.36), equation (7.37) can be written as

SP(Crie) : SP(E i) n SP(E &) n SP(E &)

(7.36)

(7.37)

(7.38)

Equation (7.36) to (7.38) may be considered to constitute a theorem of linkage

in underground chambers. They together establish three propositions linkingremovable blocks of walls, edges, and corners.

Proposition 1. If JP belongs to a remavable block of Eti, then JP belongs

to a removable block of Wt and W,.If we assume that JP is removable in Eii, then

JP c SP(E i)Introducing (7.36),

JPc(SP(et")nSP(IZ1))JP c SP(H/,) and JP c SP(|4li) (7.3e)

Proposition 2. If JP belongs to o removable block of C,io, then JP belongs

to a removable block of W,, Wr, and Wo.

Since JP belongs to a removable block of C,1p,

JP c SP(Crjk)


Introducing (7.37),

JP c (SP(W) . SP(w) n sr1nzrll

so JP c SP(ry,) and JP c SP(ft r) and JP c SP(14l*) (7.40)

Proposition 3. If JP belongs to a removable block of C,to, then JP belongsto a removable block of edges E,i, E*, and E,e.

Again, since JP belongs to a removable block of C,1p, JP c Sp(Crr*), thenintroducing (7.38),

JP c (SP(E i) n SP(Eje) n Sr1r,";and therefore

JP c SP(E, and JP c SP(Eio) and JP c SP(E o) Q.4t)Figure 7.2I diagrams the geometric connections between all walls, edges,

w6(f loor)

I

I

I

+atrq.+-E16-+C1a6+\ { ./t

| -rz-._'tl-Ll4 II r\ _i /1 |lTCr2s-Ers+Craslllll r +llrrr ,J lllw6-l l-l(ttoir)- ti'-ti,*+, -,,.Tir)*

eju**o*eou+- wo

| | | "T" I l' i" (froorr

ll* | rllI l-zgs-Ess+Cs+sI i/'"" + rt I

| -zJ.-"3.---34 Ir/ t \tCzro+Ese+Cga6

+I

I

II

I

r,[l,.r

Figure 7.21 Linkage diagram for walls, edges, and corners.


and corners. According to proposition 1, a JP that is removable in an edge is

removable in both walls contiguous to that edge. And by propositions 2 and 3,

a Jp that is removable in a corner is removable in both the edges and walls that

are contiguous to that corner. Using Fig.7.2l, all contiguous edges and corners

can be found. For example, a JP that is found to be removable in Clru must

then be removabl e in Wr, Wr, and Wu as well as in edges Eru, Erz, and E26.

Conversely, if a JP is not removable in a wall, it is not removable in any edge

or corner contiguous to that wall; if it is not removable in an edge, it is not

removable in either corner attached to that edge. For example, if a JP is not

removable in wall 4, it is also not removabl e in E1+, E+e, Ern, and En, and it isalso not removable in C, nr,Crnr,Crnu, and C1+o'If a JP is not removable in

E 3 4, it is also not removable in C3 n, and C , nu.

The relationships described above make the work easier. For if you can

choose the direction of W, and W, to avoid a particular JP, then all the edges

and corners of these walls are also safe from key blocks of that JP. In other

words, the JP cannot be part of a removable block of either the large free sur-

faces of the corners or long edges of the excavation'

Procedure for Choosing the Directionof an Underground Ghamber

1. Draw the stereographic projection of all sets of joints, as in Fi5.7.22-

2. Draw a line through the two points of intersection of a pair of joint circles :

for example, line 1 of Fig. 7.22, drawn for the intersection of joints I and

Z. Line I is the direction of a vertical wall that parallels the line of inter-

section of joint sets I and 2. Repeating for every combination of joint

planes, generate all such lines on the stereographic projection. In the

example considered, there are six such lines, radiating from the center ofthe reference circle. (In an inclined chamber, the lines are replaced by

great circles that radiate from the normal to the roof and floor.)

3. Arbitrarily denote right and left sides of each line. We then define the

right wall of line i as the wall parallel to line i and with the rock on the side

denoted as "right."4. Determine the removable blocks of the right and left walls for all i. Table

7.5 presents these results for the example (r : 1 to 6). Note that since the

right and left walls determine symmetrically opposite half-spaces, the

removable blocks of each are symmetric cousins. According to Table 6.10,

with four joint sets, none of which is repeated, there are three removable

blocks in each wall. However, the formulas in this table were derived

assuming that the free plane does not contain any line of intersection. Inthe present example, there is one less removable block in each wall parallel

to lines 1 to 6 precisely because these walls do contain the line of intersec-

tion of two joints. For example, since line I contains /1 ,, JPs 1000 and 1011

cannot determine removable blocks in the right wall, and JPs 0111 and

1111

Joint 21/\// obhr

Joint 4 /

'10111

Figwe 7.?2 Analysis of the effect of direction on the key blocks of a prismaticunderground chamber.

TABLE 7.5 Removable Blocks of All Walls Containingthe Line of Intersection of Two Sets of Joints

4et

1001

,*,\1oo0 \i

1- 7t /','),.' oom t

/" ooot""'+Hrr,,' oorl s i

LineNumber

JointSets

RightWall

LeftWall

Directionof the

Intersection(dee)

Removable Blocks

I

2

3

4

5

6

lr2

lr 3

1,4

2,3

2,4

3,4

28.4

58.1

73.0

133.2

t47.8

r77.O

1001

1101

1101

10001101

1100111011001110

010001100100

0110001000100111001000110001

00il00011011

1001

10il

230


0100 cannot determine removable blocks in the left wall. If the directionof the wall were rotated a small amount in either direction from that ofline 1, one or the other of these JPs would be added to the list of removable

blocks. (However, even then the corresponding blocks would be onlynarrow slivers in the wall.)

5. Determine the removable blocks of the right and left walls directed in each

of the angles bounded by lines i (i : 1, 6 in the example). The removable

blocks of the right and left walls are symmetric cousins. Since none ofthese walls contains the line of intersection of any pair of joints, all these

directions of walls have three removable blocks each. The results are listed

in Table 7.6.

TABLE 7.6 Removable Blocks of Walls in the Angle between

the Intersections of Two Sets of Joints

LineNumbersBoundingDirections

StartDirectionof Angle

(dee)

EndDirectionof Angle

(dee)

Removable Blocks

RightWall

LeftWall

617

1,2

213

3,4

415

516

357.O

28.4

58.1

73.0

t33.2

147.8

28.4

58.1

73.O

133.2

147.8

177.O

1001

1101

1011

1101

10001001

I101110010001110I 1001101

111001001100011001001110

0lI0001001000010011r011000100011

0ll10001001 I001000011011

001 I1001

10ll0001

The apptication of this procedure can be demonstrated by examining the

consequences of a rotation of the long walls of the example considered pre-

viously. With walls I and 3 directed to the N 28" E, as given, the key blocks are

those sketched in Fig. 7.17 .If the chamber is turned through 90o, interchangingthe directions of the long and short walls, the direction of the long wall willbelong to the angle between line 4 and line 5. Figure 7.18 shows then the maxi-mum key block of JP 0011 within the long left wall of the chamber. The pos-

sibility of having such a large, steep key block in an excavation should be avoidedat all costs.


I]UTERSECTIOIIIS OF UIUDERGROUIIID CHAMBERS

The crossing of two underground galleries creates convex edges and convex/concave corners where these edges meet the roof and the floor. In buildingtrades, these are known as 'ooutside corners" as opposed to "inside corners" ofconventional rooms. Figures 7.23,7.24, and 7.25 show horizontal sectionsthrough intersections of chambers at floor, midroom, and roof heights and thevarious edges and corners are identified. We have introduced the followingnotation:

. E'u denotes the convex edge of an exterior corner where walls i and jintersect. For example, E\2 is the edge of walls 1 and 2 where a chamberparallel to wall 1 intersects a chamber parallel to wall2 (Fig. 7.23).

$b;f7$.,,x4,/+p

Figurc 7.23 SP and EP for the inside edges of intersecting chambers.

Intersections of Underground Chambers

Figure 7.4 SP and EP for the walliwall/roof corners of intersecting chambers.

. CrtJr, denotes the exterior corner where convex edge E',t meets the roofor floor (wall k, with k : 5 or 6, respectively).

. Wt denotes the four walls involved in the intersection. In every case,

r0, is the normal unit vector of W, that points into the free space of theintersection, as shown in Fig. 7.23.

Convex Edges of Ghamber Intersections

The rock mass delimited by the convex edge E"" is L(fi,) ) L(tfi ). Remov-able blocks of a convex edge therefore have excavation pyramid EP given by

EP:L(fr,)^L(fi)r)sP: -EP:U(fr,)uU1fi)

(7.42)

(7.43)

''':A; -t--/ /\.t

lilllEP

/'\ cioe

>t',xt:


Figure 7,25 SP and EP for the wallywallTfloor corners of intersecting chambers.

The criteria for removability of a block in the edge E',t are therefore stated by

and

or

and

JP* AJPc (U(ft)vU(,ft))JP+ @

JP n (L(fr,) n r(fr)) - o

(7.44)

(7.4s)

Figure 7.23 shows the EP and SP for each of the convex edges of theintersection. Combining this with Figs. 7.L4 and 7.15, the removable blocks ofall the edges were determined for the wall directions given previously in Table7.1. For example, the removable blocks of edge E\, are those corresponding toJP's 001Q 0110,0100, 1110, 1100, and 1101. Results for all the edges of theintersection are given in Table 7.7, both for JPs lacking any repeated joints(Fig. 7.15) and for JPs with one repeated joint set (Fig. 7.14).

Since the four convex edges of the intersection have a larger SP than thatof a wall, more blocks are removable in the edges of intersections than in the

lntersections of Underground Chambers

TABLE 7.7 Removable Blocks in Convex Edges

of the Intersection of Two Underground Chambers


Position 0 Repeated Sets I Repeated Set

E's+ 1101 0010 1131 3011 03101301 1003 00301103 3001 001313lr 0003 03tl1031 00313110 0311 3001

3111 0030 30ll0130 00130113 00030310 0031

0010 il01 0030 3100 1301

E,,N

E\z

1001

10ll0001001101100111001000ll0001

011001001ll0I 1001001

1000ll0l11001110

0310 0113 I 131

0013 3110 1103

0300 ll13 1300

0130 11303001 1300 3110

3000 1131 3100

1031 11031003 1113

1301 1130

E'zt

walls of a chamber. Such edges may proye to be critical locations in an under-ground chamber.

Corners of Ghamber Intersections

The external corners Ctt,k where one chamber intersects another are com-plexly shaped. Unlike prismatic chambers, or concave surface excavations, theSP is not convex and unlike the edges of an intersection, discussed in the preced-

ing paragraph, the EP is not convex. However, each block of the external cor-ners C'a1, can be divided into two blocks BPt and BPr, where

BP1 : L(fro) n JP (7'46)

and BP2 : L(fr ')

A L(fr, ) n JP (7 .47)

Then, using Shi's theorem for nonconvex blocks, the criteria for removabilityof a block from corner C',r1, are

JP+ gand EPnJP:@or JP* @

and JP c SP

where EP : L(fr) u (L(fi,) t ^fr))

(7.48)

236 Block Theory for Underground Ghambers Chap. 7

From (7.45) and (7.46),

sP: _EP : U(fr) n *(L(fi,) n Z(fr))

or SP : U(fr) n (U(fij) u a(ti,)) (7.49)

The SPs and EPs of CI js and C',16 are shown in Figs. 7.24 and 7 .25, respectively.Using these figures with Fig. 7.L5, we can determine the removable blocks cor-responding to JPs with no repeated joint sets. And with the use of Fig. 7.I4, wecan determine the removable blocks of JPs having one repeated joint set. Theresults are listed in Table 7.8. Note that the results divide into symmetric pairs:E'r, and E|ni E'r, and, Etini C\zs and. C'3aui C'rns and C'r2u; and C'1a, and C'rro.The key blocks of all these pairs are cousins.

TABLE 7.8 Removable Blocks in Exterior Corners ofthe Intersection of Two Underground Chambers


Position 0 Repeated Sets I Repeated Set

C'st,

C\+sC\zs

C'zss

C'sqa

C\+e

C\ze

C'zta

1101

l0l l

1101

1101

0010

0010

00100100

1131 1311

1103 1031

1301 3011301 1

I 131

r 301

11031031 1301

1311 11031131

0030031000130130 03100300 001300300030 03000013 0r300310 31003100

PILLARS BETWEEN UNDERGROUND CHAMBERS

The rock separating long, parallel chambers may be called a wall pillar or arib pillar. An example is shown in Fig. 7.1. This is a critical part of any under-ground complex because its failure might bring down the roofs of the adjacentopenings. If a rib pillar is cut through on a regular pattern of cross cuts, theresulting rectangular or square column pillars are all that is left to support the

Pillars Between Underground Chambers

roof. Assuming that the stress concentrations in the pillars are insufficient tofail the rock itself, a pillar is still vulnerable to progr€ssive failure following loss

of key blocks. Key blocks of pillars can be discussed using block theory.

Wafl pillars. Figwe 7.26 shows a pillar between parallel walls ofneighboring chambers. Let ril be the normal unit vector pointing into one of the

two openings; then the EP for the rock of the pillar is determined by

EP: U(fr)nL(fi))

1t11

(7.s0)

Figure 7,26 Key blocks of a wall ("rib").

The criteria for a key block are, as usual,

tP+@and JPnEP:@The projection of the EP for the pillar is the dashed line of Figure 7.26 and theSP is all the space not touching the line. All the JP regions not touching thedashed line therefore determine blocks that are removable in the pillar. Forthis example, they are 1101, 1100, 1110, and their cousins.

Column pillars. Figure 7.27 shows several pillar cross sections. Anactual pillar will tend to have curved faces, but these can be represented by aseries of m tangent planes. Let the unit normal vector of tangent plane W,be

.." 1000

:i--fmm

001


#ffiffi

RigureT.2il Key blocks of a pillar.

tD,, pointed into the free space. The EP for the pillar is then given by the inter-section of all L(fi,):

m

EP : n (r(ltJ)J= I

(7.s1)

For a pillar of constant cross section, EP is then a line parallel to the axis of thepillar. In the case of a vertical pillar

EP:12 (7.s2)

A JP will define a removable block in the pillar only if its intersection with EPis empty. All nonempty UPs therefore define removable blocks except the twothat contain the axis of the pillar and its opposite (Fig. 7.27).

COMPARISON OF VEGTOR AITIALYSISAIUD STEREOGRAPHIC PROJECTION METHODS

Since the stereographic projection shows both EP and SP for an excavation,all the block problems of this chapter can be performed on the stereographicprojection. Not only is it easy to judge whether or not EP and JP have commonpoints (or if JP is entirely contained in SP), but also whether the boundaries ofJP and SP are close. The latter is more difficult to appreciate using vectormethods.

Comparison of Vector Analysis and Stereographic Projection Methods 239

For the calculation of key blocks of walls, edges and corners of an openingwithout intersections or reentrants, SP is convex for all elements of the excava-tion. It is then straightforward to determine, by vector methods, whether or notJP is entirely contained in SP. It is sufficient to compute the edges of JP anddetermine if each edge belongs to each of the half-spaces of SP.

For the complex blocks discussed in connection with intersections ofunderground chambers, the SP is not convex and vector methods are not con-veniently expressed in terms of SP. Instead, the EP criteria are preferred. From(7.47) the criteria for removability of a block may be written

and

(L(ft,)^L(tD)) nJP: @

L(fi) nJP- @

(7.s3)

(7.s4)

Each of the equations above represents the intersection of two convex pyramids.To determine whether or not two convex pyramids intersect, it is not sufficientto determine if the first contains any edges of the second. A calculation mustalso be made to determine if the second contains any edges of the first. Althoughall these problems can be solved with vector equations, in the case of complexblocks of intersections it seems preferable to use the computer to produce astereographic projection which is then examined interactively.

chapter 8

Block Theor>lfor Tunnolsand Shafts

Tunnels are among civilization's oldest achievements. They have been dug foraccess, transportation, defense, shelter, drainage, water supply, and mining.Except for surveying, early tunnels were probably completed with little or noengineering calculations. Even now, because the main body of rock to be pene-trated remains essentially hidden until actually encountered in the tunnel face,tunnel engineering demands on-site decisions. However, mountains of rock aredirectional with respect to excavation, as we have seen, and the directionselected for a tunnel greatly affects its excavation and support costs. Tunneldirection is usually determined before breaking ground. In this chapter weexamine exactly how block theory can be applied to choosing an optimumdirection.

We place highway and railway tunnels under a sidehill to straighten andshorten a route, and under ridges to avoid extensive surface cuts. Such tunnelstend to be safer than their cut-slope alternatives not only because the weak rockof the weathered zone is avoided, but because the span of the tunnels is farsmaller than the dip length of the slope, meaning that much smaller key blocksare involved. Moreover, tunnels have concave slopes virtually everywhere,whereas natural hillsides undercut by surface excavations often have convexslopes. In previous chapters we observed that concave excavation surfaces havesmall space pyramids and few key blocks, whereas convex excavation surfaceshave large space pyramids and many key blocks.

Temporary tunnels for mining are usually left unlined. So are long-lifetunnels for water power and water supply in good rock and for transportationin very good rock. When there is no lining, the stability analysis using block

240

Geometric Properties of Tunnels 241

theory pertains to the entire life of the tunnel. When a lining is built, for stabilityor hydraulic smootheness, block theory calculations then pertain to the perioduntil the lining is constructed. The system of blocks can then be analyzed todetermine the possible loads and load distributions on the lining under dynamicforces for example by an explosion or earthquake. The inertia force of the keyblock adds to the sliding force and the block presses against the lining. Waterforces on key blocks have to be considered for pressured water tunnels. Also,in all tunnels the state of stress acting around the tunnel can influence the

stability of key blocks. Calculations of the stability of key blocks under gravity,inertia, hydraulic, and structural loads are given in Chapter 9. In this chapter

we continue the analysis of the geometric conditions lbr the formation of remov-able blocks. What is special about tunnels in this regard is their curved surfaces,

which generate blocks with curved faces.

GEOMETRIC PROPERTIES OF TUNNETS

Tunnel directions. We use the term tunnel to describe the wholesystem by which passage is obtained through a rock mass. The commonest

system includes two portals and a horizontal cylinder.* However, vertical andinclined cylinders are also used, particularly in water power projects. Horizontaland vertical tunnels are relatively easier to excavate and to line, but the greater

design freedom to choose both a tunnel direction and inclination enables a

better choice of orientation with respect to key blocks. Therefore, the general

case of inclined tunnel systems will be treated. Often, however, inclining a tunnelis not compatible with its purpose and a horizontal cylinder is required. Thehorizontal tunnel will emerge as a special case of the general theory.

Elements of tunnels. Figure 8.1 sketches a tunnel under constructionfor a hydroelectric power project. The main part of the tunnel is the tunnelcylinder. But the portals are also very important since a difficult portaling condi-tion can delay and complicate a project. Portals are generally more difficult thanthe tunnel cylinder. Not only is weathered rock encountered, with low frictionangles along discontinuities, but the excavation surface at the portal has a largerspace pyramid than that of the tunnel cylinder. The working face of the tunnel isanother element. Since stresses are concentrated here, and the excavation surfaceis concave, key-block problems tend to be less severe than in the tunnel cylinderor the portal.

The complex cases presented by intersections, enlargements, and bends oftunnels are not discussed in this chapter. Such cases can be analyzed approxi-mately by replacing the tunnel section by a series of tangent planes. Then themethods presented in Chapter'l can be applied.

*By "cylinder" we mean a long, hollow solid of constant cross section. The shape ofthe section can conform to any continuous locus.

I ntersection

Main tdnnel cylinder

Block Theory for Tunnels and Shafts Chap. 8

Figure 8.1 Geometric elements of a tunnel.

Tunnel shapes. Figures 8.2 to 8.6 show various shapes for the tunnelcylinder. Figure 8.2 shows smooth, closed curves, without any angles. Suchshapes reduce stress concentration and diminish the number of key blocks. Butexcavation and lining may be more difficult and more costly. The shapes inFigs. 8.3, 8.4(b), and 8.5(c) represent hybrid solutions. The upper part is asmooth curve lacking angles, to minimize stress concentration and reduce the

(a) {b)

Figure 8.2 Continuously curved tunnel shapes.

Geometric Propenies of Tunnels

(a) (b)

Figure 8.3 Tunnel cylinder shapes with straight sides.

Figure 8.4 Additional shapes for the tunnel cylinder.

(b)

Polygonal and horseshoe shapes.Figure 8.5

2U Block Theory for Tunnels and Shafts Chap. I

Figure 8.6 Vaulted tunnel sections.

number of key blocks. The lower part is produced by straight-line segments, foreasier construction.

Figures 8.a(a) and (c), 8.5(a) and (b), and 8.6 show polygon shapes,occasionally found in transition sections. Ideal polygon shapes are also createdby partial collapse of sections of a tunnel as blocks drop from their initialpositions. The vaulted roof of Fig. 8.6(b), for example, could result from lossof a key block in an initially flat roof. The resulting asymmetric shape with asharp corner provides a high stress concentration which may stabilize otherblocks riding behind. Although seldom designed, asymmetric shapes can pro-vide greater stability than symmetric shapes. This is evidenced by examinationof natural tunnel sections and can be shown readily using key-block theory.

BTOGKS WITH CURVED FACES

Since the surface of a tunnel excavation is cylindrical, its intersection with asystem of joints produces blocks with a curved face. In order to apply blocktheory to this case, it will be helpful to introduce a tunnel coordinate system.

Goordinate Systems

Let the axis of the tunnel cylinder be denoted by unit vector 6,horizontalfor a horizontal tunnel, vertical for a shaft, and so on. The global coordinatesystem we have been using has i horizontal directed to the east, ! horizontal tothe north, and f directed upward. We now introduce coordinates f o, jo, and 2o

in directions attached to the tunnel. In particular,

{bt(a)

Blocks With Curved Faces 245

(8.1)

zo

xo

Yo -a

xdXxo:d1x,'(2xd)

Using the dot product of each pair of *o,9o,2o in turn, it is easily estab,lished that these coordinate vectors are mutually orthogonal. The plane pe{pen-dicular to the tunnel axis is the plane of .iofo;.f o is directed up the trace of the dipvector of this plane, while .?o is its strike, directed according to 2 x d.

As an example, assume that the plane normal to the tunnel axis has dipd : 60o and dip direction F : 0". From Q.7), the coordinates of the vector dnormal to this plane are

d: (0,0.9660,0.5000)

Equations (8.1) then

0

-0.s0000.8660

The coordinate vectors fro,io,and,20 for the example are plotted on a lower-focal-point stereographic projection in Fig. 8.7.

It will now be established that the coordinate vectors *0,90,20 defined by(8.1) are right-handed. Using these relationships, we have

(*o x 9).2o :(t x d) x @ x (2 x il).@)

N

give

/t'\ l- |

lo. l: I o

\20l \ 0

l,r*)0.5000/

Aaz

Yo

---+

Figure 8.7 Tunnel coordinate system.

246

ThensinceAxB.

(*o


C:-AxC.B,x 9)-20 : -(2 x il x (,t).Qa x Q x fi)

:(axt)x@).(AxilxG)):KAxflxAp>0

Therefore, *0,90,20 comprise a right-handed, orthogonal system.

Local Goordinate Systems for Points on the Tunnel Cylinder

Planes tangent to the tunnel cross section are all seen as edges in the fto9oplane. Consider one such tangent to the tunnel at point Q,andwhose edge makesan angle 0 with jo (measured positive in a clockwise direction) as shown in Fig.8.8. Clockwise rotation of (.f;',.f o) in the plane 2o: 0 by angle 0 about the origin

(c) {d}

Figure 8.E Position angles and local coordinate systems for points on thetunnel rylinder.

to{t0

90, t{o}

Blocks With Curved Faces

generates new vectors i(0), fr(0), tangent and normal, respectively, to the tunnelwall at point p.

(8.2)

Suppose that the shape of the tunnel section is convex (i.e., a continuousclosed curve, without reentrants). Then for any angle 0 between 0 and 360' wecan locate one point Q, ot a straight segment on the tunnel boundary such that:

l. n(q is the normal vector to the tangent plane at point Q.

2. fr(0) points out of the tunnel, into the rock.

3. t(0) is the tangent vector of the tunnel surface at point Q.

The angle d is the tunnel position angle of point Q@. In all views anddiscussions of this vector, we will adopt the convention that the vector 2o : dis pointingfrom the tunnel section toward the observer. The angle 0 is alwaysmeasured clockwise from io to i(0) or from .90 to fi(0), as shown in Fig. 8.8.

The projection plane in the projections of curved blocks will always be the planeof the tunnel cross section (fo,.f o).

Excavation Pyramids of Gurved Blocks

To analyze the curved blocks of tunnels, the boundary of the tunnelcylinder will be approximated by m tangent planes, as shown in Fig. 8.9. Firstchoose m points along the curved boundary: Q(?t), Q@),. . ., Q(0^) andconstruct a tangent line through each. The curved tunnel surface is then replacedby a multiplanar locus, seen on edge in Fig. 8.9. In Fig. 8.9(a) the tunnel curve isunsatisfactorily replaced by two tangents (i.e., m - 2). The case for m:3is in Fig. 8.9(b), while ffi - 4 is shown in Fig. 8.9(c). The latter begins to be aclose approximation to the smooth curve of the theoretical tunnel section.

Let B, be the point of intersection of tangents through Q@) and Q(0,*),so that the tunnel curve is determined by the locus

Q@)B,Q(0r)BrQ@r) .. . B^_,Q(0*)

A complex rock block between Q(0r) and Q(0^) is the union of rz convexrock blocks with the block pyramids

BP, : U(fr(0,)) n JP, i : 1, . . ., m

The excavation pyramid for the complex block is then

EP : 0 U@,@,))i=1

and a block is removable if JP + g and JP n EP : @ .

(8.3)

(8.4)

(';:1,,): (:"i ; -;: ;xi,)

248 Block Theory for Tunnels and Shafts Chap. 8

(c) (d)

Figure E.9 Representation of the tunnel cylinder, between given points, bytangent planes.

With Q@r) established at the extreme counterclockwise end of the curvedsegment, all 0, values increase uniformly with increasing t. (If angle 0 crosses

0 : 0 between 0, and 0,*r, add 360' to all 0r having index i + 1 or greater.)

Then for all i,0,*, - 0, is a small, positive angle.If 0^ - 0t is greater than or equal to 180", EP given bv (8.a) encompasses

the whole space and EP n JP : JP + @ and therefore JP cannot satisfyequation (8.4). Thus such blocks are not removable.

From Fig. 8.9(d) we can see that if 0^ - 01 is less than 180', then

{a}

EP : u U@@)) : U@(0,) u U(n(0^)) (8.5)

Equation (8.5) is true even if the number rn is so large that the tangent segments

n(02)

Types of Blocks in Tunnels 249

almost perfectly match the convex curve of the excavation boundary. This per-

mits statement of the following important proposition for curved blocks.

Proposition. The criteria of removability for blocks intersecting a curved

tunnel surface are

and

where

and

JP+@

EPNJP:QEP: u(n(0))u u@(0^))

0^- 01 < 180'

TUNNEL AXIS THEOREM

A theorem concerning the relationship between the axis of the tunnel cylinder

and the JPs of removable blocks will prove useful. This is stated here and the

proof is presented in an appendix at the end of this chapter.

Theorem. JP is a removable block of a tunnel if and only if +A + JP.

If the tunnel axis d is an element of JP, the theorem states that the JP does

not belong to any removable block of the tunnel. This is because the tunnel

axis belongs to each of the half-spaces whose union determines EP and therefore

EP n JP must contain d and accordingly is not empty'Every JP that does not contain the tunnel axis, then, has a corresponding

removable block in the tunnel. We will demonstrate how to locate this blocklater.

TYPES OF BTOCKS IN TUNNELS

In this section we discuss the numbers of blocks of different types that are

created by the intersection of a tunnel cylinder and a rock mass with n sets ofjoints. The numbers of all combinations of half-spaces are as discussed inchapters 5 and 6 (Tables 5.1 and 6.10). Also, the numbers of nonempty jointpyramids are the same as discussed in Chapter 5. The respective formulas have

been transferred to columns 1 and 2 of Table 8.1 to support the discussion here.

As previously, the number of tapered blocks is the number of all combinationsof half-spaces less the number of nonempty joint pyramids. This has been pro-

duced in column 3 of Table 8.1.

As an example, consider a case with no repeated joint sets. Then, from the

first column of Table 8.1-, the number of all combinations of joint half-spaces is

2" and the number of nonempty joint pyramids is n2 - n + 2. Accordingly, the

number of tapered blocks is 2' - (n' - n * 2).Figure 8.1"0 shows an example

of a tapered block intersecting a tunnel.

Ngro

TABLE 8.1 Numbers of Blocks of Different Types in Tunnels

Number ofRepeated Sets

Number of All Number of Non-Combinations of empty Joint

Half-Spaces Pyramids

Number ofTaperedBlocks

Number ofRemovable

Blocks

Number ofInfiniteBlocks Condition

0 repeated set1 selected repeated setAny 1 repeated set2 selected repeated sets

Any 2 repeated setsra selected repeated sets

(m>3)Any m (m> 3)

repeated sets

2"2n-rn2"-r2"-2n(n - l)2"-z2r-^

Cf;.2"-^

nz-n*22(n - r)2n(n - l),,

n(n - l)0

0

2n-(n2-n*2)2'-r -2@-l)n(2n-t -2@-1))T-2 -zn(n-1X2"-3-t)2r-^

C!.2-^

n2-n2(n - r)2n(n - l)2

n(n - l)o

0

200000

0

n2ln)2n)2n)3n)3n) m)3

n2m)3

The Maximum Key Block

Figure 8.10 Tapered block intersecting a tunnel.

The number of infinite btocks of a tunnel. From the tunnel axis

theorem, JP belongs to an infinite block of a tunnel if and only if d e JP or

-d e JP. Since d|s apoint inthe stereographic projection, there is one and only

one JP that contains d; its symmetric cousin will contain -d. Thetefore, there

are exactly two infinite blocks. In the event that dfalls exactly on the line between

Jps, it would define an infinite block of a JP having one repeated joint set. If ri

falls exactly on the line of intersection of two planes, it would define an infinite

block of a JP with two repeated joints. The probability of this happening is very

small, and therefore the number of infinite blocks with one or more repeated

joint sets is shown in Table 8.1 as equal to zero.

The number of removable blocks of a tunnel. The criteria for a

block to be removable inside the tunnel are that JP + @ and that neither d

nor -d is contained in JP. The number of removable blocks is the number ofnonempty joint pyramids minus the number of infinite blocks. For no repeated

joint sets, this means that there are nz - n removable blocks. The complete set

of formulas are given in column 4 of Table 8.1.

THE MAXIMUM KEY BLOCK

According to the proposition (8.6) for removability of blocks intersecting a

curved tunnel surface, a block is not removable if the extreme limits of itsintersection with the tunnel surface include an angle 0^ - 0, 2 180'. The angu-

lar interval over which a block intersects a tunnel depends on both the shape and

size of the block. For a given block type, there is a maximum size beyond which

blocks are no longer removable. This limiting size will be termed the *maximum

key block." The largest removable block of a given key-block type is the most

critical because it requires the largest supporting force. Further, the friction angle

mobilized on sliding surfaces generally varies inversely with the dimension of the

sliding surface. On the other hand, the statistical distribution of joint extents

dictates that the probability of actually encountering a block becomes smaller as

the size becomes larger. In this section we ignore the statistical question and

251

252 Btock Theory for Tunnels and Shafts Chap. g

always assume that the maximum key block is also the most critical key block.A formal theory will be presented first, followed by applications. proofs of thepropositions are contained in an appendix to this chapter.

Proposition on Angular Relationshipsfor an Empty Intersection

In Fig. 8.L1 a JP is projected into the plane perpendicular to d (he tunnelsection). Its edges are I, ,Ir, .. . , I, and their orthographic projections in thesection are r", lL,. ..., r! as shown. The outward normals to the extreme edgeprojections, i', and".if, enclose an angle (n, - 4) where 41 2 ur. It is establishedin the appendix to this chapter that if Id # Jp, there is a vector r0 : rrr * rnzsuch that U(mo) n JP: @ [see equation (38) in the appendix to this chapter].

fi(qr)

Figure 8.11 Projection of a Jp in the tunnel section.

This could not be so unless the angle 4t - Ut were less than 180"; therefore, theprojection of such a JP is always such that (tt, - ?,) is less than 1g0". Figure g.11establishes the following proposition :

U(ft(0)) n JP - @ if and only if 41 10 1q, (8.7)

Theorem on the Maximum Removable Area of a Tunnel section

1. Let the equation {U(fr, Q)} define the half-space containing fi whoseboundary passes through Q. Then if block .B is a removable block of Jpin a tunnel cylinder, -B belongs to the "maximum removable area" of Jp,meaning that

B c. {u(-fi(n),eQt)) ^

uen@),eeD)} (8.8)The maximum removable area is the intersection of the tunnel section andthe space outside the tunnel with the right side of (8.8) (i.e., the term insidethe [ ]).

2. There is a removable block, B, of JP such that the projection of Bin thesection perpendicular to d is exactly equal to the projection of the maxi-mum removable area of JP. This block is the maximum key block.

Computation of the Maximum Removable Area Using Vector Analysis

Figure 8.12 shows a section perpendicular to d. The directions of I'randfi(the orthographic projections of the extreme edges of a JP, as seen in the tunnel

section) are tangent to the tunnel section at points Q(rt) and Q(4). The maxi-

mum removable area is determined by the region that is between these tangents

and outside the tunnel. ,B is a removable block. It can grow larger until itsextreme edges reach almost t9 i'r,ana 1! . A similar area,l, enclosed between lines

parallel tol'r, at Q@ ) and Ii at Q@) may contain a removable block only if 0t

and g2lie between r7, and 4, (i.e.,4r 10, 10" < 4).

Figure 8.12 Maximum removable area.

A single JP yields a single angular interval regardless of the shape of the

tunnel (assuming the tunnel cylinder is convex). However, the maximum remov-

able area does depend on the shape, as shown in Figs. 8.3 to 8.6. In fact, it ispossible to choose a tunnel shape for which the maximum removable area

vanishes. Figures 8.a(a) and 8.6(b) dre examples.

The theory presented in this section can be applied usingvectorcalculations

or stereographic projections. Both methods will be developed in the following

sections.

COMPUTATION OF THE MAXIMUM REMOVABLE AREA

USING VECTOR ANALYSIS

Using vector operations, we will show how to determine angular intervals and

maximum removable areas for all key blocks of a tunnel. In the example to be

studied, we again use the sets ofjoints introduced in Chapter'7, as given in Table

8.2.

o(e2lo(4r) o(01)

Block Theory for Tunnels and Shafts Chap. g

TABLE 8.2

254

Dip Direction, fi(dee)

Dip, a;(dee)

Joint Set, i

163

243280343

7l68

4513

I23

4

Matrix for Testing the Edges of a Jp

The upward normal vector to set i is fi, given by

fi, : (sin dr sin Bs, sin d, cos Br, cos c,) (g.9)

lfrr\ | 0.2746 _0.9042 0.325s\

lo,l -f -o.azor -o.42os 0.3246l

lo,l- | -o.oslo o.t2z7 o.zo7l I\nol \-o.oosz o.ztsr o.sll;,l

The next step is to compute all values ofI'l : sign [(fi, x fi).fik\ (8.10)

The results are given in Table 8.3 (see also Table 6.6). The matrix of values of It/

TABLE 8.3 Values of If : Sign [(n1 X ry) . n&]

was used in Chapter 6 to test the finiteness of a block. Hence it will be used tocompute the edges of a JP. Let (/) be the 6 x 4 matrix of values in Table 8.3.

l; l-; -i\ L rFYl(1):l_l ; ; _?l

[-r o t ol\-1 -t o gl

1200_1_113010_114011023-100_124-l 01034-1-100


In general, (1) has m rows and n columns with m : C3 : n(n - l)12 and

n : the number ofjoint sets. Each row of (1) corresponds to the intersection fr1

of a particular pair (i andi) of joint planes.

The Edge Matrix

In Chapter 6 we denoted the JP code, D",by

Dn: (at a2 a3 ar)

For a selected D", determine a diagonal matrix (D) whose diagonal terms

correspond to a signed block code D" [as described in (6.14)].

I(a') 0 0 0

0 I(a") 0 0

::

0001(a,

(D): (8.1 l)

in which

[+t if a,: s

l-l if a,:1r(a,) :1 o if o,, :2

[+t ira,:3Corresponding to (6.15), we now form a testing matrix

(8.12)

(8.14)

(D,(1)(D), i r1-6r- '.'i '(8'13)

[Each row of (T) corresponds to a Ttr of (5.15).]

The next step is to compile a column matrix [.8] with cll-- n(n - l)l2lrows, each of which corresponds to a particular i, i pair.

{E} : [e t €2 €3 €,rn-n/zf

in which 7 indicates the "matrix transpose" and

(e),:

0 if in the ith row of (7) there are

*1 and -1, or if all of theelements of the ith row are equal tozeto

+1 if each element of the rth row of (r) (8'15)

is fl or0

-1 if each element of the ith row of (")is -1 or 0

Matrix (E) may be called the edge matrix. The test conducted by these operations

guarantees that any real edge of a JP is simultaneously in each of the half-spaces

defined by the JP code.

Gomputing the Edges

Consider D" -- (a, a2

Block Theory for Tunnels and Shafts Chap, g

, the matrix (/) was giyen in Table 8.3.

: 0000. According to (8.11),

000\I o 0lo r 0f0011

ct3 an)

ll(D):

f ;\o

Table 8.2For the joint system ofFor the JP 0000,

000l01

-1 0

-l 0

-1 -t

/,i

and {E}:ir.-l

0

II0

Similarly, for (a, e2 ao) :0101,

0 -1-l 0

-l I000lr0

(/).(D) :

(/).(D) :

(/).(r) :

as

LI il and

il and

",fi

[E]:

0

I0

I0

0

For (a, os or an) : 1031,

r0 0 +lt-

[; l.?ll ; *l\t -r o

The edge matrix {E} has been computed similarly for every JP code(at az q3 ao) and entered as a column in Table 8.4. Each row of these tablescorresponds to a particular edge Ir, (: fi, * fir) and each column to a particularJP. An element value of *1 means that the respective JP has an edge directedby the respective Iry. If the element is -1, the JP has an edge pointed in direction-lui and if the element is 0, the Jp has no edge parallel to I,r. Accordingly,

. 0000 has edges -Irr,Irn, _lrr, and -frn.' 0101 has no edges.. 0100 has edges _lrr, -I,r, and -Irr.. 1011 has edges Irr, I,r, and lrr.


TABLE 8.4(a) Edges of JPs

0001 0010 0011 0100 0l0l 01ll0110

I0

-10

-l0

I -lI -1000 -1

-1 0

-l 0

-l 00010ll0

-10-l0 0 -1-l -l -l

000-l0 -l0-lo-100

ItzIr:Irclzslz+Is+

TABLE S.4(b) Edges of JPs That Are Cousins to Those in Table 8.4(a)

1001 10001010101l11001101ll1011il

IrzItrIt+lzslz+Ir+

-l0I0

1

0

l01000100000

1 0 0 -10-10-l

-1-100101000111l1l

Table 8.4 does not show the edges for JPs with a repeated joint set, but

the method is the same. For example, the edge matrix for 1031 was shown to be

(0 1 0 I 0 0)r. Therefore, the only edges of this JP are f ,, and I23. (A JP

with one repeated joint set consists of an angle in some plane through the origin,

and therefore it has only two "edges.")Note that Table 8.4 shows no edges for 0101 and 1010. These then corre-

spond to tapered blocks.lf a, e2 a,, has edge matrix [E], its cousin has edge matrix

-{E}. This is apparent, for the example, by comparison of identical columns inTable 8.a(a) and (b). This result means that the cousin of a block pyramid is acentrosymmetric block pyramid with opposite edges.

Limiting Edges in the Tunnel Section

To determine the angular interval and maximum removable area for a.JP,

it is necessary to flnd the projections f! and fl of its extreme edges t, and t, as

seen in the tunnel section (Fig. 8.11). 1', and t; ur" orthographic projections ofI, and,f, in ttrr plane perpendicular to d,; that is, the extreme edges are projected

along d.Thelimiting edges will be established by a column matrix {.8} determined

by the following operations.

1. Compute the value of IYt defined by

I'lt : sign [(lu x A) ' lkrf

lit:ftrxfi,(8.16)

where

For the joint system of Table 8.2, the values of IIll are given in Table 8.5.

258 Block Theory for Tunnels and Shafts Chap. g

TABLE 8.5 Values of t{f : Sign t(iii x a) , tril

lu

lt+Iz+Izsrr+frrIrzhIrzrrrIt+IzsIz+fgr

-t -l11ll0-110

-1 -l

-101

-1-l-l-l

-1-l

0

-l-1-t

2. The matrix of + t, -1, and 0 of Table g.5 is denoted as (s), that is,

, 0 -1 -l -l -l l\lr o-r I t;l

(s):l I _i _l ; _l ll\_i _l _l _l _l 'l

Matrix (,S) is an antisymmetric m x m matrix, where

m:c?-n(n-l)

3. For every JP code (a, az a,), determine an edge matrix [E] asshown previously.

{E} : le, €2 e*1,

4. From [E] we define a diagonal matrix (.I) according to

.0

.0'0

.gm

.I is also an m x m matnx.Compute the product (/) (s) (/), an m x m square matrix . (Note: In thisstep, we provide the mechanism to assure that the normals to limiting edgesare computed using real edges of a given Jp.)Define a column matrix {^B} according to

(8.17)

(8. I 8)

5.

00

00

{B} : [bt bz bs b^f, (8.1e)


where b, :

if the rth row of (/XSX4 are all 0 or if the rth rowof (/)(S)(/) includes both 1 and -Lif the ith row of (/XSX4 includes only 1 or 0 and atleast one element L

if the rth row of (/)(S)(/) includes only -1 or 0 andat least one element - L

Note: In the operations above we have established a normal to an edge, I1;,

considered as a candidate to be a limiting edge. This normal was determined byIrr X d. Then, by taking the dot product with each of the other edges, we

determined if all the other edges belong to the same half-space of f,, x d. If so,

I,r X ri is the normal of one of the limiting planes and I,; is one of the limitingedges of the JP.

-1

Example(000 -1

fl:4, Q1 a2 Q3 ct4:0010; from Table 8.4, [E]:- 1)',

{B} :

000000000 *101

(B):

Example 2. at ez e3 ct4 :0111; from Table 8.4, {E'} - (1 0 -1-1 0)"

0000

-1 0

00-1 0

00

000l00000000

(j)ill?-;/

/00fo o

lo o(rxsxr):lo 0

I

[00\s0

ril

1010r0000\o o t olo o o ol'

-l 0 0 0f0006/

[j)lil

(/xsxr) :

Example 3. o1 clz o1 o4: 1031 ; using 8.11 to 8.t4,[E]: (0 1 00 0)'

0

0

0

-l0

0

(rxsx/) : [B]:


Computing the matrix [B] for each JP code produces Table 8.6, each column ofwhich is [^B] for both cousins heading the column.

TABLE 8.6 Limiting Edges of JPs

0110 0lI1l00r 1000

0011lI00

0000 00011111 1110

0100 01011011 1010

00101101At AZ A3 A4

lnIrrIrtIzslz+Is+

0000-1 0010 -1 0 0 I 0 0 0

-l 0000010r00000-1 00010000-l0t*I 00000

7. The limiting edges of a JP projected to the tunnel section correspond tothe nonzero terms of [BJ.

From the previous analysis we learned that 0101 and 1010 are tapered, becausethese JPs have no edges. The tunnel axis is in 0011 and 1100, so 0011 and 1100generate infinite blocks in the walls and have no limiting edges in the section.

In Table 8.6 the limiting edges of each JP having removable blocks arethe rows with *1 or -1. These results are summarizedin Table 8.7.

TABLE 8.7

Limiting Edges

0000,1li10001, 11100010,11010100,10ll0u0, 10010111,1000

It+,lzsIrr,Is+lzt,Is+Irz,IrrIt+,lzsItz, Iz+

Gomputing the Angular Intervals

If I!, is a limiting edge of the projection of JP to the tunnel section plane:

Ii.i X d : Ilu X 6 : tnfu) or tn@,) (8.20)

is the normal vector of a plane passing through tunnel axis 6 and the limitingedge I,r. It defines a limtt plane of the "IP (see Figs. 8.11 and 8.13). Consider aJP code having a limiting edge f,, in the kth row of Table 8.4; tbis edge is repre-sented by er (the kth term of {E}) for the JP and therefore by bo(the kth term oftB]).

Irr X d determines a vector normal to the projection in the tunnel sectionof a possible edge of JP. The term e* is f 1 when I,, is a real edge, it is - 1 when

-fry is a real edge, and it is 0 otherwise. Therefore, er(l,t x d) is a normal to theprojection of a real edge. The term bo is f 1 when all the other edges are in the


Tunnel section

Figure 8.13 Lower-focal-point stereographic projection of lines shown inFig.8.11.

same half-space of f eoQ, r x A). The term 6o is -1 when all the other edges are

in the half-space -e*(l,t x il. Then -boeo(|,, x ri) is nonzero for two vectors,each normal to a limiting plane and pointing out of the JP.

From Tables 8.4 and 8.7, for each JP we haye now determined a vector inthe tunnel section that is normal to each limiting plane and that points away

from the JP. These results are listed in Table 8.8.

Recall from (8.1) that the tunnel coordinate system is determined bycoordinate vectors fto:2 x d,/o: d x (2 x 6) and 2o: d. The vectors

.1,., fF^Dr,' oo i i "

i'" "

TABLE 8.8

Angular Interval (deg)for the JP

Outward Normals to the Limit Planesin the Tunnel Sectiono From: To:

00001ll10001r 11000101r0101001011

01101001

01111000

IlaXd-lr+Xd

l13Xd-lrt X d

lz+xd-l2a X d

-I12 X dIrz X d.

llaXd-\+xd-lrz X d

InXd

IzsXd-l23Xd

It+Xd-hrxd-l3a X d

Is+Xdl13Xd

-ltt X d

-Izs X 6lzs X d.

-Iz+ X dlz+ X d.

t71.3351.3204.1

24.185.7

265.770.2

2s0.255.6

235,6

49.r229.1

23s.65s.6

265.785.7

229.r49.r

204.124.1

171,3

351.370.2

250.2

olrt: fu x fit.


Dr : Irr x d of Table 8.8 therefore lie in the ftogo plane and

Xt:Ilr.Xo

lr:tr;'90Each vector rlr : (xr, y,) makes an angle 0, from yo (measured from !o

toward .fo). Compute 0, for each vector of Table 8.8, then, for each Jp, orderthe angles 0: 4, and 0 : q, such that 4r - 4r, measured from.to toward!0,is less than 180". The angular intervals computed for every JP code using thisprocedure are listed in Table 8.8.

The Maximum Removable Area and Maximum Blocks

Having determined the angular intervals for each removable JP code, themaximum removable areas are now fixed. These are shown in a series of figureskeyed by Table 8.9. The maximum finite blocks have also been drawn to show

TABLE 8.9

(8.21)

JPCode

Figure ShowingMaximum

RemovableArea

FigureShowing

the MaximumBlock

Summary of Removabilityand Importance

0000

0001

0010

0011

0100

0101

0110

0111

1000

100110101011

1100

1101

1110

111 I

8.14(a)

None

8.15(a)

None

8.16(a)

None

8.17(a)

None

None

8.18(a)None8.19(a)None

8.20(a)8.21(a)

8.22(a)

8.14(b)

None

8.15(b)

None

8.1(b)

None

8.17(b)

None

None

8.18(b)None8.1e(b)None

8.20(b)8.21(b)

8.22(b)

Removable stable blockunder gravity

Absent because of thetunnel shape


Infinite block; tunnelaxis belongs to JP


Tapered block; JPis empty

Key block, almost stableunder gravity



Key blockTapered block; JP is emptyKey blockInfinite block, tunnel

axis belong to the JPKey blockKey block; volume is

very smallKey block

Computation of the Maximum Removable Area Using Vector Analysis 263

their intersection with the tunnel cylinder. In these figures, the view is in thedirection of -6; that is, d rises from the paper (i.e., looking south in this case).

Visualizing the maximum blocks in relation to the tunnel permits conclu-sions to be made about which blocks are key. Table 8.9 summarizes the remov-ability and relative importance of each of the JPs. The most important blocks

ao= ?

(a) (b)

Figure 8.14 (a) Maximum removable area and (b) key block for JP 0000.

(a)



(b)

Figure 8.16 (a) Maximum removable area and (b) key block for JP 01fi).

\..'


(b)


(a)


265

Figure8.18 (Continued)

(bt



(b)


Figure 8.21 (a) Maximum removablearea and (b) key block for JP lll0.

267

Figure8.2l (Continued)

Figure 8.22 (a) Maximum removablearea and (b) key block for JP 1111.

Gomputation of the Maximum Removable Area Using Vector Analysis

(with no repeated joint sets) derive from JPs 1001, 10L1, 1101, and 1111. Thissuggests that blocks 1031, 1L31, 1301, and 1131, each having one repeated jointset, are also potential key blocks.

The Angular lnterval for a JP with One Repeated Set

Previously, w€ computed the edge matrix {E} and limiting plane matrix

[B] of JP 1031 as

{E}:(0 l 0 I 0 0)'

{B}:(0 r 0 -1 0 0)"

Using the same method, for JP 3111,

and

[E]:(1 0

[B]:(l 0

000)'000)"

-t-l

With one repeated joint set, JP is an angle in the plane of the repeated set

and a nonempty JP has only two edges. Therefore, as shown above, {,E'} has only

two nonzero terms.The normals of the limiting planes that are pointed away from the JP are

for l03l: -l* x d and lrt x d

for 3l I l: -l* x d and -Itn x d

The angular intervals are

for 1031 : 235.6" to 24.1"

for 3111: 351.3' to 70.2"

Figures 8.23(a) and 8.24(a) show the maximum area for these JPs and Figs.

8.23(b) and 8.24(b) show the maximum blocks. The angular interval for 1031 is

larger than that for 1001 or 1011 because JP 1031 contains the latter. In general,

(e)

Figure 8.2iil (a) Maximum removable area and (b) key block for JP 1031.

Figure E.23 (Continued)

Figure t.24 (a) Maximum removable area and (b) key block for JP 3111.

Computation of the Maximum Key Block Using Stereographic Projection Methods

a block with a single repeated set has alarger angular interval than one lackinga repeated set. Of course, a block with a repeated joint set has a greater resistanceto sliding than one lacking a repeated joint set.

COMPUTATION OF THE MAXIMUM KEY BTOCK USINGSTEREOGRAPHIC PROJECTION METHODS

All of the results of the preceding section can be achieved using the stereographicprojection. To demonstrate the methodology, using the system of joint sets

presented in Table 8.2, first plot the joint circles and establish the JP regions.Figure 8.25 shows this projection. Given a JP, we can draw the limit planes ofthe maximum removable area by the following procedure.

11',t1

Figure 8.25 Lower-focal-point stereographic projection ofjoint data in Table 8.2.

Constructing the Limit Planes of a Given JP

1. Plot the projection A and A' of the tunnel axis vector d, and -d; assumethat d is upward. This can be done from the given bearing and plunge ofthe axis, or from the coordinates of the axis vector d, using equations (3.2)to (3.7).

2. For each edge, f, of a JP, draw a great circle through A, A', and .L FromFig.8.26, the radius, r, of this circle is

AA'' zsind (8.22)

where, as shown in Fig. 8.26, d is the angle AIA' and AA' is the length of


AA'2

Figure 8.26 Great circles through A, A', and .I.

Figure 8.27 Limiting great circles of JP 0000.

the segment from A to A'. Figure 8.27 shows this construction applied tocorner 1 of JP 0000 for a tunnel in the direction of vector (0, 0.866, 0.5).

3. For each JP we can construct two circles through A, A', and an edge ofthe JP such that only the edge touches the circle. These two circles are

envelope planes of the JP. In Fig.8.27, the two dashed great circles were

drawn through A and A' for extreme corners of 0000. In Fig. 8.28, the twodashed circles touch extreme edges of JP 1101. The same procedure can

be used for JPs corresponding to blocks with one repeated joint set. Forexample, in Fig. 8.29, JP 1031 (an arc of circle 3) is touched at its two end

1

s'rn D-

---L- -5li- --\


------ -'z//

Figure 8.28 Limiting gteat circles of JP 1101.

/IIII

I

\\\

'il

ft

/II

K-__z II

\ .l/\I/\-.-._--,fA,//

Figure 8.29 Limiting great circles of JP 1031.


Figure 8.30 Limiting great circles of JP 311l.

points by the two dashed great circles. Similarly, the two envelope greatcircles of JP 3111 are shown in Fig. 8.30. In each of these examples, theextreme, envelope great circles are the limit planes of the JP, correspondingto the planes normal to vectors (Iri x d) discussed in the preceding section(where f,, are the two extreme edges of the JP with respect to tunnel axis d).

4. For each limit plane of the given JP (each dashed circle in Figs. 8.21 to8.30), measure the dip angle and dip direction, using methods discussed inChapter 3. From the dip and dip direction and the orientation of d, theangular intervals can be calculated. Alternatively, using the stereographicprojection entirely, measure the orientations of the traces of the limitplanes in the tunnel section. Both methods are illustrated in the followingsections.

Galculating the Angular Intervals from the Dipand Dip Direction of the Limit Planes

The angular interval for the maximum key block corresponding to a givenJP runs from 4r to r7, with the understanding that

4t-Ir<180" (8.23)

lf tt, - 4r ) L80", reyerse the subscripts. The two v€lues of q are calcu-lated from the following equation, derived in Section 4 of the appendix to thischapter,

Computation of the Maximum Key Block Using Stereographic Projection Methods 275

(8.24)4:90(l + s) * sin-1 [sin a, sin (B - F)l, ^ I I if JP is inside the limit circle

wnere " : 1-1 if JP is outside the limit circle

dr, f r : dip and dip direction of the limit plane

f : direction of d

(Note that the angular interval formula is independent of the plunge, a, of thetunnel section. However, a1 and f r were computed from limiting circles, whoseorientation does depend on both u and f, so the results do incorporate the plungeof the tunnel.)

The angular intervals were calculated using (8.24) for JPs 0000, 1101, 1031,

and 3111. The results are summarized in Table 8.10. When the tunnel cylinderaxis 6 is horizontal, (8.24) may be replaced by a simpler expression,

4 : 90(1 + s) * ur sin (p - f ,) (8.25)

The derivation of (8.25) is also presented in the appendix to this chapter.

Determining the Maximum Removable Areaby Stereographic Projection

Having found the limit planes enveloping a JP, the maximum removablearea can be determined graphically by finding the traces of the limit planes inthe tunnel section. The latter is the plane perpendicular to the tunnel axis, d.

In Fig. 8.31, for JP 0000, the limit planes, planes b and c, are seen to intersectthe tunnel section at points -B and C, respectively, 56" above horizontal from theeast and 9o above horizontal from the west in the plane of the tunnel section.The traces represented by ,B and C have been laid off in a view of the tunnelsection looking south [with the tunnel axis rising from the paper toward theobserver, as in Figs. 8.14(a) to 8.24(a)1. Since JP 0000 is inside the circle of limitplane b, the block lies above the trace of plane b. Similarly, since 0000 lies insidecircle c, the block lies above the trace of plane c. With this information, and theshape and size of the tunnel, the maximum removable area can be drawn[compare with Fig. 8.la(a).

As a second example, reconsider JP 1101, which lies between limit planesb and c in Fig. 8.32. These limit planes project to the tunnel section at points.B and C, respectively, 94' above horizontal from the west, and 49" above hori-zontal from the east in the tunnel section. The JP lies above limit plane D sinceit is inside the circle of this plane and therefore it is seen above the trace of planeb in the tunnel section (which is, again, drawn looking south). Similarly, JP 1101is outside the circle of limit plane c so that the block lies below the trace of thisplane in the tunnel section. With the known shape and size of the tunnel section,the maximum removable area is drawn [compare with Fig. 8.20(a).

li'{o)

TABLE 8.f0 1r[g;1i6nm Removable Areas

JPFigure Code

Dip ofLimit Plane

(deg)

Dip Direction ofLimit Plane

(dee)

Limits el and 41

of Angular Intervalb(deg) rt r or Ir

II

-1I

-lI

-1-l

0000

ll0l

8.29

8.30

8.27

8.28

31.160.855.586.337.860.873.O

31.1

163.025r.2246.6267.922t.9251.22s9.8163.0

t71.3235.7

49.1

26s.724.1

235.770.2

35r.3

1031

31l l

4t4t4r,lr4t4r4t4t

aWhere S: +1 if JP is inside the limit circle, -1 if JP is outside the limit circle.bThe angular interval runs from 4r to ry clockwise ; see Fig. 8.12.

t


(b)

Figure 8,31 lntersection of limit planes with the plane normal to the tunnelaxis (looking south) for JP 0000.

277

\\\Limit

\ Plane c-\- \-\--- --'\'

Plane b


(b)

Figure E.32 Intersection of limit planes with the plane normal to the tunnel axisfor JP 1101.

(a)

Removable Blocks of the Portals of Tunnels

REMOVABLE BLOCKS' OF THE PORTALS OF TUNNETS

Portals of tunnels present increased opportunities for block movement sinceeach of the tunnel interior surface planes intersects the free surface. Moreover,near the free surface the rock tends to be weathered, the discontinuities to beweaker and more numerous, and the action of surface and groundwater moretroublesome. For these reasons, portal construction is often burdensome andexpensive. Figure 8.33 shows three examples of tunnel portals. Because theintersection of the tunnel surfaces and the ground surface creates complex

(b)

Figure 8.33 Photographs of portals of tunnels: (a) in diatomite, near Lompoc,California; (b) in metamorphic roct.

(a)


(c)

Figure 8.33 (Continued) (c) In granite, after initial blast following shooting of"relief hole," for the access tunnel to Kerckhoff II underground power house.(courtesy of Pacific Gas and Electric Co.)

sP (wl)

Figure E.34 SP and EP for the floor and face of the tunnel portal.


blocks with curved edges, the general problem is difficult. For purposes of expla-nation and demonstration, we will assume a polygonal tunnel section and planarfree surfaces.

The Free Planes of the Portal

The free planes of the portal are the interior walls, roof, and floor of thetunnel and the free surfaces of the natural ground at the location of the portal.In Fig. 8.34, these have been labeled Wt through Wu, for a tunnel of rectangularshape with horizontal roof and floor. The planes are described in Table 8.11.

TABLE 8.11 Planes of the Tunnel Portal

Dip, a Dip Direction, pName (deg) (deg) Description

Wt

WzWsW+

0 0 Floor of the tunneland the portal

60 223 Face of the portal0 0 Roof of the tunnel

90 158 Northwest wall of thetunnel

Ws 90 158 Southeast wall of thetunnel

It will prove convenient to adopt a convention that the positive directionof the normal to each face is that which points into the free spqce. Let fi, be thepositive normal to plane W,.Thenby U(fi,) we mean the half-space of W, thatcontains r0r. The vectors fi, are as follows:

r?1 :(0,0, l)fr z : (sin 60' , sin 223" , sin 60' . cos 223" , cos 60")

rfl3:(0,0, -l)fir: (sin 158", cos 158", 0)

frs : (-sin 158o, -cos 158", 0)

The removable blocks of the portal of the tunnel are the blocks of the tunnelthat have portal face planes Wt or W, as additional boundary planes.

The elements of tunnel portals are:

. Free planes W, and W2 (see Fig. 8.3a)

. Edges Err, Ern, Err, and.E , (see Fig. 8.35)

. Corners Ctz+, Crrn, Crrr, and Ctr, (see Fig. 8.36)

We will compute the EP, SP, and the removable blocks of each element. Freeplane W'1is the portal base; the rock mass is in the lower side of this plane; then

EP : L(fr r)and SP : U(frt) (see Fig. 8.34)

282 Block Theory for Tunnels and Shafts Chap. I

Figure 8.35 SP and EP for edges of the tunnel portal, assuming a rectangulartunnel shape.

Free plane Wz is the face of the portal; the rock mass is in the lower sideof this plane; then

EP : L(fr r)

SP : U(fir) (see Fig. 8.34)

The Edges of the Por.tal

There are four edges in the portal-.Erzt Eztt Ey, Lfld Err-as shown inFig. 8.35.

The removable blocks of portal edge E,1are those blocks that have freeplanes W, and Wl as faces.

Each complex block of edge 812 can be divided into two types of convexblocks with excavation pyramids, respectively.

EPI : L(fr t)EP2 : L(fr r)

and

Removable Blocks of the Portals of Tunnels 283

Figure 8.36 SP and EP for the corners of the tunnel portal.

Then, using Shi's theorem for nonconvex blocks, the criteria for removabilityare

JP+ @

JPnEP:@ or JPcSPEP: EP, u EP2: L(frt)u L(1fi2)

sP: -EP:U(fi).U(fi2)

and

where

then

For Err,

and

For edge -82r,

and

Ern, Err, and E2s are convex edges, so for edge E2n,

EP: L(frr).L(t1t4)SP:U(frr)uU(lfi4)

EP: L(frr)^rL(ti)3)

SP: U(frr)uU(ti,3)

EP: L(frr).L(vDs)SP : U(fi r) u U(tt) s)

(8.26)

(8.27)

(8.28)

(8.2e)

(8.30)

(8.31)

(8.32)

(8.33)


1111

Figure 8.37 EP and removable blocks of edge .Ezr (see Fie. 8.35).

Figure 8.35 shows the EP and SP of these edges. Figure 8.37 shows the EP andthe removable blocks of edge E2r; the removable blocks are 0000, 0001, 0010,0011, 1001, 1011, and 1101.

From Fig. 8.37 we can see that the EP of edge E* is very small. Accord-ingly, the SP is very large and there are many removable blocks. The portal isa critical position of a tunnel, and edge E2, is the most critical element of theportal.

The Gorners of the Portal

There are four corners, Cr"n, Crrr, Crrn,and C2rr, in the portal as shownin Fig. 8.36. The removable blocks of portal corners Cqr arethose blocks havingfree planes W,, Wy and Wo as faces.

Each complex block of corner C23a c?flbe divided into two kinds of convexblocks, with excavation pyramids, respectively.

EPr : L(frr) n l,(fi3)

EP2:L(frr)nL(fi4)(8.34)

(8.35)

(8.36)

(8.37)

Then using Shi's theorem for nonconvex blocks, the criteria for removabilityare applicable if

EP:EP' U EPz:L(fr") n(Z(#3)UI(la4))

and sP : _EP : U(fi) v (U(ti,) n U(tn/)


Similarly, for cornet Czss we have

EP : L(frr) n (L(fi ) u L(lAr))

SP: U(frr) u (U(f,3) n U(f/s))

Corners Cr2n and C12, are different from corners C2rn and C2tt (Fig. 8.36).

Each complex block of corner Cr"n can be divided into two kinds of convex

blocks with excavation pyramids, respectively:

285

(8.38)

(8.3e)

(8.40)

(8.4r)

(8.44)

(8.45)

EPr : f(fir)EP2 : L(fr r) n L(fi,4)

Then using Shi's theorem for complex blocks, the criteria for removability

are applicable ifEP : EPl u EP2 : L(frr) u (L(fi,) n f,(fi4)) (8.42)

sP : -EP : u(fi) n (u(tir) u u(fi4)) (8.43)

Similarly, for cornet Crzs,

EP: L(fr') u (L(1fi) n L(fis))

SP: U(fr') n(U(lfir) u U(las))

Figure 8.36 shows the EP and SP of these corners. Figure 8.38 shows the EP

and removable blocks of corner Czsni the removable blocks are 0000,0010,

,fu""'

101 t

/,r,

1111

'ffi-,\

II

IItII

,ru%e

Figure 8.38 EP and removable blocks of corner Czr+ (see Fig' 8.36).

286 Block Theory for Tunnels and Shafts

1111

Chap.8

011 1

0011

Figure 8.39 EP and removable blocks of corner C12a (see Fig. 8.36).

0001, and 0011. Figure 8.39 shows the EP and removable blocks of corner Crrn.The only removable block belongs to JP 0010.

Summary of All Removable Blocks

Using formulas for the EP and SP for each element of the portal, asestablished previously, we can compute all of the removable blocks of eachposition of the portal. The removable blocks of each element are shown byTable 8.12.

TABLE 8.12 Removable Blocks of the Portal

"'l iliti

,ffi,

Position Space Pyramid (SP) Removable Blocks

lvrWzErzEzsEz+EzsCn+CnsCzt+Czts

U(frr)U(fr2)u(fr) n u(fi2)U(fii v U(frs)U(fri u U(fi,4)u(fr) u u(fis)U(fr) n (U(r?z) uu(fr) n (u(fr) vU(fr2\ u (U(fir) nu(fr) u (u(o3) n

0100,00100000,0010,000100100000, 0001, 0010, 0011, 1001,0000, 0010, 0001, 0011, 011 I0000, 0010, 0001, 1000, 1001,001000100000, 0010,0001,00110000, 0010, 0001, 1001, 1101

1011, 1101

1100, 1101

U(fr+))U(r0s))u(fi+))U(frs))

Chap. 8 Appendix 287

APPENDIX

PROOFS OF THEOREMS AND DERIVATIOIUS OF EOUATIONS

1. PROOF OF TUlIINEL AXIS THEOREM

Axis Theorem. JP is a removable block of a tunnel if and only if

+a + rP (1)

First we prove that ifde JP or -deJP (2)

then JP is not a removable block.

Proof. Suppose that there is a removable block having joint pyramid JP.

From the proposition

EP : U(fr(lJ) u U(ir(0^)) (3)

and EPnIP:@ (4)

But U(fr(O )) = *a (s)

u(fi(|-)) = ta (6)

where "=" means "contains." So

EP: U(ft(?r)) v U(n(0*)) = *a (7)

ff +A or -d e JP, then

JPnEP= 6or-d (8)

therefore JPnEP+g (9)

But this is a contradiction since this block cannot be removable.

Next we prove that if +A + JP, there is a removable block in the tunnel

with joint pyramid JP.

Proof. Suppose that

JP : hu@,) (10)

where fi, is the normal of plane f, poioliog into the joint pyramid JP. Because

a + IP: A U@,) (11)

we can find an i such that

a + u(fi)a.fu <0

(12)

( l3)

Because

-a + JP: hu@,)t:1

we can find a 7 such that

or

Choose fi* such that

-a + u@)

-d-n1 { O

k+i and k+jThe pyramid

JPo: U(fr,).U(ft).U(nk)has three edges vectors (see Fig. 8.40):

f,i : (fi, x fir).sign [(fi, x fr).firjI*r : (fio x fi,).sign [(fir x fi,).ftiI,r : (h x fe).sign [(fit x fi).fi]

SinceA x B. C : B x C . A: C x A, B, then

Itr : (fi, x fir).sign [(fi, x fi)'fir)l*t : (fir x fir).sign [(f, x fi).fir]It* : (4 x fi*).sign I(fi, x fi).fir|


(14)

(15)

(16)

(17)

(18)

(le)

(20)

(2r)

(22)

(23)

(24)

Suppose that

mr : (f, x fi,) x ,6 (25)

and r2:-e.l(firxfi,)xdl (26)

where e is a small positive number (see Fig. 8.40). Expandingr we have

mr : ftt(fi,,il - fi{fi,.A) e7)IIr2 : -e[fi{frk'il - nk(frt'il] (28)

mr.f11 :0 Q9)

"rz' rrr

:-'::;,ffiI :,Tiffi \?!,;:*r(n'

x n) nrl

From (13), we have

mr.fr1 (0 foranye>0Itrr . frr : lfi{fi,,A) - fi,(4.6\1.(ftr x fit).sign [(fi, x fi).fir[

: (fi,.illnt,(fi,, X fi,)l.sign l(fr, x fi).fir|: (fit.d)l(fi, x fi,).it*l.sign[(,0, x fi).fi1|

(30)

Chap. 8 Appendix 289

Figure 8.4,0 Projection of vectors introduced in proof of tunnel axis theorem.

From (12), we have

m1 .fps(O (31)

nr . Irr : lfit(fi,.d) * nlfrt.ill.(fit x fip).sign l(fi, x ft)'firl: -(frt'il\@t x n)-frl.sign l(ff, x fi).fio|: -(fii.il1(!, * ft,).fiel.sign l(ft, x fit).fio1

From (16) we have

m, 'fro ( 0

From (29) and (30), we have

(m,+Dz).I,r(0

(32)

(33)


Choose 6 small enough; from (31) and (32) we have, respectively,

then

Because

therefore

Recalling (25) and (26), we find that

(m,+mz).Io,(0(mr*f,2).Ii"(0

U(m, * mz) n JPo - @

JP c JPo

U(m,*mz)nJP-O

mr:(ft,Xfir)Xd12: -el(fi*xfi,)xdl

(mt*mz).d:0

(34)

(3s)

(36)

(37)

(38)

(3e)

(40)

(41)

(42)

(43)

then

We can find 01 : 0^ such that

fi(?r): fr(|-): nr * mz

(u(n(o))u u(fr(o-))) n JP: @

From the proposition there is then a removable block in the tunnel with jointpyramid JP.

2. PROOF OF PROPOSITIOIU ON ANGUTAR RELATIONSHIPSFOR AN EMPTY INTERSECTION

u(n@)) n JP - g if and only if 41 10 1rt, (B.z)

(see Fig. 8.11).(a) If 4110 aq,, then;A(0) n JP - @.If ,tr<e<q,,then there are

values a > Oand D > 0 such that

n(il: afi(q,) * bn@,)

Let the edges of the JP be lr,lr,...,r,.The extreme lines of the orthographicprojection of JP in the tunnel section are I', andIi. Accordingly,

fr(tt). Ir : 0

n(n). fr<0, i : 2,3,...,1fi(n). Ir : 0

(44)

fi(tD. Ir ( 0, i: 1,2,3,...,1- |[The spatial relationships of I,, I!, and fi(4) are shown in Fig. 8.14.] If ttr] 4,,add 360" to tl,. The rotation from Q@)to QQD is clockwise, so that

0 t1, - tlt < 180" (45)

Chap. I Appendix

Applying (44) to (43) gives us

n@). rr : afr(q') ' r, * bfi(tlt)'\: bfi(qr). rr ( 0 (46)

n(il.\: afi(Tr)'r, * bn@,) 'r, ( 0 for i: 2,3,.-.,1- | (47)

n@)' \ : afi(4,)' r, I bfi(tl)' rt: afi(q). f, ( 0 (48)

Since n(0) .I, ( 0 for all i, and JP is contained between f t and I,:

u@(0)) n JP - @ (4e)

(b) If U(ft(0)) n JP - a, then 41 10 < q,.

Suppose that 0 is not between 4t and 4t; then

n@) : aft(ry) + bfi(tt) (50)

where a:i0 or D ( 0.If a ( 0, using(44),

n(o)' \ : afi(4,)' I, + bn(o)' rt: aff(tt).I,Z 0 (51)

Similarly, if b < 0, using (44),

fi(O)' fr: ah(n) 'f, * bft(4)'It: bfi(ttr). It ) o (52)

Since JP includes /, and d,U(fr(0))nJP+@ contradiction (53)

3. PROOF OF THEOREM OtU THE MAXIMUM REMOVABLE AREA

(a) If .B is a removable block of JP in a tunnel, then ,B belongs to the maximumremovable arca of JP, namely,

B c. {u(-fi(tt), QQt)) n U(-fi(4), Q(il)} (8.8)

(b) There is a removable block .B of JP such that the projection of B along d isexactly the projection of the maximum removable area of the JP.

Proof of (a). Let

B : R (u(6,,e)) ^

TB (see Fig. 8.12) (54)

where d, is the normal to plane P,that points into the block, U(6,, Q)is thehalf-space U (6,) that passes through Qu and TB is the rock mass outside the tunnel.For any point l, such that A a B,

A e U(O,,Q,), i: 1,. . . ,fl (55)

and U(6,,A)-U(6,,Q,), i:|,...,n (56)


hu@,, A)) - (A <0,, o,t)

Let B(A): ,0 U(0,, A) n TB

then B(A) c B

[Figure 8.12 shows the relationships between B(A), B, I',, and II.lBoth B(A) and ^B have the same JP; that is,

JP: hu@,)

(s7)

(s8)

(5e)

(60)

so ^B(l{) is also a removable block of JP. In the boundary of the tunnel, if thereis a point Q(0) e B(A), by virtue of (60), JP n U(fr(0)) - g. Then by theproposition on angular relationships for an empty intersection,

41 10 l rtt (61)

Let Q(|r) and Q(?r) be limiting points on the boundary of B(A). A(l) is apyramid with apex at A and having the same shape as the JP. Then if ,,{ is anypoint in B,

B(A) c {U(-n(q ), Q(q,) n u(- fi(q), Q@D} (8.8)

Proof of (b). The projection of {U(- fr(qr), Q@)) A U(-fr(tt,), QQD)}along d is shown in Fig. 8.12 as region B(A) with apex at A. This projection isalso the projection of JP after moving its apex to A. The removable block

": (n u(fi,,,e1) n Er (62)

also has the same projection and if A is at the intersection of I', and I'1, B isexactly equal to the projection of the maximum removable area of JP.

4. DERIVATION OF EOUATIONS (8.24) and (8.25)

Recall the tunnel coordinate system (ft',9o,20),

Xo:2xdyo: dx(AxA):Qfrxt)xdzo:d'

The limit planes pass through d : ?oi therefore, their normal vectors fr, and ft,are in the *o!o plane. The angles 0 : q, and 0 : q, from !o (toward io) to fi1or fi, delimit the angular interval (see Fig. 8.12).

We have previously found the dip and dip directions of the planes normalto fi, and fi,.In order to compute the angular interval, compute

ft'.fi, and 9o.fi,as follows.

(8.1)

Chap. 8 Appendix

From (8.1),

xo : @ x d): (0, 0, 1) x (sin a sin B, sin c cos B, cos d)

: (-sin d cos B, sin a sin p,0)

so fto : (-cos /, sin 8,0) (63)

yo : d x ft,o: (sin c sin f, sin d cos p, cos a) x (-cos B, sin p,0)

so I o : (-cos a sin B, -cos d cos B, sin a) (64)

2o :4 : (sin a sin f, sin d cos B, cos a) (65)

In the following we denote both fr, and fi,by fi,. Knowing that the normal plane

of fi, has dip c, and dip direction B,, then

fir : (-.S sin o, sin B1, -S sin dl cos f ,, -S cos dr) (66)

I d t 1 if JP is inside the limit circlewnere o : 1-1 if JP is outside the limit circle

From (63) and (66),

fir'fto: S(sin a, sin p1 cos f - sin dr cos Bl sin B)

or frr. fto - -,S sin a, sin (B - f t) (67)

From (64) and (66),

fir.lo: Scos a sin p sin a, sin f , * Scos d cos B sin u, cos P,

-Ssindcosdlor fir.lo:Scosssinfircos(P-F)-Ssin&cosdl (68)

Because d is in the limit plane and fit is the normal of this plane, then

d'fi', : g

,S[sin usin p sino, sinp, * sindcospsindr cos F, * cosdcosa,] :6sincsina,cos(B - ft) *cosdcosdr:0

then cos (P - F): -ff#ffi . o

(6e)

because 0 { a ( 90", 0 { o, < 90". From (68) and (69) we know that

sign (fit'.fo) : -,S (70)

Denote the position angle of fi, in plane xoloby q;then

o : {ltl^-'(fi'.'ft,.)', ''-io ) o

[180-sin-l (fir.ft), fir.\ol-L (71)

tl :90(l - sign (,ar--fo)) * sign (fir.9).sin-l (fir.ft) (72)

Substituting (70) and (67) into (72), we get

':illl i;i;;;il-,i;lJJ#r;:i - f')l

Block Theory for Tunnels and Shafts Chap. I

The formula for the angular interval is, finally,

4:90(l + s) * sin-'lsincl.sin (P - f ')] (8.24)

A horizontal tunnel is a special case with d : 90o. When d : 90",(69) becomes

cos(B - f'):0so sin(p-F):l or -1Formula (8.24) then becomes

4: 90(1 + s) * sin-'[sin d,.sin (p - F)l- 90(l + ,S) * sin (P - B,).sin-l (sin ct)

because o ( c, =

;::::* t' * dr'sin (P - F) Q4)

sin(/-F):l or -1Assume that d is upward; then -d is downward.

Given the dip a and dip direction P of the plane normal to d, then

6 : (sin a sin f, sin d cos p, cos a)

The formula for calculating the angles to the limit planes of JPs intersecting ahorizontal tunnel is then

(8.2s)

where

4:90(I +s)*drsin(f - fr)I L if JP is inside the limit circle

o : 1-1 if JP is outside the limit circle

(73)

chapter 9

Tho Kinornaticsand Stabilityof RemovaloleBIoCKS

The previous chapters produce methods to determine the removability of ablock intersecting a free surface. Certain of the blocks created by the intersec-tion ofjoints and excavation planes prove to be finite and nontapered. We havecalled these removable. In Chapter 4 the removable blocks were divided intothree types:

I. Key blocks, which can be expected to move when the excavation is made,unless support is provided

II. Potential key blocks, which are in a position to slide or fall when theexcavation is made, but will not because the available friction on the facesin contact is sufficient for equilibrium

III. Stable blocks, which cannot slide or fall even when the angle of frictionon the faces is zero because the orientation of the resultant force promotesstability

Only removable blocks merit stability analysis. Joint blocks behind thefree face have no space into which to move until some surficial blocks vacatetheir positions. This is also true of tapered blocks (type IV). Infinite blocks(type V) intersecting the free surface can move only by becoming finite throughrock fracture. This possibility is ruled out in this book (although, as noted inChapter 1, it is a subject worthy of further consideration for weak rocks or forunderground excavations at great depth).

Having identified the removable blocks as candidates for further analysis,it is convenient to effect their subdivision into the three types on the basis of

The Kinematics and Stability of Removable Blocks Chap. 9

two kinds of analysis. First, a mode analysis is performed to distinguish stableblocks (type III) from potential key blocks, or real key blocks (types II and I).The direction of the resultant force must be specified but no joint-strength prop-erties are needed. Finally, a sliding equilibrium stability analysis is performed,with friction angles input for each joint surface in contact, in order to separatepotential key blocks (II) from real key blocks (I). The results of the modeanalysis guide the stability analysis.

The kinematic mode analysis discussed in the next section has not beenpresented previously. Limit equilibrium stabilityanalyses have been publishedfortetrahedral blocks by Wittke (1965), Londe (1965), John (1968), Goodman(1976), Shi (1981), and others. We shall extend this established technique toblocks with any number of faces.

MODES OF SLIDIIUG

In this section we establish relationships connecting the direction of the resultantforce, on an incipiently sliding block, and the direction of sliding. Coupled withother kinematic constraints and a specific direction for the resultant force, theserules will permit us to establish which, if any, sliding mode is applicable to eachJP.

We shall denote a specific removable block by B.Ignoring rotation, everypart of -B undergoes motion described by the same vector. The unit vector ofthe direction of sliding shall be termed .f. The discussion will imagine a stateof limiting equilibrium, in which motion ensues without acceleration.

Under a given set of forces, .B cannot be expected to be exactly in a con-dition of limiting equilibrium. To bring B to a limiting state, we add a ficticiousforce -FS, as shown in Fig.9.1. When Fis positive, the block tends to slideunless artificial support is added. Conversely, a negative value of F implies thatthe block B is safe from sliding. Thus F can be used as a vehicle to discuss thelimiting conditions.

Forces Acting on B

There are three contributions to the forces acting on block .8.

1. The resultant (N) of the normal components of the reaction forces on thesliding planes. LetA, be a unit vector normal to joint plane /, directed intoblock .8. Then the normal reactions are

N:EN,o,

We assume that there is no tensile strength across the joint, so

&>0

(e.1)


Figure 9.1 Explanation of ficticiousforce F and sliding direction vector ^i.

2. The resultant T of the tangential frictional forces:

Tr : - N' tan $'3

and the resultant of this and the ficticious force ts

-Ts : -t tr/, tan 0$ - Fs (9.3)

so T:ElI,tan6,*F (9.4)

For a potential or a real keyblock, by design, sliding will occur if f, : g.

Since positive F implies sliding, Z > 0.

3. The resultant r of all other forces acting on block B, including weight,seepage forces or external water pressures, inertia forces, and supportforces from rock bolts or cables. The force r will be termed the activeresultant.

The condition of equilibrium for a potential or real key block ^B is

t+FN,0,-Li:0 (9.5)

with f>0 and I/,>0From the theorem of removability, Chapter 4, the sliding direction ^f of remov-able block B belongs to the JP of block ,8, that is,

(e.2)

:l

.icJP (e.6)


Figure 9.2 shows a key block translating freely from its home, openingfrom each face. We shall call such a mode lifting. Since no joint plane remainsin contact, ,f cannot be contained in any joint plane.

Figure 9.2 Lifting.

Since no joint is in contact, N, : 0 and (9.5) becomes

t:TSFor a key block or potential key block, T > 0. Therefore,

8:f (e.7)

The joint pyramids have been defined as closed sets, meaning that a JP includesnot only the space inside the pyramid but the boundary faces and edges as well.The condition for lifting is that ,f be contained inside JP but not in its boundary.

Proposition on Lifting (Free Translation). If S is not parallel to any planeof a JP, then the necessary and sfficient condition for B to satisfy the equilibriumequation (9.5) ,r that S : i.

Sliding on a Single Face

Figure 9.3 shows an example of a block sliding along one of its faces.

In this case, ,f is parallel to only one plane of B, plane i, and the sliding direction,f is the orthographic projection of r on plane f.

^f:jl

Sr:9t#+&

I

where (e.8)


Figure 9.3 Single-plane sliding.

and frt is the upward normal vector to plane i, determined by (2.5) and (2.7).

In this case, all of the joint planes except plane i will open and i c JP ) P,,

where P, represents plane i. This is stated formally in the following proposition,

which is proved in the appendix to this chapter.

Proposition on Single-Face Sliding. If the sliding direction S lies withinonly one plane, P,, the sfficient and necessary conditions that the removable

block B satisfy the equilibrium equation (9.5) is i : S, and At' t < 0, where S,

is the orthographic projection of s on plane Pt[given by (9.8)1.

Sliding on Two Faces

Figures 9.4 and 9.5 show examples of blocks sliding along two planes Pt

and P1 simultaneously (i.e., along their line of intersection) since that is the

only direction common to both planes. The sliding direction ^i is the directionalong the line of intersection that makes an acute angle with the direction ofthe active resultant r.

3 : ,fir : #ihsign ((fi, x fi,) . r) (e.e)

Furthermore, the sliding direction is an edge of the JP formed by the intersectionof planes i and j.

/.a\\/ \s t.-\\

^feJPnP,APt (e.10)

300 The Kinematics and Stability of Removable Blocks Chap. 9

Figure 9.4 Double-plane sliding.

Figure 9.5 Double-plane sliding.

The Silding Force

Proposition on Double-Face Sliding. If the sliding direction is simultane-ously in two planes, P, and Pr, the necessary and sfficient conditions that blockB satisfy the equilibrium equation (9.5) are

ir.^f, < 0a,.0, a o

and s: ,4'x *,,sign((fi, x fir). r)"-1ft1 xry:where 3, and 31 are the orthographic projections of r on planes P, and P1, r€sp€c-

tively, as given by (9.8).

The proof is given in the appendix to this chapter.

THE STIDING FORCE

The equilibrium equations for free translation, single-face sliding, and double-face sliding have been given in the preceding section. Using these equations,we can compute corresponding equations for the sliding force FJ.

Lifting

In this case, Nt :0 for all joint planes and equation (9.4) becomes

F:Tand the equilibrium equation is

TS:rand combining with (9.11) gives us

r: lrl (9.13)

When gravity is the only contributor to the active resultant, F is simply theweight of the block.

Single-Face Sliding

Since there is only one plane in contact, (9.4) becomes

T: Nttan6, * F

(e.11)

(e.r2)

(e.14)

(e.15)

(e.16)

(e.r7)

It is shown in Section 1 of the appendix to this chapter [equation (2)] that theequilibrium equation (9.5) leads to

TS: (0, x r) x I,so T:li,xtlAlso, equation (3) of the same section leads to

Nt:-l'At


Substituting (9.16) and (9.L7) into (9.14), we have

F:l0,xrl *0,.itanQ,where At : -frt sign (fir . r)

Since from the proposition for single-face sliding,

fr'r(0Substituting (9.19) into (9.18) gives

F:\fi, x rl - \fr,. rltan f,

(e.18)

(e.le)

(e.20)

(9.20) is the formula for the net sliding force of single-face sliding along plane

Pi.When the active resultant r is given only by gravity,

r:(0,0,-W), W>O

Suppose that the dip and dip direction of plane P, are a, and p,,

fi, : (sin a, sin fr, sin dr cos B,, cos a,),

1fi, x rl : leW sin c, cos 8,, W sinc, sin Ft,O)l: ((W' sin2 &,. cos2 f , * W2 sinz d,. sin2 f )r/2

- W sin o,

\ft, . r,l : | -W cos q+l : W cos u,

From (9.20), we have

F: W sin a, - W cos d, tan$, (e.21)

(g.2L)expresses the net sliding force when the active resultant r is due to gravity

alone.

Double-Face Sliding

In this case, removable block B slides along planes of joint sets f and 7and all the other joint planes open. The normal reaction force Nr :0 for allI + i or7 and (9.4) becomes

T : NttanS, + il, tanQt + F, T>0, Nd>0, Nr>O

so F:T- Nr tan4, - Nrtanfl

From equation (5) in Section 1 of the appendix to this chapter,

N,@, x dr).@' x 0t): -(r x 0)'(0' x 0r)

(e.22)

-(r x 0).(0, x 0i)mglvlng N,: (e.23)

The Sliding Force

Since & > 0, (9.23) may be written

303

(e.24)A.r _ l(r x fit).(fi, x n)lrrr-W

This follows from the proposition yr . ir < 0, as shown in the appendix to thischapter. Similarly,

Nt: lG x,A)'@' I ft)l1fi, x fitlz

then from (9) in Section 1 of the appendix to this chapter,

T: r..fFor a key block or potential key block, 7 > 0, so

(e.2s)

,- lr.(fi,xfi)l1fi, x firl (e.26)

Substituting (9.24), (9.25), and (9.26) in (9.22) gives, finally,

1.,F:Wfuilr. (fi, x fit)lln,x ntl- l(r x ft)-(fi, x fit)ltanf,

- l(r x fr).(fu * fir) ltan Srl e.27)

Equation (9.21) gives the net sliding force for a key block or potential key blocksliding on plane P, and P1.

An Example of Sliding-Force Calculations

Table 9.1 gives the dip, dip directions, and friction angles of four sets ofjoints. Given active forces,

rr:Wir, fr:(0,0,-1)and t2 : Wi r, i z : (0, 0.866, 0.500)

TABLE 9.1

Dip, a Dip Direction, p Friction AngleJoint Set (deg) (deg) (dee)

Using formulas (9.13), (9.20), and (9.27), we have calculated the net slidingforces along all of the sliding directions. The results are presented in Table 9.2.

| 7r 163 15

2 68 243 203 45 280 404 13 343 30

SlidingPIanes

The Kinematics and Stability of Removable Blocks Chap- 9

TABLE 9.2 Net Sliding Forces (F)

Net Sliding Forces (F)

forrr : VI/ir fOr rz : Wfz

NoneI23

4

l.ool4/0.86W -o:l9wO.IIW

*o.34Wo.72W

-o.nw*0.6rw-o.45W*o.36W

-o.45W

t.oow0.62W0.92W *-o.50wo.35Wo.62Wo.l6w

-0.36W-t.72W-o.o3w

o.l9w

lr2l, 3

1,42132143r4

From Table 9.2 we can see that different resultant directions generate

different net sliding forces. If F is positive, the block slides if it is removable,

meaning that it may be a key block (type I). If F is negative, the block is safe,

meaning it may be a potential key block (type II). However, as we shall see, inthe next section a given excavation configuration does not yield removable

blocks corresponding to every mode.

KIITEMATIC CONDITIONS FOR TIFTIIUG AIUD STIDING

In this section we prove that there is only one joint pyramid corresponding to agiven sliding direction and then establish procedures to identify it. A "modeanalysis" determines a complete list of JPs corresponding to all the slidingdirections. This will be demonstrated in subsequent sections -----/

The Gomplete List of Sliding Directions, Given r

When the direction of the active resultant r has been determined, we can

compute all of the sliding directions.f belonging to a given system ofjoint planes.

. For lifting modes, S : f .

. For sliding on a single joint faces

determined by (9.8).

. For sliding on two faces (i and j),by (e.e).

(f), the sliding directions .f, are

the sliding directions sry are given

Table 9.3 gives the results of a sample calculation, corresponding to thejoint system of Table 9.1 with r : (0, 0.866, 0.500). ',\

Kinematic Conditions for Lifting and Sliding 305

TABLE 9.3 Coordinates of Vectors of Joint Sets

and Sliding Directions for r : (0,0.866,0.500)

Vector

ftrftzfts

fiti.ir3z

.ir

.f+

3tzjtsStc3zs

3zt3s+

0,2764

-0.8261*0.6963

-0.065700.2r86

-0.14870.36060.05990.2w70.68110.9563

-0.5587-0.5196-0.0500

-0.9042-4.4209

o.12270.21510.86600.38900.80410.91160.97550.3873o.4233o.29230.5256o.8262o.9745

0.325so.17460.7071o.97430.s0000.89490.57550.1968

-0.2113o.89770.59730

-0.6415-0.2174-0.2185

According to the definition of 'oremovability" given in Fig. 4.t6(a), every

removable block has a sliding direction along which it can be moved withoutcolliding with the adjacent rock mass. It follows then that for such a block

i e JP (9.28)

Lifting. When a block is lifting, ^f is not parallel to any plane P,. Then

since d, is the normal directed into the block, for each plane / of the block

i.a, > 0 (e.2e)

Single-plane sliding. For the case of sliding in plane P,, i is parallelto P,. Then with (9.28) and the proposition on single-face sliding,

and

i'4>0 foralll,l+ir.61{-0

(e.30)

(e.31)

(e.32)

(e.33)

(e.34)

Two-plane sliding. For sliding on planes i and7, .i is parallel to planes

i and7. Using (9.28) and the proposition on double-face sliding,

,f.4>0 foralll, l+iorjst.6r < 0

ij.0r < 0

If none of the vectors r,31,31, and.i,, are parallel, the ( signs in (9.31),

(9.33), and (9.34) can be replaced by <. Then knowing.i and r, (9.29) to (9.34)provide sufficient information for identifying all of the corresponding d,, and


therefore identifying the JP. An example will be worked out, using first vectorcalculations, and, subsequently, stereographic projection. The joint systemfor the example is determined in Table 9.2. Two cases are considered, corre-sponding to r, : (0,0, -W) and r, : (0,0.866W,0.5WW).

VECTOR SOLUTIOII FOR THE JP GORRESPOIIDI]TG TOA GIVEIT SLIDITTG DIRECTIO]T

Lifting. Since from (9.29) 3.A, ) 0, 4 : fu, if S.fi, ) 0 and 0t: -firif S.fu ( 0. More succinctly,

0r : sigt (s'fi)fu for all / (e.3s)

For lifting, i: f. Using the fi, yalues given in Table 9.3 with r: rz (0,0.866W,0.500H/) yields

U=OL=l

-..7,'t-i

6t: -fir 0s: fit

0z: -fi2 0t: fit

4 : sieR (Su.fit)fu, I + i, jdr : -sign (ji'fi,)ft,

0t: -sign (St'fi,)fit

Accordingly, the JP in the lifting mode under the action of r, is (1100).

Single-face sliding on plane f. In this case, .f : ir. From (9.30)S.A, ) 0 for all I # f. Therefore,

dr : sign (s'fr)& fot I * i

Also from (9.31), | . 6t < 0; accordingly,

0r : -sign (r ' fit)fit

(e.36)

(e.37)

(e.38)

(e.3e)

(e.40)

For example, for sliding on plane L (i:1), using the data of Table 9.3 and, :,,,

0t:ftt0z: -fi20t: fis

6+: fi+

So the JP that slides on plane 1 is (0100).

Double-face sliding on planes i and j. From (9.32), (9.33), and(9.34), following the logic of the previous cases, 6, is determined by

and

Stereographic Projection for the JP Corresponding to a Given Sliding Direction 3O7

Consider, for example, sliding on planes I and 2.Table 9.3 gives 3rz: (-0.3005,0.24M,0.9229) when r : (0,0.866,0.500). Considering each of fib as given inTable 9.3, yields

6r:fit6z : -fi26s: fis

and 8n: fi+

So the JP that tends to slide on planes 1 and 2 is (0100).

Table 9.4 gives the JPs for all modes of failure both for r : (0, 0.866W,0.500W) and for r : (0, 0, -W).

TABLE 9.4 JPs Corresponding to Each Potential Sliding Mode

Sliding Direction r : (0, o, -W) r : (0, O.866W,O5OOW)

r.frSz

,ir.f+

3rzsrrjr+3zt3z+

Ss+

1111

01111011

1001

1100001 I000101101101

10001110

1100010010001 110

1 111

000001100111001 I1001

1101

STEREOGRAPHIC PROJECTION FOR THE JP CORRESPONDINGTO A GIVEN STIDING DIREGTION

All of the preceding analysis can be performed graphically. Figure 9.6 shows

the projections of the four joint sets previously considered. Recall that .f, is theorthographic projection of f on plane i. (The stereographic procedure for find-ing "the orthographic projection of a vector on a plane" was discussed in Chap-ter 3.) Also, ,f,, is one of the two points where circles i andT intersect; it is theone closest to f, that iso the intersection inclined less than 90' with l.

Figure 9.7 shows all sliding directions for the joint sets of Fig. 9.6 whenthe resultant is due to weight alone I r : (0, 0, - W).In this case, .f, correspondto the dip vectors in each joint plane, and ^i' are the lower-hemisphere inter-sections of P, and, Pt.

Figure 9.8 shows all sliding directions for the resultant r : (0, 0.866W,0.500W). The .f, directions are no longer dip vectors.


Figure 9.6 Projection of the jont data from Table 9.1.

Lifting along r. The JP that tends to lift is the one that contains l.(If the resultant force is contained in a plane, there is no lifting mode.) In Figs.9.7 and 9.8, the JP that contains f has been labeled by a "0."

Single-face sliding on plane i. When sliding is in plane r, .ir e(JP n P,) meaning that one side of the JP is the circular arc length that containsSr. Moreover, from the proposition on single-face sliding, 6t . t < 0. Thisrequires that the JP be on the side of plane i that does not contain i. These twoconditions describe a unique JP.

Double-face sliding on planes i and 7. One corner of the JP is.f,7. There are usually four JPs having this corner. The additional conditionsare given in (9.33) and (9.34). The former, S,.Ai { 0, requires that the JP be

on the side of plane j that does not contain ^f,; that is, if .i, is inside the circlefor plane j, the JP must be outside of the circle for plane 7. Similarly, (9.34)requires that the JP be on the side of plane i that does not contain ,ir.

In Figs. 9.7 and 9.8, the JPs corresponding to each mode have been

tlll

Figure 9.7 All sliding directions and modes when r : (0,0, -W).

Figure 9.8 All sliding directions and modes when r : (0,0.866W,0.500W).

310 The Kinematics and Stability of Removabte Blocks Chap. g

identified for both resultant force directions of the previous examples. Com-parison with Table 9.4 shows that the vector and graphical methods giveidentical results.

COMPARISON OF REMOVABILITY AIUD MODE ANALYSES

A keyblock must be removable and have a sliding mode. Overlaying the twoanalyses then shortens the list of potential key blocks. For example, Fig. 9.7examines the same joint pyramids as in the underground chamber case studiedin Chapter 7. The reference circle is the projection of the roof and floor of thechamber. In particular, removable blocks in the roof are those whose JPs areentirely outside the reference circle and removable blocks of the floor are thosewhose JPs are entirely inside the reference circle. Comparison of Figs. 7.1I,9.7, and 9.8 gives the sliding directions for each removable block. Table 9.5summarizes the comparison.

TABLE 9.5

Removable Blocks:

Mode of Slidingunder Gravityr : (0, O, -W)

Mode ofBlock I Sliding with BlockType r : (0,0.866, 0.500)W Type

Of the rooftOl1

1101

Of the floor01000010

Single face along3z

Double face along3zt

NoneNone

IorII

IorII

UIIII

Double face along I or II3rz

Double face along I or IIjr+

None IIIDouble face along I or II

3zs

In order to be a keyblock (I) or potential key block (II), a JP needs to beremovable and have a sliding mode. Removable blocks lacking a sliding modeare stable, (type III). (Roman numerals refer to the blctck classification givenin Table 4.1.)

FI]UDING THE STIDING DIRECTION FOR A GIVEN JP

In the preceding section we determined a unique JP given the resultant force r.Here we treat the inverse problem. Recall that only certain JPs have a slidingdirection. Those that are removable but lack a sliding direction correspond tostable, type III blocks. In the following section we establish a criterion for aremovable block to be stable.

Finding the Sliding Direction for a Given JP

The Nearest Vector of a JP with Respect to r

Given the active resultant r and a joint pyramid JP, for any vector f e JPwe denote by (f , g) the angle between f and f. If there is a vector $ e JP suchthat(i,f) is less than or equal to the angle between i and anyvector of JP, $is "a nearest vector of JP with respect to f ." The angle Q, g shall be called"the smallest angle between JP and i." The following three propositions deter-mine the nearest vectors of JPs for all cases. The proofs are presented in theappendix to this chapter.

Proposition 1. If there is a vector rt e JP such that (i,n1 <90", then

there is one and only one nearest vector of JP wtth respect to t.

Proposition 2. If 3 is the sliding direction of IP under active resultant r,then S is the nearest vector of JP with respect to r and r . i > 0.

Proposition 3. If g is a nearest vector of JP with respect to r, and I . r ) 0,

then $ is the sliding direction (i.e., I - i).

Griteria for Stable Blocks

These propositions generate criteria to judge whether or not a JP ccrre-sponds to a stable block.

Criterion 1. A JP corresponds to a stable block if for uny fr. € JP,

fr.r<o (e.41)

Criterion 2. A JP corresponds to a stable block if r, all ,f,, and all ,f,, arenot contained in JP:

i +JPir + JP for all i

.i,, d JP for all ij

(e.42)

(e.43)

(e.44)

The second criterion permits computation of stable JPs. Any JP satisfying(9.42), (9.43), and (9.44) is seen on the stereographic projection as a sphericalpolygon lacking any sliding directions; for example, when r is (0,0,-W),(0000), (0100), and (0010) satisfy criterion 2 (Fig. 9.7). Under active resultantr : (0,0.866, 0.500), Fig. 9.8 shows that (0001), (1011), and (0011) satisfycriterion 2.

Example: Gomputation of Sliding Direction and Mode

By using the propositions about the nearest vectors, we can compute thesliding direction and the sliding mode of a given JP. This is demonstrated bythe following example, based on the same joint system as discussed previouslyin this chapter. For computation, r : (0, 0.866,0.500).


1. Compute the sliding directions using equations (9.8) and (9.9). Forconvenience the coordinates of all .f, and 3r, are repeated in Table 9.6.

TABLE 9.6 Angle (3, f) for f : (0,0.866, 0.500)

SlidingDirection X

Angle with rz (dee)

.ir3z

^ir

^f+

3tzSrrjrcSzs

3z+

sr+

0.2186

-o.14870.36060.0599o.20970.68110.9563

-0.5587-o.5196-0.0500

0.38900.804r0.91t6o.97550.38730.4233o.29230.52s60.8262o.9745

0.89490.57550.1968

-o.2l13o.89770.59730

-0.6415-o.2r74-0.2185

38.336to.2l27.3842.3438.33948.3075.33382.2752.@42;t2

Now compute their angles with r, using (2.26). The results are given in theright column of Table 9.6.

Now for a given JP, calculate which sliding directions are contained inthe JP. Recall that each JP corresponds to the solution set of a system ofinequalities. A sliding direction belongs to a JP if and only if it alsosatisfies all these inequalities. Table 9.7 lists the sliding directions that

TABLE 9.7

Sliding Directions Contained.in JPNearest Vector;Sliding Mode

000000010010o0ll0100010101100111tmo1001l0l01011

1100

1101

l110111 I

3rzStableStableStable.irTapered,$r r.fr+Sz

Sz+

Tapered,fzs

r3s+

^ig

.ic

srz

,f r, .it z, ,it g

$rs, jr+Sr+

Stz, Sz, Sz+

izs,3z*

3zs

f,31,32r il, itz, Sts'ize,.frc32s,324,33431, 3+, ,f r +, .f r l, .ig +

3+, St+,32s, h+

Mode and Stability Analysis With Varying Direction for the Active Resultant Force

are contained in each JP. Each makes an angle with r that was previouslyreported in Table 9.6. The sliding mode corresponds to the sliding direc-

tion that makes the smallest angle with r. The sliding modes are listed

for each JP in the right column of Table 9.7.

MODE AND STABILITY ANALYSIS WITH VARYING DIREGTION

FOR THE ACTIVE RESULTANT FORCE

In the preceding discussion, we assumed that the direction of the active resultant,

f, was constant. This simplified the analysis of sliding modes in the different

blocks. Indeed, problems of block sliding under gravity develop with unchanged

orientation for the resultant force. However, there are other practical problems

in which the resultant force does vary in direction. One class of problems ofthis type concerns water forces on a block in the abutment of a dam; the

magnitude and direction of the water forces on the faces of the block change as

the reservoir level fluctuates. Another class of problem with variable direction

for the actual resultant is posed by analysis of resistance to earthquake forces;

since such analyses must be performed without sure knowledge of the direction

of the inertia forces, parametric studies are performed to determine the most

critical direction.Let us now confine our attention to a single joint pyramid. As the direction

of the active resultant shifts, different modes of sliding and different magnitudes

of the sliding force will be calculated. If i is a unit vector of any orientation,

the range of its stereographic projection is the entire projection plane. We willsubdivide the whole projection plane into regions within which f can move

without changing the mode of failure. The "equilibrium regions" will be labeled

as follows. A region is called "0" if the sliding mode is lifting when I plots inside

the region. The symbol i identifies a region for i in which the mode of failure

for the JP is sliding on planei. Similarly, the symbol y establishes the applicable

mode as sliding on planes i and j, along their line of intersection. When the wholeprojection plane has been subdivided in this w&y, there will be an additionalclosed region lacking any sliding mode. This region will be labeled S, meaning

safe, as sliding is impossible when the resultant force plots therein, even if the

friction angle is zero.Within each equilibrium region, it is possible to construct contours of

friction angle required for equilibrium. The magnitude of a contour at a certain

point within a region will indicate the value of the friction angles on the slidingplanes in order to produce a net sliding force equal to zero. This is the frictionangle required for limiting equilibrium when the resultant force acts in the direc-

tion corresponding to the point in question.

We will first establish a construction method to bound and contour the

equilibrium regions. Subsequently, we will prove that these regions and contours

derive from the equilibrium equations.

313

314 The Kinematics and Stability of Removable Blocks Chap, g

Construction of Equilibrium Regions for a Jp Consistingof One Joint Set

Consider first a case of sliding with only one set of joints. Let fi,,6,, and,rD, be unit normal vectors of plane P, such that fi, is upward, 6, points into theJP, and r0, points out of the JP. The steps in the construction of the equilibriumregions are as follows.

1. Draw the great circle for plane 1. The region inside this circle is theprojection of the JP. Let 0t : (-X, -f , -Z). The radius of the greatcircle for plane 1 and the coordinates of the center for this circle are givenby (3.13) as

R| : -----7-

lzl.A RFu*: Z

and

where R is the radius of the reference circle.

2. Denote the region corresponding to the JP as "0." The region outside ofthe JP is then denoted "1."

3. Assume that the friction angle of the joints of set I is f1. If the activeresultant is inclined exactly Qt from the normal to plane P.-itwill producea net sliding force equal exactly to zero. The locus of points making afixed angle with a given line is a cone; this projects as a small circle aboutthe normal pointed out of the JP (i.e., making an angle of d, with fir).The radius (r) of the required small circle and the coordinates (C", Cr)of its center are given by (3.20) and (3.22). Figure 9.9 shows a family ofsuch small circles, drawn as dashed circles, for a plane (p,) with 6,: L3"and B : 343o. In this case JP is the upper half-space, so

(e.45)

rt RYl-r, : --=-'z

and

fir : (sin a sin p, sin a cos B, cos a)

6t:fir

frt : -fir

Gonstruction of Equilibrium Regions for a Jp Gonsistingof lr Joints (z 2 2)

Consider the system of four joint sets listed in Table 9.1. The stereographicprojection of these joints is shown in Fig. 9.10 and the JPs are labeled. Considerone particular JP, (ararasat). The method for constructing the equilibriumregions for (arararar) is as follows.

-at' ---- ta

./' \r. \/\\! / ------- \-I t' -/- --- '\. \

l/t/ tt\ \

IIII/I//

tiffi\\!\\\iiruT;l i\t,

_\ .\\:__:::_-- ---- ,rr, ,r,

'.*-

\\--=

--a' ./'--... -\-\-- -----r-// ,r'----

-_--t ,r,\-\-'---------

-------"/'/' ,ra{r/=zo"-t-

Figure 9.9 Contours of required friction angle and equilibrium regions for one

joint set.

Figure 9.10 Lower-focal-point stereographic projection of joint data ofTable 9.1.

1111

316

316

1.


Project the JP by the usual stereographic projection procedure and labeleach of its sides as P,; establish the clockwise order of the sides. Forexample, for JP 1100, the clockwise order is p, , pr, pn,p, (Fig. 9.11).

"6Dr;

Figure 9.ll Lower-focal-point stereographic projection of equilibrium regionsfor JP I100 of the joint system of Table 9.1, with friction angles as given in ttrattable.

2. Compute the edge vector of the JP and find its projection. We can computethe edge vectors and their coordinates using an 'oedge matrix," as estab-lished in Chapter 8. The joint data of Table 9.1 arc the same as used inChapter 8 and the edge vectors of JP 1100 are given in Table g.4(a) as

-lrr, -f , r, Irn, and, frn, whereltt : fi, X fir. On the stereographic piojec-tion, Fig. 9.11, the four corners of JP 1100 are labeled. Cu,with the valuesof the indexes i andT determined by the common sides. The corners are,in clockwise order, crz, crn, crn, and c'3. comparing with the calculatededge vectors,

e ,, : -1rr, e ,, : -frr, e ,o: Irn, e ,o: lrn(tn the above, C,i is the stereographic projection of i,r.;

3. compute the projections (vr,) of normal vectors Qi,,) and construct great-circle segments connectin g w, and, C,r. This can be accomplishec as follows.For JP (arararan), equatipn (8.I2) established a function I(a,). Then

wzwl

Mode and Stability Analysis With Varying Direction for the Active Resultant Force 317

fit: -I(a,)fi, (9'46)

1100), fi r : fit, fr, : fir, ftt : -fir, and fi) 4 :

frr: (X, Y,2)

wi : (Xo, Yo)

where the values of -tr' and y0 are giYen by (3.a). Plot w,, as shown inFig. 9.11 for JP 1100. (In this figure, wn plots off the drawing.)

Using (3.13), compute r, C*, and Cy for the projection circle of the

plane whose normal vector is rD, x i,r. This great circle will pass through

points lr', and C,r. Erase the portion of the circle that is greater than 180".

If rD, x i,, is upward, draw this circular segment counterclockwise fromw, to Cs.;; otherwise, draw it clockwise. Using the same approach, drawgreat-circle segments connecting w, and C,i. Finally, draw the great-circle

segments connectin g wt and wr, proceeding in the order of the sides of the

JP. For JP 1100, connect wrto wr,thenwrto wn, then woto r'3' and'

flnally, w, to wr. For JP 1100, the great-circle segments constructedaccord-ing to this procedure a;te w1C1zi wzCtzi wzCz+i w+Cz+i wtCs.; wsCt+',

wsCrt, wrCrr; wrwzi wzla+i wtws; and wtwr.

4. Identify the sliding mode of each region established in the preceding

construction.

Lifting, mode 0, is defined by the region inside the JP.

Single-plane sliding along plane i in direction ,f, is delimited by the

spherical triangle whose corners are Ct,, wt, and C,o. This is termed

"mode i."Double-face sliding on planes i and 7 along direction ^i,, is delimited

by the spherical triangle whose corners are wi, C11, efld wr. This is termed

"mode ij."No sliding can occur, even with zero friction on the joints, when the

resultant plots inside the spherical polygon whose corners lta tuiT w11

wk,...,wn. This region is denoted,S.

In the example of Fig. 9.I1, the regions are as follows: lifting occurs

in the polygon between Crr,Crn, Crn, and C1r. Single-face slidingoccurson all four faces. Double-face sliding occurs in modes 12,24,34, and 13.

It is important to observe that some modes of double-face sliding cannot

occur; in this example, modes 23 and L4 do not occur.

5. Draw the circle of zero sliding force for modes i corresponding to a given

value of the friction angle f,. This will be constructed as a dashed circle.

Assume that the corners of region i ate w,, C11, Lftd C,o. Because tD, is the

normal to plane i and both e ,, and e ,o ar" in plane i, fi, is perpendicular

to both e ,, and i,o. D.fine t,lin the plane of fi, and e ,,by

For JP Il00 (ataztja+:-fr* Suppose that

Then its projection is

tu : cos (t',)rD, * sin (6)e ,, (e.47)

318 The Kinematics and Stability of Removable Blocks Chap. g

The vector i,, is inclined f, from r0, in the plane fr,e ,,.Similarly, vector trkgiven by

t,r,: cos (rf,)rO, * sin (6)e ,u

is inclined f, from fi, in plane fr,e ,o.Now construct a small circle from t11to lr*. This represents the cone

of constant angle {r with the normal fi, and is therefore the locus of zeronet sliding force bn plane i. The dashed circles in regions l,z,and 3 areshown in Fig. 9.11 (their continuations in region 4 are offthe drawing).In region 1, the dashed line connects points /,, and r1s, both f, fromwr. In region 2, the dashed locus connects tu and t2n, both of which makean angle of 6, with wr. And in region 3, the small circle runs from131 to tro, each of which are fi, from wr. The points tq Lte plotted usingequations (3.4) and the small circle radii and centers are located usingequations (3.20) and (3.22).

Finally, draw the circle of zero net sliding force for modes 1,7'. The cornersof regions f,7 are defined by ,r, Crr, and wr. The point r,, is in the greatcircle wtCtt. Similarly, the point /r, is in the great circle wtC,t. The vectorsf ,1 und f;l w€r€ calculated from (9.47) and t, and t 1, are thlir stereographicprojections. The required locus is a great circle connecting t,, und tr,.To construct it, we calculated the normal to the plane repreiented bythis great circle. That normal, fi,, is computed by

ft,:##:(n,r,z) (e.48)

Project the plane perpendicular to fi, using equations (3.13). This greatcircle will intersect circle wtCtt at t11and will intersect circle wtCu at t11.In Fig. 9.11, the great circle segments tz'trzare drawn with a dashed linein region 12; great circle tz+t+z is drawn in region 24; great-circle seg-ment t+sttr is drawn as a dashed line in region 34; and segment tsrtttruns through region 13. These circles have validity only within theappropriate regions and therefore are erased from all points outside theappropriate equilibrium regions.

Examples

We have examined in detail the example posed by construction of theequilibrium regions fcrr JP 1100, with the joint system of Fig. 9.10 and Table 9.1.The friction angles for the four joint sets stated in that table were used in theconstructions.

Figure 9.12 shows additional contours of zero net sliding force for 10oincrements in friction angle. In the i,7 regions, the contours can apply only if6t : f1' This type of diagram can be useful because it shows, at a glance, whatfriction angles are required to achieve equilibrium corresponding to any posi-tion of the active resultant.

Figure 9.13 examines the equilibrium regions for JP 1101, with the joint

6.

4

Figure 9.12 Equilibrium regions and contours of required friction for JP 1100,

assuming equal friction angle on each plane.

Figure9.l3 Equilibrium regions for JP 1101, with friction angles as given inTable 9.1.

t.rt\r \ i

X\;'\ i'i\i}. 1

\ i 11. I. | | :'

41ii.ry?:,i:'.--"


system and friction angles given in Table 9.1. The position of 1101 relative tothe other joint pyramids is shown on Fig. 9.10. JP 1101 has onlythree cornersand three sides. Even though there are four joint sets, the particular JP has onlythree faces. The sliding regions that exist for JP 1101 are 2, 3, and 4, and 23, 34,and 24. Figure 9.14 shows the contours of zero net sliding force correspondingto a range in friction angles in 10' increments.

70" /qo"r 60o

3

'\\

I

\\\\

r\'r \\\\\\\

23

Figure 9.14 Equilibrium regions and contours of required friction for JP l l0l.

Special Gase: Equilibrium Regions for a Jpwith Only Two Sets of Joints

A JP with two sides presents an important special case of the previousconstruction procedures. Consider the joint data given in Table 9.8. The jointpyramid (arat): (10) is shadedin Fig. 9.15. This Jp has twoedges, e ,rand

TABLE 9.8

0= 10o

Vrt ---\r/./ ,/ ,/ ,.-/ / ,/ ,,' _---4,///,.' /-

/,/ /./ ,,/ --/// / ,/'ll;jt+1'-

Joint Set

Dip, a(dee)


Friction Angle(dee)

20

40

243280

I2

6845


Figure 9.15 Equilibrium regions for JP 10 in the two-joint system defined inTable 9.8.

-e ,r, whose projections are C* and Cl2, respectively. The vectors rD 1 and li/2

are outward normals to planes L and 2, respectively. Vectors fi1and fi, arc bothperpendicular to e ,, and -e ,r.Point 1,, is the projection of t tz; i, is in the

plane of r0, and, e t2 and makes an angle f , with r0r. Point t'12is the projection

of vector t\, andis in the plane of r0, and *dr , (and e ,r), inclined {, from rD1

toward -e ,". Point /r,, similarly, is the projection of vector ir,, which makes

an angle 6, wittr i,rinthe plane fr"e ,L; and t'1ris inclined $, from r0, toward

-e ,r. The regions are as follows:

. Mode 0 is in the projection of the JP.

. Mode 1, representing single-face sliding along plane 1 in direction .i,,is in the region between great circle l and great circle CrrwrC'1r; this

region is contiguous to the JP.

. Mode 2, representing single-face sliding along plane 2 in direction ,ir,is in the region between great circle 2 and great circle C12w2C'1r. It iscontiguous to the JP.

. Mode l2 is in the spherical triangle between great-circle segments

Ctzwt, Ctzwz, and ur,wr. It represents sliding on planes 1 and 2 in direc-li.on e ,".

. Mode -12 is in the region remaining that is between great-circle seg-

322 The Kinematics and Stability of Removable Blocks Chap. g

ments c'rzwt, c\zwz, and wrwz. This mode represents sliding on pranes1 and 2 in direction -i, r.

' Mode ^S,

which is safe from sliding with zero joint friction, occurs alongthe great circle wrwz between regions !2 and -12.

Figure 9.15 shows the equilibrium regions for the joint data of Table 9.8and Fig. 9.16 shows the contour values for these regions corresponding to 10"increments of friction angle. In the regions of modes 12 and -tZ, these contoursapply only if 0, : Qz.

600 5oo7o"l--- \ i

5oo i A -\\\\\ \\ ,

40"

4oo \ \ \i\o' -\r \i, \ \ \i\ 1++zz>>_-'rr.'.r. \ i

\ \ "\t.. \t '\r

'*i\ I I-.,r. trr'.\----y{_'Y))h\i i ir,30o---_- --r-rt.=l\--=__::;wA] i i'..i I--\------. ----iYt-.-----i^t:l@i l/ i i-\i.----,tr\.""\ZZ i ! ! Ii I_;,_____ __ :_..\1.-a\i\t),,V

i i i f i--u1..2\\Lllr\jiJlii / /,' /

-------__

/'---ii0=10"

d= 10"

Figure 9.16 Equilibrium regions and contours of required friction angle forJP 10 assuming equal friction angle on each plane.

Proofs of the Validity of the Gonstructionsfor Equilibrium Regions

We now proceed to show, formally, the connections between the equi-librium equations and the constructions above. Denote by ,othe projection ofthe active resultant r.

l. Region 0. lf r" is in region 0, then r c Jp. From the proposition onlifting, then, the sliding mode is lifting.

-----\rY!-/--..--\

\.\

;r^i,\s*i.r'i'll)N


2. Regioni. Assume that rolies in region l, having corners wr, Cii, and C*.Since wt L C,1, w, L C,o, and r belongs to this region, the projection of I toplane P1,

(fi, x i) x rfi,: 3,

lies between Cil and Ctk. When I is between fit and ir, there are values of Nr ) 0

and,T)0suchthat

and

r:Ntfit*Tit

r * l[rdr - Tft: g

This is the limit equilibrium condition for sliding along .f, [given as equation (1)

in the appendix to this chapterl.The friction angle of P, i. 6,. From (9.3), -73, - -N' tanf,.i, - FS,.

Substituting this value of -2,f, into (9.49) gives

r f N,d, - /Yi tan frsr - F3, : g

givingr : i/,(r}, f tan di3r) + Fi,

In the case of limiting equilibriun, F: 0 and (9.50) becomes

(e.50)

323

then

(e.4e)

(e.51)

r : -{+ (cos d,ro, * sin f,s,)cos 0j

i : cos 6,fi, * sin {,.i,

which means that i lies between lil, and ir at an angle f, from r0,. The projection

ro of f is in the small circle with rD, as axis and angle {,.3. Region y. Assume that r" lies in the region U, which has cornets rr121rrit

and C,1. Because region y is the stereographic projection of a pyramid withthree edge vectors, tD,, fi,, and C;1,ltrd r lies in this pyramid, there are values

ofNd>0, l/j>0, and T>0

such thatr:Nrrit*Ntfit*TC,t (e.s2)

Because

e ,, L to,, 0,, L fi,,

and ro belongs to region fli.e,,> o

so e ,,: 3tl- Equation (9.52) becomes

r I N,A,* Nfit - 73,,:g (9.53)

which is the limit equilibrium equation [equation (4) of the appendix to this

The Kinettratics and Stability of Removable Blocks Chap. 9

chapterl of double-face sliding along .i,r. so the sliding mode issliding along ,f;;, from the proposition of (9.13).

From (9.22), sliding parallel to .f,, requires that

-73tt : *.1[. tan Q,3,, - N j tan Qf ,1 - FS.J

Substituting this expression for -fitt into (9.53), we have

r { N,0, + N jOr - ff, tanQ,S,, - lfr tanQ,S,1 - FSu:

double-face

0

There are

such that

For any vector

giving

Then

or r : tr/,(rli * tan i,3,) + Nrli,t * tan 0,3,,) * Ff,t (e.s4)

In the case of limit equilibriun, F: 0 and (9.54) becomes

r : i[,(fi, * tan fi,S,i) * lfi(rAi * tan 6$,)r:-*=-(cosf,,r0r*sini F \, N, ,-

cos yr, fi' s'i) t ;ffi<"os 6fit * sin 6t3')

Finally,

r: & r.r*I': d?; t,t * ffitt,, ffr > 0, N, > 0 (9.55)

Equation (9.55) shows that r lies between f tr and f ,,. The projection of I is in theprojection of the plane of f ,, and f;,. The normal vector of this plane is f,, x ir,.

Then ro lies in the great circle connecting t,, and, t1, in the reglon ij.4. Region

^S. Assume that ro lies in region,S. Denote the corners of region

^S as

W17W1tWkr.,.

lr,>0, Nj>0, lrk>0,

r : N;fii * Ntfrt * Noti,o+ ... (e.56)

SeJPif the angle between g and I is less than 90", then

i.E > 0There is a vector r0, such that

g'(l[lf/,) > 0

8'fr > 0

8'6, < 0

g+JPThis is a contradiction. So for any g € JP, the angle between f and f is greaterthan or equal to 90'. Then there is no direction in the region that can satisfythe propositions for lifting, sliding on one face, or sliding on two faces. Theblock cannot slide even if the friction angle of each face is zero.

Since

then

and

Chap. 9 Appendix

APPENDIX

PROOFS OF PROPOSITIONS

1. PROOFS OF PROPOSITIONS ON SINGLE-AND DOUBLE-FACED STIDING

Single-Face Sliding

Since .i lies only within plane P,, the normal component of the reactionforces on the block reduce to N,6,, with N, ) 0; accordingly, the equilibriumequation (9.5) becomes

r*i[iOr-T,i:0, N,>0, f>0 (1)

Taking the vector cross product of equation (1) with 6, on the left and then 6,

on the right,(6, x r) x 0, - T(6, x ^i) x 6t : O

(6, x ^i) x dr : S(6t.0') - A,($,.i):.i(6rxr)xd':TS

^_(0rxr)xd," - l(6;x r)T-

ftt: 6t ot fi, - -6o so the above can be written

o_(fi,xr)xfi,_o" -l(ri x r)T--"'

Taking the dot product of d, with equation (1) gives

0t. t * N,(0,.0,) : 0, Nd > 0 (3)

Then since N, ) 0, 0, . r ( 0.

Double-Face Sliding

Since .i is contained in P, and P1, end,B is removable, all of the jointplanes except those of set i and 7 will open. The normal components of thereaction forces on the sliding planes are therefore

1**, : N,6, ! Np1, ,nrr > 0, .lri > 0

The equilibrium equation, (9.5), becomes

rf.N,0,*Nfit-73:0, .lfj>0, If/>0, T>O (4)

Taking first the cross product of (a) with 0, and then the dot product with(dr x Dr) gives

(r x d).(6, x 6t) + N,(0, * 6t).(0, x 6)

- f(.i x dr). (6' x dr) : 0 (5)

(2)


Since ^f is parallel to (6, x 6r), the last term : 0. Also,

(r x 6r).(0, x 6,) : (0t x 6r).(r x dr) : 6t x (r x 0).6,

so (5) reduces to : -(r * 6t) x 6t'6': (0t x r) x 6''6'

[(6r x r) x dr].6, * N,(0, * 0,).(!, x 61) : 0, & >0 (6)

Therefore,

[(0txr)xd/.d,{0or I@t x r) x n).u, ( 0

So i,.di(O (7)

Similarly, multiplying (4) by 0, and then taking the dot product with (d, x 6,)gives

.f,.0, < 0 (g)

Taking the dot product of (a) with 3 gives

r.,f-f(^f.i)-0 (9)

Since T > 0 for a key block or potential key block,

r..r)0 (10)

Since the sliding direction is parallel to fi4 x fi1, when (ff, x fi) . r > 0, (10)requires s:fitx fiiand when (ft,x fi,)., <0, (10) requires s: -fir * fr,;therefore,

s : (i' x \t)r. Sign ((fi, x ft) . r)" lfu x fi1:

2. PROPOSITIONS OIU THE NEAREST VECTOR

Proposition 1. If there is a vector fi - JP such that

C,n) < 90"

then there is one and only one nearest vector of JP with respect to f .

Proof, Suppose that there are two nearest vectors gr and fr; then

i'8r: i'8r) O

Denotei:(X,y,Z)

Er : (At, Bb C)8z : (A2, 82, C2)

Let Etz:#+€h (see Fig. s.tl)

Figure 9.17

In the force triangle of gr , 52, and 8r * Pr, the sum of the lengths of the twosides S, and $, equals 2. Therefore, the third side's length is less than 2.

18'*8rl<18'l*l9rl:2Err.i : ,€'! 8,, . i >(8' -r-9)'i :gr.i

lgr -t gzl 2

or 9rz- r ) Er. r

so (i,|'r) < (f,8')$, and $rare not the nearest vectors of JP to i; this is contradiction.

Proposition 2. If .i is the sliding direction of JP under active resultantr, then.f is the nearest vector of JP to f, and f .i > 0.

Proof. .i is either i, 3r, or ,i,r.

1. If 3: i, we haveS.i : L

and .f is the nearest vector of JP to f .

2. If i : jr, then ^i, is the orthographic projection of f on plane P,.

From Fig. 9.18 it can be seen that

OB: OAcosu

-N: O.Bcos p:mcosdcosBa/a

COS ? : L: COS LCOS B'oA'cos (f, fr) : cos (1, .f,) cos G,, rt)

where fr is any vector of plane P,. From (11) it can be seen that

cos (f , fi) < cos (f, 3,)

(1 1)

(12)

l

328 The Kinematics and Stability of Removable Blocks

Figure 9.18

for any rt e P, andfr#3i then (l,s) is smaller than e,fr), meaningthat^f,is the nearest vector of plane P,to f. As before, letfl,be the normal vector of P,pointing into the JP.

when i-6,1o, i + u(flt); the nearest vector of the half-space u(d)withrespect to I must therefore be on the boundary plane, P,. By (12), therefore, gis the nearest direction of half-space U(a) to f. SinceJP c U(6,)and,f, € JP,.f, is also the nearest direction of the JP to l.

The dot product of equation (1) with S shows that

i.3:f>0and (1, f) < 90'. By Proposition 1, the nearest direction is unique and equals g.

3. If .i : ,is.l, from (9), i. .f > 0.

Denote by 6t and 6t the normal vectors of P, and Pl such that 6, and d, pointinto the JP.

consider u@,) n u(6); its boundary is the union of two half-planes:

(U(6) . U(0)) ^

(P, u p,) : lu(o) n U(6) n p,l ulU(6,) ^

U(6r). Prl : (U@t) n PJ u (U(d,) n Pr)

The last step follows from

U(0) A Pt : Pt and U(0,) n Pr : Pt

For any h e (U(0,) n 4), from (11):

cos (f, fi) : cos (f,.ir) cos (5r,fi)

cos (i, i,r) : cos (f , ,ii) cos (^fi, iri)

Chap.9]

(13)

(14)and

Chap. 9 Appendix

For double-face sliding, (9.34) gives $,-Si <0, so Si + U@,). In partic-ular, ,f, 4 U(6,) A Pi. In Fig. 9.19 the region U(0,) ) 4îs shaded and î,is outside of it. We denote any vector in this half-plane by h. From the figure,

G,,h) ( (ii, i,i)Comparing (13) and (14) gives us

cos (f, i,i) ) cos (f , fr) (see Fig. 9.19)

Figure 9.19

Then.i,, is the nearest vector of U(6') o Pt to i. Similarly, ^f,, is the nearest

vector of U(6r) ) P, to r. Therefore, .irr is the nearest vector of (U(6J n Pr)U (U(0) n P,) to l. Because

i+U(6,)Û@j)the nearest vector of U(6,) ) U@ j) to I is in its boundary; then.f,, is the nearest

vector of U(6,) ) U(A) of i.iii € JP, and JP c U(6,)

^ U(A) shows that î,y is the nearest vector of

JP to f.

Proposition 3. If f is the nearest vector of JP to the active resultant fand $.i > 0, then $ is the sliding direction i E -: 3.

Praof. Because g-i > 0, $ is unique.

1. If $is notin the boundary of the JP, theng e JP. Therefore, i: $andI : j is the sliding direction; the mode is lifting.

2. lf $ is in the boundary of the JP and belongs to plane r but not to anyother plane of the JP, then 6,-i 10.^Consider another vector ft in planeP, and suppose that $ # 3,.When ft moves from $ toward ,i,, that is,

when the angle (fr, s,) becomes smaller, the angle e,b calculated by (11)

also becomes smaller. This is impossible because $ is the nearest vector ofJP n P,. Therefore, $ : Sr, meaning that g is the sliding direction andthe sliding mode is single-face sliding along J,.

u(0i) n Pj

I

1

3.


If $ e P, n P7, then I : 3,r.As assumed before,6rand,0, are the normalvectors of P, a;ad Pp respectively, such that 6, and 0r point into JP.

JP n P, is an angle in P,, the boundary of which is j' and .f,o asshown in Fig. 9.20.

Figure 9.20

When fi e Prmoves from E : S,ttoward i,, (i,, fr; becomes smaller,and, from (11), (i, fi) also becomes smallel. Because g is the nearest vectorto f of JP n P,, Fig. 9.20 shows that

(J,r, s,) ) (i,r, i,)Also, since $ is the nearest vector of JP A Pr,

soas shown in Fig. 9.20

Similarly, we have

,f, # u(o) n P,

6r.J, < 0

6r.s, < 0

Then by (9.33) and (9.34), JP has the mode of double-face sliding along.f,1.

\\.\\\\\\

3', fi\

Referoncos

AnNoto, A. 8., R. P. Brsro, D. G. TflvnEs, and A. O. Wrr.sou (1972). "Case historiesof three tunnel support failures, California Aqueduct," Bulletin of the Associationof Engineering Geologisls, Vol. IX, No. 3, pp. 265-300.

Barcnrn, G. 8., and N. A. L.q,rrNBy (1978). "Sampling for joint persistence," Proc. I9thU.S. Symposium on Rock Mechanics, Mackay School of Mines, University ofNevada, Reno, pp. 5G65.

BnEcuEn, G. B., N. A. LaNwry, and H. H. ErNsrsru (1977). "Statistical descriptions ofrock joints and sampling," Proc. 18th U.S. Symposium on Rock Mechanics, ColoradoSchool of Mines, Golden, Colo. (Johnson Publishing Company, Golden, Colo.),pp. 5Cl-8.

Bentou, N. (1977). Proc. I6th U.S. Symltosium an Rock Mechanics (ASCE, NewYork), p.238.

BnnNarx, J. (1966). "Contribution d l'6tude de la stabilitd des appuis de barrages-6tude g6otechnique de la roche de Malpasset," Ph.D. thesis, Ecole Polytechnique,Paris (Dunod, Paris).

Btsuot, A.W., and N. Moncsr,rsrnnrv (1960). "Stability coefficients for earth slopes,"Geotechnique, Yol. 10, pp. 129-150.

Bnrrxr, T.L. (1972). "Shotcrete in hard rock tunneling," Bulletin of the Associationaf Engineering Geologisfs, Vol. IX, No. 3, pp. 241-264.

BRocH, E., and L. Opnc^q,ano (1980). "Storing water is safe and cheap," SubsurfaceSpace, pp.259-266.

CHIN, L.-Y., and R. E. GoonuaN (1983). "Prediction of support requirements forhard rock excavations using keyblock theory and joint statistics," Proc. 24th U.S.Symposium on Rock Mechanics (Association of Engineering Geologists and TexasA&M University, College Station), pp. 557-576.

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Rererences I

332

coor, J.8., J. w. LBny, and J. T. Mlnrrr (1962). "Kemano tunnel, operation andmaintenance," The Engineering Journal, Vol. 45, pp. 39-86.

cononrc, E.J., A. J. Hnr.rDRoN, Jn., and D. u. Dnnne (1971). "Rock engineering forunderground caverns," in Underground Rock Chambers (ASCE, New york), p. 567.

CnuorN, D. M. (1977). "Describing the size of discontinuities," International Journalof Rock Mechanics and Mining Sciences & Geomechanics Abstracts, Vol. 14, pp.133-137.

Cuunllr, P. (1971). "A computer model for simulating progressive, large scale move-ments in blocky rock systems," Proc. International Symposium on Rock Fractures,Nancy, France (International Society for Rock Mechanics).

CuNnrtr, P., M. Vorcrtn, and C. F.lrnnunsr (1977). "Computerized design of rockslopes using interactive graphics for the input and output of geometrical data," Proc.I6th U.S. Symposium on Rock Mechanics (ASCE, New York), pp. 5-14.

Fuvacnru, E. (1978). "Geomechanical models-notes on the state of the art," Proc.International Symposium on Rock Mechanics Related to Dam Foundations, Rio deJaneiro (reprinted by ISMES, Bergamo, Publ. 109).

GoopnaeN, R.E. (1976). Methods of Geological Engineering in Discontinuous Rocks,West Publishing Co., St. Paul, Minn.

Goonuarq, R.E. (1980). Introduction to Rock Mechanics, Wiley, New york.GoopuaN, R.E,, and J. w. Bnl,v (1976). "Toppling of rock slopes," proc. specialty

Conference on Rock Engineering for Foundations and Slopes, ASCE, Boulder, Colo.,Yol.2, pp.2Ol-234.

GoonulN, R. 8., and Gerq-nuA SHI (1982). "Geology and rock slope stability-appli-cation of the key block concept for rock slopes," Proc. 3rd International Conferenceon Stability in Surface Mining (AIME/SME, New York), pp. 347-373.

Gooontex, R.E., Grm-nua Snr, and W. Bovu (1982). "Calculation of support forhard, jointed rock using the keyblock principle," Proc. 23rd U.S. Symposium onRock Mechanics (AIME/SME, New York), pp. 883-898.

IfAv, G. E. (1953). Vector and Tensor Calculus, Dover, New York.HrrrrNcrn, M. (1978). "Numerical analysis of topping failures in jointed rock," Ph.D.

thesis, University of California, Department of Civil Engineering.llonx, E., and J. Bnav (1977). Rock Slope Engineering,2nd ed. (Inst. of Mining and

Metallurgy, London) 402 pages.

Honr E., and E. T. BnowN (1980). Underground Excavations in Rock (Inst. of Miningand Metallurgy, London), 527 pages.

Hunsox, J.A., and S. D. Pnrnsr (1979). "Discontinuities and rock mass geornetry"International Journal of Rock Mechanics and Mining Sciences & GeomechanicsAbstracts, Vol. 16, pp.339-362.

Jotttt, K. W. (1968). "Graphical analysis of slopes in jointed rock," Journal of the SoilMechanics and Foundations Division, ASCE, Vol. 94, No. SM2, pp.497-526.

Lowor, P. (1965). "LJne m6thode d'analyse ir trois dimensions de la stabilitd d'une riverocheuse,o' Annales des ponts et chaussdes, No 7, p.37.

LoNon, P., and B. Tanorru (1977). "Fractical rock foundation design for dams," Proc.16th U.S. Symposium on Rock Mechanics (ASCE, New York), pp. 115-138.

Luz.qoBnn, W. (1952). Fundamentals of Engineering Drawing, Prentice-Hall, Engle-wood Cliffs, N.J.

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McM.q,HoN, B. (1974). "Design of rock slopes against sliding on pre-existing surface,"Proc. 3rd Congress of the International Society for Rock Mechanics, Denver, Colo.(National Academy of Sciences, Washington, D.C.), Vol. IIB, pp. 803-808.

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Indox

A

Angles between lines and planes

by stereographic projection, 82

by vectors, 40,41,49Angles between planes

stereographic projection, 95

Angular relationships for an emptyintersection, 290

Arnold, A., 8

Assumptions, 9-11

B

Bamaix, J., 8

Barton, N., 19

Bending, 162

Blasting, 125,237Block codes

for omitted joint sets, 129

with repeated faces, 130

without repeated faces, 107

334

Block pyrarnid, 105

computed using vectors 4l-43, l15defined, 4l

Block typesnumbers of different classes ofjoint blocks,

139numbers of different classes in surficial

excavations, 186

numbers of different classes in tunnels. 250Blocks

corner coordinates, 29

with curved faces,247faces, determination using vector analysis,

28-40.49joint blocks, 125

key blocks, 98with repeated joint sets, in a surface

excavation, 187

stable, 295

types,98, 100

united, l2l 'volume, 3l-38

BP (see Block pyramid)

Brekke. T'.. 5

Buckling, 6, 7

c

Castaic dam, 8

Caving, 4Chan, L.Y., 164

Clay, 5Complementary set, 172, 2O8

Complex blocks (see United blocks)

Concave slopes, 173

Contacts, 3Convex block, 113

Convex slopes, l7lCook, J.8., 8

Coordinate axes, 25

for tunnels, 246Corners

of chamber intersections. 235

of a portal, 284of an underground excavation, 219, 22'7

Cousins, 147

Crack growth, 6,7,9, 162Cundall, 14

Curved faces,244Cylinder, 241

D

Designlocations for reinforcement in undersround

excavations, 220slope angle, 190

slope direction, 194

Dip and dip direction of a plane, 25

stereographic projection, 85

Directionchoice for a chamber,225,229choice for a slope, 194

Discontinuities, 3

Distinct (or discrete) element analysis, 14, 15

Don Pedro spillway, 2

E

Edges

of a portal, 282of a pyramid, 82, 254of an underground chamber, 218, 22'l

Engineering judgment, l5EP (see Excavation pyramid)

335

Equal angle net (see Stereonets)

Equilibrium regions, 313

for a JP with only 2 joints, 320

Excavation pyramid

for curved blocks, 247

defined, 105

Excavationsdefined, 3modes of failure, 4-9, 157-62purposes, 157, 2O3

stresses, 4

at the surface,157tunnels, 240types, 3

underground openings, 203

F

Finite element analysis, 1l-14Finiteness, theorem, 101

Fissures, 3

Forces

acting on blocks, 296vector operations involving, 43-45, 54

Foundations (see Stability analysis)

G

Geological maps (see Trace maps)

Great circle, 67-70, 80center of projection, 80

stereographic projection is a true circle-proof, 92

H

Half space

is a convex block, 113

description by stereographic projection, 75

description by vectors, 30

Hittinger, M., 13

I

lnfinite block, 103

lntersection line, 28, 49

Intersections of underground excavations, 232

336

J

Joint blocks, 125

with repeated faces, l3lshapes, 127

types in three dimensions, 139types in two dimensions, 135

Joint properties

extent, l0spacing, 150

Joint pyramidsliding direction when JP is given

by stereographic projection, 75-78, 88Joint spacing, 150Joint trace map, 152

Joints, 3

JP (see Joint pyramid)

K

Kemano tunnel, 8

Kerckhoff II underground power house, 281Key blocks

analysis, 164

cousins, 147

in dam foundation, 20,21defined, 98

in underground openings, 19

versus joint blocks, 128, 162, l&Kinematic conditions for lifting and sliding,

304Kinematics, 295

L

Lifting mode, 45

Limit equilibrium analysis, 16, 17

Londe, P., 16,296

M

Microfaults, 3

Minesunstable vs. stable, 4

Mode analysis, 310

example, 311

Modelsnumerical, ll-17physical, 17, 18

lndex

Modes of failurelifting, 45, 83, 298,32Osingle face sliding, 46,54,55, 83, 159, 298toppling, 6-8, 16ltwo face sliding, 47,48,55, 84, 158, 299

Multiple blocks, 158, 161

N

Nearest vector of a JP with respect to r, 3ll,326

o

Orientationchoice for excavations, 2lterms to describe, 24.25

Orthographic projection, 57 -60of a vector on a plane in the stereographic

projection, 81, 82

P

Physical models, 17, 18

Pillars between underground chambers, 236Portals, 279-86Potential key block, 99,295

R

Removable area of a tunnel, 253

Removable blocks, 98, 100

in inside edges of underground excavations,206

in corners of underground excavations, 21 Iin pillars, 236in the roof, 205

in a surface excavation, 168

of a tunnel, 251

in the walls of an underground excavation,204

Rigidity of blocks, l0Rock

discontinuous, Ihard versus soft, 2

Roof of an underground gallery, 205

s

Shi's theorem, l2lapplied to concave slopes, 173

lndex

applied to nonconvex blocks in undergoundexcavations, 208,2I1

applied to portals, 283-86Sliding direction

example of calculation using vectors, 54, 55

under gravity alone, 47, 48

Sliding forcedouble face sliding, 302example of computation, 303

lifting, 301

single face sliding, 301

Slope designfind direction, given angle, 194

find safe slope, given direction, 190

Small circle, 'll, '14, 87

stereographic projection is a true circle,proof, 93

SP (see Space pyramid)

Space pyramid

defined, 105

Spacing, 150

Squeezing rock, 5Stability analysis, 295

Stable blocks.295Statistics, l0Stereographic projection

compared with vector analysis methods, 238

comparison with vector analysis forunderground excavations, 238

of a cone, 71

description, 61-64

equilibrium regions, 314-20

finding the JP for a given sliding direction,310

finiteness theorem, 105, 106

half spaces, 75

intersection of planes, lO,7ljoint blocks, 138

joint pyramid,l5-78key blocks in a surface excavation, 169

maximum key blocks of a tunnel, 271

normal to a plane, 78

of opposite to a vector, 66, 67

of a plane, 67-70removability theorem, 1 l0sliding direction, 91, 92upper versus lower focal point, 63

upper versus lower hemisphere projections,64

of vectors, 67

Stereonets, 74

construction, 87

Stillwater damsite, 126

Stresses. 4

337

Strike and dip,74, 25

Surface excavations, 157

design considerations, 165

effect of slope curvature, l7l-74Symmetry, l2O, I47

T

Tapered blocks, 99, 100, 102

in a surface excavation. 168

Theoremsfiniteness. 101

maximum removable area. 252. 291removability, 108

removability of united blocks, 122

Shi's theorem. l2ltunnel axis.249. 287

Toppling, 6-8, 161

Trace maps, 152

finding removable blocks, 198,272Tunnels. 240

cylinder.240direction,24lface,24lportals, 279

shape, 242

U

Underground chambers

example of key block analysis, 214

underground space, 203

united blocks, l2l

v

Vector analysis

angles between lines and planes, 40, 4I, 51 ,

52

of block pyramid, 4l-43, 52, 53

compared with stereographic projection

methods. 238

determining block corners. 29, 49-50emptiness of a pyramid, 53, 54, 147

equation of a line, 25

equation of a plane, 27, 48

faces of a block, 38-40

to find slope angle, 193

half spaces, 30, 50

intersections of lines and planes, 28, 49

of sliding direction, 45-48slope direction, 196

I ndex

Vector analysis (contd.) Wsurface excavations. 179-83

vectors wall Pillars' 237Walls. 205

normal,27 WaErradius, 26 5stereographic projection, 85 *"|?ltXh, ng (seeModes of failure, two facetn\t,27 sliding)

Voegele, M., 15 Wittke, W., 296

t

block theory and its application to rock engineering

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