Analytical solutions of the mechanical behaviour of rock ... · The rock mass was assumed to be a continuous, homogeneous and isotropic medium obeying Hooke's law, and the time-dependent

FI9700152

S T U K • Y M O • TR 1 3 6 October 1997

Analytical solutions ofthe mechanical behaviourof rock with applications toa repository for spentnuclear fuelPetri JussilaSTUK, Nuclear Waste and Materials Regulation

The study was supervised byEsko Eloranta

The conclusions presented in the report are those of the authorand do not represent the official position of STUK.

I 2 9 - 0 1STUK • SATEILYTURVAKESKUS • STRALSAKERHETSCENTRALEN

RADIATION AND NUCLEAR SAFETY AUTHORITY

ISBN 951-712-230-6ISSN 0785-9325

Oy Edita Ab, Helsinki 1997

STUK-YMO-TR 136 RADIATION AND NUCLEAR SAFETY AUTHORITY

JUSSILA, Petri. Analytical solutions of the mechanical behaviour of rock with applications to arepository for spent nuclear fuel. STUK-YMO-TR 136. Helsinki 1997. 38 pp. + Appendices 17 pp.

ISBN 951-712-230-6ISSN 0785-9325

Keywords: rock mechanics, spent fuel disposal, mathematical models, analytical models,Kirsch equations

ABSTRACT

Mechanical behaviour of rock is one of the main research areas of disposal of spent nuclear fuel.Various calculation programs designed to solve the problem are utilized by the planners of spentfuel disposal. The results of the numerical approaches can be validated by comparing them to theanalytical solution of a simplified problem.

As a basis for this study, we used Kirsch's equations, which give the planar solution to the stressfield developing around a horizontal tunnel of a circular cross section. Kirsch's equations and thecorresponding deformation (strain) and displacement fields were derived for a tunnel with andwithout a circular support. The rock mass was assumed to be a continuous, homogeneous andisotropic medium obeying Hooke's law, and the time-dependent aspects of the fields were takeninto account by applying the Poynting-Thomson model to the interaction between rock and a sup-port. The Matlab® code was used to calculate and visualize the results.

As a result of this study, we state that the tangential, tensile stresses at the sides of the tunnel are ofthe same order of magnitude as the tensile strength of rock, which influences the fracturing. With asupport, the effects can be moderately reduced.

RADIATION AND NUCLEAR SAFETY AUTHORITY STUK-YMO-TR 136

JUSSILA, Petri. Kallion mekaanisen käyttäytymisen analyyttisia ratkaisuja ja niidensovelluksia käytetyn ydinpolttoaineen loppusijoitustilaan. STUK-YMO-TR 136.Helsinki 1997. 38 s. + liitteet 17 s.

ISBN 951-712-230-6ISSN 0785-9325

Avainsanat: kalliomekaniikka, käytetyn polttoaineen loppusijoitus, matemaattiset mallit,analyyttiset mallit, Kirschin yhtälöt

TIIVISTELMÄ

Kallion mekaaninen käyttäytyminen on eräs käytetyn ydinpolttoaineen loppusijoitukseen liittyvis-tä tutkimusalueista. Käytetyn polttoaineen loppusijoituksen suunnittelijat käyttävät tutkimuksis-saan monien eri tahojen tekemiä ohjelmistoja, joiden antamien tulosten oikeellisuutta ja suuruus-luokkia voidaan arvioida analyyttisten ratkaisujen perusteella. Työn pohjana ovat 1800-luvunlopulla johdetut Kirschin yhtälöt, jotka antavat tasoratkaisun ympyräpoikkileikkauksisen, kahtaprimaarijännityskomponenttia vastaan kohtisuorassa olevan, äärettömän pitkän tunnelin ympärillemuodostuvalle jännityskentälle.

Työssä johdettiin Kirschin yhtälöt sekä niitä vastaavat muodonmuutokset ja siirtymät sekä tyhjälleettä sisältä tuetulle tunnelille ja sovellettiin näitä tuloksia suunnitteilla olevan loppusijoitustilanvaakasuoraan sijoitustunneliin.

Kallion oletettiin olevan jatkuva, homogeeninen ja isotrooppinen väliaine ja noudattavan Hookenlakia. Mekaanisen käyttäytymisen aikariippuvuus otettiin huomioon soveltamalla kallion ja tuenväliseen vuorovaikutukseen Poynting-Thomsonin mallia. Tulosten laskemiseksi ja visualisoimi-seksi käytettiin Matlab®-ohjelmistoa.

Työn tuloksena tangentiaaliset vetojännitykset tunnelin seinissä ovat vetolujuuden suuruusluok-kaa, mikä vaikuttaa rakoiluun. Muodonmuutokset ja siirtymät ovat pieniä tunnelin säteeseenverrattuna ja rajoittuvat tunnelin välittömään läheisyyteen. Tuella voidaan rajoitetusti vähentäävaikutuksia tunnelin lähiympäristössä.


CONTENTS

ABSTRACT PageTIIVISTELMA 4PREFACE 6

1 INTRODUCTION 7

2 THE MECHANICAL FIELD 82.1 The field variables 82.2 The fundamental equations 92.3 The biharmonic equation 12

3 THE FIELD AROUND A CIRCULAR CAVITY 143.1 Primary state 143.2 Secondary state 153.3 Tertiary state 19

3.3.1 The time factor in the tertiary field 213.3.2 Values of the derived parameters 233.3.3 Comparison of the secondary and tertiary fields 233.3.4 The time-dependent variation of the tertiary field 243.3.5 The final deformation of the cross section 25

4 THE REPOSITORY OF SPENT NUCLEAR FUEL 274.1 The parameters of tunnel and rock 274.2 The parameters of support 28

5 RESULTS 305.1 Secondary state 305.2 Tertiary state 32

6 DISCUSSION 346.1 Secondary state 34

6.2 Tertiary state 35

7 CONCLUSION AND COMPARISON TO THE ACTUAL SITUATION 36

REFERENCES 38

APPENDICES 1-17


PREFACE

The characterization and behaviour of rock mass in and around a spent fuel repository are some ofthe main research areas in the geological disposal of spent nuclear fuel. The Radiation and NuclearSafety Authority of Finland (STUK) reviews all the research work done by Posiva Oy, the organi-zation for planning and later implementation of spent fuel disposal.

In rock mechanics one important research area is numerical simulations made by such sophistica-ted computer codes as UDEC, 3DEC and FLAC. From the authority's point of view it is essentialthat we can make independent checking calculations to confirm that the numerical results of the so-phisticated codes are sound and valid, especially in the orders of magnitude. To achieve this goalthe authority has actually two possibilities. The first possibility is to use the same numerical codesand the same data as the planning organization and to repeat all the calculations in detail. Thispossibility is of course the one which could be used if we had great resources and much time andmoney to do the work. Unfortunately, we have only limited resources at present time, and actuallyit may make little sense to do such an imitation. The second possibility is to use robust, classicalanalytical solutions which require, however, even stronger idealizations and simplifications of thereal world situations than numerical ones. This possibility offers us, however, an easy, quick,cheap, and direct way to make checking calculations.

It was almost one hundred years ago, in 1898, when the German engineer G. Kirsch presented hisfamous equations concerning the stresses in an infinite plate with a circular hole. This analyticalsolution has been widely used in different rock mechanical problems, and it is much treated also inthe elementary text books on rock mechanics and elasticity theory. This solution is exact androbust. Of course, the real situations are rarely such that they conform exactly to the analytical case.But as a first approximation this is better than nothing. The solutions are also got for the strain anddisplacement fields. Using the equations and their solutions we can gain an insight into and a deepunderstanding of the physics related to rock mechanical issues.

This report is based on the work done at STUK in summer 1997 by Mr. Petri Jussila, a student intechnical physics. He derived the Kirsch equations from the principles of continuum mechanics andprogrammed them and their extensions to the cases of supporting structures using the Matlab® codein a PC (Pentium) environment. He also made comparisons with the published results got by nu-merical codes.

Esko Eloranta


1 INTRODUCTION

The mechanical behaviour of rock is one of themajor research areas of the disposal of spentnuclear fuel. The analytical solutions of themechanical state serve as a basis for the ad-vanced numerical approaches.

There are two main ways to describe rock mass:• as a solid and continuous medium by means

of continuum mechanics,• statistically, as a fractured, discontinuous

medium.

In sufficiently small scale (e.g. at the atomiclevel) matter is always discontinuous, and theconcept of continuity is relative. Rock mass ischaracteristically a fractured medium contai-ning cracks of a wide range of size. In this studywe use the continuum approach assuming theonly fractures occurring in rock to be smallcompared to typical distances in rock, e.g. thespace between the ground surface and theunderground tunnel.

The primary stresses, i.e. the stresses originallypresent in the crust, are due to the own weightof rock, to geological and structural influencescontrolled by the processes of formation of therocks and by local crustal history. The openingof cavities will upset an established equilibriumof stresses, triggering a variety of displacementsand deformations. The course, intensity andnature of any such process depend on a varietyof reasons, e.g. the natural, petrological andexcavational conditions, the form and positionof the cavity, the form, multitude and positionof the neighbouring cavities, etc.

In order to be able to successfully derive ana-lytical solutions, various simplifications have to

be made, the most fundamental of which are:• The rock mass is a continuous, homogene-

ous, isotropic, rheological medium of well-defined properties.

• Excavation of a cavity does not affect theparameters of rock.

• The cavity has a smooth regular surface, theprofile of which is a mathematically knownplane curve (a circle or an ellipse).

• The cavity extends to infinity in the rock fill-ing the infinite half-space.

• There are not any neighbouring cavities.

In this study we derive the analytical solutionsfor the stress, deformation and displacementfields developing around a long horizontal cir-cular cavity and the effect of a support on thosefields. The solutions are applied to the condi-tions measured at the Finnish candidate sites forthe planned repository of spent nuclear fuel.

The application of the elasticity theory to des-cribe rock mechanics suggests that rock is amedium of linear elasticity or a so calledHookean solid. One of the disadvantages of thisis that the linear elastic analysis is incapable ofhandling time-dependent aspects of rock me-chanics like deformation and displacement pro-cesses. A more advanced approach used for thispurpose is the Poynting-Thomson model thathas homogeneous-linear equations of state.

In this study the rock mass is characterized bythe Hookean model, while the time-dependencyis taken into account by applying the Poynting-Thomson model to the interaction between rockand a support.


2 THE MECHANICAL FIELD

2.1 The field variables

The mechanical rheological field can be regarded as adequately specified when three mechanicalfunctions corresponding to the stress, deformation and displacement fields, or 15 correspondingvariables are known [1]. Those quantities are:

• the stress field

= F(r,r) =

a. o. -a.

0

0

0

a0

0

0

0

(1)

ay = ay (r, t) = ay (x, y, z, t), Tyz = ryz (r, t) = ryz (x, y, z,t) , r^ = xyz ,

where<7x, C7 and a_ are the normal stresses in the x-, y- and z-directions respectivelyT..(i, j = x, y, z) is the shear stress in a plane perpendicular to the i-axis and in the direction ofthe j-axis

<y0 — ~(®x + ay +<Jz) is m e average normal stress

T is the stress deviator tensorTo is the spherical stress tensorI is the identity matrix

• the deformation (or strain) field

1 1F — y — v

x « / xy * * xz

l l-Jyx £y -Yyz

1 12 ^ 2 7 ^ £z

1 1

0

0

0

e0

0

0

0

e0

= E(r, t) + Eo (r, t) = E(r, t) + e0 (r, t)l,

ex =ex(r,t) = e1[{x,y,z,t), rx,=7xy(r,t) = 7xy(x,y,z,t), yyx =yxy ,

ey = ey (r, t) = ey (x, y, z, t), yyz = yyz (r, t) - yyz (x, y, z, t) ,yzy= yyz,

(2)


where£x, e and £. are the deformations (strains) in the x-, y- and z-directions respectivelyy is the shear deformation (strain) or the angular distortion

£o ~z(£x+ Ey + ez) is the average relative deformation (strain)

E is the strain deviator tensorEo is the spherical strain tensorI is the identity matrix

• the displacement field

u = u(r, t)

ux=ux{r,t) = ux(x,y,z,t),

uy = uy (r, t) = uy (x, y, z, t), (3)

uz = uz (r, t) = uz (x, y, z, t) .

whereux, u and uz are the displacements in the x-, y- and z-directions respectively

2.2 The fundamental equations

The relationship between the mechanical variables can be derived from the three fundamentalequations of rheology, which are the equations of equilibrium, the geometry equations and thematerial or the state equations.

The equations of equilibrium [2]

do or ox—— h ——— -\ — t A — U ,

dx dy dz

driy do^ dr^

dx dy dz ^dT~ dr- <9<T- z = odx dy dz

are derived straight from Newton's second axiom and give the dynamical equilibrium of forcesacting on an infinitesimal 3-dimensional volume element in rectangular x, y, z-coordinates. HereX, Y and Z are volume forces (e.g. the gravitational force) acting in x-, y- and z-directionsrespectively.


The geometry equations are essentially a definition giving the geometrical relationships betweendeformations and displacements [2]:

du. dux

dy

duy

dzdu.

dx

' dx

du.

dy~

dz

G

Tyz

G

Tz*

G

x dx ' '

dy ' y- dz dy G (5)

du.£ - = - * ' ^

where G is the shear modulus of elasticity or the modulus of rigidity.

The state equations or the material equations of a deformed solid considered to be materiallycontinuous can be derived in terms of stress tensor T and strain tensor E, rate of change of T and Eand higher order time derivatives of them. By the hypothesis of continuity of matter, anymechanical change taking place in a volume bounded by a closed surface of the infinite half-spacemay be expressed in terms of continuous functions of these variables, and the mentioned functionsmay serve as equations of state.

At a first approximation, in a homogeneous and isotropic medium within a range of sufficientlysmall deformations, the functional relationships

/ 0 (T 0 ,T 0 > . . . ,E 0 ,E 0 , . . . ) = O

among the mechanical variables can be considered linear and time derivatives of higher order thanone can be neglected [1]. Consequently, six constants are needed to specify linear state equations

f = C0 +C1T + C2T + C3E + C4E = 0

f0 = C'5T0 + C6E0 = To + C5E0 = 0 . ( 7 )

This can be rewritten in the form

T = Co + 2GE + 277E - i t ,

T0=3KE0, f o = 3 K E o , ( 8 )

by using constantsC, = - 1 , C2 = -T, C3 = 2G, C4 = 27J, C5=3K, (9)

the symbols in which are shear modulus of elasticity G, viscosity coefficient r\, time of relaxationT, a constant tensor Co, and modulus of compressibility K.

As material equations the derived expressions (8), corresponding to the so called Hohenemser-Prager model are rather general. Numerous material equations can be reduced from them by

10


evaluating certain constants to be zero. For example an equation without time-dependency (timederivatives) and without the constant tensor C(),

T = 2GE (10)

is the generalized Hooke's law, which expressed in scalars and in the rectangular xyz-coordinatesis

ve \ 1

1 + v l A \-2vJ ' • G

1 / . vv £ ( ve \ 1£ , = — C,. - V I C +<T. ) . <J= £.. + . 7 = —

Ey • v - ; ; • 1 + v l 1 - 2 v J -v G

1 / / w E ( ve1 / / u E ( v

= — \o.-v{o.+a..)\, a.= \£.+ —

where [3]? = e + e + f. is the relative change of volume,E is the modulus of elasticity,v is the Poisson's ratio.

For the constants above also holds [3]

r E E

In this context we are dealing with phenomena of relatively small deformations occurring belowthe plastic limit (yield stress), where the rock mass remains absolutely homogeneous and isotropic.Thus, we can assume, that in an undeformed state there is no stress and without stress there is nodeformation and Co = 0. The result is the Poynting-Thomson model with homogeneous-linear stateequations:

T = 2GE + 2rjE - ft,

T0=3/ttv fo=3*E,, ° 3 )

which in scalar form are

ox-a0=2G{ex -£ ( ) ) + 277(e, - £ 0 ) - T ( C T V -<x())

CTV - ff „ = 2G(fv - £0) + 2ri(e, - £„) - r(cxv - <r0)

a. -<T{)=2G(£: -eo) + 2r}(e: -E0)-T(O: -&a).

11


2.3 The biharmonic equation

An additional tool for solving planar problems is the hiharmoiiic equation, which relates differentstress variables with a help of a so called Airy's stress function <t>.

The equations of equilibrium for a 3-dimensional volume element in rectangular x, y. z-coordinates (4) can be simplified by assuming the body forces X, Y. Z = 0. and considering theproblem in the xy-plane, i.e. making o = r = T = T. = 0. Thus, we obtain simplified equations ofequilibrium

do drv,,

IT ~dx~ 'dr do (15)— - + — - = 0 .

dx dy

which are valid for plane stresses and strains. By differentiating the geometry equations (5) we get

so called compatibil i ty equations.

d\-2

dz2

- +

- +

d

d2

d~

V"

e.

f,

dxdy

d2y,z

dydz (16)

dx2 ' rt: rtdx

which relate the different deformations and express the continuity of matter. In the xy-plane thecompatibility equations (16) reduce to

dy' dx' dxdy

By substituting Hooke's law (1 1) to (17) we get a new compatibility equation

O"(J (JO O O o~O O X

dy' (h' dx' dx' dxdy

By differentiating (15) we get

32a, d'o, d2rx.

dx2 dx' dxdx

IX)

19)

which is satisfied by introducing Airy's stress function <t> = <t>(.\. y) and by denoting the stresscomponents of (19) to be

( 2 0 )

12


Substitution of (20) to (18) yields the hiharmonic equation

d4® „ d4Q> d4® „* n

^ T - + 2 y ^ T + -T l - = V < I > = 0 'ax ox ay ay

which is a fundamental equation in the theory of elasticity and valid both for plane stress andplane strain for a material obeying Hooke's law [2].

The biharmonic equation in polar coordinates has the form

(d2 \d i d2 Yd2® \d® l d2®) n

dr' r dr r d<p' ){ dr~ r dr r dip(22)

the general solution of which (for a ring or a solid circle) is [4]

o() I n r + bor2 + c(]r

2

— r(— r(p sin (p

dor2(p + a'0(p +

+ b[ In;) cos <p - /,'/• In/-) sin <jp

13


3 THE FIELD AROUND A CIRCULARCAVITY

In this chapter we will derive the stress, deformation and displacement fields developing around anopening of a circular cross section in a medium with a prescribed planar primary stressdistribution. Although the solution is suitable only for describing fields around circular and elliptictunnels in the idealized conditions defined below, it can be used as a source of information of theorders of magnitude of the field variables in actual situations.

Definitions:• Primary state corresponds to the field in the medium without a cavity.• Secondary stale corresponds to the field in the medium with a cavity.• Tertiary state corresponds to the field in the medium with a cavity supported from inside.

Simplifyint; assumptions:• The cavity extends to infinity in the z-direction.• The primary stress distribution near the cavity is independent of the distance from the cavity.• The rock mass is homogeneous, isotropic and characterized by the Hookean model (11).

3.1 Primary state

The principal primary stresses/? and/) set a frame of reference with the z-axis coinciding with theaxis of the tunnel (Fig. 1). In the cylindrical coordinates r, (p. and z the general stress field (1) bymeans of rectangular x-. y-. and z-components has the form [ 1 ]

a, = —(o\ + o \ ) + —(<r( - (7v)cos2<p + rx>sin2<p,

°<r = ^ K ^ a v ) - - ( < 7 l -<T,)cos2<p-rX )sin24?,

a. =a. .

*rv =--{<*> -Ov)sm2(p + r^cos2(p. (24)

V- =r v cos«>- r / v sirup,

r:, = r w sinv>+r

In the planar state of stress described in Fig. 1, these equations (24) reduce to

P, =^(P, + / ' , ) + - ( / ' . , - / \ ) c o s 2 < p .

/ ' « , .= - ( / ' , + / ' . , ) - - ( / ' , -pv)cos2(p.

',•«, = - - ( / > < -/>Jsin2<p,

14


where the primary normal stresses have been denoted with p (instead of a) and the primary shearstress with / (instead of T). Here the signs and directions of the quantities have been chosen to be:• Positive values of radial stress pr correspond to compressive and negative values to tensile

radial stress.• Positive values of tangential stress p correspond to compressive and negative values to tensile

tangential stress.• Positive values of shear stress t correspond to stress on a tangential plane in the direction of

decreasing angle <p, negative values of shear stress correspond to stress on a tangential plane inthe direction of increasing angle (p.

3.2 Secondary state

By assuming that the stress function <J> has the same form as the primary stresses in (25), we try tofind it in the form

O(/\ <p, /) = f(rj) + g(i\ t) cos 2<p, (26)

which can be seen as a special case of (23). By substituting (26) in (22) we get differentialequations for functions/and g to be

d\f(rj) . 2 d'f(r.t) 1 d2f(rj) , 1 #(i-./)_n*—Tl— Tl ^ T̂ 5 - i— ~ '

OT' /" dr r" OT"~ r ' dr

d4?>(rj) | 2<9-\«?(r,Q 9 d2f>(rj) | 9 ^(/-,Q = Q

dr4 r dry r2 dr2 r3 ^ -

which have the general solutions

f(rj) = A\nr + Br2 ln;- + O- : + D

(27)

(28)

Figure 1. A horizontal, circular tunnel in planar primary stress field.

15


The equations of equilibrium (4) in cylindrical coordinates without the volume forces X. Y and Zhave the form 11)

do, 1 dz dr.,. 0,--°? n

— - + - +—— + - = 0 ,

dr r dip dz r

dziv | i dav ^ dr:v, | 2zlip =

dr r dip dz r ( 29)

dr r dip dz r

which in the xy-plane reduce to

do, 1 dr o , - o+ h = 0 ,

dr r dip r

dxlV i do, 2r (30)

dr r dip r

Similarly, the components of the stress function (20) in cylindrical coordinates can be determinedto be [ 1 ]

1 c*t> 1 <9:<t> i9:O

/• di' r~ dip' * dr'

d ( 1 d<t>\ 1 dQ> 1 (9 2 O ( 3 1 )llp dr{rdipj r1 dip r di'dip

Now by substituting (26. 28) into (31) we get an expression for the secondary stress field to be

/•" V r~ r

2 In ;-) + 2C-f 2A' + ̂ - I c o s 2tp .\ r ) ( 3 2 )

r,r =\2A - - 3 - + — p sin2^ ./ • / •

In infinity, the primary condition is conserved, thus

'•ma, = p, =-(/», +l\) + -(/\ ~p,)cos2ip, (33)

and from (33) and (32) we see that B = 0 and further

Now. by introducing new factors/?, g'and @. which are constants or functions of time, as

2,4 _ (1 2)B' C

*~~P^1\' S = J^7/ 7^71' (35)

16


the stress field of (32) can be rewritten in the form

(36)

When excavating a tunnel, the rock mass around the cavity receives an extra load previouslycarried by the removed rock mass. Thus the secondary stress field can be expressed as the sum ofthe primary stresses and opening counterstresses. The deformational stress field using (25) is

a]=a,-p, = 2(PX + 3

- Py2 ^ + 3

)A

(37)

The deformation or strain field in polar coordinates from Hooke 's law ( 1 1 . 12) is

2GJ21

2G 2

Yry QT"P1

2G

cosltp(38)

The geometry equations (5) in cylindrical coordinates are [1]

e *'' dr

£<f r dip

du

1 ,

r

du,,, 1V — 4.Ira i '

dr r\du.

;• dtp

du. di

dur Uy

dtp r

duv

dz '

t..£. =

dz

(39)

17


which in the xy-plane reduce to

At..e, =

r>

^ /• Dip r

_ du,,, 1 du, "pI<P dr r dip r

(40)

The radial and tangential displacements can be derived from the first two equations of (40) byintegration and by assuming both the displacements to be zero as r approaches infinity:

cos2<p \.

2{l-2v)—-±r\s\n2<p .r r (41

By assuming the radial and shear stresses to vanish at the surface of the cavity we can expect theboundary conditions

(42)

and by substituting (42) to (36) we can obtain factors/#, g'and £ to be

From (43) and (36) we obtain the components of the .secondary stress field

(43)

o, = — 1-1*

1 + 31

R

' R

r )

cos 2<p.

cos 2(p.

(44)

I /sin2(p.

which are known as the Kirscli equations for the stresses around a cavity of circular cross sectionin a Hookean medium.

Also from (38). (41) and (43) we get the secondary deformation or strain field

•*• 2 G 2

R

'(TT4 v | - | - 3 | * cos 2(45)

STUK-YMO-TR136 RADIATION AND NUCLEAR SAFETY AUTHORITY

and the secondary displacement field

1 1M, =

2G 2, , R R

4(1-v) rcos2(p k

(46)

Here and from now on, the directions of the displacements are chosen to be against thecoordinates, i.e.:• Positive values of ur and u^ correspond to displacements in the direction of decreasing values of

r and <p, respectively.• Negative values of ur and u^ correspond to displacements in the direction of increasing values

of r and <p, respectively.

3.3 Tertiary state

In this section we will study the effects of a support of circular symmetry (Fig. 1) and linearcharacteristics on the form of the secondary solution starting from the equations (36, 38, 41). Thecontact between rock and support is supposed to be ideal and independent of the position. Theemphasis is on the radial displacement, which is the same for the support and the rock at the pointsof their contact and assumed to be the only displacement occurring in the support. This assumptionhas practical justifications for a thin support according to [ 1 ]. In order to define the three factors /?,S and @ in this situation and to yield an advanced solution, we will make the followingassumption.

Radial and shear stresses on the support are proportional to the displacements of the peripheralpoints:

(47)

Here the factors r4, S and C are sought for in the form [1 ]

(48)

which after substituting (48), (36) and (41) to (47) yields the new factors a and \jfXo be

a =

v = 2G •3-4v

(49)

19


and the proportionality factors in (47) to be

_ 3-4v

** 5-4v(50)

We can now obtain from (36, 48) the tertiary stress field components in a medium obeying Hooke'slaw to be

ar=^(Px+Py)

R

:-i(p.+p, M rR

4 - - 3 -R

(\-y/)\cos2(p,

cos2<p,

RY_JR (\-y/)\sin2<p,

(51)

where a and \y are as in (49). By substituting (48) in (38) and by eliminating G:s by (49) thecomponents of the corresponding tertiary deformation field can be obtained to be

1 Ry 1j - v ) | - | - 3 1 - cos

a (R)2 1 A I R\2 *(R^

4v — I - 3 |V r

cos2<p,

(52)

21 i l 1 - ^ 'sin2(jo .

Similarly, from (41, 48, 49) we get the components of the corresponding tertiary displacementfield to be

\ R 1 14(1 -v)^-

r V rcos2<p,

r \rsin2(p .

(53)

20


3.3.1 The time factor in the tertiary field

The equations derived so far are of static nature. In practice, however, the deformations anddisplacements are well known to be time-dependent presenting an elastic behaviour—the shape ofa hole excavated in rock deforms in the course of long periods of time due to the primary stressesin the bedrock. This is the moment for us to apply our material equations, the Poynting-Thomsonmodel.

The time-dependency can be obtained by considering our new parameters a and \j/io be functionsof time

a = a(t) and y = ij/(t), (54)

and determining them by subtracting the second of the material equations (14) for the stress fieldin the polar form

ar-a0=2G(er-e0) + 2T](Er-e0)-r(&r-a0),

p-e0)-r(aip-d0), (

from the first one, resulting in

<rr -av +r(ar -&v) = 2G(er -ev) + 2r][er -£„) , (56)

the constants in which (as in (9)) are shear modulus of elasticity G, viscosity coefficient rj and timeof relaxation T. Substitution of (51) and (52) into (56) gives two differential equations

2G ^ , n

W H ! 1 L , o (57)dt

The initial conditions ur(t = 0) = u (t = 0) = 0 as in (53) yield 5(0) = i^(0)= 0. and the solutions of(57) become

5(0 = a(l-«-

where

8

2G '3-4v

h^T^7 (59)

3-4v

21


Now, we are ready to conclude the tertiary fields obeying the Poynting-Thomson model from (51-53) and (58):

The tertiary stress field:

4 (/>,-/>,)>- 4 1 *r

R

-wy«-'i

cos 2^),

T..n = - - sin2

(60)

tertiary deformation or strain field:

3 - 4v V r

R

"3-4v

(61.

tertiary displacement field:

1

3-4v

1

4 ( 1 - v ) - -

3-4v/? V

(62)


3.3.2 Values of the derived parameters

In equations (60-62) there are parameters a, /3, 8 and yr defined in (59) as

3-4v2G

3-4v

These parameters depend on• the parametric constants of rock (G, v, r\, T),• the dimensions of the cavity (R),• the characteristics of the support (^).

The support parameter % can be estimated to be [1].

Z~E,d R\ (63)

where Es is the modulus of elasticity of the support and d is the thickness of the support.

The role of the parameters a and y/ is that they determine the load on the support and themagnitudes of the final deformations and displacements, whereas the parameters j8 and Sdetermine the rate of change of the field components. The solutions for the physical problem statedin this chapter are thus not barely functions of the natural conditions but they also depend on thedesign of the support in a manner not easily seen from the solution above (59-63). However, it isrational to assume that increasing the rigidity of the support, i.e. choosing a thicker support whichis of a material that has a larger modulus of elasticity, reduces the deformations and displacementsin the rock mass near the surface of the cavity.

In the forthcoming chapters we will emphasize the initial and final states of the problem thusavoiding the need of the constants /3, 5, Tand TJ. When the common relation (12) between theparameters G, E, K and v of materials is known, only two of them are needed from references. Es

can also be found from references when the support material is chosen. Then, by giving thesupport thickness d and the tunnel radius R some practical values, a and y^can be calculated fromequations (63, 59).

3.3.3 Comparison of the secondary and tertiary fields

The tertiary results (60-62) can be compared to the secondary ones (44-46) by making the supportthickness d and consequently the parameter £ to go to zero. From (59, 63) can be observed, thatwhen d —» 0, also the parameters a, y/ —> 0. Consequently, when d —> 0, the tertiary stress field ofthe Poynting-Thomson model (60) coincides with the secondary stress field given by the Hookeanmodel (44) at any value of time t. Furthermore, from (59)

,. a R .. w 3 -4vhm—= — , \imJ- =£ 2 %, \im R

2G S->o % 2G

, hm72G «-»O^? 2G

,. a 1 ,. y 3 -4v (64)hm— = hm

23


Substitution of (64) to (61, 62) reveals, that when d —» 0, the tertiary deformation anddisplacement fields of the Poynting-Thomson model (61-62) approach the form predicted by theHookean model (45—46) as the time approaches infinity.

3.3.4 The time-dependent variation of the tertiary field

In the formulation of the problem, we assumed that there exist two principal primary stresses px

and pv in the bedrock. To simplify the notation, we can reduce the number of parameters byintroducing a quasi-Poisson number k to be

k - P * + 1

and denoting

P v = P ' Px =

(65)

(66)

For example, in the hydrostatic primary stress distribution (px = p) the quasi-Poisson number getsthe value k = 2, and in the case of no primary vertical stress (pv = 0) k —» °°. In the forthcomingchapters, we will study the situation in a constant depth, in which case also the primary verticalstress /?v is constant.

The variation of the tertiary stress field

At the instant of opening the cavity (t = 0) the tertiary stress field (60) coincides with the(Hookean) secondary field (44), which, expressed by means of k (65-66), is

1 -R

jf(*-2) 1-41-1 +31

1+1*r

R

cos2(p,

cos2<p,

1 + 2R R

sin2(p.

As the time approaches infinity, the final stresses get the form

(67)

l - - | 0 - « )r

(l-y/Hcos2<p,

l + 3[ -r\ (l-v^)|cos2(p,

R2 - - 3 -

R (l - ) pin 2(p.

(68)

24


As mentioned in the previous section, in the secondary case a, y —»0. This means, that without asupport the stress field is the same in the initial (67) and the final (68) states.

The variation of the tertiary deformation or strain field

As physically reasonable, the deformations at the initial state {t = 0) are zero as can readily be seenfrom (61). The deformations change with time approaching the values

1 , a2 i * S

;_Mt_2)^2r«Y-3t; 3 - 4 v [rj ir

(69)

in infinity.

The variation of the tertiary displacement field

As can be seen from (62), the tertiary displacements are also zero at t = 0. In infinity they approachthe values

, , 1 , a R

M.->->-j*j7+

(70)

3.3.5 The final deformation of the cross section

The final displacements of the points at the tunnel surface (r = R) can be calculated from (70) to be

/ „ 1 . cc

(71)

25

RADIATION AND NUCLEAR SAFETY AUTHORITY STUK-YMO-TR136

By introducing constants a and b as

1 ak

(72)

(71) takes the form

ur{r- Rj-^°°) = a

uv(r= R,t —> °°) = -bsi (73)

which represents an ellipse, another form of which is

(R-a-bf {R-a + bfwhere

y = (R-a-b)cos(p .

(74)

(75)

In the secondary case, by (63) the parameters a and b as in (72) reduce to a' and b'

' l i Ra = — pk ,2 2G

2G

(76)

A schematic picture of the final cross section against increasing values of k is plotted in Fig. 2,where the dashed lines represent the initial circular cross section. The correctness of this pictureand the values of parameters in the actual situation will be discussed later.

Figure 2. Schematic change of the form of the final tunnel cross section.

26


4 THE REPOSITORY OF SPENTNUCLEAR FUEL

In this chapter we will apply the derived analyt-ical solutions to the conditions of a depositiontunnel of the planned repository of spent nuclearfuel. The deposition tunnel (Fig. 3) is situated ina horizontal position above various verticaldeposition holes where the canisters of spentnuclear fuel will be placed. The tunnel is loca-ted in the depth of h = 500 m [5].

The area of thetunnel is A = 14.a circle is /?, =perimeter of thetunnel is L = 14.of a circle is R2 =that we will uselent radii, i.e.

cross section of the deposition01 m2. The equivalent radius of~[ATK= 2.11m. Similarly, thecross section of the deposition38 m, and the equivalent radius• LI2K= 2.29 m. The radius R,is the average of these equiva-

*2V2 = 2.20 m4.1 The parameters of tunnel and * = (rock

The parameters of the rock mass are [6]In order to apply the derived solutions, the cross E = 65 GPa, v = 0.15, p = 2700 kg/m3

section of the deposition tunnel will be approxi-mated by an equivalent circle in the following where p is the density. From (12), G gets themanner. value

G = 28.3 GPa.

aI

The depositiontunnel

The depositionbole

330 m

• *

The equivalent circle

The canister

L50m

Figure 3. Cross section of the deposition tunnel and the equivalent circle.

27


The primary vertical stress p is assumed to beconstant and calculated from the lithostaticcondition analogous to the hydrostatic pressure:p=phg= 13.2 MPa,where ggravity.

= 9.81 m/s2 is the acceleration of

According to [6], at the candidate sites of therepository the largest primary horizontal stress-es at the depth of h - 500 m vary in the rangePx = 20- 45 MPa.

The secondary stress, deformation and displace-ment fields (44-46) are plotted in App. 1-8against the values px = 20, 30, 40 and 45 MPa ofthe primary horizontal stress, which correspondto the values k = 2.51, 3.27, 4.02 and 4.40 of thequasi-Poisson number (65), respectively. Aspreviously noticed, these secondary fields areequivalent to the initial stresses (67) with orwithout a support and the final deformations(69) and the final displacements (70) without asupport.

4.2 The parameters of support

To study the effects of a support to the fieldvalues, i.e. the tertiary state, we have chosenconcrete and steel to be two support materials

and given the support thickness d values in therange of 0.01-0.5 m. The values of elasticmodulus of support [3] are £ = 0.20-10" N/m2

(a chosen value from a wide range) for concreteand Es = 0.20-1012 N/m2 for steel. The equiv-alent values of £, a and y/ for concrete and steelfrom (59, 63) and from the values of previoussection are collected in Table I. A picture of aand \(f against d for concrete and steel is plottedin Fig. 4.

Here we have plotted pairs of pictures of thefields both in the case of concrete and steelsupport:• along a constant angle, at which maximum

positive or negative values occur, against thedistance r (m) from the center of the tunnel,

• along the constant radius r = R = 2.20 m, i.e.on the surface of the tunnel, against theangle <p (rad).

Because a concrete support has smaller effectsthan a steel support on the fields, we have usedlarger and fewer values of support thickness dfor concrete (d = 0.5, 0.2, 0.1 m) than for steel(d = 0.3, 0.2, 0.1, 0.05, 0.02 m) in order to beable to get sensible pictures. The plots areshown in App. 9-16.

28


Table I. Values of support parameters £ (Pa/m), a (-) and \ff (-); the minimum tangential stress<T (MPa) (k = 4.40) and the maximum radial displacement urmu (mm) (k = 4.40), against the valuesofd (m)for concrete and steel.

«m)

0.010.020.050.10.20.30.40.5

aiO-NAn1)

0.0040.0080.0210.0410.0830.1240.1650.207

<*•)

0.0020.0030.0080.0160.0310.0460.0600.074

Concretev<-)

0.0040.0080.0190.0370.0720.1040.1340.162

u r — (mm)

2.612.602.582.542.482.412.352.29

.̂JMPa)

-5.1-5.0-4.6-4.0-2.8-1.7-0.70.3

^ " M / m " )

0.0040.0080.0210.0410.0830.1240.1650.207

oir)

0.020.030.070.140.240.330.390.45

Steelvi-)

0.040.070.160.280.440.540.610.66

ulmM (mm)

2.542.482.292.051.691.451.271.13

o^(MPa)

-4.0-2.80.34.08.410.812.213.1

COQ .

alph

a (

—

0.6

0.5

0.4

0.3

0.2

0.1

0

- : :

- : : :

: • / • :

- • / : • - • ' :

- • / : • • • . . . . . . . . •

0.05 0.1 0.15 0.2 0d

25(m)

steel

0.3 0.35

^ ^

• • • - ' -

• • • • • • -

• • - - . -

concrete

0.4 0.45

Figure 4. Support parameters a and y/ against support thickness d (m)for concrete and steel.

29


5 RESULTS

A Matlab program was created to aid in calcu-lations and illustrations of the results. For eachsurface plot in App. 1-8 a 51 x 51 matrix ofresult values was calculated in the range x, ye [-5R, 5R] = [-11.0 m, 11.0 m], which corres-ponds to the grid spacing of 27.5 cm (thecoordinates in appendices are as in Fig. 1, page15). For each curve in App. 9-16 was calculateda 51x1 matrix of result values. Because of theirgreat multitude, these values are not presentedhere in their totality. Further study of the valuesnear some specific points demanded even morecalculation points, the values at which can notbe seen in the plots but are mentioned in thetext below.

5.1 Secondary state

The secondary (or initial tertiary) radialstress (App. 1): The angularly sinusoidal fieldis positive (compressive) almost everywhereand can be seen to consist of two distinctregions inside and outside the distance r = 3.81m at which the field is constant against theangle <p.

At the inner region the values at the surface ofthe runnel are zero but increase rapidly with r(except for small areas of decreasing valuesnear both sides of the tunnel for large values ofit). The largest values against (p occur on the y-axis both near the ceiling (<p = Ttll = 1.57 rad)and the floor (<p = (3l2)n = 4.71 rad) of thetunnel, while the smallest values occur on the x-axis near both of sides (<p = 0, n), where thefield can also be negative (tensile) at suffici-ently large values of k. When k = 4.40 (px = 45MPa), the minima a = -0.38105 Pa occur at

v r. mm

0.032 m from both sides of the tunnel.

A phase shift of n/1 occurs on the borderline ofthe two regions, in which case the largest valuesat the outer region occur on the x-axis and thesmallest on the y-axis, which resembles theprimary state. The field approaches the primarystress distribution in infinity but remains small-er at every other point. At the distance of r = 5/?= 11.0 m (the plot range) the field values arealready very close to the primary stresses.

The secondary (or initial tertiary) tangentialstress (App. 2): The field values near the cavityrepresent great sinusoidal variation against theangle cp, and get their maxima and minima atthe surface of the tunnel. The maxima occur onthe y-axis both on the ceiling (<p = nil = 1.57rad) and on the floor (cp = (3/2)TT= 4.71 rad) ofthe tunnel. The maxima are positive(compressive) stresses almost three times thatof the primary horizontal stress px and locatedin a very narrow area The minima or themaximum negative (tensile) values occur onboth the sides of the tunnel (<p = 0, TZ) on the x-axis. In the case k = 4.40 (px = 0.45 MPa) themaxima are a = 0.12-109 Pa and the minima

(p. max

a =-O.531O7Pa.(p. mm

As in the radial case, the field values approachthe primary stresses in infinity being very closeto them already at the distance of r = 5R.

The secondary (or initial tertiary) shearstress (App. 3): The field is zero on the x- andy-axes and on the surface of the tunnel, whilethe maxima occur at the angles (p = (3/4) TT =2.36 rad and <p = (7/4)TT = 5.50 rad and the

minima at the angles q> = {\IA)n = 0.79 rad and<p = {5/4)K= 3.93 rad. The angularly sinusoidalshape of the solution is independent of the

30


value of k and both the maxima and minimaoccur at the distance of r = 3.81 m. In the case it= 4.40 the maxima are T^ = 0.21 108 Pa and theminima x = -0.21108 Pa.

The apparent antisymmetry of the solution rela-tive to the axes is explained when we recall thatthe positive values of shear stress refer to thedirection of decreasing angle q> and negativevalues to the increasing values of <p on atangential surface. In other words, the field inevery quadrant points away from the x-axis andtowards the y-axis resulting in a physicallyreasonable situation of a symmetric field rela-tive to the origin.

As in the radial and tangential cases, the fieldvalues approach the primary stresses in infinitybeing very close to them already at the distanceof r = 5R.

The secondary radial deformation (App. 4):The field is negative almost everywhere in thevicinity of the tunnel corresponding to de-creasing values of radial displacement as thevalue of r increases. The minimum (the maxi-mum negative) values occur on the x-axis onboth sides of the tunnel surface (q> = 0, *) beinge = -0.63 10"4 in the case k = 4.40. As the

K nun

primary horizontal stress px and consequentlythe quasi-Poisson number k increase, the fieldgets more heterogeneous and small areas ofpositive deformation can be noticed on the y-axis (<p = ±*/2) near the ceiling and the floor ofthe cavity. When k = 4.40 the maxima are e

•* r, max

= 0.57-10"1 at r = 4.40 m (= 2R).

The field represents everywhere in the plotrange a clearly noticeable sinusoidal variationagainst <p. With the results above, this agreeswith our earlier derivation of the final form ofthe cross section to be an ellipse, whose eccen-tricity increases with the value of it (Fig. 2, page26). The field approaches zero in infinity cor-responding to the occurence of the deformationsonly in the vicinity of the tunnel.

The secondary tangential deformation (App.5): The form of the field resembles the tangen-tial stress distribution, and major changes to the

primary zero-state occur in the vicinity of thetunnel surface. The maxima occur on the y-axisboth on the ceiling and on the floor of thetunnel being e = 0.1210"2 in the case k =

O in max

4.40. The minima occur on both the sides of thetunnel and turn from positive to negative as px

and k increase. At large values of k this corres-ponds to tangential compression on the ceilingand on the floor and tangential strain on bothsides of the tunnel. When k = 4.40, the minimaaree . =-0.16-10"3.

ip, nun

As in the radial case, the form of the field iscompatible with the elliptic form of the finalcross section.

The secondary shear deformation (App. 6):Due to the finite number of calculation points,the extreme values at the tunnel surface are notshown in App. 6. All the values at the surfacecan be seen in dashed curves for constant radiusin App. 14. Sinusoidal variation of the field islarge at the tunnel surface. The field is anti-symmetrical relative to the x- and y-axes onwhich the field is zero. The maxima occur onthe surface at the angles q> = (1/4)* = 0.79 radand q> = (5/4)* = 3.93 rad and the minima alsoon the surface at the angles <p = (3/4)* = 2.36rad and <p = (7/4)* = 5.50 rad. When k = 4.40,the maxima are y = 0.56-10~3 and the

I rip, max

minima y = -0.56-10~3. The form of the' rip, mm

solution is independent of k.

Other special points occur at the distance r =3.81 m, where there are negative peaks at theangles <p = (1/4)* and <p = (5/4)* and positivepeaks at the angles (p = (3/4)* and <p = (7/4)*.When k = 4.40, the values at the peaks are y =±0.1910-3.

Accordingly, the field at and near the tunnelsurface points away from the x-axis and to-wards the y-axis, decays as r increases andturns opposite at the distance r = 2.69 m. Theopposite field reaches its maximum angularvariation at r = 3.81 m and dies away as rapproaches infinity.

The form of the field suggests considerablecontraction of the tunnel cross section on the x-

31


axis and expansion on the y-axis. This alsorefers to the elliptic form of the final crosssection.

The secondary radial displacement (App. 7):The largest positive values exist on the x-axis,and the maxima occur on both sides of thetunnel surface being u = 0.26-102 m in the

° K max

case k = 4.40. The smallest values occur on they-axis near both the ceiling and the floor of thetunnel, where the field turns negative as px andk increase. The minimum (largest negative)values occur at the distance r = 3.30 m (i.e. 1.10m = R/2 m from the surface) being u . =-0.47-10 3 m in the case k = 4.40.

The form of the field corresponds to inwardmovement of the points near the sides of thetunnel and outward movement of the pointsnear the floor and the ceiling. This agrees withthe final form of the cross section to be anellipse as in Fig. 2.

The secondary tangential displacement (A pp.8): The field is antisymmetrical relative to thex- and y-axes, on which the field is zero. Thismeans, that the field changes direction from onequadrant to another pointing everywhere awayfrom the x-axis and towards the y-axis. Thisagrees with the previous result of compressivetangential stresses occurring on the ceiling andtensile tangential stresses occurring on the floorof the tunnel. The largest positive values ((p =(3/4)n= 2.36 rad and <p=(7/4)n= 5.50 rad) andthe largest negative values (cp = (l/4)?r = 0.79rad and <p = (5/4)7r. = 3.93 rad) occur at thetunnel surface. In the case k = 4.40 the maximaare u = 0.15-10~2 m and the minima w . =

(p, max (p, mm

-0.15-10 2m.

The form of the field is again in full agreementwith the elliptic form of the final cross section.

5.2 Tertiary state

In App. 9-16 the dashed lines represent thesecondary or initial tertiary state and deviationfrom that state increases with d, i.e. the solidcurves being closest to the dashed curves cor-respond to the smallest value of d etc.

From now on, the expression 'the tunnel sur-face' refers to the surface of rock, i.e. thepoints, where rock and support contact.

The final radial stress (App. 9): At theconstant angle (JD = 0 a support can be seen tohave a small effect on the stresses and the effectdecreases as the distance r increases. The thick-er the support the larger the final stress, whichis clearly noticeable in both plots.

Compared to the secondary or initial tertiarystate, in addition to the changes in the magni-tude of the solution, there is a phase shift of nilfor the values of the final tertiary state near thetunnel surface. This effect depends greatly onthe parameters of the support. For d - 0.01 m ofconcrete the phase is shifted inside the range of0.21 cm from the surface, while in the case ofthe thickest steel supports the region of phaseshift reaches the outer region of the field men-tioned above in the secondary state. In thelatter case, the largest values against q> alwaysoccur on the x-axis and the smallest values onthe y-axis.

It is important to notice that as a boundarycondition the initial radial stress at the tunnelsurface is zero (the dashed line), which is alsothe case for the values of the tertiary state at theinner surface of the support A support takessome of the load originally carried by the rockmass and, in a sort of way, the field is movedinwards the distance d. This accounts for therelatively large effect of the support thickness don the field on the tunnel surface.

The final tangential stress (App. 10): A sup-port reduces the final tangential stresses. How-ever, especially in the case of concrete, theeffect of a support on the field at the constantangle (p = nil = 1.57 rad is hardly noticeable.On the tunnel surface a support tends to flattenpeaks, and consequently the largest negativevalues on the sides of the tunnel turn positive asd increases. Values of minimum (maximumnegative) final tangential stress a min (MPa)against d for concrete and steel have beencollected in Table I (page 29).

32


The final shear stress (App. 11): A supportreduces the absolute values of the field andflattens the peaks, which can be seen in the plotof constant angle q> = (3/4)TT = 2.36 rad. On thetunnel surface, however, as in the radial case,the final shear stress increases strongly with thesupport thickness.

The final radial deformation (App. 12): Theplot of the constant angle (p = 0 shows, that theeffect of a support occurs mainly near thesurface and decreases as r increases. A supportreduces the absolute values of the field. Theeffect is rather small, and the field remainsnegative everywhere.

The final tangential deformation (App. 13):A support reduces the absolute values of thefield in a way similar to that of the tangentialstress. At (p=nJ2 = 1.57 rad the effect ispractically unnoticeable except on the tunnelsurface. The thickest steel supports {d > 0.2 m)make the largest negative values to turn posi-tive.

The final shear deformation (App. 14): Asupport reduces the absolute values of the fieldand flattens the peaks. At (p = n IA = 0.79 radthe effect can be noticed near the surface andaround the peaks.

The final radial displacement (App. 15): Asupport reduces the absolute values of the fieldand flattens the peaks. At <jp = 0 the effect is thelargest near the surface. On the surface thethickest steel supports (d > 0.2 m) make thelargest negative values to turn positive. Valuesof maximum final radial displacement ur max

(mm) against d for concrete and steel have beencollected in Table I.

The final tangential displacement (App. 16):A support reduces the absolute values of thefield and flattens the peaks. At (p = (3/4)7t =2.36 rad the effect is the largest near the surface.

33


6 DISCUSSION

6.1 Secondary state

All the changes to the primary state occur in thevicinity of the tunnel surface, and in infinity allthe field values approach those of the primarystate, that is, secondary (initial) stress distri-bution approach the primary stress distributionand (final) deformations and displacementsapproach zero. At the edge of the plot range of5R = 11.0 m the values of the primary state arealmost reached. Most of the quantities studiedhere reach their maximum positive and negativevalues on the tunnel surface or very close to itwhere there can exist peaks at which the valuescan be manifold compared to those of theprimary state. Radial stress is an exceptionwhose values have minima on the tunnel surfaceand never exceed those of the primary state.More attention has to be paid to the tangentialstress, which gets larger positive and negativevalues.

Comparison of the compressive (-250 MPa)and tensile (-5-10 MPa) strengths of rock [6]reveals that more attention has to be paid to thenegative values of stresses. When k = 4.40, thelargest positive value of (tangential) stress isa«>max = 0.12-109 Pa and the largest negativevalue a = -0.53-107 Pa The maximum

<p, min

positive value, even though almost three timesthat of the largest primary stress (p = 45 MPa),remains below 50 % of the compressivestrength of rock, whereas the maximum nega-tive value is of the same order of magnitude asthe tensile strength.

One of the earliest assumptions that we madewas that our model is valid only for small

deformations, i.e. when the stresses are clearlybelow the plastic limit and do not affect thecharacteristics of the rock mass. Now, when asa result of our study the tensile tangential stresson the sides of the tunnel reaches the yieldstress of rock, we can not be sure of the actualvalues of the fields occurring there. At least wecan say that considerable disturbances to thehomogeneity, continuity and isotropicity willmost probably be encountered at and aroundthose points. In other words, rock becomesfractured and the stress distribution around thefractured region does not obey our model anymore.

All of the deformations and displacements referto the same final situation: the points near thesides of the tunnel move inwards and, at suffi-ciently large values of k, the points near theceiling and the floor move outwards. Thus, thefinal form of the originally circular cross sec-tion is an ellipse as in Fig. 2 (page 26). Thedisplacements are so small, however, that theelliptic form can not be perceived in practice.When k = 4.40, the largest inward radial dis-placements on the sides of the tunnel are 0.26cm and the largest outward radial displacementson the floor and the ceiling are 0.47 mm.

All the notable deformations and displacementsoccur in the vicinity of the tunnel, inside arange of approximately 2 m (= R) from thesurface. Distributions of the secondary defor-mations (k = 4.40) by means of 8 contour linesare shown in Fig. 5. The lines are the closest toeach other at the regions where the variation ofthe fields against the coordinates is the greatest,i.e. near the surface of the tunnel.

34


6.2 Tertiary state

Because of their relatively small values, thevisible effects of a support on the secondarydeformations and displacements may be diffi-cult to observe in practice. The main function ofa support is to reduce local extreme values nearthe tunnel surface. This is most important in thecase of negative valued (tensile) stress peaks atwhich the values can even be turned positivewith sufficiently rigid supports. In App. 10 andTable I it can be seen that when k = 4.40, toreduce the negative values of initial tangentialstress on both the sides of the tunnel to zerorequires d = ~5 cm of steel or d = ~50 cm ofconcrete. Only half of those thicknesses aresufficient to reduce the peaks considerably be-low the tensile stress of rock.

Our assumption of the interaction between rockand a support was that tangential stresses do notdevelop in the support This means, that the

support has to be relatively thin, and as thethickness d increases, the validity of our modelgets worse. At least, the results for the largestconsidered thicknesses (d > 0.2 m or so) shouldnot be taken for granted.

The difference between secondary and tertiaryfields depends on the values of the rock-sup-port-interaction parameters a and iff, whichessentially appear in coefficient factors (1-a)and (\-\ff) in the tertiary solutions. When a andy/ are small, the mentioned coefficients areclose to unity and the tertiary solution is closeto the secondary one. In Table I and Fig. 4 (page29) it can be seen, that the values of a and J^fordifferent materials can differ distinctly. For thinsupports a, i// « 1 , especially for materials ofsmall elastic modulus. Thus, because our modelis the most valid for small values of d, the valideffects of a support on the solution shouldgenerally be considered limited.

10

5

1 °-5

-10

Radial

(dH) •

Tangential Shear

10

5

1 0

-5

-10-10 0

x(m)10 -10 0

x(m)

-1010 -10 0

x(m)10

Figure 5. Contour lines of the final secondary deformations, k = 4.40.

35


7 CONCLUSION AND COMPARISONTO THE ACTUAL SITUATION

Compared to the real situation, our study wasstrikingly simplified:

• Rock is always a fractured, discontinuousmedium and the excavation of a tunnelincreases fissuring especially in the vicinityof the cavity.

• The excavation of a long tunnel takes time,during which the primary stresses can affectthe initial conditions of the problem. In ourstudy the cavity was assumed to emerge at t= 0.

• A tunnel does not extend to infinity, but thelength of the tunnel, ~ 100 m in the case ofthe deposition tunnel, is large compared tothe displacements (~ 1 mm) occurring inrock.

• The problem was studied in a plane withonly two components of primary stressestaken into account.

• The primary stress distribution near the cavi-ty was supposed to be independent of thedistance from the cavity, because the tunnelradius is small compared to the depth of therepository and the typical distances in thebedrock.

• In the actual case of a repository of spentnuclear fuel there are supposed to be numer-ous other cavities affecting the fields devel-oping around each other, the situation ofwhich is most difficult to cover even withnumerical approaches.

• The shape of the cross-section of the deposi-tion tunnel (Fig. 3, page 27) was approxi-mated by a circle, which definitely alters theshape of the solution.

• Some additional features of the actual situ-ation that we have not taken into account inthis study, are for example, the effects ofsand and bentonite filling, ground waterflow and production of heat by the nuclearwaste.

As a result from this study we conclude that, inthe unsupported situation, depending on thevalue of the primary horizontal stress, the pointsat the sides of the deposition tunnel moveinwards 0.10-0.26 cm, while the displacementof the points on the floor and the ceiling of thetunnel varies from 0.33 mm inwards to 0.35mm outwards. The compressive stresses remaineverywhere below the strength of rock, whilethe tensile tangential stresses tend to break therock on the sides of the tunnel. The forms of thesolutions suggest that the distribution of theprimary stress field to its components affectsthe secondary and tertiary field values, andconsequently the breaking of rock, substantiallymore than the magnitude of the primary stress.

With a support of sufficient rigidity, the extremevalues of the fields near the tunnel can bereduced. The deformations and displacementsare so small already in the secondary case, thatthe main function of a support seems to be inreducing the tensile stresses and consequentlythe fissuring of rock, as well as inhibiting themovements of pieces of rock and groundwater.

36


This is an interesting feature, because some ofthe most crucial factors in the safety analysis ofnuclear waste disposal are fractures and ground-water flow near the repository.

In [6], the various numerical results to theproblem are of the same tendency and order ofmagnitude. The largest inward displacements,0.8—4.5 mm, occur on the sides of the disposaltunnel where the tangential tensile stress mayexceed the strength of rock and all the con-siderable deformations are bounded inside therange of 2-3 m from the tunnel.

The international DECOVALEX-project [7] hasbeen founded to establish collected and verifiedinformation of coupled thermo-hydro-mecha-nical processes of fractured rock to contributeto the research of spent nuclear fuel disposal. InJune 1997, some of the participants of thisproject visited the Underground Research Labo-ratory (URL) in Pinawa, Manitoba, Canada,

where tunnels have been excavated in con-ditions of dense and dry rock and exceptionallylarge horizontal stresses. In the depth of 420 mthe largest lateral stress is ~ 60 MPa. In thephotographs in App. 17 it can be seen that inabout five years the originally circular crosssection of the tunnel has deformed into anirregular shape, the largest displacements beingin the order of magnitude of 0.1-1 m.

The purpose of this study, as well as the purposeof modelling in general, was to give an idea ofthe shape and the orders of magnitude of thesolutions to a practical problem. Although thesolutions given by this highly simplified app-roach proved to be similar to those of otherstudies with different approaches, practice dem-onstrates apparent dissimilarities. The mechan-ical behaviour of rock seems to be a com-plicated area of research even at its simplest,without the additional problems presented bythe disposal of spent nuclear fuel.

37


REFERENCES

[ 1 ] C. Asszonyi & R. Richter, The ContinuumTheory of Rock Mechanics, Trans TechPublications, USA, 1979.

[2] K. Ikonen, Levy-, laatta- ja kuoriteoria, 874,Otatieto Oy, Helsinki 1990.

[3] A.Ylinen, Kimmo- ja lujuusoppi I, Valtionteknillinen tutkimuslaitos, julkaisu 10,Werner Soderstrom Oy, Porvoo 1948.

[4] L.E. Malvern, Introduction to the Mechanicsof a Continuous Medium, Prentice-Hall Inc.,Englewood Cliffs, N.J., USA, 1969.

[5] T. Vieno, A. Hautojarvi, L. Koskinen & H.Nordman, TVO-92 Safety analysis of spentfuel disposal, Report YJT-92-33 E, NuclearWaste Commission of Finnish Power Com-panies, Helsinki 1993.

[6] E. Johansson & M. Hakala, Loppusijoitus-tiloja ymparoivan kiteisen kallion mekaa-ninen, terminen ja hydraulinen kayttayty-minen, Raportti YJT-92-17, Voimayhtioidenydinjatetoimikunta, Helsinki 1992.

[7] Coupled Thermo-Hydro-Mechanical Pro-cesses of Fractured Media, Mathematical andExperimental Studies, Recent Developmentsof DECOVALEX Project for RadioactiveWaste Repositories, Edited by Ove Stephans-son, Lanru Jing & Chin-Fu Tsang, Develop-ments in Geotechnical Engineering, 79,Elsevier, Amsterdam-Lausanne-New York-Oxford-Shannon-Tokyo 1996.

38


THE SECONDARY OR INITIAL TERTIARY RADIAL STRESS O~ APPENDIX 1

x10°k = 2.51

y(m) -10x ( m )

x10k = 3.27

y(m) -10 -100

x(m)

x10k = 4.02

y(m) -10 -100

x(m)

x10k = 4.40

y(m) -10 -100x(m)


APPENDIX 2 THE SECONDARY OR INITIAL TERTIARY TANGENTIAL STRESS CJ

CT5Q_

x 10k = 2.51

-100x(m)

03 X 10= 3.27

-10x ( m )

TOC L

x10k = 4.02

y(m) -10 -100

x(m)

X10k-4.40

y(m) -10 -100x(m)


THE SECONDARY OR INITIAL TERTIARY SHEAR STRESS X APPENDIX 3

k = 2.51 k = 3.27x10

y(m) -10 -10 x ( m )

k = 4.02x107

x 1 0

-10 x(m) -10x ( m )


APPENDIX 4 THE SECONDARY OR FINAL TERTIARY RADIAL DEFORMATION e

= 2.51 k = 3.27

y(m) -100

x(m) y(m)

x 10—4

= 4.02

y(m) -10 -100

x(m)

x 10-4

k = 4.40

0y(m) ~10 -10 x ( m )


THE SECONDARY OR FINAL TERTIARY TANGENTIAL DEFORMATION e APPENDIX 5

k = 2.51 k = 3.27x10

-4x10

-4

x10-4

k = 4.02

-10 -100

x(m)

k = 4.40x10~

y(m) -10 -100

x(m)


APPENDIX 6 THE SECONDARY OR FINAL TERTIARY SHEAR DEFORMATION yup

k = 2.51 k = 3.27x 10"

y (m) -10 -100

x(m)

k = 4.02x 10

- 4

0

y(m) -10 -

x 1 0-4

x(m) -10 x ( m )


THE SECONDARY OR FINAL TERTIARY RADIAL DISPLACEMENT U APPENDIX 7

k = 2.51x10

-4k = 3.27

x10"

y(m) -10 x ( m ) y(m) -10 -100

x(m)

x10'k = 4.02

I f x10-3

k = 4.40

y(m) -10 -100

x(m) y(m) -10x ( m )


APPENDIX 8 THE SECONDARY OR FINAL TERTIARY TANGENTIAL DISPLACEMENT U

x 10-4

k = 2.51x10

-4= 3.27

y(m) -10 -10

k = 4.40x 10

-3

y(m) -10x ( m ) y(m)


FINAL RADIAL STRESS o r AGAINST SUPPORT THICKNESS d

R = 2 .20m,k = 4.40,px = 45 MPadashed lines represent the secondary or initial tertiary state

APPENDIX 9

Concrete support

constant angle, 0 rad. constant radius, 2.2 m.

2 4fii(rad)

E = 20-109 N/m2, d = 0.5, 0.2, 0.1 m

Steel support

X 1 Q 7 constant angle, 0 rad. x 1 Q6 constant radius, 2.2 m.

-10 -5 0 5 10 0 2 4r(m) fii(rad)

E = 200109 N/m2, d = 0.3, 0.2, 0.1 0.05, 0.02 m


APPENDIX 10 FINAL TANGENTIAL STRESS a AGAINST SUPPORT THICKNESS d

R = 2.20 m , k = 4.40, px = 45 MPadashed lines represent the secondary or initial tertiary state

Concrete support

x -|Q7 constant angle, 1.6 rad. x 10? c o n s t a n t radius, 2.2 m.

-10

12

Pa)

I 1 0

"Si 9To

tangent

-NI

OO

1 6

5

1 1

/ V/ V0

-5 0r(m)

10 2 4fii(rad)

£ = 20-10" N/m2, J = 0.5, 0.2, 0.1 m

Steel support

x -|Q7 constant angle, 1.6 rad. x -|Q7 constant radius, 2.2 m.

12

CO

COCO

0

IBU

I

11

10

9

8

7

6

10 0 2 4fii(rad)

E = 200-109 N/m2, d = 0.3, 0.2, 0.1 0.05, 0.02 m


FINAL SHEAR STRESS T AGAINST SUPPORT THICKNESS drip

APPENDIX 11

R = 2.20 m , k = 4.40, px = 45 MPadashed lines represent the secondary or initial tertiary state

Concrete support

-|Q7 constant angle, 2.4 rad. x 1 06 constant radius, 2.2 m.

-10 -5

CL-

coCO

£tocc

CO

2

1

0

-1

CVI

IX • Y \/ \ / \/ -/"^V \ / •/*"*\ A"/ / \ \ / / \ \/ / \ \ / / \ \/// \ \ \ /// \ \ \

\ \ / / \ \ / / ^

\ / \ / ^

0 5 10 0 2 4r(m) fii(rad)

£ = 20-109 N/m2, rf = 0.5, 0.2, 0.1 m

Steel support

CO

9*1.5COCO£to« 1CD

CO

ISS0.5

constant angle, 2.4 rad.

i

constant radius, 2.2 m.

-10 -5 0 5 10 0 2 4r(m) fii(rad)

£ = 200-109 N/m2, rf = 0.3, 0.2, 0.1 0.05, 0.02 m

RADIATION AND NUCLEAR SAFETY AUTHORITY STUK-YMO-TR136

APPENDIX 12 FINAL RADIAL DEFORMATION e AGAINST SUPPORT THICKNESS dr

R = 2.20 m , k = 4.40, px = 45 MPadashed lines represent the secondary (supportless) state

x IQ- 4 constant angle, 0 rad.

gto

X!

HIDCO

"cCc

-6

-10

Concrete support

-4

co••§ - 4 - 5E

Q-4constant radius, 2.2 m.

CD -5CO

2-5.5COc

-6

/

/

J 1w /

r1

i0

E =20-109N/m2, ^ = 0.5,0.2, 0.1 m

2 4fii(rad)

Steel support

-6

constant angle, 0 rad.

-10 -5


0 5 10 0 2 4r(m) fii(rad)

E = 200-109 N/m2, d = 0.3, 0.2, 0.1 0.05, 0.02 m


FINAL TANGENTIAL DEFORMATION e AGAINST SUPPORT THICKNESS d APPENDIX 13

R = 2 .20m,k = 4.40,px = 45 MPadashed lines represent the secondary (supportless) state

Concrete support

x -|Q-4 constant angle, 1.6 rad.

gCD

1 0

8

CD _

Z 6

CD 4

CO

-10


10

E = 20-109 N/m2, d = 0.5, 0.2, 0.1 m

2 4fii(rad)

Steel support

x ^Q-4 constant angle, 1.6 rad. x 1Q-4constant radius, 2.2 m.

c 1 0

g

IsCD

CO

CD 4

CO

-10 0 5 10 0 2 4r(m) fii(rad)

E = 200109 N/m2, rf = 0.3, 0.2, 0.1 0.05, 0.02 m

final shear deformation

rv> c*s -&• cn


o ->• N> co -^ en

p

p

pb

pbto

Ifinal shear deformation

o

3 o

t/2

ce

T3O

top

%a.IIp

p cn


o

nonte suppor

cn

ft

133§ 3

§ P

>3IItotoo

1313O3

g

^

C/3

a0

o

zon

inC

3H

n

C/3

JO

HO2

a2ancn

on>

cn

ojoH

on

2oHJO

ON


FINAL RADIAL DISPLACEMENT ur AGAINST SUPPORT THICKNESS d APPENDIX 15

R = 2.20 m , k = 4.40, px = 45 MPadashed lines represent the secondary (supportless) state

Concrete support

X 1 Q - 3 constant angle, 0 rad. -3 constant radius, 2.2 m.

10

E = 20-109 N/m2, d = 0.5, 0.2, 0.1 m

2 4fii(rad)

Steel support

E,

ent

EoCOQ.COT3

laT3JOCOc

2.

1.

0.

5

2

5

1

5

x 1C

%

(-3 constant angle, 0 rad.• i • \ • ;

ll \ 'I ... 1 :..; //// ; \\\\ ;' i \\\v

/ / / / / \ \ \ \ \//vv, , \ v | •

////If \\ Vw/III \\ \\^

//// 1 1 \ \ \w////// W \\SJ

///' \V\\V;

-3 constant radius, 2.2 m.

-10 -5 0 5 10 0 2 4r(m) fii(rad)

E = 200109 N/m2, d = 0.3, 0.2, 0.1 0.05, 0.02 m


APPENDIX 16 FINAL TANGENTIAL DISPLACEMENT U^ AGAINSTSUPPORT THICKNESS D

R = 2.20 m,k = 4.40, px = 45 MPadashed lines represent the secondary (supportless) state

Concrete support

cg-co

CO0

W

-10 -5

5

4

3

2

1

0

1

x i Q-4constant angle, 0.79 rad.

0r(m)

-4constant radius, 2.2 m.

10

E = 20-109 N/m2, d = 0.5, 0.2, 0.1 mfii(rad)

Steel support

10-4constant angle, 0.79 rad.

co

"•3

(!)T3k_

C00

CO

-1

-10

-4constant radius, 2.2 m.

2 4fii(rad)

= 0.3, 0.2, 0.1 0.05, 0.02 m


DEFORMATION OF A TUNNEL CROSS SECTION,URL, PINAWA, MANITOBA, CANADA

APPENDIX 17

Documents

Analytical solutions of the mechanical behaviour of rock ... · The rock mass was assumed to be a continuous, homogeneous and isotropic medium obeying Hooke's law, and the time-dependent