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Finite element analysis of jointed rock masses and engineering application
Int. J. Rock Mech. Min. Sci. & Geomech. Abstr. Vol. 30, No. 5, pp. 537-544, 1993
Printed in Great Britain. All rights reserved
0148-9062/93 $6.00 + 0.00 Copyright © 1993 Pergamon Press Ltd
Finite Element Analysis of Jointed Rock Masses and Engineering Application w. zI-iU% P. WANGf In this paper, regularly jointed rock masses are regarded as composed of isotropic rock elements and equivalent anisotropic jointed rock elements. On the basis of analyzing the deformation and strength of jointed rock elements, fundamental formulae are derived for equivalent deformation and equivalent strength: thus, an equivalent continuum model is presented. By non-linear FEM analysis the simulation is performed for the mechanical properties of jointed rock masses. The validity of the model is verified by means of physical model tests. This numerical model has been applied in for strength prediction and stability analysis for two typical engineering jointed rock masses, with satisfactory results.
method, discrete element method and so on. However, these methods have their limitations, for example, in compatibility of joint elements with adjacent continuum elements, the scale effect in equivalence, definition of the damage tensor, equation of damage evolution and the discretion error. Our purpose lies in developing a new numerical simulation for jointed rock masses and providing an efficient technique for rapid and exact prediction of their mechanical behaviour.
INTRODUCTION
It is difficult to quantitatively predict mechanical properties of jointed masses due to their complexity, even by expensive, large-scale in situ tests. The parameters can be obtained, as a rule, only for joint planes or rock blocks of very limited scale (with a test area smaller than 1 m, in general). The parameters in the macroscopic or average sense for masses of more extended range are not available through direct measurement. Prediction of properties of jointed rock masses by computer simulation is convenient for a number of cases such as different loadings, variation in the kinds of materials, and the simulation of joint distribution. Therefore, it is promising in its future application. Naturally, the creditability of this simulation should be supported by physical simulation tests. For jointed rock masses, various kinds of numerical simulations have been developed in the world. For rocks containing a small number of joints, there are in a joint element method [1] and that in which the concept of a discontinuous plane [2] is accepted. The simulation of densely jointed rock masses is mainly realized by two methods. One is the approach of continuum mechanics, or the equivalence approach, such as the following methods: material parameter equivalence, energy equivalence [3], deformation equivalence [4], composite equivalence [5], fracture mechanics [6] and damage mechanics [7]. The other approach is to take the rock blocks as particles of a discontinuum, to study the mechanical properties of the assembly of these particles: stress, strain, stability in light of discontinuum theory, and to derive mechanical properties for the blocks. Examples of this approach are the rigid block
AN E Q U I V A L E N T C O N T I N U U M M O D E L F O R JOINTED ROCK MASSES
As shown in Fig. 1, two sets of joints are spaced regularly in the surrounding rock of the cavity. Typical elements can be found, which are large enough to involve these two sets and their connection character [8] (Fig. 2). Of course, it is desired that the size of the element be sufficiently small compared to the engineering range. Numerical analysis of these typical joint-containing elements leads to the strength deformation property, which is then to be taken as a constitutive relation for the jointed rock mass and in stability analysis. According to papers already published, the basic idea of all the "equivalence methods" is as follows: to have an overall consideration of the influence of joints on properties of a rock mass on the basis of some equivalence principles, to take the jointed rock mass as uniform and continuous so as to derive a set of constitutive relations and then to obtain mechanical properties for the jointed rock mass by means of numerical analysis. These kinds of equivalence models are mostly elastic: discussion of elasto-plastic constitutive relations are seldom encountered in the published papers. The method proposed here is characterized by the following: firstly, to discretize typical elements of the jointed rock
flnstitute of Rock & Soil Mechanics, Academia Sinica, Wuhan, People's Republic of China. 537
538
ZHU and WANG: FEA OF JOINTED ROCK
Typical element
Table I. Principle of the equivalent continuum model Prototype Equivalent continuummodel Geometry [] [] mode discontinuous equivalent continuous body Deformation Total deformation is Deformationequivalence: character the sum of deformation deformationequal to that of of rock and that of jointedrock mass under joints the same loading Strength Joint strength is used Strengthequivalence: character when failure is along failuretakes place the joint plane, rock simultaneouslywith strength is used when the prototype failure takes place in the intact rock
Fig. 1. Jointed rocks around the cavity. mass into rock elements and jointed rock elements (that is, rock elements with joints), to establish equivalent constitutive relations through detailed study of the deformation-strength properties for the jointed rock elements, to perform FEM analysis of the typical elements and obtain mechanical properties of the elements. In the treatment of joints, this method lies between the "generality" of the equivalence method and the "particularity" of the joint element method; therefore, it has significance in engineering. Table 1 gives the principle of the equivalent continuum model.
Formula of deformation equivalence The principle of deformation equivalence assumes equal deformation for the equivalent continuous body and the jointed rock elements under the same loading, and hence derives the relation of material constants between equivalent elements and jointed rock elements. This equivalent body can be assumed, according to different circumstances, as isotropic, orthogonally anisotropic, transversely isotropic and generally anisotropic. Here, the equivalent body is regarded as generally anisotropic. Assume the joint strike is in the same direction as the elastic principal direction (axis z). In the plane-stress condition, an anisotropic material has a stress-strain relation as follows:
Because c~ = cji(i, j = 1, 2, 3), there are 6 independent constants. Elastic FE calculation will not be difficult provided these 6 parameters can be determined. However, their in situ measurement is not easy. Since this jointed rock mass can be regarded as composed of isotropic rock and joints the deformation parameters of the equivalent continuous body can be obtained according to deformation equivalence through measuring rock elastic constants Er, v~, the constants of joints with thickness, Ej, vj, or joint stiffness coefficients Kn, Ks, and the joint geometrical parameters (spacing, persistance inclination angle, width, etc.).
Deformation equivalence formula with no dilation of joints. Assume that the rock is generally anisotropic. The elastic constitutive relation can be written in the form:
~xy
\st31
Sr32 sr33/
r.KV
The joints have no thickness and penetrate the whole element. Forces are applied to the jointed rock element, as shown in Fig. 3. The stress on the joint plane can be obtained from the equilibrium equation:
a~ = ax sin 2 ct + ay cos 2 ~t - ~y sin 2~ r=aysin~tcosct -a~sin~cosu+z~ycos2ct~"
(1)
Deformation superposition leads to: "Cxy
\C31
C32
C33/
~xy
o1
~x=Sll
n T "~ r a x d + s l 2 ¢rY y d + s tr3 z x v d + O" - - s in ~ - - - c o s o t d
"
kn
k,
-~.
r ~n T . 6y = s~ trxl + s~2ayl + s23Zx~ I + -- cos ~ + -- sm ct I k, k, J
(2)
7
I (the 1st set)
Ye
Xe 2 (the 2nd set)
,i
Fig. 2. Typical element.
02
After substituting (1) into (2) and rearranging, we have: 6 x = ( s ~ t d + ~ s i n 2 ° t s i n ~ + l c °as 2 ~ s i nk ~ ) s I c°s2 ~t sin ~t --ks I c°s2 ~t sin ~t)tre + ( s~2d+k,
(
,
)j
+ s~3 d - 1 sin 2e sin e - .-- cos 2e cos at ~.,, k~ ks •
1
ZHU and WANG: FEA OF JOINTED ROCK
dir-----/~S211+ ~ 1s i n
2
well. As for the case of multiple joints, the derivation is similar.
~t cos • - L ~ ksin2 . cos=jax
+(s~l
• 2 ot cos ~)o'y + 1k. cos2 ct cos~t + ~1 sin /
+ s~fl
lsin2ecose - k~
x
~
539
Deformation equivalence formula with dilation of joint. Let the joint dilation deformation (normal defor-
~. (3)
mation caused by shear deformation) be 6a, the dilation angle i, then did = z/k, tan i. With dilation considered, (2) becomes:
c o s 2 e s i n e xx~
On . T T ~t fix = s~t dtr~ + S~2 day + s~3 dxxy+ .-=-sm ~t - -- cos • -_-- tan i sin ~t =
,
+
l . y + s . l x, +
On
Let the jointed rock element be equivalent to a generally anisotropic body. Then its elastic constitutive relation is:
~X)
/Sll
S12 S13~
(axy)
~y =Is2, s22 s23]x ay ~xy \S31 S32 S33/ T
6~ = sll oxd + sl2ayd + sl3xxyd] dicy s2~oxi+ s~oyl + s23"cxyl ]"
sl2 = sI2 +
S13 = S I 3 - ( +
ksd
sin
-
T
(6) tan i cos
From the two terms on the right-handed side in each of the expressions of (6), we can obtain, respectively: T T T - k ~ c°s ct - k~ "-- tan i sin ~ = -7-7 ks c°s °t sin
z sin ~t - --z tan i cos ~t = 77;,, ks gs
where k" = ks/(1 + tan i tan ~) ]
(4)
c°s2 = sin =
(7)
? ks'
ks/(1 - tan i cotan ~ ) ]
With K~ replacing Ks in the first 3 equations of (5), and K~' replacing Ks in the last 3 equations, then sH, s~2, sl3, s2~, s~ and s23 can be obtained. In addition, as an approximation, we take s33 = s~3. In this way, the deformation equivalence formula with dilation considered is established.
Formula of strength equivalence
cos 2 ~t - k ~ c°s2 ~t sin at
COS 20¢ COS0~ +lk.d sin2~ sin~
,. (5) s2t = s[i +
T
ks
According to the above deformation equivalence principle, namely: fix = 6~,, 6y = 6~, we have: k ~ sin2 = + ~
+
•
Under the same loading, the deformation of the equivalent body is:
sn = sll +
cos
sin 2 ~t - ~ / s i n 2 ot cos
Assume the strength of rock, joint plane and equivalent body follow the Mohr-Coulomb criterion, the parameters being (Cr, ~b,), (q, ~bj), (c,, ~=) respectively. The strength of jointed rock elements consists of the strength of rock and that of the joint plane• The joint rock element may undergo two kinds of failure: the failure of rock and that of the joint plane•
1 2 S22=S[2"~(+COS20g + ~ f l s i n ~)cos=
I
~ From Fig. 3, we have l = d tan e and then s~ = s2~ can be obtained. From the symmetry of the constitutive relation, it is easy to obtain s3t = s~3 h2 = s23. Because the derivation of s33 is difficult, we take s33 = s~3 as a simplification. Therefore, all of the six parameters needed for the constitutive relations are obtained. For the case of an element with two joints, the s o for one joint can be obtained from (5) at first, then by replacing s,~ with s o for one joint and repeating the calculation, (~ being the inclination of the second joint) in (5) as before, the s o and the constitutive relations of the equivalent body with two joints can be obtained as
XYx ~xy
~t
Fig. 3
Ox
540
ZHU and WANG: FEA OF JOINTED ROCK
Strength equivalence in the case of a single joint plane. The strength conditions of the rock element and equivalent continuous body are respectively: 0-~ - a3 _ 0-~ + a3 sin ~or = cr cos (/Or, 2 2 0-1 - - 0"3
O'l "~- 0"3 •
- 2
2
sm ~Oe= ce cos ~0e,
I
CJI
(8)
(9)
03
03
and it is evident
c,=cr,
q~ = tp~.
(10)
For jointed rock elements,//is the angle of the joint plane with the plane of the maximum principal stress• Now, we discuss two cases with different values o f / / : When ////r,~, the strength of jointed rock elements is dominated by that of the rock, having the same equivalent formula as (lO). When //n~n ~
-a3sin(2fl-~bj)
at+a3sin~bJ2
qcos%.
(11)
When //~n ~
0-1 "~ 0"3
2
2
sin ~bj cos ~bj s i n ( 2 / / - tpj) = cj sin(2fl - ~bj)'
tpe = sln Ce
Cj .
-
t
sin (pj sin(2//-- %) cos % l
.
.
the sliding trace is determined, the sliding continues on this trace. Therefore, what is important is which joint is dominant in respect of strength. The equivalent strength should be dependent on the strength of the dominant joint. Let the strength criterion of the joint plane be written in another form: 2cj + 2 t a n ~j0-3 "~-a3. 0-1 -- (1 -- tan ~0~cotan fl)sin 2fl
(13)
1
1
"
(12)
.
s i n ( 2 / / - 4~j)cos ¢po
~
oI
/
Fig. 5. Model of two joint sets.
At = (1 - tan % cotan//i )sin 2//~'
]
Therefore, when the strength of the joint plane is dominant, the strength of the jointed rock elements will vary with the joint direction, a fact which is different from the case of rock strength. Strength equivalence in the case of two joint planes. Figure 5 shows the mechanical model of two joint sets. What is significant is along which joint the sliding begins. This is dependent on the geometry of the combination of joint sets within the rock mass, the stress distribution, the shear strength and the anisotropy. Once
o3
o
Assume both fl~ and r2 satisfy//mi, ~
and comparing it to (9) will lead to: •
T
A2=
(1 - tan % cotan//2)sin 2//2"
If A t < A2, the first joint is dominant in strength. If A, >A2, the second joint is dominant in strength. It is easy to obtain the equivalent strength with a similar derivation to that used for (12). If an element contains three joints, the parameters of equivalent strength can be determined in a similar way. Hoek and Brown [9] have pointed out that for strength and deformation properties, rock elements with 4 joint sets or more can be considered isotropic. Analysis of tensile failure. When a jointed rock element is under tension, its tensile strength is, to a greater extent, dependent on that of joint planes. Non-tension analysis should be adopted in this circumstance• The constitutive relations can be established in the light of incremental plasticity. For elements with two sets of joints, non-tension analysis should be performed with the dominant joint plane fully considered.
03
IMPLEMENTATION, VERIFICATION AND APPLICATION OF THE EQUIVALENT CONTINUUM MODEL I Ol
Fig. 4. M o d e l o f a single joint set.
Treatment of elements with non-persistent joints In a jointed rock element already discretized, it is possible that joints are not persistent. Therefore,
541
ZHU and WANG: FEA OF JOINTED ROCK
(a)
(b)
I
(c)
(d)
(e)
(g)
(f)
(h)
(i)
Fig. 6. Distribution of joint length.
Fig. 7. Distribution of joint angle.
modification should be made for the fundamental formula of deformation and strength equivalence. For simplicity, let us define the projection length ratios in the x- and y-directions, respectively, as:
discussed (Fig. 6). Calculation results are listed in Table 2. Variation of joint angles. Assume there is a joint with a length of 1.0 crn at the centre of the element. The values of its angle are as follows: (g) a = 0°; (h) • = 45°; (i) ~ = 71.56 °. The distribution of the joint is shown in Fig. 7. Results are given in Table 3. Scale effect of an element. The form of joint is the same as (d) in (1); cc = 45 °, Ax = Ay = 0.6. The side length of jointed rock element I has 6 values. Table 4 gives the results. Calculation results show that when the size of the jointed rock element is greater than 10cm, the error decreases to 5% or less. In discretion of engineering the jointed rock mass, the scale of the finite elements is generally much greater than 10 cm. Therefore, the treatment of the deformation equivalence with (5) and its modified version is to a certain extent precise and is also convenient in computer implementation.
Ax = Lj~
A, =
Ljy
(14)
where L~ and Lie are the joint projection lengths in the x- and y-directions. Lx and Ly are the lengths in the x- and y-directions. It is apparent that deformation of elements is positively correlative to A~, Ay; thus, the following modification should be made for equation (5). When 0 ~
= cjn% -k Cr(l -- n°~), ~j = ~jn °//o -~- ¢~r(1 -- n O/0).
This is a working assumption for the treatment of element properties, c~ and ~ of the equivalent body can be derived from ~j, ~j and expression (12). Now we begin to verify the validity of the modified formula in terms of the following: joint length, joint inclination and the scale effect of jointed rock elements. First, deformation 61 (or strain El) in the direction ~l can be calculated using FEM (Goodman joint element used). With ~l (El) regarded as the equivalent deformation, we have elastic constant sn = E1/a~, DN is defined as s~]/sH. Alternatively, s~, DH can be derived from the modification of (5) according to the joint distribution, and then the two can be compared to each other. The calculation of the relative error DH which comes from FEM, is regarded as exact. Variation of joint length. Assume the jointed rock element is a square in shape with a size of 5 × 5 era, joint inclination cc = 45 °, E, ffi 68 MPa, Vr = 0.25, K, = 75.0MPa/cm, K~=5.0MPa/cm. Six cases will be
Case a b c d e f
Table 2. Influenceof joint length on Du Ax Ay Du(analytical) Dn(FEM) Relativeerror (%) 0.0 0.2 0.4 0.6 0.8 1.0
0.0 0.2 0.4 0.6 0.8 1.0
1.00 0.83 0.71 0.62 0.55 0.49
1.00 0.94 0.83 0.72 0.62 0.49
0 12 14 14 11 0
Verification of the numerical model by means of a physical model The global mechanical properties of jointed rock masses can be reflected if the jointed rock elements proposed here are put into structural analysis as one kind of basic FE elements. To guarantee rationability and reliability, strict verification should be performed by means of a physical model test proposed by previous scholars [10]. The size of the joint model is 50 x 50 x 7 cm, containing two sets of orthogonal joints with the same spacing and length of 5 crn. Figure 8 shows 3 kinds of joint distribution and their FE meshes. Due to the difficulty in the physical model test, comparisons cover only strength deformation properties under the uniaxial stress state of 3 kinds of jointed rock and under the biaxial stress state of joints of type (a), (Figs 9, 10). In Fig. 10, E*, ¢* are the strains in the direction of ¢~ and cr2 . Figures 9 and 10 indicate that the stress-strain relations under the uniaxial stress state coincide with each other. In the biaxial stress state, the maximum principal stress-principal strain relation, peak
Case g h i
Ax 0.2 0.2 0.2
Table 3. Influenceof joint angle on Du A~ Dn(analytical) Dn(FEM) Relativeerror (%) 0.0 0.96 0.90 7 0.2 0.83 0.88 6 0.3 0.81 0.93 13
Table 4. The scale effect(a = 45°, dx = dy = 0.6) Side length (¢m) 5 I0 20 30 40 50 D,, (analytical) 0.62 0.77 0.87 0.90 0.93 0.94 Du (FEM) 0.72 0.81 0.88 0.91 0.93 0.94 Relative error (%) 14 5 1 1 0 0
542
Z H U and W A N G :
(a)
FEA OF JOINTED ROCK
(b)
"4 .~,
fig im
lli
?b 7. :
IK
ill im ill ilR llli im Mill im Ill ill
,
\
(,
\ f
- . ) ") ' "
',
,d
i ',,
I
/
/ \
i
~
"q/N/\/\/IN/N/IN L/ / / / / / ~q/N/N/N~/N/N
,,,), " -. ) ' ' × \
/'/'/'/'
x
\
\
\ \
( , I
/
,
/'/'/
\
/ \
(c)
x
/
/
/
/
v v
N/K/\/\V'\/\/]'N ~"~/ \ / \ / \ L / \ / \
\
/\/'/~'/'/'/\/"
,<
/
/
/\-,/~-,/,-\-, ) , ) , )~ ) , / )\ ),/~ ), /
/ / / / \// \ / /\ / \ , /< \ //\ \/\/\/\.~\/\
/
/
/
/
~/ /IN
/i
/
/
V
,
\
i
/~ /
\
)\ ;, )\ ~
'
"
/
/
/
/
/
/
Fig. 8. M o d e l o f joint distribution a n d F E m e s h .
strength and strain have good coincidence. But the maximum, principal stress-lateral strain relations have a poor coincidence with each other, which may be caused by the fact that the measured deformation is on the small side due to the boundary friction in the model test.
of the rock mass contains about l0 joints, which guarantees the "mechanical representativeness". The distribution of joints and the FE meshes can be seen from Fig. I 1. There are 625 elements, 242 of which are jointed rock elements. The number of nodes is 676. The mode of joint distribution is type (a), as described above. The mechanical parameters of rock are obtained from the laboratory tests on the rock of grade A. The parameters of rock and joints are in Table 5. For lack of data on K,, Ks, only strength equivalence is considered here. FE analysis involves 3 kinds of element (Fig. I 1)" rock elements, and two kinds of jointed rock elements with a size of 71.2 x 71.2 cm. The equivalence of strength property is automatically treated by a program. By our numerical simulation, loading simulated by the computer can be used to give the yield values of th with respect to different lateral pressures a2; then c, ~b, the strength parameters of masses for the whole calculation range can be obtained by inverse analysis, with the Drucker-Prager criterion used. The above results arc given in Fig. 12 which shows envelopes of different strength.
ENGINEERING APPLICATIONS Strength prediction o f jointed rock masses
The authors have used the above method to predict the strength of a rock mass at the Ertan electric station of China, and received good results. According to the engineering geological reports of the Ertan project, in the rock mass of grade A-C, there are 3 major joint sets, two of which are steep and the other gentle. The two orthogonal main joint sets are considered in the calculation. The length of the joints is between 4 and 7 m, and taken in the calculation as 5 m, with a spacing of I m. These values are typical according to geological data. The joint persistence is 50 and 30%, respectively. The calculation range of 17.8 × 17.8 m satisfies the demand that each side (a)
(b)
o (MPa)
o (MPa)
- 0.4
--0.4 f~
0.2
/ I I~2 20
I 10
I 0
f
10
I
J
I
E2 20
10
10
t0-~
20 ~1
o (MPa)
(c)
Physical ....
0.2
Numerical
f
10-3 82
20
10
0
10
20 E 1
Fig. 9. C o m p a r i s o n o f a - ~ relations u n d e r uniaxial loading.
I 10-3 20 e 1
ZHU and WANG:
FEA OF JOINTED ROCK
543 01 - 02 (MPa)
(b)
(a)
"--
//r~
I
01 - 02 (MPa)
02 = 0.339
I x-~ X
--
\
0.4 / / / ~ " " ~ -
0;4 /
t
\
,~i/~/f
/
0 2 = 0.15
-
"ll}"2
/
- 0.2
I
I E 2 -- E2*
10
o
I 20
10
_
10-3
E l -- E l
E2-
E2
I 10
01 - 02 (MPa) (C)- . . . . .
~'..
f'
I I I
I
E2 - E2* 10
Numerical
o~ = o.2s
[
0
E 1 -- E l
Physical
/ ~ ~-- ~-
'i ~ - o.4,'" 'I I
10 -3
I 10
0
10
[
. 10 -3
20 El - EI
Fig. 10. Comparison of o ~ relations under biaxial loading of jointed rock of type (a). / 1
3
15
2
5
0
5
10
15
O (MPa) Fig. 12. Comparison of results with proposed values: (1) intact rock; (2) joint plane; and (3) numerical simulation. Fig. 1i, Distribution of joints and meshing of jointed rock.
Stability analysis of jointed rock masses
1
Now a stability analysis was performed on the jointed rock surrounding an underground power station. It is situated 400 m underground. The surrounding rock is granite. The relative distance of the power house to the transformer hall, their relative size, and the excavation order are shown in Fig. 13. During excavation, measurement with extensometers was performed on several sections of the two cavities. The distribution of two Table 5. Parameters of rocks and joints Material parameters Intact rock Joint plane
E(MPa) 35,000 --
v 0.30 --
c(MPa) 14.58 0.5
VI ~ (degrees) 65.20 36.89
I
II
II
III
If!
IV
IV
V
T r a n s f o r m e r hall
[
Power
house
Fig. 13. The excavation order of the power house and the transformer hall.
544
ZHU and WANG: FEA OF JOINTED ROCK
Materials Rock Joint
E(MPa) 37,000 1
Table 6. Mechanical parameters of materials v c(MPa) ~b( ° ) R,(MPa) KnfMPa/cm) 0.20 22.27 48.1 5.0 -__ 0.5 35 0.0 600,000
Table 7. The predicted displacement convergence values (mm) Measurement lines 1 2 3 4 5 6 7 8 9 11.4 18.5 19.5 6.0 6.9 7.3 10.8 7.4 6.9 9.7 15.5 16.3 5.0 5.8 6.1 9.0 6.3 5.8
m a j o r joint sets and 3 faults is shown in Fig. 14. D e f o r m a t i o n properties o f the jointed rock, and then the constitutive relations can be obtained with the deformation equivalence method. Assume the rock element is generally anisotropic and the joints have zero thickness spaced over all the element. According to statistical data, the persistence rate o f the jointed rock mass is taken as 50%. Constitutive relations o f the rock mass with two joint sets can be derived with the above method. Mechanical parameters are listed in Table 6. The calculation range was 525 x 320 m, with 654 elements and 630 nodes. The c o m p o n e n t s o f the initial stressfield, ax, Oy, "rxy, were obtained from the displacement data o f the fourth excavation stage o f the power house and the data o f the second stage o f the transformer hall, by means o f inverse analysis. Optimization calculation was performed with the simplex accelerated method in inverse analysis. The object function is F = ~ [ui(x) - u°(x)] 2, i=1
where u~(x) indicates calculated relative displacement, u ° ( x ) is measured relative displacement, n is the n u m b e r o f relative measurement points. The c o m p o n e n t s o f g r o u n d stress field, which are received from inverse analysis, are trx = - 10.19 MPa, t r y = - 8 . 5 1 MPa, Txy = - 1.12 MPa.
Ks(MPa/cm) -7500
Remarks Equivalent Elastic
Using the above g r o u n d stresses and mechanical parameters, non-linear F E M calculation has been performed with respect to the ultimate excavation state o f the two cavities. The D r u c k e r - P r a g e r criterion was adopted. The range o f failure zone a r o u n d the cavities and some convergence values was obtained (Fig. 14). Table 7 gives the comparison o f the convergence values o f jointed rock analysis and elastic analysis with no joint considered. CONCLUSION In this paper, the analysis m e t h o d o f mechanical properties o f rock masses with regularly spaced interrupted (or coherent) joints is developed on the basis o f the equivalent c o n t i n u u m model for the jointed rock elements. The physical model tests and engineering application have shown the feasibility and convenience o f this method. It is applicable to most cases with a regular distribution o f joints and especially to the stability analysis o f hydroengineering structures, in which circumstances significant deformation and d a m a g e are not allowed. Because the equivalence is in the macroscopic and average sense, further research is needed for developing methods which can deal with interaction and the failure mechanism o f joints. Acknowledgement--The authors are grateful for the financial support
from the National Natural Science Foundation of China. Accepted for publication 8 March 1993.
Fig. 14. Overstress areas of surrounding rocks.
REFERENCES 1. Goodman R. E. Methods of geological engineering in discontinuous rocks. West, St Paul, MN (1976). 2. Ren W. Advance in Plastic Mechanics. The Publishing House of Chinese Railway (1988). 3. Gerrard C. M. Equivalent elastic moduli of a rock mass consisting of orthorhombic layers. Int. J. Rock Mech. Min. Sei. & Geomech. Abstr. 19, 15 (1982). 4. Fossum A. F. Effective elastic properties for a randomly jointed rock mass. Int. J. Rock Mech. Min. Sci. & Geomech Abstr. 22, 467 (1985). 5. Wu C. and Xianhong C. Elastic models of jointed rock masses. Chinese J. Geotech. Engng No. 4 (1987). 6. Ping C. The theory of anistropic rock mechanics and its application. Doctorial Dissertation of Zhongnan Industry University, Changsha, China (1990). 7. Kawamoto T. et al. Deformation and fracturing behaviour of discontinuous rock mass and damage mechanics theory. Int. J. Numer. Analyt. Meth. Geomech. 12, (1988). 8. Zhu W. et aL Research on strength behaviour of jointed rock mass by numerical and physical simulation. Proc. Int. Conf. on Mechanics of Jointed and Faulted Rock, Vienna (1990). 9. Hoek E. and Brown E. T. Underground Excavation in Rock. The Institution of Mining and Metallurgy (1980). 10. Lian Z. Geomechanical model test of mechanical behaviour of jointed rock masses. Research Report of Wuhan Inst. of Rock and Soil Mechanics (1990).
Printed in Great Britain. All rights reserved
0148-9062/93 $6.00 + 0.00 Copyright © 1993 Pergamon Press Ltd
Finite Element Analysis of Jointed Rock Masses and Engineering Application w. zI-iU% P. WANGf In this paper, regularly jointed rock masses are regarded as composed of isotropic rock elements and equivalent anisotropic jointed rock elements. On the basis of analyzing the deformation and strength of jointed rock elements, fundamental formulae are derived for equivalent deformation and equivalent strength: thus, an equivalent continuum model is presented. By non-linear FEM analysis the simulation is performed for the mechanical properties of jointed rock masses. The validity of the model is verified by means of physical model tests. This numerical model has been applied in for strength prediction and stability analysis for two typical engineering jointed rock masses, with satisfactory results.
method, discrete element method and so on. However, these methods have their limitations, for example, in compatibility of joint elements with adjacent continuum elements, the scale effect in equivalence, definition of the damage tensor, equation of damage evolution and the discretion error. Our purpose lies in developing a new numerical simulation for jointed rock masses and providing an efficient technique for rapid and exact prediction of their mechanical behaviour.
INTRODUCTION
It is difficult to quantitatively predict mechanical properties of jointed masses due to their complexity, even by expensive, large-scale in situ tests. The parameters can be obtained, as a rule, only for joint planes or rock blocks of very limited scale (with a test area smaller than 1 m, in general). The parameters in the macroscopic or average sense for masses of more extended range are not available through direct measurement. Prediction of properties of jointed rock masses by computer simulation is convenient for a number of cases such as different loadings, variation in the kinds of materials, and the simulation of joint distribution. Therefore, it is promising in its future application. Naturally, the creditability of this simulation should be supported by physical simulation tests. For jointed rock masses, various kinds of numerical simulations have been developed in the world. For rocks containing a small number of joints, there are in a joint element method [1] and that in which the concept of a discontinuous plane [2] is accepted. The simulation of densely jointed rock masses is mainly realized by two methods. One is the approach of continuum mechanics, or the equivalence approach, such as the following methods: material parameter equivalence, energy equivalence [3], deformation equivalence [4], composite equivalence [5], fracture mechanics [6] and damage mechanics [7]. The other approach is to take the rock blocks as particles of a discontinuum, to study the mechanical properties of the assembly of these particles: stress, strain, stability in light of discontinuum theory, and to derive mechanical properties for the blocks. Examples of this approach are the rigid block
AN E Q U I V A L E N T C O N T I N U U M M O D E L F O R JOINTED ROCK MASSES
As shown in Fig. 1, two sets of joints are spaced regularly in the surrounding rock of the cavity. Typical elements can be found, which are large enough to involve these two sets and their connection character [8] (Fig. 2). Of course, it is desired that the size of the element be sufficiently small compared to the engineering range. Numerical analysis of these typical joint-containing elements leads to the strength deformation property, which is then to be taken as a constitutive relation for the jointed rock mass and in stability analysis. According to papers already published, the basic idea of all the "equivalence methods" is as follows: to have an overall consideration of the influence of joints on properties of a rock mass on the basis of some equivalence principles, to take the jointed rock mass as uniform and continuous so as to derive a set of constitutive relations and then to obtain mechanical properties for the jointed rock mass by means of numerical analysis. These kinds of equivalence models are mostly elastic: discussion of elasto-plastic constitutive relations are seldom encountered in the published papers. The method proposed here is characterized by the following: firstly, to discretize typical elements of the jointed rock
flnstitute of Rock & Soil Mechanics, Academia Sinica, Wuhan, People's Republic of China. 537
538
ZHU and WANG: FEA OF JOINTED ROCK
Typical element
Table I. Principle of the equivalent continuum model Prototype Equivalent continuummodel Geometry [] [] mode discontinuous equivalent continuous body Deformation Total deformation is Deformationequivalence: character the sum of deformation deformationequal to that of of rock and that of jointedrock mass under joints the same loading Strength Joint strength is used Strengthequivalence: character when failure is along failuretakes place the joint plane, rock simultaneouslywith strength is used when the prototype failure takes place in the intact rock
Fig. 1. Jointed rocks around the cavity. mass into rock elements and jointed rock elements (that is, rock elements with joints), to establish equivalent constitutive relations through detailed study of the deformation-strength properties for the jointed rock elements, to perform FEM analysis of the typical elements and obtain mechanical properties of the elements. In the treatment of joints, this method lies between the "generality" of the equivalence method and the "particularity" of the joint element method; therefore, it has significance in engineering. Table 1 gives the principle of the equivalent continuum model.
Formula of deformation equivalence The principle of deformation equivalence assumes equal deformation for the equivalent continuous body and the jointed rock elements under the same loading, and hence derives the relation of material constants between equivalent elements and jointed rock elements. This equivalent body can be assumed, according to different circumstances, as isotropic, orthogonally anisotropic, transversely isotropic and generally anisotropic. Here, the equivalent body is regarded as generally anisotropic. Assume the joint strike is in the same direction as the elastic principal direction (axis z). In the plane-stress condition, an anisotropic material has a stress-strain relation as follows:
Because c~ = cji(i, j = 1, 2, 3), there are 6 independent constants. Elastic FE calculation will not be difficult provided these 6 parameters can be determined. However, their in situ measurement is not easy. Since this jointed rock mass can be regarded as composed of isotropic rock and joints the deformation parameters of the equivalent continuous body can be obtained according to deformation equivalence through measuring rock elastic constants Er, v~, the constants of joints with thickness, Ej, vj, or joint stiffness coefficients Kn, Ks, and the joint geometrical parameters (spacing, persistance inclination angle, width, etc.).
Deformation equivalence formula with no dilation of joints. Assume that the rock is generally anisotropic. The elastic constitutive relation can be written in the form:
~xy
\st31
Sr32 sr33/
r.KV
The joints have no thickness and penetrate the whole element. Forces are applied to the jointed rock element, as shown in Fig. 3. The stress on the joint plane can be obtained from the equilibrium equation:
a~ = ax sin 2 ct + ay cos 2 ~t - ~y sin 2~ r=aysin~tcosct -a~sin~cosu+z~ycos2ct~"
(1)
Deformation superposition leads to: "Cxy
\C31
C32
C33/
~xy
o1
~x=Sll
n T "~ r a x d + s l 2 ¢rY y d + s tr3 z x v d + O" - - s in ~ - - - c o s o t d
"
kn
k,
-~.
r ~n T . 6y = s~ trxl + s~2ayl + s23Zx~ I + -- cos ~ + -- sm ct I k, k, J
(2)
7
I (the 1st set)
Ye
Xe 2 (the 2nd set)
,i
Fig. 2. Typical element.
02
After substituting (1) into (2) and rearranging, we have: 6 x = ( s ~ t d + ~ s i n 2 ° t s i n ~ + l c °as 2 ~ s i nk ~ ) s I c°s2 ~t sin ~t --ks I c°s2 ~t sin ~t)tre + ( s~2d+k,
(
,
)j
+ s~3 d - 1 sin 2e sin e - .-- cos 2e cos at ~.,, k~ ks •
1
ZHU and WANG: FEA OF JOINTED ROCK
dir-----/~S211+ ~ 1s i n
2
well. As for the case of multiple joints, the derivation is similar.
~t cos • - L ~ ksin2 . cos=jax
+(s~l
• 2 ot cos ~)o'y + 1k. cos2 ct cos~t + ~1 sin /
+ s~fl
lsin2ecose - k~
x
~
539
Deformation equivalence formula with dilation of joint. Let the joint dilation deformation (normal defor-
~. (3)
mation caused by shear deformation) be 6a, the dilation angle i, then did = z/k, tan i. With dilation considered, (2) becomes:
c o s 2 e s i n e xx~
On . T T ~t fix = s~t dtr~ + S~2 day + s~3 dxxy+ .-=-sm ~t - -- cos • -_-- tan i sin ~t =
,
+
l . y + s . l x, +
On
Let the jointed rock element be equivalent to a generally anisotropic body. Then its elastic constitutive relation is:
~X)
/Sll
S12 S13~
(axy)
~y =Is2, s22 s23]x ay ~xy \S31 S32 S33/ T
6~ = sll oxd + sl2ayd + sl3xxyd] dicy s2~oxi+ s~oyl + s23"cxyl ]"
sl2 = sI2 +
S13 = S I 3 - ( +
ksd
sin
-
T
(6) tan i cos
From the two terms on the right-handed side in each of the expressions of (6), we can obtain, respectively: T T T - k ~ c°s ct - k~ "-- tan i sin ~ = -7-7 ks c°s °t sin
z sin ~t - --z tan i cos ~t = 77;,, ks gs
where k" = ks/(1 + tan i tan ~) ]
(4)
c°s2 = sin =
(7)
? ks'
ks/(1 - tan i cotan ~ ) ]
With K~ replacing Ks in the first 3 equations of (5), and K~' replacing Ks in the last 3 equations, then sH, s~2, sl3, s2~, s~ and s23 can be obtained. In addition, as an approximation, we take s33 = s~3. In this way, the deformation equivalence formula with dilation considered is established.
Formula of strength equivalence
cos 2 ~t - k ~ c°s2 ~t sin at
COS 20¢ COS0~ +lk.d sin2~ sin~
,. (5) s2t = s[i +
T
ks
According to the above deformation equivalence principle, namely: fix = 6~,, 6y = 6~, we have: k ~ sin2 = + ~
+
•
Under the same loading, the deformation of the equivalent body is:
sn = sll +
cos
sin 2 ~t - ~ / s i n 2 ot cos
Assume the strength of rock, joint plane and equivalent body follow the Mohr-Coulomb criterion, the parameters being (Cr, ~b,), (q, ~bj), (c,, ~=) respectively. The strength of jointed rock elements consists of the strength of rock and that of the joint plane• The joint rock element may undergo two kinds of failure: the failure of rock and that of the joint plane•
1 2 S22=S[2"~(+COS20g + ~ f l s i n ~)cos=
I
~ From Fig. 3, we have l = d tan e and then s~ = s2~ can be obtained. From the symmetry of the constitutive relation, it is easy to obtain s3t = s~3 h2 = s23. Because the derivation of s33 is difficult, we take s33 = s~3 as a simplification. Therefore, all of the six parameters needed for the constitutive relations are obtained. For the case of an element with two joints, the s o for one joint can be obtained from (5) at first, then by replacing s,~ with s o for one joint and repeating the calculation, (~ being the inclination of the second joint) in (5) as before, the s o and the constitutive relations of the equivalent body with two joints can be obtained as
XYx ~xy
~t
Fig. 3
Ox
540
ZHU and WANG: FEA OF JOINTED ROCK
Strength equivalence in the case of a single joint plane. The strength conditions of the rock element and equivalent continuous body are respectively: 0-~ - a3 _ 0-~ + a3 sin ~or = cr cos (/Or, 2 2 0-1 - - 0"3
O'l "~- 0"3 •
- 2
2
sm ~Oe= ce cos ~0e,
I
CJI
(8)
(9)
03
03
and it is evident
c,=cr,
q~ = tp~.
(10)
For jointed rock elements,//is the angle of the joint plane with the plane of the maximum principal stress• Now, we discuss two cases with different values o f / / : When ////r,~, the strength of jointed rock elements is dominated by that of the rock, having the same equivalent formula as (lO). When //n~n ~
-a3sin(2fl-~bj)
at+a3sin~bJ2
qcos%.
(11)
When //~n ~
0-1 "~ 0"3
2
2
sin ~bj cos ~bj s i n ( 2 / / - tpj) = cj sin(2fl - ~bj)'
tpe = sln Ce
Cj .
-
t
sin (pj sin(2//-- %) cos % l
.
.
the sliding trace is determined, the sliding continues on this trace. Therefore, what is important is which joint is dominant in respect of strength. The equivalent strength should be dependent on the strength of the dominant joint. Let the strength criterion of the joint plane be written in another form: 2cj + 2 t a n ~j0-3 "~-a3. 0-1 -- (1 -- tan ~0~cotan fl)sin 2fl
(13)
1
1
"
(12)
.
s i n ( 2 / / - 4~j)cos ¢po
~
oI
/
Fig. 5. Model of two joint sets.
At = (1 - tan % cotan//i )sin 2//~'
]
Therefore, when the strength of the joint plane is dominant, the strength of the jointed rock elements will vary with the joint direction, a fact which is different from the case of rock strength. Strength equivalence in the case of two joint planes. Figure 5 shows the mechanical model of two joint sets. What is significant is along which joint the sliding begins. This is dependent on the geometry of the combination of joint sets within the rock mass, the stress distribution, the shear strength and the anisotropy. Once
o3
o
Assume both fl~ and r2 satisfy//mi, ~
and comparing it to (9) will lead to: •
T
A2=
(1 - tan % cotan//2)sin 2//2"
If A t < A2, the first joint is dominant in strength. If A, >A2, the second joint is dominant in strength. It is easy to obtain the equivalent strength with a similar derivation to that used for (12). If an element contains three joints, the parameters of equivalent strength can be determined in a similar way. Hoek and Brown [9] have pointed out that for strength and deformation properties, rock elements with 4 joint sets or more can be considered isotropic. Analysis of tensile failure. When a jointed rock element is under tension, its tensile strength is, to a greater extent, dependent on that of joint planes. Non-tension analysis should be adopted in this circumstance• The constitutive relations can be established in the light of incremental plasticity. For elements with two sets of joints, non-tension analysis should be performed with the dominant joint plane fully considered.
03
IMPLEMENTATION, VERIFICATION AND APPLICATION OF THE EQUIVALENT CONTINUUM MODEL I Ol
Fig. 4. M o d e l o f a single joint set.
Treatment of elements with non-persistent joints In a jointed rock element already discretized, it is possible that joints are not persistent. Therefore,
541
ZHU and WANG: FEA OF JOINTED ROCK
(a)
(b)
I
(c)
(d)
(e)
(g)
(f)
(h)
(i)
Fig. 6. Distribution of joint length.
Fig. 7. Distribution of joint angle.
modification should be made for the fundamental formula of deformation and strength equivalence. For simplicity, let us define the projection length ratios in the x- and y-directions, respectively, as:
discussed (Fig. 6). Calculation results are listed in Table 2. Variation of joint angles. Assume there is a joint with a length of 1.0 crn at the centre of the element. The values of its angle are as follows: (g) a = 0°; (h) • = 45°; (i) ~ = 71.56 °. The distribution of the joint is shown in Fig. 7. Results are given in Table 3. Scale effect of an element. The form of joint is the same as (d) in (1); cc = 45 °, Ax = Ay = 0.6. The side length of jointed rock element I has 6 values. Table 4 gives the results. Calculation results show that when the size of the jointed rock element is greater than 10cm, the error decreases to 5% or less. In discretion of engineering the jointed rock mass, the scale of the finite elements is generally much greater than 10 cm. Therefore, the treatment of the deformation equivalence with (5) and its modified version is to a certain extent precise and is also convenient in computer implementation.
Ax = Lj~
A, =
Ljy
(14)
where L~ and Lie are the joint projection lengths in the x- and y-directions. Lx and Ly are the lengths in the x- and y-directions. It is apparent that deformation of elements is positively correlative to A~, Ay; thus, the following modification should be made for equation (5). When 0 ~
= cjn% -k Cr(l -- n°~), ~j = ~jn °//o -~- ¢~r(1 -- n O/0).
This is a working assumption for the treatment of element properties, c~ and ~ of the equivalent body can be derived from ~j, ~j and expression (12). Now we begin to verify the validity of the modified formula in terms of the following: joint length, joint inclination and the scale effect of jointed rock elements. First, deformation 61 (or strain El) in the direction ~l can be calculated using FEM (Goodman joint element used). With ~l (El) regarded as the equivalent deformation, we have elastic constant sn = E1/a~, DN is defined as s~]/sH. Alternatively, s~, DH can be derived from the modification of (5) according to the joint distribution, and then the two can be compared to each other. The calculation of the relative error DH which comes from FEM, is regarded as exact. Variation of joint length. Assume the jointed rock element is a square in shape with a size of 5 × 5 era, joint inclination cc = 45 °, E, ffi 68 MPa, Vr = 0.25, K, = 75.0MPa/cm, K~=5.0MPa/cm. Six cases will be
Case a b c d e f
Table 2. Influenceof joint length on Du Ax Ay Du(analytical) Dn(FEM) Relativeerror (%) 0.0 0.2 0.4 0.6 0.8 1.0
0.0 0.2 0.4 0.6 0.8 1.0
1.00 0.83 0.71 0.62 0.55 0.49
1.00 0.94 0.83 0.72 0.62 0.49
0 12 14 14 11 0
Verification of the numerical model by means of a physical model The global mechanical properties of jointed rock masses can be reflected if the jointed rock elements proposed here are put into structural analysis as one kind of basic FE elements. To guarantee rationability and reliability, strict verification should be performed by means of a physical model test proposed by previous scholars [10]. The size of the joint model is 50 x 50 x 7 cm, containing two sets of orthogonal joints with the same spacing and length of 5 crn. Figure 8 shows 3 kinds of joint distribution and their FE meshes. Due to the difficulty in the physical model test, comparisons cover only strength deformation properties under the uniaxial stress state of 3 kinds of jointed rock and under the biaxial stress state of joints of type (a), (Figs 9, 10). In Fig. 10, E*, ¢* are the strains in the direction of ¢~ and cr2 . Figures 9 and 10 indicate that the stress-strain relations under the uniaxial stress state coincide with each other. In the biaxial stress state, the maximum principal stress-principal strain relation, peak
Case g h i
Ax 0.2 0.2 0.2
Table 3. Influenceof joint angle on Du A~ Dn(analytical) Dn(FEM) Relativeerror (%) 0.0 0.96 0.90 7 0.2 0.83 0.88 6 0.3 0.81 0.93 13
Table 4. The scale effect(a = 45°, dx = dy = 0.6) Side length (¢m) 5 I0 20 30 40 50 D,, (analytical) 0.62 0.77 0.87 0.90 0.93 0.94 Du (FEM) 0.72 0.81 0.88 0.91 0.93 0.94 Relative error (%) 14 5 1 1 0 0
542
Z H U and W A N G :
(a)
FEA OF JOINTED ROCK
(b)
"4 .~,
fig im
lli
?b 7. :
IK
ill im ill ilR llli im Mill im Ill ill
,
\
(,
\ f
- . ) ") ' "
',
,d
i ',,
I
/
/ \
i
~
"q/N/\/\/IN/N/IN L/ / / / / / ~q/N/N/N~/N/N
,,,), " -. ) ' ' × \
/'/'/'/'
x
\
\
\ \
( , I
/
,
/'/'/
\
/ \
(c)
x
/
/
/
/
v v
N/K/\/\V'\/\/]'N ~"~/ \ / \ / \ L / \ / \
\
/\/'/~'/'/'/\/"
,<
/
/
/\-,/~-,/,-\-, ) , ) , )~ ) , / )\ ),/~ ), /
/ / / / \// \ / /\ / \ , /< \ //\ \/\/\/\.~\/\
/
/
/
/
~/ /IN
/i
/
/
V
,
\
i
/~ /
\
)\ ;, )\ ~
'
"
/
/
/
/
/
/
Fig. 8. M o d e l o f joint distribution a n d F E m e s h .
strength and strain have good coincidence. But the maximum, principal stress-lateral strain relations have a poor coincidence with each other, which may be caused by the fact that the measured deformation is on the small side due to the boundary friction in the model test.
of the rock mass contains about l0 joints, which guarantees the "mechanical representativeness". The distribution of joints and the FE meshes can be seen from Fig. I 1. There are 625 elements, 242 of which are jointed rock elements. The number of nodes is 676. The mode of joint distribution is type (a), as described above. The mechanical parameters of rock are obtained from the laboratory tests on the rock of grade A. The parameters of rock and joints are in Table 5. For lack of data on K,, Ks, only strength equivalence is considered here. FE analysis involves 3 kinds of element (Fig. I 1)" rock elements, and two kinds of jointed rock elements with a size of 71.2 x 71.2 cm. The equivalence of strength property is automatically treated by a program. By our numerical simulation, loading simulated by the computer can be used to give the yield values of th with respect to different lateral pressures a2; then c, ~b, the strength parameters of masses for the whole calculation range can be obtained by inverse analysis, with the Drucker-Prager criterion used. The above results arc given in Fig. 12 which shows envelopes of different strength.
ENGINEERING APPLICATIONS Strength prediction o f jointed rock masses
The authors have used the above method to predict the strength of a rock mass at the Ertan electric station of China, and received good results. According to the engineering geological reports of the Ertan project, in the rock mass of grade A-C, there are 3 major joint sets, two of which are steep and the other gentle. The two orthogonal main joint sets are considered in the calculation. The length of the joints is between 4 and 7 m, and taken in the calculation as 5 m, with a spacing of I m. These values are typical according to geological data. The joint persistence is 50 and 30%, respectively. The calculation range of 17.8 × 17.8 m satisfies the demand that each side (a)
(b)
o (MPa)
o (MPa)
- 0.4
--0.4 f~
0.2
/ I I~2 20
I 10
I 0
f
10
I
J
I
E2 20
10
10
t0-~
20 ~1
o (MPa)
(c)
Physical ....
0.2
Numerical
f
10-3 82
20
10
0
10
20 E 1
Fig. 9. C o m p a r i s o n o f a - ~ relations u n d e r uniaxial loading.
I 10-3 20 e 1
ZHU and WANG:
FEA OF JOINTED ROCK
543 01 - 02 (MPa)
(b)
(a)
"--
//r~
I
01 - 02 (MPa)
02 = 0.339
I x-~ X
--
\
0.4 / / / ~ " " ~ -
0;4 /
t
\
,~i/~/f
/
0 2 = 0.15
-
"ll}"2
/
- 0.2
I
I E 2 -- E2*
10
o
I 20
10
_
10-3
E l -- E l
E2-
E2
I 10
01 - 02 (MPa) (C)- . . . . .
~'..
f'
I I I
I
E2 - E2* 10
Numerical
o~ = o.2s
[
0
E 1 -- E l
Physical
/ ~ ~-- ~-
'i ~ - o.4,'" 'I I
10 -3
I 10
0
10
[
. 10 -3
20 El - EI
Fig. 10. Comparison of o ~ relations under biaxial loading of jointed rock of type (a). / 1
3
15
2
5
0
5
10
15
O (MPa) Fig. 12. Comparison of results with proposed values: (1) intact rock; (2) joint plane; and (3) numerical simulation. Fig. 1i, Distribution of joints and meshing of jointed rock.
Stability analysis of jointed rock masses
1
Now a stability analysis was performed on the jointed rock surrounding an underground power station. It is situated 400 m underground. The surrounding rock is granite. The relative distance of the power house to the transformer hall, their relative size, and the excavation order are shown in Fig. 13. During excavation, measurement with extensometers was performed on several sections of the two cavities. The distribution of two Table 5. Parameters of rocks and joints Material parameters Intact rock Joint plane
E(MPa) 35,000 --
v 0.30 --
c(MPa) 14.58 0.5
VI ~ (degrees) 65.20 36.89
I
II
II
III
If!
IV
IV
V
T r a n s f o r m e r hall
[
Power
house
Fig. 13. The excavation order of the power house and the transformer hall.
544
ZHU and WANG: FEA OF JOINTED ROCK
Materials Rock Joint
E(MPa) 37,000 1
Table 6. Mechanical parameters of materials v c(MPa) ~b( ° ) R,(MPa) KnfMPa/cm) 0.20 22.27 48.1 5.0 -__ 0.5 35 0.0 600,000
Table 7. The predicted displacement convergence values (mm) Measurement lines 1 2 3 4 5 6 7 8 9 11.4 18.5 19.5 6.0 6.9 7.3 10.8 7.4 6.9 9.7 15.5 16.3 5.0 5.8 6.1 9.0 6.3 5.8
m a j o r joint sets and 3 faults is shown in Fig. 14. D e f o r m a t i o n properties o f the jointed rock, and then the constitutive relations can be obtained with the deformation equivalence method. Assume the rock element is generally anisotropic and the joints have zero thickness spaced over all the element. According to statistical data, the persistence rate o f the jointed rock mass is taken as 50%. Constitutive relations o f the rock mass with two joint sets can be derived with the above method. Mechanical parameters are listed in Table 6. The calculation range was 525 x 320 m, with 654 elements and 630 nodes. The c o m p o n e n t s o f the initial stressfield, ax, Oy, "rxy, were obtained from the displacement data o f the fourth excavation stage o f the power house and the data o f the second stage o f the transformer hall, by means o f inverse analysis. Optimization calculation was performed with the simplex accelerated method in inverse analysis. The object function is F = ~ [ui(x) - u°(x)] 2, i=1
where u~(x) indicates calculated relative displacement, u ° ( x ) is measured relative displacement, n is the n u m b e r o f relative measurement points. The c o m p o n e n t s o f g r o u n d stress field, which are received from inverse analysis, are trx = - 10.19 MPa, t r y = - 8 . 5 1 MPa, Txy = - 1.12 MPa.
Ks(MPa/cm) -7500
Remarks Equivalent Elastic
Using the above g r o u n d stresses and mechanical parameters, non-linear F E M calculation has been performed with respect to the ultimate excavation state o f the two cavities. The D r u c k e r - P r a g e r criterion was adopted. The range o f failure zone a r o u n d the cavities and some convergence values was obtained (Fig. 14). Table 7 gives the comparison o f the convergence values o f jointed rock analysis and elastic analysis with no joint considered. CONCLUSION In this paper, the analysis m e t h o d o f mechanical properties o f rock masses with regularly spaced interrupted (or coherent) joints is developed on the basis o f the equivalent c o n t i n u u m model for the jointed rock elements. The physical model tests and engineering application have shown the feasibility and convenience o f this method. It is applicable to most cases with a regular distribution o f joints and especially to the stability analysis o f hydroengineering structures, in which circumstances significant deformation and d a m a g e are not allowed. Because the equivalence is in the macroscopic and average sense, further research is needed for developing methods which can deal with interaction and the failure mechanism o f joints. Acknowledgement--The authors are grateful for the financial support
from the National Natural Science Foundation of China. Accepted for publication 8 March 1993.
Fig. 14. Overstress areas of surrounding rocks.
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