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Analytical solution for a circular opening in an elastic–brittle–plastic rock
ARTICLE IN PRESS
International Journal of Rock Mechanics & Mining Sciences 43 (2006) 616–622 www.elsevier.com/locate/ijrmms
Analytical solution for a circular opening in an elastic–brittle–plastic rock Kyung-Ho Parka,, Yong-Jin Kimb a
School of Civil Engineering, Asian Institute of Technology, P.O. Box 4, Klong Luang, Pathumthani 12120, Thailand b NOWENG Co. Ltd, Sucho-dong 1461-1, Hallagisan B/D 3F, Sucho-gu, Seoul, Korea Accepted 3 November 2005 Available online 19 December 2005
Abstract This paper deals with the analytical solutions for the prediction of displacements around a circular opening in an elastic–brittle–plastic rock mass compatible with a linear Mohr–Coulomb or a nonlinear Hoek–Brown yield criterion. Three different cases of definitions for elastic strains in the plastic region, used in the existing solutions, are mentioned. The closed-form analytical solutions for the displacement in the plastic region are derived on a theoretically consistent way for all the cases by employing a non-associated flow rule. The results of the dimensionless displacements are compared using the data of the soft and hard rocks to investigate the effect of different definitions for elastic strains with the dilation angle. r 2005 Elsevier Ltd. All rights reserved. Keywords: Analytical solution; Brittle plastic rock; Hoek–Brown yield criterion; Mohr–Coulomb yield criterion; Circular opening
1. Introduction Prediction of the stresses and displacements around a circular opening in the rock mass at great depth is an important problem in geotechnical, petroleum and mining engineering such as the design of tunnels, boreholes and mine shafts. A large number of analytical solutions for this axisymmetric opening problem have been presented by considering different models of material behavior, such as the elastic-perfectly plastic, elastic–brittle–plastic and elastic-strain softening models, with the different yield criteria, like the linear Mohr–Coulomb (M–C) and nonlinear Hoek–Brown (H–B) criteria [1]. For an elastic–brittle–plastic model, Brown et al. [1] presented closed-form solutions for the stresses and displacements of a circular opening in the H–B media. These solutions included a simplifying assumption for elastic strains in the formulation for displacement in the plastic region. Later Wang [2] mentioned the errors in the solutions of Brown et al. [1] and presented a numerical Corresponding author. Tel.: +66 2 524 5508; fax: +66 2 524 5509.
E-mail address: [email protected] (K.-H. Park). 1365-1609/$ - see front matter r 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.ijrmms.2005.11.004
solution for the correct constitutive equations. Recently Sharan [3,4] pointed out the errors in the solutions by Brown et al. [1] and Wang [2], and presented simple closed-form solution for the displacement in the plastic region in the H–B media. On the other hand, Ogawa and Lo [5], Reed [6] and Yu [7] introduced different solutions for the displacements in the M–C and H–B media by employing a non-associated flow rule. Thus there is some confusion in the solution for the displacement in the plastic region. This study deals with the closed-form analytical solutions for the displacement within the plastically deforming region in the elastic–brittle–plastic analysis of a circular opening. Three different cases of definitions for elastic strains, used in the existing solutions, are discussed. The closed-form solutions for all the cases are derived on a theoretically consistent way using a non-associated flow rule with two different (linear M–C and nonlinear H–B) yield criteria. In order to investigate the effect of different definitions for elastic strains on the displacement, the results of the dimensionless radial displacements obtained from the solutions are compared for the hard and soft rocks, by using the data used by Sharan [3] and Ogawa and
ARTICLE IN PRESS K.-H. Park, Y.-J. Kim / International Journal of Rock Mechanics & Mining Sciences 43 (2006) 616–622
617
Lo [5]. The effects of dilatancy and yield criterion on the displacement are also discussed.
2. Definition of the problem Fig. 1 shows a circular opening being excavated in a continuous, homogeneous, isotropic, initially elastic rock mass subjected to a hydrostatic stress po. The opening surface is subjected to an internal pressure pi. As pi is gradually reduced, the radial displacement occurs and a plastic region develops around the opening when pi is less than p1y, i.e. the initial yielding stress. After yielding, the strength of rock suddenly drops and follows the post-yield softening behavior. The material behavior of elastic–brittle–plastic model used in this study is shown in Fig. 2. It is required to solve for the stresses and displacements in the plastic region. Two most commonly used yield criteria for the rock are considered in this study: the nonlinear H–B yield criterion, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s1 ¼ s3 þ msc s3 þ ss2c (1) and the linear M–C yield criterion, s1 ¼ as3 þ Y ,
(2)
where s1 ¼ the major principal stress at failure, s3 ¼ the minor principal stress at failure, sc ¼ the uniaxial compressive strength of the intact rock material, m and s ¼ material constants which depend on the properties of the rock and on the extent to which it has been broken before being subject to the stresses (s ¼ 1 for intact rock and so1 for previously broken rock), a ¼ ð1 þ sin fÞ= ð1 sin fÞ, Y ¼ 2c cos f=ð1 sin fÞ, c ¼ the cohesion of the rock, and f ¼ the friction angle of the rock.
Fig. 2. Material behavior model used in this study.
Because of the axial symmetry of the problem, the radial and tangential stresses, sr and sy, in the rock mass will be principal stresses, such as s1 ¼ sy and s3 ¼ sr . Then Eqs. (1) and (2) can be expressed as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sy ¼ sr þ msc sr þ ss2c for peak strength, (3) sy ¼ sr þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mr sc sr þ sr s2c for postpeak strength
(4)
or sy ¼ asr þ Y sy ¼ ar sr þ Y r
for peak strength, for postpeak strength,
(5) (6)
where mr and sr ¼ the residual values of H–B constants for the yielded rock, ar ¼ ð1 þ sin fr Þ=ð1 sin fr Þ, Y r ¼ 2cr cos fr =ð1 sin fr Þ, and cr and fr ¼ the residual values of cohesion and friction angle of the rock, respectively. 3. Analytical solution for stresses in the plastic region The differential equation of equilibrium for the axisymmetric problem can be expressed as Fig. 1. A circular opening in an infinite medium.
dsr sr sy þ ¼ 0. dr r
(7)
ARTICLE IN PRESS K.-H. Park, Y.-J. Kim / International Journal of Rock Mechanics & Mining Sciences 43 (2006) 616–622
618
By substituting Eq. (4) into Eq. (7) and using the boundary condition sr ¼ pi at r ¼ a, the stresses in the plastic region can be obtained for the H–B yield criterion r r þ BH2B ‘n2 , (8) sr ¼ pi þ AH2B ‘n a a r sy ¼ pi þ AH2B þ ðAH2B þ 2BH2B Þ‘n a H2B 2 r þB ‘n , ð9Þ a where qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mr sc (10) AHB ¼ mr sc pi þ sr s2c ; BH2B ¼ 4 and the superscript H–B represents the H–B yield criterion. The radius of the plastic region c can be obtained by considering the continuity of the radial stress at the elastic–plastic interface, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9 8 <AH2B þ AH2B2 4BH2B ðpi pH2B Þ= c 1y ¼ exp , (11) : ; a 2BH2B where pH2B 1y
ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 mpo m ¼ p o sc þ þs 2 4 8 sc
(12)
which is the initial yielding stress. In the same way, using Eq. (6), the stresses in the plastic region and the radius of the plastic region can be obtained for the M–C criterion r ar 1 , (13) sr ¼ AM2C þ BM2C a r ar 1 , (14) sy ¼ AM2C þ BM2C ar a c ¼ a
(
pM2C AM2C 1y BM2C
)
1 ar 1
,
(15)
where ¼ pM2C 1y
2po Y , aþ1
AM2C ¼
Yr ; ar 1
(16) BM2C ¼ pi þ
Yr ar 1
(17)
and the superscript M–C represents the M–C yield criterion. 4. Analytical solution for displacements in the plastic region In the plastic region, total radial and tangential strains, er and ey, can be decomposed into elastic and plastic parts as er ¼ eer þ epr ,
(18)
ey ¼ eey þ epy .
(19)
For small strain problem, the strains can be expressed in terms of inward radial displacement u: er ¼
du , dr
u ey ¼ . r
(20) (21)
In addition, in order to determine the displacement field in the plastic region, a plastic flow rule is needed. By assuming that the elastic strains are relatively small in comparison to the plastic strains and that a non-associated flow rule is valid, the plastic parts of radial and tangential strains may be related for the plane strain condition (s3os2os1) epr þ bepy ¼ 0,
(22)
where b ¼ ð1 þ sin cÞ=ð1 sin cÞ, c ¼ the dilation angle of the rock. From Eqs. (18)–(22) the differential equation for the radial displacement can be expressed as du u þ b ¼ f ðrÞ, dr r
(23)
where f ðrÞ ¼ eer þ beey .
(24)
By using the following boundary condition for the radial displacement uc at the elastic–plastic interface: uc ¼
c ðp p1y Þ 2G o
(25)
in which G ¼ the shear modulus of the rock mass, the solution for Eq. (23) can be obtained as Z cb 1 r b u¼ b r f ðrÞ dr þ uc . (26) r c r In order to evaluate the integral in Eq. (26), expressions for elastic strains are needed. If the function f(r) is expressed in simple function of r, the integration can be carried out analytically. Generally three different definitions for elastic strains can be found: (1) Case 1: With the assumption of the constant elastic strain in the plastic region such as that at the elastic–plastic interface, the elastic strains can be expressed as [1] eer ¼ eey ¼
1 ðp p1y Þ, 2G o
1 ðp p1y Þ. 2G o
(27)
(28)
So, the function f(r) is f ðrÞ ¼
1 ðb 1Þðpo p1y Þ. 2G
(29)
(2) Case 2: By considering the plastic region as the thickwalled cylinder subjected to the inner stress, pipo at r ¼ a, and the outer stress, p1ypo at r ¼ c, the elastic strains can
ARTICLE IN PRESS K.-H. Park, Y.-J. Kim / International Journal of Rock Mechanics & Mining Sciences 43 (2006) 616–622
be obtained as [3] 1 D ð1 2nÞC þ 2 , eer ¼ 2G r 1 D ð1 2nÞC 2 , eey ¼ 2G r
where (30)
(31)
Then the integration in Eq. (26) can be made analytically and the solution for the radial displacement can be simplified as u 1 1 ¼ ½D1 f 1 ðrÞ þ D2 f 2 ðrÞ þ D3 f 3 ðrÞ r 2G rbþ1 þ 2Guc cb D1 f 1 ðcÞ D2 f 2 ðcÞ D3 f 3 ðcÞ,
C¼
and n ¼the Poisson’s ratio. So, the function f(r) is 1 D ð1 2nÞð1 þ bÞC þ ð1 bÞ 2 . f ðrÞ ¼ 2G r
¼ ð1 þ bÞð1 2nÞðAM2C po Þ, DM2C 1 DM2C ¼ fð1 n bnÞ þ ar ðb nb nÞgBM2C . 2
where ðp1y po Þc2 ðpi po Þa2 , c 2 a2 a2 c2 ðpi p1y Þ D¼ c 2 a2
ð32Þ
Z
(33)
1 ½ð1 nÞðsr po Þ nðsy po Þ, 2G
(34)
ðrÞ ¼ f M2C 2
eey ¼
1 ½ð1 nÞðsy po Þ nðsr po Þ. 2G
(35)
f H2B ðrÞ ¼ 3
1 ½ð1 n bnÞsr þ ðb bn nÞsy 2G ð1 2nÞð1 þ bÞpo .
ð36Þ
It should be noted that the functions f(r) in Cases 1 and 2 are simple and the same forms can be used for two different yield criteria. Thus the integration in Eq. (26) and the displacements in the plastic region can be obtained in the same form for different yield criteria. However Case 3 needs stresses in the plastic region for the derivation of displacement. By using Eqs. (8) and (9) for the H–B yield criterion and Eqs. (13) and (14) for the M–C yield criterion, the function f(r) can be expressed in terms of r: for the H–B yield criterion, r i 1 h H2B H2B 2 r f ðrÞ ¼ D1 þ DH2B ‘n ‘n þ D , (37) 2 3 2G a a where ¼ ðb nb nÞAH2B þ ð1 þ bÞð1 2nÞðpi po Þ, DH2B 1 DH2B ¼ ð1 þ bÞð1 2nÞAH2B þ 2ðb nb nÞBH2B , 2 DH2B ¼ ð1 þ bÞð1 2nÞBH2B 3 and for the M–C yield criterion, 1 r ar 1 D1 f ðrÞ ¼ þ D2 , 2G a
(38)
rb dr ¼
rbþ1 , bþ1
Z
r rb ‘n dr a rbþ1 r 1 ‘n , ¼ a bþ1 bþ1
f H2B ðrÞ ¼ 2
eer ¼
So, the function f(r) is
ð39Þ
where f 1 ðrÞ ¼
(3) Case 3: By using the elastic stress–strain relationship with the consideration of initial hydrostatic stress, the elastic strains can be expressed as [6,7]
f ðrÞ ¼
619
Z
rb
r ar 1 a
dr ¼
1 aar 1
rbþar , b þ ar
Z
r rb ‘n2 dr a r rbþ1 2 2 2 r ‘n ‘n þ . ¼ a bþ1 a bþ1 ð b þ 1Þ 2
Table 1 summarizes the constants, D1, D2 and D3, and functions, f1(r), f2(r) and f3(r), for all the cases with the H–B and M–C yield criteria. It is of interest to note that the solutions by Cases 1 and 2 are the same as those of Brown et al. [1] and Sharan [3] respectively, for the H–B yield criterion, while the solutions by Case 3 are the same as those of Reed [6] and Yu [7] for the M–C and H–B yield criteria, respectively. In addition, the solutions presented in this paper have been obtained by using the deformation theory of plasticity [8,9] and equivalent results can be found using the incremental theory of plasticity [10–13] with the correct form of elastic strain increments [12]. 5. Comparison of solutions In order to investigate the effect of different definitions for elastic strains on the displacement, the dimensionless radial displacements in the plastic region are compared by using the typical cases of soft and hard rocks, used by Sharan [3] and Ogawa and Lo [5]. Tables 2 and 3 show the properties of the rocks. It should be noted that Ogawa and Lo [5] determined c and f from m, s and sc to compare the results with equivalent parameter values for two different yield criteria. The dimensionless radial displacement, uE/poa, in the plastic region is plotted in Figs. 3–8 for cases of c ¼ 0 (no plastic volume change) and c ¼ 301 (plastic volume
ARTICLE IN PRESS K.-H. Park, Y.-J. Kim / International Journal of Rock Mechanics & Mining Sciences 43 (2006) 616–622
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Table 1 Summary of the constants and functions for Eq. (39) Case 1
Case 2
Case 3 (HB)
Case 3 (MC)
D1 D2 D3 f1(r)
ðb 1Þðpo p1y Þ 0 0 rbþ1 bþ1
ð1 2nÞð1 þ bÞC ð1 bÞD 0 rbþ1 bþ1
ðb nb nÞA þ ð1 2nÞð1 þ bÞðpi po Þ ð1 2nÞð1 þ bÞA þ 2ðb nb nÞB ð1 2nÞð1 þ bÞB rbþ1 bþ1
ð1 2nÞð1 þ bÞðA po Þ ð1 n bnÞ þ ar ðb bn nÞ B 0 rbþ1 bþ1
f2(r)
0
rb1 b1
rbþ1 r 1 ln a bþ1 bþ1
1 rbþar aar 1 b þ ar
f3(r)
p1y
c/a
Case 3 ðH2BÞ :
r rbþ1 r 2 2 ln ln2 ; Others : 0 ðzeroÞ þ 2 a bþ1 a bþ1 ðb þ 1Þ
MC 2po Y aþ1
HB rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 mpo m po sc þ þs 2 4 8 sc
1 p1y A ar 1 B
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8 ffi9 > > 2 > = <A þ A 4B pi p1y > exp > > 2B > > ; :
Yr ar 1
A
B
pi þ
uc
c ðp p1y Þ 2G o
Yr ar 1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mr sc pi þ sr s2c mr sc 4 c ðp p1y Þ 2G o
Table 2 Data set 1 (Sharan [3])
Table 3 Data set 2 (Ogawa and Lo [5]) Hard rock
Soft rock
Radius of opening, a (m) Initial stress, po (MPa) Internal pressure, pi (MPa) Young’s modulus, E (MPa) Poisson’s ratio, n Shear modulus, G (MPa)
4 108 0 40 000 0.2 16 667
5 30 5 5500 0.25 2200
H–B m s mr sr sc (MPa)
7.5 0.1 1.0 0.01 300
1.7 0.0039 1.0 0 30
increase). While the high value chosen for the dilation angle may not be usually encountered in practice, the use of the high value in this study is to illustrate the maximum possible effect of dilatancy on the predicted displacements. In Figs. 3–8, the solid line represents the results for the case of c ¼ 0, while the dotted line for c ¼ 301.
Hard rock
Soft rock
Radius of opening, a (m) Initial stress, po (MPa) Internal pressure, pi (MPa) Young’s modulus, E (MPa) Poisson’s ratio, n Shear modulus, G (MPa)
1 1 0 50 000 0.2 20 833
1 1 0 5000 0.2 2083
H–B m s mr sr sc (MPa)
0.5 0.0001 0.3 0.00001 75
0.2 0.0001 0.05 0.00001 50
M–C c (MPa) f (deg) cr (MPa) fr (deg)
0.173 55 0.061 52
0.276 35 0.055 30
(1) Effect of different definitions for elastic strains with dilation angle. Figs. 3 and 4 show the displacements for data set 1. It can be seen that Case 1 gives the smallest
ARTICLE IN PRESS K.-H. Park, Y.-J. Kim / International Journal of Rock Mechanics & Mining Sciences 43 (2006) 616–622
621
10.0
4.0
Case 1
Case 1
8.0
Case 2
3.0
Case 2 Case 3
uE/poa
uE/poa
Case 3
2.0
1.0
0.0
6.0 4.0 2.0 0.0
1
1.05
1.1
1.15
1.2
1.25
1.3
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
r/a
r/a
Fig. 3. Dimensionless radial displacement (data set 1, hard rock, H–B).
Fig. 6. Dimensionless radial displacement (data set 2, soft rock, H–B).
2.5
10.0 Case 1
8.0
2.0
Case 2
uE/poa
uE/poa
Case 3
6.0 4.0
1.5 1.0 Case 1 Case 2
0.5
2.0
Case 3
0.0
0.0 1
1.25
1.5
1.75
2
1
1.05
r/a
1.1
1.15
r/a
Fig. 4. Dimensionless radial displacement (data set 1, soft rock, H–B).
Fig. 7. Dimensionless radial displacement (data set 2, hard rock, M–C).
16.0
2.0
Case 1 Case 2
12.0
1.5 uE/poa
uE/poa
Case 3
1.0
8.0
Case 1
0.5
4.0
Case 2 Case 3
0.0
0.0
1
1.02
1.04
1.06 r/a
1.08
1.1
1.12
1
1.2
1.4
1.6
1.8
r/a
Fig. 5. Dimensionless radial displacement (data set 2, hard rock, H–B).
Fig. 8. Dimensionless radial displacement (data set 2, soft rock, M–C).
displacements. While the difference between results from Case 1 and the others is relatively small for the case of c ¼ 0, it increases with increasing dilation angle. So the assumption for elastic strains of Eqs. (27) and (28) leads to an underestimate of the displacement for both the hard and soft rocks. For Cases 2 and 3, the displacements are almost the same when c ¼ 0. The errors due to the approximation involved in Case 2 are relatively smaller than those due to the
approximation involved in Case 1 [4]. But the difference in displacements between Cases 2 and 3 increases with increasing dilation angle, and becomes larger for the soft rock. So the dilation angle seems to have more impact on the displacement for the soft rock rather than for the hard rock. In addition the displacements by Case 3 are smaller than those by Case 2. (2) Effect of the use of equivalent parameter values. Figs. 5 and 6 show the displacements for data set 2 with the H–B
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622
consistent way. The difference in the definitions of elastic strains among the existing solutions is investigated and the results of the dimensionless radial displacement are compared for the hard and soft rocks. The following main conclusions can be drawn:
2.0
uE/poa
1.5
1.0 Case 1
0.5
Case 2 Case 3
0.0
1
1.01
1.02
1.03
1.04
1.05
1.06
r/a
Fig. 9. Dimensionless radial displacement (data set 2, hard rock, perfectly-plastic). 2.0
uE/poa
1.5
1.0 Case 1
0.5
Case 2
(1) The radial displacement in the plastic region can be simplified as Eq. (39) and Table 1 for three different definitions of elastic strains with two different (linear M–C and nonlinear H–B) yield criteria by employing the non-associated flow rule. (2) Case 1 gives the smallest displacements and seems to give an underestimate of the displacements. While Cases 2 and 3 give almost the same results of displacements when c ¼ 0, the difference of the displacements between Cases 2 and 3 increases as the dilation angle increases. (3) In all the cases, the use of equivalent parameter values for two different yield criteria may lead to different displacements. (4) Even in the case of using the elastic-perfectly plastic model with the M–C yield criterion, the displacements obtained from the solutions by Cases 2 and 3 with c ¼ f (associated flow) are slightly different.
Case 3
0.0
1
1.02
1.04
1.06
1.08
1.1
1.12
1.14
1.16
1.18
References
r/a
Fig. 10. Dimensionless radial displacement (data set 2, soft rock, perfectly-plastic).
yield criterion, while Figs. 7 and 8 with the M–C yield criterion. The same tendency as data set 1 can be seen. Comparison of Figs. 5, 6 with 7,8 indicate that the use of equivalent parameter values for two different yield criteria may lead to different displacement, while Ogawa and Lo [5] mentioned the similar results of the displacements for two different yield criteria. (3) Effect of different definitions for elastic strains in the elastic-perfectly plastic model. The effect of different definitions for elastic strains in the elastic-perfectly plastic model is investigated by calculating the dimensionless radial displacement with the peak strength parameter values. Figs. 9 and 10 shows the displacements using the data set 2 with the M–C yield criterion. In these figures, the solid line represents the case of c ¼ 0 (non-associated flow), while the dotted line for c ¼ f (associated flow). The same tendency mentioned the above can be seen. Even in the perfectly plastic model, the displacements by Cases 2 and 3 are still slightly different for c ¼ f. 6. Conclusion The closed-form analytical solutions of the displacement in the plastic region for the elastic–brittle–plastic analysis of a circular opening have been presented in a theoretically
[1] Brown ET, Bray JW, Ladanyi B, Hoek E. Ground response curves for rock tunnels. J Geotech Eng 1983;109:15–39. [2] Wang Y. Ground response of circular tunnel in poorly consolidated rock. J Geotech Eng 1996;122:703–8. [3] Sharan SK. Elastic–brittle–plastic analysis of circular openings in Hoek–Brown media. Int J Rock Mech Min Sci 2003;40:817–24. [4] Sharan SK. Exact and approximate solutions for displacements around circular openings in elastic–brittle–plastic Hoek–Brown rock. Int J Rock Mech Min Sci 2005;42:542–9. [5] Ogawa T, Lo KY. Effects of dilatancy and yield criteria on displacements around tunnels. Can Geotech J 1987;24:100–13. [6] Reed MB. Stresses and displacements around a cylindrical cavity in soft rock. IMA J Appl Math 1986;36:223–45. [7] Yu HS. Cavity expansion methods in geomechanics. Dordrecht: Kluwer Academic Publishers; 2000. [8] Duncan Fama ME. Numerical modeling of yield zones in weak rock. In: Hudson JA, editor. Comprehensive rock engineering, 2. Oxford: Pergamon; 1993. p. 49–75. [9] Panet M. Understanding deformations in tunnels. In: Hudson JA, editor. Comprehensive rock engineering, 1. Oxford: Pergamon; 1993. p. 663–90. [10] Salenc- con J. Contraction quasi-statique d’une cavite´ a´ syme´trie sphe´rique ou cylindrique dans un milieu e´lastoplastique. Annl Ponts Chauss 1969;4:231–6. [11] Detournay E. Elastoplastic model of a deep tunnel for a rock with variable dilatancy. Rock Mech Rock Eng 1986;19:99–108. [12] Carranza-Torres C, Fairhurst C. The elasto-plastic response of underground excavations in rock masses that satisfy the Hoek– Brown failure criterion. Int J Rock Mech Min Sci 1999;36:777–809. [13] Carranza-Torres C. Elasto-plastic solution of tunnel problems using the generalized form of the Hoek–Brown failure criterion. In: Hudson JA, Xia-Ting F, editors. Proceeding of the ISRM SINOROCK 2004 symposium, China, 2004. Int J Rock Mech Min Sci 2004;41(3): 480–1.
International Journal of Rock Mechanics & Mining Sciences 43 (2006) 616–622 www.elsevier.com/locate/ijrmms
Analytical solution for a circular opening in an elastic–brittle–plastic rock Kyung-Ho Parka,, Yong-Jin Kimb a
School of Civil Engineering, Asian Institute of Technology, P.O. Box 4, Klong Luang, Pathumthani 12120, Thailand b NOWENG Co. Ltd, Sucho-dong 1461-1, Hallagisan B/D 3F, Sucho-gu, Seoul, Korea Accepted 3 November 2005 Available online 19 December 2005
Abstract This paper deals with the analytical solutions for the prediction of displacements around a circular opening in an elastic–brittle–plastic rock mass compatible with a linear Mohr–Coulomb or a nonlinear Hoek–Brown yield criterion. Three different cases of definitions for elastic strains in the plastic region, used in the existing solutions, are mentioned. The closed-form analytical solutions for the displacement in the plastic region are derived on a theoretically consistent way for all the cases by employing a non-associated flow rule. The results of the dimensionless displacements are compared using the data of the soft and hard rocks to investigate the effect of different definitions for elastic strains with the dilation angle. r 2005 Elsevier Ltd. All rights reserved. Keywords: Analytical solution; Brittle plastic rock; Hoek–Brown yield criterion; Mohr–Coulomb yield criterion; Circular opening
1. Introduction Prediction of the stresses and displacements around a circular opening in the rock mass at great depth is an important problem in geotechnical, petroleum and mining engineering such as the design of tunnels, boreholes and mine shafts. A large number of analytical solutions for this axisymmetric opening problem have been presented by considering different models of material behavior, such as the elastic-perfectly plastic, elastic–brittle–plastic and elastic-strain softening models, with the different yield criteria, like the linear Mohr–Coulomb (M–C) and nonlinear Hoek–Brown (H–B) criteria [1]. For an elastic–brittle–plastic model, Brown et al. [1] presented closed-form solutions for the stresses and displacements of a circular opening in the H–B media. These solutions included a simplifying assumption for elastic strains in the formulation for displacement in the plastic region. Later Wang [2] mentioned the errors in the solutions of Brown et al. [1] and presented a numerical Corresponding author. Tel.: +66 2 524 5508; fax: +66 2 524 5509.
E-mail address: [email protected] (K.-H. Park). 1365-1609/$ - see front matter r 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.ijrmms.2005.11.004
solution for the correct constitutive equations. Recently Sharan [3,4] pointed out the errors in the solutions by Brown et al. [1] and Wang [2], and presented simple closed-form solution for the displacement in the plastic region in the H–B media. On the other hand, Ogawa and Lo [5], Reed [6] and Yu [7] introduced different solutions for the displacements in the M–C and H–B media by employing a non-associated flow rule. Thus there is some confusion in the solution for the displacement in the plastic region. This study deals with the closed-form analytical solutions for the displacement within the plastically deforming region in the elastic–brittle–plastic analysis of a circular opening. Three different cases of definitions for elastic strains, used in the existing solutions, are discussed. The closed-form solutions for all the cases are derived on a theoretically consistent way using a non-associated flow rule with two different (linear M–C and nonlinear H–B) yield criteria. In order to investigate the effect of different definitions for elastic strains on the displacement, the results of the dimensionless radial displacements obtained from the solutions are compared for the hard and soft rocks, by using the data used by Sharan [3] and Ogawa and
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Lo [5]. The effects of dilatancy and yield criterion on the displacement are also discussed.
2. Definition of the problem Fig. 1 shows a circular opening being excavated in a continuous, homogeneous, isotropic, initially elastic rock mass subjected to a hydrostatic stress po. The opening surface is subjected to an internal pressure pi. As pi is gradually reduced, the radial displacement occurs and a plastic region develops around the opening when pi is less than p1y, i.e. the initial yielding stress. After yielding, the strength of rock suddenly drops and follows the post-yield softening behavior. The material behavior of elastic–brittle–plastic model used in this study is shown in Fig. 2. It is required to solve for the stresses and displacements in the plastic region. Two most commonly used yield criteria for the rock are considered in this study: the nonlinear H–B yield criterion, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s1 ¼ s3 þ msc s3 þ ss2c (1) and the linear M–C yield criterion, s1 ¼ as3 þ Y ,
(2)
where s1 ¼ the major principal stress at failure, s3 ¼ the minor principal stress at failure, sc ¼ the uniaxial compressive strength of the intact rock material, m and s ¼ material constants which depend on the properties of the rock and on the extent to which it has been broken before being subject to the stresses (s ¼ 1 for intact rock and so1 for previously broken rock), a ¼ ð1 þ sin fÞ= ð1 sin fÞ, Y ¼ 2c cos f=ð1 sin fÞ, c ¼ the cohesion of the rock, and f ¼ the friction angle of the rock.
Fig. 2. Material behavior model used in this study.
Because of the axial symmetry of the problem, the radial and tangential stresses, sr and sy, in the rock mass will be principal stresses, such as s1 ¼ sy and s3 ¼ sr . Then Eqs. (1) and (2) can be expressed as qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sy ¼ sr þ msc sr þ ss2c for peak strength, (3) sy ¼ sr þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mr sc sr þ sr s2c for postpeak strength
(4)
or sy ¼ asr þ Y sy ¼ ar sr þ Y r
for peak strength, for postpeak strength,
(5) (6)
where mr and sr ¼ the residual values of H–B constants for the yielded rock, ar ¼ ð1 þ sin fr Þ=ð1 sin fr Þ, Y r ¼ 2cr cos fr =ð1 sin fr Þ, and cr and fr ¼ the residual values of cohesion and friction angle of the rock, respectively. 3. Analytical solution for stresses in the plastic region The differential equation of equilibrium for the axisymmetric problem can be expressed as Fig. 1. A circular opening in an infinite medium.
dsr sr sy þ ¼ 0. dr r
(7)
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By substituting Eq. (4) into Eq. (7) and using the boundary condition sr ¼ pi at r ¼ a, the stresses in the plastic region can be obtained for the H–B yield criterion r r þ BH2B ‘n2 , (8) sr ¼ pi þ AH2B ‘n a a r sy ¼ pi þ AH2B þ ðAH2B þ 2BH2B Þ‘n a H2B 2 r þB ‘n , ð9Þ a where qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mr sc (10) AHB ¼ mr sc pi þ sr s2c ; BH2B ¼ 4 and the superscript H–B represents the H–B yield criterion. The radius of the plastic region c can be obtained by considering the continuity of the radial stress at the elastic–plastic interface, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9 8 <AH2B þ AH2B2 4BH2B ðpi pH2B Þ= c 1y ¼ exp , (11) : ; a 2BH2B where pH2B 1y
ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 mpo m ¼ p o sc þ þs 2 4 8 sc
(12)
which is the initial yielding stress. In the same way, using Eq. (6), the stresses in the plastic region and the radius of the plastic region can be obtained for the M–C criterion r ar 1 , (13) sr ¼ AM2C þ BM2C a r ar 1 , (14) sy ¼ AM2C þ BM2C ar a c ¼ a
(
pM2C AM2C 1y BM2C
)
1 ar 1
,
(15)
where ¼ pM2C 1y
2po Y , aþ1
AM2C ¼
Yr ; ar 1
(16) BM2C ¼ pi þ
Yr ar 1
(17)
and the superscript M–C represents the M–C yield criterion. 4. Analytical solution for displacements in the plastic region In the plastic region, total radial and tangential strains, er and ey, can be decomposed into elastic and plastic parts as er ¼ eer þ epr ,
(18)
ey ¼ eey þ epy .
(19)
For small strain problem, the strains can be expressed in terms of inward radial displacement u: er ¼
du , dr
u ey ¼ . r
(20) (21)
In addition, in order to determine the displacement field in the plastic region, a plastic flow rule is needed. By assuming that the elastic strains are relatively small in comparison to the plastic strains and that a non-associated flow rule is valid, the plastic parts of radial and tangential strains may be related for the plane strain condition (s3os2os1) epr þ bepy ¼ 0,
(22)
where b ¼ ð1 þ sin cÞ=ð1 sin cÞ, c ¼ the dilation angle of the rock. From Eqs. (18)–(22) the differential equation for the radial displacement can be expressed as du u þ b ¼ f ðrÞ, dr r
(23)
where f ðrÞ ¼ eer þ beey .
(24)
By using the following boundary condition for the radial displacement uc at the elastic–plastic interface: uc ¼
c ðp p1y Þ 2G o
(25)
in which G ¼ the shear modulus of the rock mass, the solution for Eq. (23) can be obtained as Z cb 1 r b u¼ b r f ðrÞ dr þ uc . (26) r c r In order to evaluate the integral in Eq. (26), expressions for elastic strains are needed. If the function f(r) is expressed in simple function of r, the integration can be carried out analytically. Generally three different definitions for elastic strains can be found: (1) Case 1: With the assumption of the constant elastic strain in the plastic region such as that at the elastic–plastic interface, the elastic strains can be expressed as [1] eer ¼ eey ¼
1 ðp p1y Þ, 2G o
1 ðp p1y Þ. 2G o
(27)
(28)
So, the function f(r) is f ðrÞ ¼
1 ðb 1Þðpo p1y Þ. 2G
(29)
(2) Case 2: By considering the plastic region as the thickwalled cylinder subjected to the inner stress, pipo at r ¼ a, and the outer stress, p1ypo at r ¼ c, the elastic strains can
ARTICLE IN PRESS K.-H. Park, Y.-J. Kim / International Journal of Rock Mechanics & Mining Sciences 43 (2006) 616–622
be obtained as [3] 1 D ð1 2nÞC þ 2 , eer ¼ 2G r 1 D ð1 2nÞC 2 , eey ¼ 2G r
where (30)
(31)
Then the integration in Eq. (26) can be made analytically and the solution for the radial displacement can be simplified as u 1 1 ¼ ½D1 f 1 ðrÞ þ D2 f 2 ðrÞ þ D3 f 3 ðrÞ r 2G rbþ1 þ 2Guc cb D1 f 1 ðcÞ D2 f 2 ðcÞ D3 f 3 ðcÞ,
C¼
and n ¼the Poisson’s ratio. So, the function f(r) is 1 D ð1 2nÞð1 þ bÞC þ ð1 bÞ 2 . f ðrÞ ¼ 2G r
¼ ð1 þ bÞð1 2nÞðAM2C po Þ, DM2C 1 DM2C ¼ fð1 n bnÞ þ ar ðb nb nÞgBM2C . 2
where ðp1y po Þc2 ðpi po Þa2 , c 2 a2 a2 c2 ðpi p1y Þ D¼ c 2 a2
ð32Þ
Z
(33)
1 ½ð1 nÞðsr po Þ nðsy po Þ, 2G
(34)
ðrÞ ¼ f M2C 2
eey ¼
1 ½ð1 nÞðsy po Þ nðsr po Þ. 2G
(35)
f H2B ðrÞ ¼ 3
1 ½ð1 n bnÞsr þ ðb bn nÞsy 2G ð1 2nÞð1 þ bÞpo .
ð36Þ
It should be noted that the functions f(r) in Cases 1 and 2 are simple and the same forms can be used for two different yield criteria. Thus the integration in Eq. (26) and the displacements in the plastic region can be obtained in the same form for different yield criteria. However Case 3 needs stresses in the plastic region for the derivation of displacement. By using Eqs. (8) and (9) for the H–B yield criterion and Eqs. (13) and (14) for the M–C yield criterion, the function f(r) can be expressed in terms of r: for the H–B yield criterion, r i 1 h H2B H2B 2 r f ðrÞ ¼ D1 þ DH2B ‘n ‘n þ D , (37) 2 3 2G a a where ¼ ðb nb nÞAH2B þ ð1 þ bÞð1 2nÞðpi po Þ, DH2B 1 DH2B ¼ ð1 þ bÞð1 2nÞAH2B þ 2ðb nb nÞBH2B , 2 DH2B ¼ ð1 þ bÞð1 2nÞBH2B 3 and for the M–C yield criterion, 1 r ar 1 D1 f ðrÞ ¼ þ D2 , 2G a
(38)
rb dr ¼
rbþ1 , bþ1
Z
r rb ‘n dr a rbþ1 r 1 ‘n , ¼ a bþ1 bþ1
f H2B ðrÞ ¼ 2
eer ¼
So, the function f(r) is
ð39Þ
where f 1 ðrÞ ¼
(3) Case 3: By using the elastic stress–strain relationship with the consideration of initial hydrostatic stress, the elastic strains can be expressed as [6,7]
f ðrÞ ¼
619
Z
rb
r ar 1 a
dr ¼
1 aar 1
rbþar , b þ ar
Z
r rb ‘n2 dr a r rbþ1 2 2 2 r ‘n ‘n þ . ¼ a bþ1 a bþ1 ð b þ 1Þ 2
Table 1 summarizes the constants, D1, D2 and D3, and functions, f1(r), f2(r) and f3(r), for all the cases with the H–B and M–C yield criteria. It is of interest to note that the solutions by Cases 1 and 2 are the same as those of Brown et al. [1] and Sharan [3] respectively, for the H–B yield criterion, while the solutions by Case 3 are the same as those of Reed [6] and Yu [7] for the M–C and H–B yield criteria, respectively. In addition, the solutions presented in this paper have been obtained by using the deformation theory of plasticity [8,9] and equivalent results can be found using the incremental theory of plasticity [10–13] with the correct form of elastic strain increments [12]. 5. Comparison of solutions In order to investigate the effect of different definitions for elastic strains on the displacement, the dimensionless radial displacements in the plastic region are compared by using the typical cases of soft and hard rocks, used by Sharan [3] and Ogawa and Lo [5]. Tables 2 and 3 show the properties of the rocks. It should be noted that Ogawa and Lo [5] determined c and f from m, s and sc to compare the results with equivalent parameter values for two different yield criteria. The dimensionless radial displacement, uE/poa, in the plastic region is plotted in Figs. 3–8 for cases of c ¼ 0 (no plastic volume change) and c ¼ 301 (plastic volume
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Table 1 Summary of the constants and functions for Eq. (39) Case 1
Case 2
Case 3 (HB)
Case 3 (MC)
D1 D2 D3 f1(r)
ðb 1Þðpo p1y Þ 0 0 rbþ1 bþ1
ð1 2nÞð1 þ bÞC ð1 bÞD 0 rbþ1 bþ1
ðb nb nÞA þ ð1 2nÞð1 þ bÞðpi po Þ ð1 2nÞð1 þ bÞA þ 2ðb nb nÞB ð1 2nÞð1 þ bÞB rbþ1 bþ1
ð1 2nÞð1 þ bÞðA po Þ ð1 n bnÞ þ ar ðb bn nÞ B 0 rbþ1 bþ1
f2(r)
0
rb1 b1
rbþ1 r 1 ln a bþ1 bþ1
1 rbþar aar 1 b þ ar
f3(r)
p1y
c/a
Case 3 ðH2BÞ :
r rbþ1 r 2 2 ln ln2 ; Others : 0 ðzeroÞ þ 2 a bþ1 a bþ1 ðb þ 1Þ
MC 2po Y aþ1
HB rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m2 mpo m po sc þ þs 2 4 8 sc
1 p1y A ar 1 B
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8 ffi9 > > 2 > = <A þ A 4B pi p1y > exp > > 2B > > ; :
Yr ar 1
A
B
pi þ
uc
c ðp p1y Þ 2G o
Yr ar 1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mr sc pi þ sr s2c mr sc 4 c ðp p1y Þ 2G o
Table 2 Data set 1 (Sharan [3])
Table 3 Data set 2 (Ogawa and Lo [5]) Hard rock
Soft rock
Radius of opening, a (m) Initial stress, po (MPa) Internal pressure, pi (MPa) Young’s modulus, E (MPa) Poisson’s ratio, n Shear modulus, G (MPa)
4 108 0 40 000 0.2 16 667
5 30 5 5500 0.25 2200
H–B m s mr sr sc (MPa)
7.5 0.1 1.0 0.01 300
1.7 0.0039 1.0 0 30
increase). While the high value chosen for the dilation angle may not be usually encountered in practice, the use of the high value in this study is to illustrate the maximum possible effect of dilatancy on the predicted displacements. In Figs. 3–8, the solid line represents the results for the case of c ¼ 0, while the dotted line for c ¼ 301.
Hard rock
Soft rock
Radius of opening, a (m) Initial stress, po (MPa) Internal pressure, pi (MPa) Young’s modulus, E (MPa) Poisson’s ratio, n Shear modulus, G (MPa)
1 1 0 50 000 0.2 20 833
1 1 0 5000 0.2 2083
H–B m s mr sr sc (MPa)
0.5 0.0001 0.3 0.00001 75
0.2 0.0001 0.05 0.00001 50
M–C c (MPa) f (deg) cr (MPa) fr (deg)
0.173 55 0.061 52
0.276 35 0.055 30
(1) Effect of different definitions for elastic strains with dilation angle. Figs. 3 and 4 show the displacements for data set 1. It can be seen that Case 1 gives the smallest
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621
10.0
4.0
Case 1
Case 1
8.0
Case 2
3.0
Case 2 Case 3
uE/poa
uE/poa
Case 3
2.0
1.0
0.0
6.0 4.0 2.0 0.0
1
1.05
1.1
1.15
1.2
1.25
1.3
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
r/a
r/a
Fig. 3. Dimensionless radial displacement (data set 1, hard rock, H–B).
Fig. 6. Dimensionless radial displacement (data set 2, soft rock, H–B).
2.5
10.0 Case 1
8.0
2.0
Case 2
uE/poa
uE/poa
Case 3
6.0 4.0
1.5 1.0 Case 1 Case 2
0.5
2.0
Case 3
0.0
0.0 1
1.25
1.5
1.75
2
1
1.05
r/a
1.1
1.15
r/a
Fig. 4. Dimensionless radial displacement (data set 1, soft rock, H–B).
Fig. 7. Dimensionless radial displacement (data set 2, hard rock, M–C).
16.0
2.0
Case 1 Case 2
12.0
1.5 uE/poa
uE/poa
Case 3
1.0
8.0
Case 1
0.5
4.0
Case 2 Case 3
0.0
0.0
1
1.02
1.04
1.06 r/a
1.08
1.1
1.12
1
1.2
1.4
1.6
1.8
r/a
Fig. 5. Dimensionless radial displacement (data set 2, hard rock, H–B).
Fig. 8. Dimensionless radial displacement (data set 2, soft rock, M–C).
displacements. While the difference between results from Case 1 and the others is relatively small for the case of c ¼ 0, it increases with increasing dilation angle. So the assumption for elastic strains of Eqs. (27) and (28) leads to an underestimate of the displacement for both the hard and soft rocks. For Cases 2 and 3, the displacements are almost the same when c ¼ 0. The errors due to the approximation involved in Case 2 are relatively smaller than those due to the
approximation involved in Case 1 [4]. But the difference in displacements between Cases 2 and 3 increases with increasing dilation angle, and becomes larger for the soft rock. So the dilation angle seems to have more impact on the displacement for the soft rock rather than for the hard rock. In addition the displacements by Case 3 are smaller than those by Case 2. (2) Effect of the use of equivalent parameter values. Figs. 5 and 6 show the displacements for data set 2 with the H–B
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622
consistent way. The difference in the definitions of elastic strains among the existing solutions is investigated and the results of the dimensionless radial displacement are compared for the hard and soft rocks. The following main conclusions can be drawn:
2.0
uE/poa
1.5
1.0 Case 1
0.5
Case 2 Case 3
0.0
1
1.01
1.02
1.03
1.04
1.05
1.06
r/a
Fig. 9. Dimensionless radial displacement (data set 2, hard rock, perfectly-plastic). 2.0
uE/poa
1.5
1.0 Case 1
0.5
Case 2
(1) The radial displacement in the plastic region can be simplified as Eq. (39) and Table 1 for three different definitions of elastic strains with two different (linear M–C and nonlinear H–B) yield criteria by employing the non-associated flow rule. (2) Case 1 gives the smallest displacements and seems to give an underestimate of the displacements. While Cases 2 and 3 give almost the same results of displacements when c ¼ 0, the difference of the displacements between Cases 2 and 3 increases as the dilation angle increases. (3) In all the cases, the use of equivalent parameter values for two different yield criteria may lead to different displacements. (4) Even in the case of using the elastic-perfectly plastic model with the M–C yield criterion, the displacements obtained from the solutions by Cases 2 and 3 with c ¼ f (associated flow) are slightly different.
Case 3
0.0
1
1.02
1.04
1.06
1.08
1.1
1.12
1.14
1.16
1.18
References
r/a
Fig. 10. Dimensionless radial displacement (data set 2, soft rock, perfectly-plastic).
yield criterion, while Figs. 7 and 8 with the M–C yield criterion. The same tendency as data set 1 can be seen. Comparison of Figs. 5, 6 with 7,8 indicate that the use of equivalent parameter values for two different yield criteria may lead to different displacement, while Ogawa and Lo [5] mentioned the similar results of the displacements for two different yield criteria. (3) Effect of different definitions for elastic strains in the elastic-perfectly plastic model. The effect of different definitions for elastic strains in the elastic-perfectly plastic model is investigated by calculating the dimensionless radial displacement with the peak strength parameter values. Figs. 9 and 10 shows the displacements using the data set 2 with the M–C yield criterion. In these figures, the solid line represents the case of c ¼ 0 (non-associated flow), while the dotted line for c ¼ f (associated flow). The same tendency mentioned the above can be seen. Even in the perfectly plastic model, the displacements by Cases 2 and 3 are still slightly different for c ¼ f. 6. Conclusion The closed-form analytical solutions of the displacement in the plastic region for the elastic–brittle–plastic analysis of a circular opening have been presented in a theoretically
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