APPLIED MATHEMATICS AND MECHANICS (ENGLISH ......The general solution of elastic stress can be obtained using Eq.(4), ⎧ ⎨ ⎩ σr = C1t λ−1 +C 2t −λ−1, σθ = C1λt λ−1

Appl. Math. Mech. -Engl. Ed., 41(12), 1847–1860 (2020)

APPLIED MATHEMATICS AND MECHANICS (ENGLISH EDITION)

https://doi.org/10.1007/s10483-020-2665-9

Displacement of surrounding rock in a deep circular hole consideringdouble moduli and strength-stiffness degradation∗

Zenghui ZHAO1,2, Wei SUN1,2, Shaojie CHEN1,2,†,Yuanhui FENG1,2, Weiming WANG3

1. College of Energy and Mining Engineering, Shandong University of Science

and Technology, Qingdao 266590, Shandong Province, China;

2. State Key Laboratory of Mining Disaster Prevention and Control Co-founded by

Shandong Province and the Ministry of Science and Technology,

Qingdao 266590, Shandong Province, China;

3. College of Civil Engineering and Architecture, Shandong University of Science and

Technology, Qingdao 266590, Shandong Province, China

(Received Mar. 20, 2020 / Revised Jul. 17, 2020)

Abstract The problem of cavity stability widely exists in deep underground engineer-ing and energy exploitation. First, the stress field of the surrounding rock under theuniform stress field is deduced based on a post-peak strength drop model consideringthe rock’s characteristics of constant modulus and double moduli. Then, the orthogonalnon-associative flow rule is used to establish the displacement of the surrounding rockunder constant modulus and double moduli, respectively, considering the stiffness degra-dation and dilatancy effects in the plastic region and assuming that the elastic strainin the plastic region satisfies the elastic constitutive relationship. Finally, the evolutionof the displacement in the surrounding rock is analyzed under the effects of the doublemodulus characteristics, the strength drop, the stiffness degradation, and the dilatancy.The results show that the displacement solutions of the surrounding rock under constantmodulus and double moduli have a unified expression. The coefficients of the expressionare related to the stress field of the original rock, the elastic constant of the surroundingrock, the strength parameters, and the dilatancy angle. The strength drop, the stiffnessdegradation, and the dilatancy effects all have effects on the displacement. The effectscan be characterized by quantitative relationships.

Key words deep rock, double moduli, strength-stiffness degradation, circular hole,displacement solution

∗ Citation: ZHAO, Z. H., SUN, W., CHEN, S. J., FENG, Y. H., and WANG, W. M. Displace-ment of surrounding rock in a deep circular hole considering double moduli and strength-stiffnessdegradation. Applied Mathematics and Mechanics (English Edition), 41(12), 1847–1860 (2020)https://doi.org/10.1007/s10483-020-2665-9

† Corresponding author, E-mail: [email protected] supported by the National Natural Science Foundation of China and Shandong ProvinceJoint Program (No. U1806209), the National Natural Science Foundation of China (Nos. 51774196and 51774194), and Shandong University of Science and Technology (SDUST) Research Fund(No. 2019TDJH101)

c©The Author(s) 2020

1848 Zenghui ZHAO, Wei SUN, Shaojie CHEN, Yuanhui FENG, and Weiming WANG

Chinese Library Classification O344.32010 Mathematics Subject Classification 74L10

1 Introduction

Circular holes are widely involved in deep underground energy exploitation, nuclear wastestorage, and underground space development, such as circular chambers in mines, water con-veyance tunnels, mine borehole pressure relief, oil and gas development, and columnar holes incoalbed methane mining[1–5]. In addition to the original rock stress, this type of holes is alsosubject to different internal pressures. The instability of the chamber is manifested as shrink-age or expansion, which is related to the plastic failure and displacement of the surroundingrock. The instability of the surrounding rock of the hole has brought many challenges to theexcavation of deep underground engineering, such as drill hole inclination, hole collapse in softrock, and jamming of a drilling tool. Therefore, understanding the mechanical response of thesurrounding rock in deep circular holes under different combinations of pressures is of greatsignificance for maintaining the stability of deep engineering.

Researchers have done a lot of work in predicting the stress and displacement fields of circularholes. For example, based on the Hoek-Brown failure criterion, Brown et al.[6] analyzed thetheoretical solutions for the stress and displacement of circular holes in two models, i.e., elastic-plastic and elastic-brittle-plastic. Sharan[7–8] analyzed the displacement of the cavern basedon the Hoek-Brown failure criterion in an elastoplastic model. Jiang and Shen[9] studied theelastic, brittle, and plastic expansion of cylindrical holes based on the Drucker-Prager (D-P) criterion and Mohr-Column (M-C) criterion. Wen et al.[10] used the twin-shear strengththeory to obtain analytical solutions for the stress and displacement of the failure zone, theplastic zone, and the elastic zone of the circular chamber. For deep well seepage problems,some scholars considered the seepage force as radial force and provided analytical solutions forunderwater chambers[11] and porous elastic-brittle-plastic rock[12–13] based on the elastic-brittle-plastic model and M-C criterion. The tunnel boring machine (TBM) method or mine blastingmethod causes initial damage to the surrounding rock of a chamber. Thus, in many reports, theanalytical solutions of the stress and displacement of a circular hole were obtained consideringthe initial damage caused by blasting loads[14–17]. From the existing results, researchers haveestablished analytical solutions for various circular holes based on different working conditionsusing different constitutive models. In these solutions, the displacement in the plastic zone wasdeveloped based on the following three assumptions. First, the elastic strain in the plastic zonewas assumed to be constant and equal to the strain at the elastoplastic interface[18–20]. Second,the plastic region was considered as an elastic thick-walled cylinder to obtain the solution of thestrain[7]. Third, the elastic displacement of the plastic region was assumed to meet the law ofelasticity, and thus the elastic constitutive equation and the stress in the plastic region can beused to obtain the elastic strains[21–22]. Park and Kim[23] compared the calculated displacementin the plastic zone using the above three methods. Based on the comparison results, it wasfound that the first method would underestimate the displacement value. When the dilatancyangle was zero, the results from the second and third methods were similar, but the differencebetween the two methods increased as the dilatancy angle increased.

The relevant conclusions in the above reports are of great significance for understanding theinstability of deep circular holes. However, rock is an inhomogeneous material with internalcracks. Therefore, it has double moduli in tension and compression. After the chamber wasexcavated, the surrounding rock in the plastic zone experiences not only the strength drop butalso the modulus damage, which was rarely studied in the present achievements. Therefore, thispaper focuses on the solution of the displacement field of surrounding rocks under the dual-modulus properties and strength-stiffness degradation, instead of the influence of multi-fieldcoupling or the selection of strength criteria.

Displacement of surrounding rock in a deep circular hole considering double moduli 1849

2 Definition of problem

2.1 Description of problemAs shown in Fig. 1(a), the deep circular chamber with a radius a is subject to the hydrostatic

pressure p0. The inner wall of the chamber is subject to an internal pressure pi. At this time,three stresses of the surrounding rock σr, σθ, and σz are all principal stresses. pi can be regardedas the virtual internal supporting force provided by the tunnel face during the excavation ofthe roadway. As the tunnel face moves away from the section, pi gradually decreases. Whenit decreases to a certain value, the surrounding rock close to the free face of the roadwayenters the plastic zone, and then strength drop and modulus damage occur. At this time, thethree principal stresses are all compressive stresses. In addition, pi can also be regarded as theinternal pressure, the drilling pressure, or the high internal water pressure of the water and oilpipelines. When the force is greater than a certain value, the surrounding rock of the roadwayalso undergoes the plastic deformation, and the strength and modulus of the plastic zone drop.At this time, σr and σz are still compressive stresses, while the tangential stress σθ changesfrom the tensile force to the compressive force from the inner boundary to the outer boundaryof the hole.

Let the compressive stress be positive and the tensile stress be negative. The dimensionlessradial coordinates are defined as t = r/a (t � 1), the boundary of the plastic zone of thesurrounding rock is defined as t = tp, and the position for the tangential stress to change signis t = ts. The deformation of the surrounding rock can be divided into the following zones (seeFigs. 1(b) and 1(c)).

Case A Ω1I zone: 1 � t � tp, σr > 0, σθ > 0, σz > 0, the plastic damage zone (PDZ).Ω2I zone: t > tp, σr > 0, σθ > 0, σz > 0, the elastic compression zone (ECZ).Case B Ω1Π zone: 1 � t � tp, σr > 0, σθ < 0, σz > 0, the PDZ.Ω2Π zone: tp � t � ts, σr > 0, σθ < 0, σz > 0, the elastic tension and compression zone

(ETCZ).Ω3Π zone: t � ts, σr > 0, σθ > 0, σz > 0, the ECZ.According to the theory of different elastic moduli in tension and compression[24], Ω1Π zone

and Ω2Π zone should be analyzed by the dual-modulus theory, and the other zones can beanalyzed by the constant modulus.

Fig. 1 Analytical model for circular hole under deep hydrostatic pressure (color online)

2.2 Elastic constitutive and deterministic equations of rock considering doublemoduli

Assume that the elastic moduli in compression and tension of the surrounding rock are E+

and E−, respectively, and the transverse deformation coefficients are μ+ and μ−, respectively.According to the theory of different elastic moduli in tension and compression[24–25], the plane


strain elasticity constitutive is {εr = b11σr + b12σθ,

εθ = b21σr + b22σθ,(1)

where

b11 =a11a33 − a13a31

a33, b12 =

a12a33 − a13a32a33

,

b21 =a21a33 − a23a31

a33, b22 =

a22a33 − a23a32a33

.

For deep surrounding rocks, σr > 0, and σz > 0. Therefore, a11 = a33 = 1/E+, anda21 = a31 = a13 = a23 = −μ+/E+. a12, a22, and a32 are set based on the compressiveand tensile signs of the tangential stress σθ. The plane differential equation for axisymmetricproblems can be simplified as

dσrdr

+σr − σθ

r= 0. (2)

The geometric equation is

εr =dudr, εθ =

u

r. (3)

Let σr = Φ/r, and σθ = dΦ/dr. The deterministic control equations of stress can beobtained by combining Eqs. (2) and (3),

t2d2Φdt2

+ tdΦdt

− λ2Φ = 0, (4)

where λ =√

b11+b12−b21b22

, which is related to the double moduli of the rock.

The general solution of elastic stress can be obtained using Eq. (4),⎧⎨⎩σr = C1tλ−1 + C2t−λ−1,

σθ = C1λtλ−1 − C2λt−λ−1,(5)

where C1 and C2 can be determined by the stress boundary conditions.2.3 Yield and damage behaviors of surrounding rock

The M-C criterion is used to describe the strength characteristics of the surrounding rocks.Set the initial yield criterion to

σ1 = ησ3 + ξ. (6)

After the strength drop, the subsequent yield criterion is

σ1 = ηrσ3 + ξr, (7)

where

η =1 + sinϕ1 − sinϕ , ξ =

2c cosϕ1 − sinϕ , ηr =

1 + sinϕr1 − sinϕr , ξr =

2cr cosϕr1 − sinϕr ,


in which c and ϕ are the initial cohesion and internal friction angle of the surrounding rock,respectively, and cr and ϕr are the cohesion and internal friction angle of the surrounding rockin the residual stage after the strength drop, respectively.

Assume that the elastic strain is smaller than the plastic strain and that the strain inthe plastic region obeys the non-associated flow rule, as shown in Fig. 2. Then, the followingequation can be satisfied:

εp3p + Θεp1p = 0, (8)

where the superscript p represents the plastic region, and the subscript p represents the plasticstrain. Considering the modulus damage of the surrounding rock in the plastic zone, andassuming that the elastic strain in the plastic zone satisfies the law of elasticity, the elasticstrain in the plastic zone can satisfy the following equation:{

εr = br11σr + br12σθ,

εθ = br21σr + br22σθ,

(9)

where the superscript r indicates the modulus after damage.

Fig. 2 Elastic-brittle-ideal plastic behavior of rocks and non-associated flow laws (color online)

3 Analytical solutions of surrounding rock in circle opening

3.1 Analytical solutions in Case A3.1.1 Stress solutions

In this case, in the general solution of the elastic stress, i.e., Eq. (5), λ = 1. Assume that thestress in the axial direction of the chamber is the intermediate principal stress. Then, σ1 = σθ,and σ3 = σr. Combining Eqs. (2) and (7) and considering the boundary conditions of t = 1 andσr = pi, the stress solutions of the plastic zone can be obtained as follows:⎧⎪⎪⎨

⎪⎪⎩σpr =

(pi +

ξrηr − 1

)tηr−1 − ξr

ηr − 1 ,

σpθ = ηr(pi +

ξrηr − 1

)tηr−1 − ξr

ηr − 1 .(10)

The elastoplastic interface stress satisfies σr+σθ = 2p0. Substituting this equation intoEq. (6), the radial stress at the elastoplastic interface can be obtained as follows:

σtpr =2p0 − ξη + 1

. (11)

Based on σpr (tp) = σtpr , the radius of the plastic zone can be obtained as follows:

tp =( (ηr − 1)(2p0 − ξ) + (η + 1)ξr

(η + 1)((ηr − 1)pi + ξr)) 1

ηr−1. (12)


Using Eq. (5) and the boundary conditions of t = tp, σer(tp) = σtpr and t→ ∞, σer = p0, the

stress solutions of the elastic zone can be obtained as follows:⎧⎪⎪⎨⎪⎪⎩σer =

(1 −

( tpt

)2)p0 +

( tpt

)2σtpr ,

σeθ =(1 +

( tpt

)2)p0 −

( tpt

)2σtpr .

(13)

3.1.2 Displacement solutionsBoth elastic and plastic deformations exist in the plastic zone, and the total strains can be

expressed as follows: {εpr = ε

pre + ε

prp,

εpθ = εpθe + ε

pθp.

(14)

According to Eq. (8), the following equation can be obtained for this condition:

εprp + Θεpθp = 0, (15)

where Θ = (1 + sinψ)/(1 − sinψ), in which ψ is the dilatancy angle of the surrounding rock.If only the displacement caused by excavation is considered, the deformation caused by the

original rock stress before excavation should be removed, and the elastic strains should meetthe following equation: {

εpre = br11(σ

pr − p0) + br12(σpθ − p0),

εpθe = br21(σ

pr − p0) + br22(σpθ − p0).

(16)

For plane axisymmetric problems, the surrounding rock has only the radial displacement u.Combining Eqs. (3), (15), and (16), the differential equations for the displacement of the plasticzone can be obtained as follows[23]:

dupdt

+ Θupt

= af(t), (17)

where f(t) = (br11 + Θbr21)σpr + (br12 + Θbr22)σpθ − (br11 + br12 + Θbr21 + Θbr22)p0.

The displacement at the elastoplastic interface is

utp = atpεθ = atp(b22 − b21)(p0 − σtpr ). (18)Combine Eqs. (17) and (18). Then, the displacement of the plastic zone can be obtained as

follows:

up =a

tΘ

∫ ttp

tΘf(t)dt+ utp( tpt

)Θ. (19)

The displacement of the plastic zone can be obtained by integrating the above equation,

upr

=1

tΘ+1((Ar1B

r1 +A

r2B

r3)(t

Θ+ηr − tΘ+ηrp )− (Ar1Br2 +Ar2Br2 +Ar3B0)(tΘ+1 − tΘ+1p ) + utptΘp ), (20)

where ⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Ar1 = (br11 + Θb

r21), A

r2 = (b

r12 + Θb

r22), A

r3 = b

r11 + b

r12 + Θb

r21 + Θb

r22,

Br1 =(ηr − 1)pi + ξr

(Θ + ηr)(ηr − 1) , Br2 =

ξr(Θ + 1)(ηr − 1) ,

Br3 =ηr((ηr − 1)pi + ξr)(Θ + ηr)(ηr − 1) , B0 =

p0Θ+1

.


Combining Eqs. (1) and (13), the displacement of the elastic zone can be obtained as follows:

ue = a(b22 − b21)(p0 − σtpr )t2pt. (21)

3.2 Analytical solutions in Case B3.2.1 Stress solutions

(i) Stress in Ω1Π zoneSince σr > 0, σθ < 0, σ1 = σr, and σ3 = σθ, combining Eqs. (2) and (6), the stress solutions

of the plastic zone can be obtained,

⎧⎪⎪⎨⎪⎪⎩σpr =

(pi +

ξrηr − 1

)t

1−ηrηr − ξr

ηr − 1 ,

σpθ = −1ηr

(pi +

ξrηr − 1

)t

1−ηrηr − ξr

ηr − 1 .(22)

The stress at the elastoplastic interface is

σtpr =2p0η + ξ

1 + η. (23)

The radius of the plastic zone can be obtained by σpr (tp) = σtpr and Eq. (12),

tp =( (ηr − 1)(2p0η + ξ) + (1 + η)ξr

(1 + η)((ηr − 1)pi+ξr)) ηr

1−ηr. (24)

(ii) Stress in Ω2Π zoneThe boundary conditions of the Ω2Π zone are σr = σ

tpr when t = tp, and σθ = 0 when t = ts.

The stress solutions can be obtained by Eq. (5),⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩σr =

(tλ−1 +

t2λstλ+1

) tλ+1pt2λp + t2λs

σtpr ,

σθ =(tλ−1 − t

2λs

tλ+1


λσtpr .

(25)

(iii) Stress in Ω3Π zoneIn this zone, a11 = a22, and b11 = b22. Thus, λ = 1, and the boundary conditions are σθ = 0

when t = ts, and σr = p0 when t→ ∞.The stress solutions can be obtained by Eq. (5),

⎧⎪⎪⎨⎪⎪⎩σr =

(1 +

t2st2

)p0,

σθ =(1 − t

2s

t2

)p0.

(26)

According to the stress continuity condition (σr)t=s− = (σr)t=s+ , the transcendental equa-tion of s can be obtained,

t2λs −σ

tpr

p0tλ+1p t

λ−1s + t

2λp = 0. (27)


3.2.2 Displacement solutions(i) Displacement in Ω1Π zone In the plastic zone, the following equation is obtained:

εpθp + Θεprp = 0. (28)

The elastic strain still satisfies Eq. (9), and the differential equation for displacement controlis rewritten as

Θdupdt

+upt

= ag(t), (29)

where g(t) = (br21+Θbr11)σ

pr + (b

r22+Θb

r12)σ

pθ − (br21 + br22 + Θbr11 + Θbr12)p0.

The displacement at the elastoplastic interface can still be obtained by Eq. (19),

up =a

Θt1Θ

∫ ttp

g(t)t1Θ dt+ utp

( tpt

) 1Θ. (30)

The displacement of the plastic zone can be obtained by integrating the above equation,

upr

=1

t1Θ+1

((Cr1D

r1 + C

r2D

r2)

(t

1Θ+

1ηr − t

1Θ+

1ηr

p

)− (Cr1Dr3 − Cr2Dr3 + Cr3D0)

(t

Θ+1Θ − t

Θ+1Θ

p

)+ utp(tp)

1Θ

), (31)

where {Cr1 = (b

r21 + Θb

r11), Θ

r2 = (b

r22 + Θb

r12), C

r3 = b

r21 + b

r22 + Θb

r11 + Θb

r12,

Dr1 = Br1ηr, D

r2 = B

r1, D

r3 = B

r2, D0 = B0.

(ii) Displacement in Ω2Π zoneThe displacement in this zone can be obtained by combining Eqs. (1) and (2),

uΩ2Πe = a

(((b21 + λb22)tλ + (b21 − λb22) t

2λs

tλ


σtpr − t(b21 + b22)p0). (32)

(iii) Displacement in Ω3Π zoneThe displacement in this zone can be obtained by combining Eqs. (1) and (26),

uΩ3Πe = a(b21 − b22) t

2s

tp0. (33)

3.3 Unified solution of displacement in two casesFrom the above analysis, the displacement of the plastic zone can be written in a unified

form for both cases,

upr

=1

tα1+1

(K1(tα2 − tα2p ) +K2(tα3 − tα3p ) + utp(tp)α1

). (34)

The expression of each parameter is summarized in Table 1.


4 Analysis and comparison of solution

The tensile and compressive elastic modulus ratio is defined as E = E+/E−, the ratio ofPoisson’s ratios is defined as μ = μ+/μ−, and the residual strength coefficient in the plasticstrength drop zone is defined as λc = cr/c and λϕ = ϕr/ϕ. Considering the modulus damage inthe strength drop zone, the residual modulus coefficient is set to be DE = E+r /E

+ = E−r /E−;

that is, the tensile and compressive moduli are considered to be degraded at the same level, andthe transverse deformation coefficient damage is not considered. The values of the parameterin the two cases in Eq. (34) are shown in Table 2.

Table 1 Values of variables and constants in unified solution of displacement

Variable Case A Case B

K1 Ar1Br1 + A

r2B

r3 C

r1D

r1 + C

r2D

r2

K2 −(Ar1Br2 + Ar2Br2+Ar3B0) −(Cr1Dr3 − Cr2Dr3 + Cr3D0)

utp atp(b22 − b21)(p0 − σtpr ) a““

(b21 + λb22)tλp + (b21 − λb22) t2λstλp

”tλ+1p

t2λp +t2λsσ

tpr − tp(b21 + b22)p0

”

σtpr

2p0−ξη+1

2p0η+ξ1+η

tp“

(ηr−1)(2p0−ξ)+(η+1)ξr(η+1)((ηr−1)pi+ξr)

” 1ηr−1

“(ηr−1)(2p0η+ξ)+(1+η)ξr

(1+η)((ηr−1)pi+ξr)” ηr

1−ηr

α1 Θ1Θ

α2 Θ + ηr1Θ

+ 1ηr

α3 Θ + 1Θ+1Θ

Table 2 Values of each modulus in unified solution of displacement

Modulus Case A Case B

br111

E+r(1 − μ+μ+) 1

E+r(1 − μ+μ+)

br21μ+

E+r(−1 − μ+) μ+

E+r(−1 − μ+)

br12μ+

E+r(−1 − μ+) μ−

E−r(−1 − μ+)

br221

E+r(1 − μ+μ+) 1

E−r(1 − μ+μ−)

The values of the basic calculation parameters in both cases are shown in Table 3.

Table 3 Values of basic calculation parameters

Original Internal Initial ResidualModulus

Modulus Damage DilatancyType stress stress strength strength ratio coefficient angle

p0/MPa pi/MPa c/MPa ϕ/(◦) λc λϕ E−/GPa μ− �E �μ DE ψ

Case A 60 3 30 25 0.3 0.3 30 0.25 0.8 0.6 0.6 5Case B 30 80 30 25 0.3 0.3 30 0.25 0.8 0.6 0.6 5

4.1 Influence of dual-modulus characteristics on displacementFigure 3 shows the influence of compressive and tensile dual-modulus characteristics on

the displacement of the surrounding rock. In Case A, the elastic constants in tension andcompression of the surrounding rock are the same; that is, the tensile and compressive elasticmodulus ratio E = 1.0, and the ratio of Poisson’s ratios μ = 1.0. The displacement of thesurrounding rock is not affected by the difference of modulus. In Case B, the shear stresses,radial stresses, and axial stresses in Ω1Π and Ω

2Π have opposite signs. Therefore, the difference

between compressive and tensile moduli has a great impact on the displacement. Figure 3(a)shows the evolution law of the displacement of the surrounding rock when the tensile andcompressive elastic modulus ratio E is different. At smaller E , the displacement of the


surrounding rock in the plastic zone is larger. As E increases, the displacement of the plasticzone decreases. When E < 1.0, the effect of the difference between tensile and compressiveelastic moduli on the displacement of the plastic zone is more significant than that at E > 1.0.The influence of E on the displacement in the Ω3Π elastic zone exhibits the same trend as thatin the Ω1Π zone. In comparison, the effect of E on the displacement in the Ω

3Π elastic zone is

less significant. In addition, E has no effect on the relative radius tp of the plastic zone, but ithas a greater effect on the relative radius ts of the Ω2Π zone. As E increases, ts also continuesto expand outward. From Fig. 3(b), the influence of μ on the displacement of the surroundingrock is just the opposite to the influence of E . With the increase in μ, the displacements ofthe surrounding rock in Ω1Π and Ω

2Π zones increase continuously and have the same trend. μ

has no effect either on the relative radius of the plastic zone tp. The relative radius of the Ω2Πzone decreases with the increase in μ, but the change is not significant.

µ

µ

µ

µ

µ

µ

’ µ

Fig. 3 Influence of double modulus characteristics on displacement of surrounding rock (Case B)(color online)

4.2 Influence of strength drop on surrounding rocksFigures 4 and 5 show the influence of the strength drop characteristics of the plastic zone

on the displacement of the surrounding rock in two cases. With the increase in the strengthdrop, the roadway displacements are continuously increased. In the case of constant modulus(Case A), the radius of the plastic zone of the surrounding rock and the displacement at theelastoplastic interface both increase with the increase in the strength drop. In comparison, theeffect of cohesive drop is more significant. When λc is dropped to 10% of the original value,the maximum displacement of the surrounding rock is increased by 223%. On the other hand,

ϕ

ϕ

ϕ

ϕ

ϕ

ϕ

ϕ

Fig. 4 Influence of strength drop characteristics on displacement of surrounding rock (Case A) (coloronline)


Fig. 5 Influence of strength drop characteristics on displacement of surrounding rock (Case B) (coloronline)

when λϕ drops to 10% of its original value, the maximum displacement of the surrounding rockis increased by only 32.3%.

In the case of double moduli (Case B), with the drop of cohesion and friction angle, thedisplacement of the surrounding rock increases significantly, especially in the plastic zone. Un-like Case A, the radius of the surrounding plastic zone increases with the drop of the strengthparameter, but the displacement at the elastoplastic interface is not changed. In addition, thefriction angle drop has a more significant effect on the displacement of the surrounding rockthan the cohesive force drop. When λϕ is dropped to 10% of the original value, the maximumdisplacement of the surrounding rock increases by 148%. With the same drop of cohesion, themaximum displacement of the surrounding rock increases by only 63%.4.3 Influence of dilatancy effect

The dilatancy effect has great influence on the displacement of the surrounding rock in theplastic zone as shown in Fig. 6. Under different dilatancy angles, the displacements of the sur-rounding rock in Case A and Case B show different characteristics. In Case A, the displacementof the surrounding rock shrinks inward. As the dilatancy angle increases, the convergent dis-placement of the surrounding rock increases significantly, showing a dilatation effect. In CaseB, the displacement of the surrounding rock expands outward. As the dilatancy angle increases,the displacement of the surrounding rock tends to decrease, showing a compacting effect on theplastic zone due to the large pressure in the roadway. In comparison, the dilatancy effect hasmore significant influence in Case A.

°°°°

°°°°

Fig. 6 Influence of dilatancy angle on displacement of surrounding rock (color online)


4.4 Influence of stiffness degradation on displacementAfter excavation of the chamber, in addition to the strength drop of the surrounding rock

in the plastic zone, the stiffness is also significantly degraded. Figure 7 shows the effect of theresidual modulus coefficient on the displacement of the surrounding rock in both cases. For theconvenience of analysis, it is assumed that the elastic moduli of tension and compression areattenuated by the same degree. Under the conditions of constant modulus and double moduli,the displacement of the surrounding rock increases obviously with the deterioration of stiffness.As the modulus gets more degraded, the displacement of the surrounding rock in the plasticzone is increased faster.

Fig. 7 Influence of modulus damage on displacement of surrounding rock (color online)

The above analysis considers the influence of the tension-compression double-modulus char-acteristics of the surrounding rock, the strength of the plastic zone, the stiffness degradation,and the dilatancy effect on the displacement of the surrounding rock. From the perspectiveof the displacement of the surrounding rock, Park and Kim[23] analyzed in detail the validityof three different calculation methods of elastic strain in the plastic zone and the equivalentreplacement of M-C and Hoek-Brown parameters. If the double-modulus characteristics of therock and the stiffness degradation of the plastic zone are not considered, the solution in thestudy is consistent with the solution proposed by Park and Kim.

5 Conclusions

In this paper, a unified analytical solution for the displacement of the surrounding rockis obtained for the problem of circular holes in deep underground engineering, consideringdifferent characteristics of the tensile and compressive elastic moduli of the surrounding rock,the strength-stiffness degradation, and the dilatancy behavior in the plastic zone. In addition,the effects of the tensile and compressive elastic modulus ratio, the residual strength, the residualstiffness, and the dilatancy angle on the displacement of the surrounding rock are analyzed.The main conclusions are as follows:

(i) According to the sign change of tangential stress, the surrounding rocks under a constantmodulus can be divided into two zones, i.e., the PDZ and the ECZ. On the other hand, thesurrounding rocks under double moduli can be divided into three zones, i.e., the PDZ, theETCZ, and the ECZ.

(ii) Considering the strength drop, the stiffness degradation, and the dilatancy effect, theanalytical solutions of displacement of the surrounding rock in the plastic zone under the con-ditions of constant modulus and dual moduli using non-associated flow laws and the M-C yieldcriterion can adopt a unified expression.


(iii) Under the dual moduli, the tensile and compressive elastic constant ratio has a sig-nificant effect on the displacement of the surrounding rock. With the increase in the tensileand compressive elastic modulus ratio E , the displacement of the plastic zone is decreased.However, with the continuous increase in the ratio of Poisson’s ratio μ, the displacement ofthe surrounding rock appears to continuously increase. In comparison, the impact of E ismore significant. Although E has no effect on the relative radius of the plastic zone, it has agreat effect on the boundary between the two zones of the ETCZ and the ECZ.

(iv) The strength drop, the stiffness degradation, and the dilatancy effect in the plasticzone affect the displacement of the surrounding rock in both cases. As the strength drop of thesurrounding rock is greater and the stiffness degradation is more severe, the displacement inthe plastic zone is greater. However, the dilatancy characteristics have a different effect. Thedilatancy angles have opposite influence on the displacement of the surrounding rock underthe constant modulus and double moduli.

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APPLIED MATHEMATICS AND MECHANICS (ENGLISH ......The general solution of elastic stress can be obtained using Eq.(4), ⎧ ⎨ ⎩ σr = C1t λ−1 +C 2t −λ−1, σθ = C1λt λ−1