Analytical solutions for the stress of a lined non-circular tunnel under full-slip contact conditions

Analytical solutions for the stress of a lined non-circular tunnel under full-slip contact conditions

International Journal of Rock Mechanics & Mining Sciences 79 (2015) 183–192 Contents lists available at ScienceDirect International Journal of Rock ...

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International Journal of Rock Mechanics & Mining Sciences 79 (2015) 183–192

Contents lists available at ScienceDirect

International Journal of Rock Mechanics & Mining Sciences journal homepage: www.elsevier.com/locate/ijrmms

Analytical solutions for the stress of a lined non-circular tunnel under full-slip contact conditions Ai-zhong Lu n, Ning Zhang, Yuan Qin Institute of Hydroelectric and Geotechnical Engineering, North China Electric Power University, Beijing 102206, China

art ic l e i nf o

a b s t r a c t

Article history: Received 24 November 2014 Received in revised form 7 May 2015 Accepted 10 August 2015

Stress fields of a lined non-circular tunnel subjected to in situ stress are derived based on the complex variable method and on the assumption that the interface between the liner and surrounding rock is fullslip. The basic equations for solving the stress solutions are obtained according to the stress boundary condition along the inner boundary of the lining and the stress and normal displacement continuity conditions along the rock-lining interface. In the solving process, the support delay is also considered. The basic equations can be solved by the power series method, and the stresses in the surrounding rock mass and lining can be calculated. The distributions of the tangential stresses (also known as the circumferential stresses) along the excavation boundary and the inner boundary of the lining and the contact stresses along the rock-lining interface are analysed. An example demonstrates that the results are significantly affected by the number of terms in the power series. When the number of terms is greater than 100, the boundary conditions can be well satisfied, and the results of the stresses and displacements are highly accurate. The tangential stress results along the inner boundary of the lining for the full-slip condition are compared with those for the perfect bond condition, and the analysis indicates that the maximum value of the tangential stress for the full-slip condition is smaller than that for the perfect bond condition, which gives that the full-slip condition is superior to the perfect bond condition. Thus, the carrying capacity of the lining can be increased if sliding materials are installed between the lining and the surrounding rock mass. The analytic solutions are verified using computer simulation software. & 2015 Elsevier Ltd. All rights reserved.

Keywords: Lined non-circular tunnel Full-slip contact Conformal transformation Power series method Analytical solutions of stress

1. Introduction Underground tunnels are widely used in hydropower, traffic, mining and military engineering. To ensure the safety of the tunnels, a concrete lining is applied in these tunnels. The complex variable method developed by Muskhelishvili can be used to calculate the stresses and displacements in the lining and in the surrounding rock mass.[1] This technique is a highly accurate analytical method and is particularly suitable for solving underground tunnel problems.[2] Analytical method has been of high interest for determining stress distribution within lining and surrounding rock mass with high level of accuracy. Recent studies have focused on problems with deep unlined tunnels.[3–6] However, the interaction between the support and the rock mass has been rarely considered in most literatures. Problems involving lined tunnels cannot be solved easily using n Correspondence to: No. 2 Beinong Road Huilongguan, Changping District, 102206 Beijing, China. Fax: þ 86 10 61772234. E-mail address: [email protected] (A.-z. Lu).

http://dx.doi.org/10.1016/j.ijrmms.2015.08.008 1365-1609/& 2015 Elsevier Ltd. All rights reserved.

the complex variable method. Prior to 2014, only the plane strain problem associated with a single circular tunnel with a ring lining in an infinite domain had been studied in detail.[7–9] Two different regions must be considered when the lining is included, which increases the complexity of the problem.[10,11] Until 2014, the stress and displacement solutions for a closed lined non-circular tunnel were obtained using the Cauchy integration method.[11,12] In the studies described above, the lining and surrounding rock mass are assumed to have perfect bond contact, i.e., along the rock-lining interface, the stresses and displacements in the normal direction and the shear stresses and displacements in the tangential direction are continuous. However, the contact between the lining and rock cannot be expected to completely satisfy perfect bonding conditions because slippage occurs between the lining and rock. Additionally, the tangential displacements in the lining and rock are also discontinuous because of the slippage. In this paper, we utilise full-slip contact, i.e., along the interface, the normal stresses and normal displacements are continuous, and the shear stresses are continuous and equal to zero. However, the tangential displacements can be discontinuous. The stress vectors

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Fig. 1. (A) Lined non-circular tunnel under an initial stress field. (B) Ring-shaped region in the ζ plane.

and normal displacements along the interface satisfy the continuity conditions, but slippage can occur between the lining and rock. The full-slip contact condition has been analysed in in-plane and anti-plane problems involving a circular hole with two linings composed of different materials.[13] Some deformation occurs in the surrounding rock mass before the lining is applied. We assume that this portion of the displacement is given.[12] The stress–displacement solutions for a non-circular tunnel with closed support under full-slip boundary conditions can be obtained by the power series method. Similar to the studies mentioned above, the following assumptions are made: (1) the surrounding rock mass and lining and the interaction between them behave in a linear elastic manner under the in situ stresses; and (2) the tunnel is deep enough that the case can be simplified as a plane strain problem in an infinite domain (Fig. 1A).

[15,16]

obtain

2G1(uρR1 + iuθR1) =

ζ¯ ω‵(¯ ζ ) ω(ζ ) ¯ φ‵1(ζ ) − ψ1¯(ζ )] [κ1φ1(ζ ) − ρ ω‵(ζ ) ω‵(¯ ζ )

(2)

We apply the displacement components in orthogonal curvilinear coordinates rather than in rectangular coordinates considered in the literature.[11,12] In this way, the curvilinear coordinate line overlaps with the contact boundary and the equations for full-slip case can be found. The conformal mapping function ω(ζ ) and analytic functions φ1(ζ ) and ψ1(ζ ) in Eqs. (1) and (2) are given by the following equations [2,12]: n

z = ω(ζ ) = R(ζ +

∑ ckζ −k)

(3)

k=0 n

φ1(ζ ) =

2.1. Analytical expressions of displacements in the surrounding rock mass and lining Because the non-circular tunnel is supported closely and the surrounding rock mass and lining are perfect bond, the interaction can be treated as a typically elastic contact problem. Assuming that the tunnel is sufficiently deep, the problem of a monolithic lining can be treated as a ring-shaped elastomer that is embedded within an infinite domain. The stresses can be analysed using the conformal transformation method of complex functions. The transformation z = ω(ζ ) is introduced here to transform the complicated support across the section in the z plane (Fig. 1A) into a simpler ring-shaped region in the ζ plane (Fig. 1B).[2,14] For simplicity, we set the inner and outer radii to R0 (to be determined) and 1, respectively. Let γ1 and γ2 denote the inner and outer circles, respectively. If the tunnel is unlined, the maximum displacements in the surrounding rock mass, u1R and v1R , can be written as [2,12]

ω(ζ ) ¯ φ‵1(ζ ) − ψ1¯(ζ ) ω‵(¯ ζ )

(4)

k=1

2. Fundamental theories and equations

2G1(u1R + iv1R ) = κ1φ1(ζ ) −

∑ ak ζ − k

(1)

where κ1 = 3 − 4μ1, G1 = E1/[2(1 + μ1)]; E1 and μ1 are the Young's modulus and Poisson's ratio of the surrounding rock mass, respectively; and u1R and V1R are the displacement components in the x and y directions, respectively. Let uρR1 and uθR1 denote the normal and tangential displacement components, respectively, in orthogonal curvilinear coordinates in the z plane. From Eq. (1), we

ψ1(ζ ) =−

ω¯(1/ζ ) φ‵1(ζ ) ω‵(ζ ) n− 2

+

n

∑ Skζ k + S‵0 − k=1

pR pR (1 + λ )ζ −1 + (1 − λ ) ∑ ckζ −k 2 2 k=1

(5)

where λ is the lateral pressure coefficient and λ ¼Q/P. The coefficients ck , ak , Sk and S′0 in Eqs. (3)–(5) are real numbers if the tunnel is symmetric about the x-axis. If the support is installed when the displacement is η ( 0 ≤ η ≤ 1) times the total displacement uρR1 + iuθR1, the displacement that occurs before the support installation is η(uρR1 + iuθR1). The surrounding rock mass and lining interact with each other after the lining is installed. The interaction will restrict the displacement in the surrounding rock mass. This displacement reduced due to the interaction uρR2 + iuθR2 can be derived using the following equation:

2G1(uρR2 + iuθR2) =

ζ¯ ω‵(¯ ζ ) ω(ζ ) ¯ [κ1φ2(ζ ) − φ‵2 (ζ ) − ψ2¯(ζ )] ρ ω‵(ζ ) ω‵(¯ ζ )

(6)

where φ2(ζ ) and ψ2(ζ ) are the analytic functions in the surrounding rock mass caused by the lining support. The power series of these functions can be expressed as ∞

φ2(ζ ) = b0 +

∑ bkζ −k k=1

(7)

A.-z. Lu et al. / International Journal of Rock Mechanics & Mining Sciences 79 (2015) 183–192 ∞

ψ2(ζ ) = d0 +

condition is

∑ dkζ −k

(8)

k=1

where b0 , d0 , bk and dk are constants to be determined. After the lining is installed, the displacements uρL and uθL at any point on the lining under the action of the surrounding rock mass can be expressed as

2G2(uρL + iuθL ) =

ζ¯ ω‵(¯ ζ ) ω(ζ ) ¯ φ‵3(ζ ) − ψ3¯(ζ )] [κ2φ3(ζ ) − ρ ω‵(ζ ) ω‵(¯ ζ )

(9)

where κ2 = 3 − 4μ2; G2 = E2/[2(1 + μ2)]; E2 and μ2 are the Young's modulus and Poisson's ratio of the lining, respectively; and φ3(ζ ) and ψ3(ζ ) are the analytic functions in the ring-shaped region corresponding to the lining after it is installed, the power series of which can be written as ∞ ∞ φ3(ζ ) = p0 + ∑k = 1 ekζ −k + ∑k = 1 fk ζ k ∞

φ3(ζ ) = p0 +

ψ3(ζ ) = q0 +



∑ ekζ −k + ∑ fk ζ k k=1

k=1





∑ gkζ −k + ∑ hkζ k k=1

k=1

(10)

(11)

where p0 , q0 , ek , fk , gk and hk are constants to be determined. If the lining is symmetric about the x-axis, the coefficients b0 , d0 , p0 , q0 , bk , dk , ek , fk , gk and hk are real constants to be determined. These coefficients can be calculated based on the stress boundary conditions along L1 and the full-slip contact conditions along L2. 2.2. Basic equations for solving φ2(ζ ), ψ2(ζ ), φ3(ζ ) and ψ3(ζ ) The basic equations for solving φ2(ζ ), ψ2(ζ ), φ3(ζ ) and ψ3(ζ ) can be determined according to the stress boundary condition along L1 and the full-slip contact conditions along L2. Because there is no load acting on the inner boundary L1, the stress boundary condition along γ1 in the ζ plane can be expressed as

φ3(σ1) +

ω(σ1) ¯ φ‵3(σ1) + ψ3(¯σ1) = 0 ω‵(¯σ1)

φ3(R 0σ ) +

ω(R 0σ ) ¯ φ‵3(R 0/σ ) + ψ¯3(R 0/σ ) = 0 ω¯ ‵(R 0/σ )

(13)

The analytic functions that correspond to the interaction between the surrounding rock mass and the lining are φ2(ζ ), ψ2(ζ ) and φ3(ζ ), ψ3(ζ ), respectively. Thus, the stress vector continuity condition along the full-slip interface L2, corresponding to γ2 in the ζ plane, can be expressed as

φ2(σ ) +

ω(σ ) ¯ ω(σ ) ¯ φ‵2 (σ ) + ψ2¯(σ ) = φ3(σ ) + φ‵3(σ ) + ψ3¯(σ ) ω‵(¯σ ) ω‵(¯σ )

(14)

where σ = eiθ and θ is the polar angle in the ζ plane (Fig. 1B). Because the stress vector is continuous along L2, Eq. (14) indicates that the normal stress in the surrounding rock mass is equal to that in the lining and the tangential shear stress in the surrounding rock mass is equal to that in the lining. The expressions for the shear stress along L2 can be derived from φ2(σ ) and ψ2(σ ) or from φ3(σ ) and ψ3(σ ). If we let the expressions of the shear stress be equal to zero, two equations can be obtained. However, the two equations are not independent of Eq. (14); therefore, only one equation can be used. Eqs. (7), (8), (10) and (11) indicate that the expressions for φ2(ζ ) and Ψ2(ζ ) are simpler than those for φ3(ζ ) and ψ3(ζ ). Herein, the expression for the shear stress along the full-slip interface is expressed using φ2(ζ ) and ψ2(ζ ). The other expression of the stress boundary

[1,15]

σρ − iτρθ = 2Re[Φ(σ )] −

σ2 [ω(¯σ )Φ‵(σ ) + ψ ‵(σ )] ρ ω‵(¯σ ) 2

(15)

where Φ(σ ) = φ′(σ ) /ω′(σ ) and . We substitute Φ′(σ ) and ρ = 1 into Eq. (15) and then replace φ(σ ) and ψ (σ ) by φ2(σ ) and ψ2(σ ), respectively. The imaginary part of Eq. (15) is zero, which yields

Im{

ω ‵‵(σ ) 1 [σ 2ω(¯σ )φ‵‵2 (σ ) − σ 2ω(¯σ ) φ‵2 (σ ) + σ 2ω‵(σ )ψ ‵2 ω‵(¯σ )ω‵(σ ) ω‵(σ ) (σ )]} = 0

(16)

where ω′(σ )ω′(σ ) is a real number. Thus, Eq. (16) can be rewritten as

Im{σ 2ω(¯σ )φ‵‵2 (σ ) − σ 2ω(¯σ )

ω ‵‵(σ ) φ‵2 (σ ) + σ 2ω‵(σ )ψ ‵2 (σ )} = 0 ω‵(σ )

(17)

Eq. (17) is the expression of the zero-value shear stress along the full-slip interface. After the lining is installed, the normal displacements in the surrounding rock mass and lining are equal to each other along the interface L2. We have

uρR1 − ηuρR1 + uρR2 = uρL

(18)

Substituting the real parts of Eqs. (2), (6) and (9) into Eq. (18) gives

1 − η σω ω(σ ) ¯ ¯ ‵(¯σ ) φ‵1(σ ) − ψ1(¯σ )]}+ [κ1φ1(σ ) − 2G1 ω‵(σ ) ω‵(¯σ ) 1 σω ω(σ ) ¯ ¯ ‵(¯σ ) Re{ φ‵2 (σ ) − ψ2¯(σ )]} [κ1φ2(σ ) − 2G1 ω‵(σ ) ω‵(¯σ ) 1 σω ω(σ ) ¯ ¯ ‵(¯σ ) φ‵3(σ ) − ψ3¯(σ )]} = Re{ [κ2φ3(σ ) − 2G2 ω‵(σ ) ω‵(¯σ )

Re{

(19)

In Eq. (19), Re{} denotes the real part of {}. ω′(σ ) is a real number, thus Eq. (19) can be rewritten as

Re{

1−η [κ1σω′(σ )φ1(σ ) − σω(σ )φ′1(σ ) − σω′(σ )ψ1(σ )]}+ G1

Re{

1 [κ1σω′(σ )φ2(σ ) − σω(σ )φ′2 (σ ) − σω′(σ )ψ2(σ )]} G1

(12)

where σ1 = R0σ . We also have

185

= Re{

1 [κ2σω′(σ )φ3(σ ) G2

− σω(σ )φ′3(σ ) − σω′(σ )ψ3(σ )]}

(20)

Equations (13), (14), (17) and (20) are the basic equations for solving φ2(ζ ), ψ2(ζ ), φ3(ζ ) and ψ3(ζ ).

3. Process of solving for φ2(ζ ), ψ2(ζ ), φ3(ζ ) and ψ3(ζ ) To solve φ2(ζ ), Ψ2(ζ ), φ3(ζ ) and ψ3(ζ ) using the power series method, we take a finite number of terms. The example demonstrates that the result can be extremely accurate if the number of terms is sufficiently large. We take the highest powers of ζ corresponding to bk , dk , ek , fk , gk and hk as Nb , Nd , Ne , Nf , Ng and Nh , respectively. For simplicity of calculation, we take Nb = Ne = Nh and Nd = Nf = Ng = Nb + 2. Let bk = xk , dL = x Nb+ L , ek = x Nb+ Nd+ k , fL = x2Nb+ Nd+ L , gL = x2Nb+ 2Nd+ L , hk = x2Nb+ 3Nd+ k , b0 = x3Nb+ 3Nd+ 1, d0 = x3Nb+ 3Nd+ 2, p0 = x3Nb+ 3Nd+ 3 and d0 = x3Nb+ 3Nd+ 4 , where k = 1, .. . , Nb , L = 1, . . . , Nd . There are a total of 3Nb + 3Nd + 4 undetermined variables. From Eq. (13), we have

ω′(R 0/σ )φ3(R 0σ ) + ω(R 0σ ) φ′3(R 0/σ ) + ω′(R 0/σ ) ψ3(R 0/σ ) = 0

(21)

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The left side of Eq. (21) can be expressed as a power series of σ . The highest negative power term of ω′(R0/σ )φ3(R0σ ),

ω(R0σ ) φ′3(R0/σ ) and ω′(R0/σ ) ψ3(R0/σ ), are σ −Ne , σ −(Nf + n − 1) and σ −Nh ,

yields ω (σ ) − ω ′ (σ ) φ′1(σ ) − ψ1(σ ) n−2

= − ∑k = 1 Skσ −k − S′0 +

pR (1 2

+ λ )σ −

pR (1 2

n

− λ ) ∑k = 1 ckσ k

respectively. The highest positive powers are σ Nf + n + 1, σ Ne+ 2 and σ Ng + n + 1, respectively. For an arbitrary value of θ (σ = eiθ ), Eq. (21) should be satisfied, and thus, the constant and the coefficients of the negative and positive power items are equal to zero. During the derivation, we take the highest negative power of σ −Ne , σ −(Nf + n − 1) and σ −Nh as −Nb and the highest positive power of σ Nf + n + 1, σ Ne+ 2 and σ Ng + n + 1 as Nb + 2, i.e., the higher power terms of σ are ignored. We then have the following linear equations:

= u0 + here uk = − sk , u0 = − s′0 k = 1, …, n − 2, and vk = − PR(1 − λ )ck/2, v1 = PR(1 + λ ) /2 − PR(1 − λ )c1/2, k = 1, … , n. Thus, Eq. (20) can be rewritten as

Ej(Nb + Nd + k)x Nb + Nd + k + Ej(2Nb + Nd + L)x2Nb + Nd + L

Re{

+ Ej(2Nb + 2Nd + L)x2Nb + 2Nd + L + (22)

where j = 1, … , 2Nb + 3, k = 1, … , Nb , L = 1, … , Nd , and M = 1, … , 4 . A total of 2Nb + 3 equations are included in Eq. (22), and their coefficients are provided in Appendix A. From Eq. (14), we have

ω‵(¯σ )φ2(σ ) + ω(σ )φ‵2¯(σ ) + ω‵(¯σ )ψ2¯(σ ) = ω‵(¯σ )φ3(σ ) + ω(σ )φ‵3¯(σ ) + ω‵(¯σ )ψ3¯(σ )

(23)

Similar to the discussion for Eq. (21), the equality of the constant and the coefficients of the negative power terms σ −1, … , σ −Nb and the positive power terms σ 1, … , σ Nb+ 2 of both sides of Eq. (23) yields

Re{−κ1σω′(σ )φ2(σ ) + σω(σ )φ′2 (σ ) + σω′(σ )ψ2(σ )}+ G1 [κ2σω′(σ )φ3(σ ) − σω(σ )φ′3(σ ) − σω′(σ )ψ3(σ )]} G2 (27)

Both sides of Eq. (27) can be expressed as a power series of σ . The highest positive powers of σω′(σ )φ2(σ ), σω(σ )φ′2 (σ ), σω′(σ )ψ2(σ ), σω′(σ )φ3(σ ), σω(σ )φ′3(σ ), σω′(σ )ψ3(σ ), σω′(σ )φ1(σ ) and σω′(σ )ψ4(σ ) areσ n , σ Nb+ 1, σ Nd+ n , σ Nf + n , σ Ne+ 1, σ Ng + n , σ n − 1 and σ 2n , respectively. The highest negative powers are σ −(Nb+ 1) , σ −(n − 1) , σ −1, σ −(Ne+ 1), σ −(Nf + n), σ −(Nh+ 1), σ −(n + 1) and σ −(n − 1) , respectively. The real parts of σ j and σ −j are equal to each other and have a value of cos(jθ ). Moreover, Eq. (27) should be satisfied for an arbitrary value of θ ; hence, we take the highest positive and negative powers of σ to be equal to each other at a value of Nb + 1. The coefficients of cos(jθ ) are equal to each other, which yields the following linear equations:

+ Cj(2Nb + Nd + L)x2Nb + Nd + L +

+ Dj(2Nb + Nd + L)x2Nb + Nd + L +

Cj(2Nb + 2Nd + L)x2Nb + 2Nd + L + Cj(2Nb + 3Nd + k)x2Nb + 3Nd + k

Dj(2Nb + 2Nd + L)x2Nb + 2Nd + L + Dj(2Nb + 3Nd + k)x2Nb + 3Nd + k + Dj(3Nb + 3Nd + M )x3Nb + 3Nd + M = 0



ω(σ )ω″(σ )/ω′(σ ) = R[ ∑ wkσ k+ ∑ w′k σ −k + 1] k=1

+ Cj(3Nb + 3Nd + M)x3Nb + 3Nd + M = Hj

(24)

where j = 1, … , 2Nb + 3, k = 1, … , Nb , L = 1, … , Nd , and M = 1, … , 4 . A total of 2Nb + 3 equations are included in Eq. (24), and their coefficients are provided in Appendix B. We now discuss Eq. (17). For simplicity, we first discuss the term ω(¯σ )ω ‵‵(σ ) /ω‵(σ ), which can be expressed as a power series as

(25)

where wk and w′k can be derived. The parameters wk and wk′ are provided in Appendix C. The left side of Eq. (17) can be expressed as a power series of σ . The highest negative and positive powers of σ 2ω(σ )φ″2 (σ ) are σ −(Nb+ 1) and σ n − 1, respectively, and σ 2ω′(σ )ψ ′2 (σ ) has only negative power terms of σ , with the highest power being σ −(Nd+ n). The ω ″ (σ ) highest positive power of σ 2ω(σ ) ω ′ (σ ) φ′2 (σ ) is σ n − 3, and the highest negative power is infinite. The imaginary parts of σ j and σ −j are Im{σ j} = sin(jθ ) and Im{σ −j} = − sin(jθ ), respectively. Moreover, Eq. (17) should be satisfied for an arbitrary value of θ ; hence, we take the highest positive and negative powers of σ to be equal to each other with a value of Nb + 1. The coefficients of sin(jθ ) are equal to zero, which yields the following linear equations:

Fjkxk + Fj(Nb + L)x Nb + L = 0

n

− λ ) ∑k = 1 ckσ k ,

Cjkxk + Cj(Nb + L)x Nb + L + Cj(Nb + Nd + k)x Nb + Nd + k

Djk xk + Dj(Nb + L)x Nb + L + Dj(Nb + Nd + k)x Nb + Nd + k

k=1

λ )σ −

pR (1 2

= Re{(1 − η)κ1σω′(σ )φ1(σ ) + (1 − η)σω′(σ )ψ4(σ )}

Ej(2Nb + 3Nd + k)x2Nb + 3Nd + k + Ej(3Nb + 3Nd + M )x3Nb + 3Nd + M = 0

n− 3

Let ψ4(σ ) = −

pR − S′0 + 2 (1 + n−2 n ∑k = 1 uk σ −k + ∑k = 1 vkσ k

n−2 ∑k = 1 Skσ −k

(26)

where j = 1, … , Nb + 1, k = 1, … , Nb , and L = 1, … , Nd . A total of Nb + 1 equations are included in Eq. (26). The coefficients of the undetermined variables in Eq. (26) can be obtained based on those of the corresponding items of Im{σ j, σ −j}(j = 1, ... , Nb), whose coefficients are given in Appendix D. Before Eq. (20) is discussed, taking the conjugate of Eq. (5)

(28)

where j = 1, ... , Nb + 2, k = 1, . . . , Nb , L = 1, . . . , Nd , and M = 1, . . . , 4 . A total of Nb + 2 equations are included in Eq. (28). The coefficients of the undetermined variables of Eq. (28) can be solved according to the following process: let I1 = 1 − η , I2 = κ1(1 − η), I3 = G1/G2, and I3 = κ2G1/G2. The coefficients and Hj of Equation Set (27) are provided in Appendix E. A total of 6Nb + 9 equations are included in Eqs. (22), (24), (26) and (28), but 6Nb + 10 variables must be determined. Hence, one additional equation is required. The effect of the support on the surrounding rock mass decreases with increasing ζ ; thus, uρR2 → 0 and uθR2 → 0 as ζ → ∞, which corresponds to infinity on the z plane. Combining this result with Eq. (6) yields

lim {2G1(uρR2 + iuθR2)}

ζ →∞

= lim { ζ →∞

ζ ω′(ζ ) ω(ζ ) φ′2 (ζ ) − ψ2(ζ )]} = 0 [κ1φ2(ζ ) − ρ ω′(ζ ) ω′(ζ )

Because lim { ζρ ζ →∞

ω ′ (ζ ) ω ′ (ζ )

(29)

ω (ζ ) } = 1, lim [φ2(ζ )] = b0 , lim [ ω ′ (ζ ) φ′2 (ζ )] = 0,

ζ→∞

ζ →∞

and lim [ψ2(ζ )] = d0 , combining these expressions with Eq. (29) ζ→∞

yields

κ1b0 − d0 = 0

(30)

Thus, we have 6Nb + 10 equations in Eqs. (22), (24), (26), (28) and (30). 6Nb + 10 variables need to be determined. To solve these using a computer, we rewrite Eqs. (22), (24), (26), (28) and (30) in the following uniform form

Pijxj = Q i, i, j = 1, …, 6Nb + 10

(31)

where Pij and Oi can be listed as follows based on the derivation

A.-z. Lu et al. / International Journal of Rock Mechanics & Mining Sciences 79 (2015) 183–192

process described above: Pij = Eij , i = 1, … , 2Nb + 3; j= 1, … , 6Nb + 10 P(2Nb+ 3 + i)j = Dij , i = 1, … , 2Nb + 3; j= 1, … , 6Nb + 10 P(4Nb+ 6 + i)j = Fij , i = 1, … , Nb + 1; j= 1, … , 2Nb + 2 P(5Nb+ 7 + i)j = Cij , i = 1, … , Nb + 2; j= 1, … , 6Nb + 10 P(6Nb+ 10)(6Nb+ 7) = κ1 P(6Nb+ 10)(6Nb+ 8) = − 1 Q 5Nb+ 7 + i = Hi , i = 1, … , Nb + 2 The remaining zero terms, including Pij = 0 and Q i = 0, are not listed. The basic undetermined variables, b0 , d0 , p0 , q0 , b1, … , bNb ,

d1, … , d Nd , e1, … , eNe , f1 , … , fN , g1, … , g N , and h1, … , h Nh , can be f

g

solved easily using a computer. Thus, the analytic functions φ2(ζ ), ψ2(ζ ), φ3(ζ ) and ψ3(ζ ) can be determined.

2G1(uρR + iuθ R ) =

187

ζ ω′(ζ ) ω(ζ ) φ′1(ζ ) − ψ1(ζ )] (1 − η)[κ1φ1(ζ ) − ρ ω′(ζ ) ω′(ζ ) +

ζ ω′(ζ ) ω(ζ ) φ′2 (ζ ) − ψ2(ζ )} {κ1φ2(ζ ) − ρ ω′(ζ ) ω′(ζ )

(38)

5. Example and discussion We consider an inverted U-shaped tunnel with 1-metre-thick lining supported (Fig. 2).[12] Take n ¼8, the transformation can be given as

z = ω(ζ ) = 7.54561(ζ − 0.07499 + 0.12076ζ −1 + 0.04536ζ −2 − 0.06513ζ −3+ 0.02083ζ −4 + 0.00118ζ −5 − 0.00285ζ −6

4. Stress–displacement solutions in the lining and surrounding rock mass Because the analytic functions φ2(ζ ), ψ2(ζ ), φ3(ζ ) and ψ3(ζ ) have been derived, the stress–displacement solutions in the lining and surrounding rock mass can be obtained by substituting the analytic functions into the following expressions. The stress components at any point in the lining can be solved using the following two equations [12]:

σρ + σθ = 4Re[φ‵3(ζ )/ω‵(ζ )]

− 0.00096ζ −7 − 0.00221ζ −8)

(39)

From Eq. (39), we note that R = 7.54561, c0 = − 0.07499 , c1,…, c8 are the coefficients of the negative powers of ζ , and the inner radius R0 = 0.88569. We use the same parameters as in reference [12]: E1 = 20, 000 MPa , E2 = 30, 000 MPa , μ1 = μ2 = 0.2, P = 10 MPa , and Q = 5 MPa . The in situ stresses P and Q are compressive; if the values of the in situ stresses are positive, positive values of σρ and σθ denote compressive stresses, and negative values denote tensile stresses.

(32) 5.1. Discussion of the accuracy

σθ − σρ + 2iτρθ 2

=

φ‵‵ (ζ )ω‵(ζ ) − φ‵3(ζ )ω ‵‵(ζ ) 2ζ 1 {ω(¯ζ ) 3 + ψ ‵3(ζ )} ρ2 ω‵(¯ ζ ) [ω‵(ζ )]2

(33)

where σρ , σθ and τρθ are the stress components in orthogonal curvilinear coordinates in the z plane. The line ρ = 1 denotes the interface between the lining and surrounding rock mass L2, and σρ and σθ are the normal and tangential stresses along the interface, respectively. The stress components at any point in the surrounding rock mass can be solved using the following two equations:

σρ + σθ = 4Re[φ′(ζ )/ω′(ζ )]

σθ − σρ + 2iτρθ =

2ζ 2 1 φ″(ζ )ω′(ζ ) − φ′(ζ )ω″(ζ ) {ω(ζ ) + ψ ′(ζ )} ρ2 ω′(ζ ) [ω′(ζ )]2

The results are significantly affected by the number of terms in the power series expressions of φ2(ζ ), ψ2(ζ ), φ3(ζ ) and ψ3(ζ ). Theoretically, the stress continuity condition Eq. (14), the normal displacement continuity condition Eq. (20), the zero-value shear stress along L2 Eq. (17), and the stress boundary condition along L1 Eq. (13) will be properly satisfied if the number of terms is infinite; the results are exact solutions. If the number of terms is finite, some of the equations are ignored because the linear equations are found; thus, Eqs. (13), (14), (17) and (20) cannot be solved

(34)

(35)

where

φ(ζ ) = Γω(ζ ) + φ1(ζ ) + φ2(ζ )

(36)

ψ (ζ ) = Γ ′ω(ζ ) + ψ1(ζ ) + ψ2(ζ )

(37)

where Γ = P (1 + λ ) /4 and Γ ′ = P (λ − 1) /2. The first term of Eqs. (36) and (37) represent the corresponding complex potential function before excavation, and the remaining terms reflect the effects of the excavation and the lining on the surrounding rock mass, respectively. The equations for the axial stresses σz , for both lining and surrounding rock mass, are the same, i.e., σz = μ(σρ + σθ ). Once σρ and σθ are calculated, σz can be obtained. After the lining is installed, the displacement components uρL and uθL in the lining can be derived from Eq. (9). The displacement components uρR and uθR in the surrounding rock mass can be derived from the following equation:

Fig. 2. An inverted U-shaped tunnel.

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accurately, and the results are approximate solutions. The accuracy of the boundary conditions along L1 and L2 for different values of Nb if the values of the given parameters remain unchanged and if η = 0 is discussed below. We take η = 0 because for the case of η = 1, no action occurs along the interface between the lining and the surrounding rock mass; all of the boundary conditions can be satisfied, and thus, we do not need to discuss this case. A smaller value of η leads to a larger interaction and thus the results should be more representative. Therefore, we take η = 0. Along the inner boundary of the lining L1, ρ = R0 ; thus, if the results based on Eqs. (32) and (33) can satisfy σρ = 0 and τρθ = 0, the stress boundary conditions along L1 can be satisfied. Along the L denote the values of σρ and τρθ interface L2, ρ = 1. Let σρL and τρθ R based on Eqs. (32) and (33), respectively, and let σρR and τρθ denote

the values based on Eqs. (34) and (35), respectively. If σρL = σρR and L R , the stress continuity conditions along L2 can be satisfied. τρθ = τρθ Along the interface L2, ρ = 1; thus, the normal displacement continuity conditions can be satisfied if the displacement based on Eq. (9) (i.e., uρL ) is equal to the displacement based on Eq. (38) (i.e., uρR ). Because both the in situ stresses and the tunnel are symmetric about the x-axis, the left half of Fig. 2, which corresponds to the range θ ∈ [0∘ , 180∘] in the ζ plane, will be discussed here. The value θ = 0∘ corresponds to the x-axis above the midpoint of the vault, and θ = 180∘ corresponds to the x-axis below the midpoint of the floor. For Nb = 30, the values for σρ and τρθ along L1, σρL and σρR along L2,

L R and τρθ along L2, and uρL and uρR along L2 are illustrated in Figs. 3– τρθ 6, respectively. As Nb = 30, the stress and normal displacement continuity conditions along L2 can be satisfied well. In particular, the normal displacement continuity conditions can be satisfied, whereas the results along L1 are poor, especially at θ ¼138°, the corner of the tunnel. The stress and normal displacement continuity conditions along L2 can be better satisfied by increasing Nb . The example verifies this conclusion as well. We now discuss the problem of how large Nb should be as the the stress boundary conditions along L1 are satisfied basically. The distributions of σρ and τρθ along L1 for Nb = 100 are shown in Fig. 7. The maximum absolute value of

Fig. 4. Values of σρL and σρR along L2 ( Nb = 30 ).

L R Fig. 5. Values of τρθ and τρθ along L2 ( Nb = 30 ).

σρ and τρθ , max{[ σρ , τρθ ]}, is less than 0.025. Thus, for Nb = 100, the values of σρ and τρθ along L1 are small, which means that the accuracy is high.

Fig. 6. Values of uρL and uRρ along L2 ( Nb = 30 ).

5.2. Discussion of the stresses in the lining and surrounding rock mass

Fig. 3. Values of σρ and τρθ along L1( Nb = 30 ).

The results demonstrate that the tangential stress concentration occurs along the excavation boundary and that the stress concentrations decrease to the level of the in situ stress away from

A.-z. Lu et al. / International Journal of Rock Mechanics & Mining Sciences 79 (2015) 183–192

Fig. 7. Values of σρ and τρθ along L1 ( Nb = 100 ).

the boundary. Thus, for the stresses in the surrounding rock mass, only the tangential stress along the excavation boundary will be discussed. From the results, it also follows that, the greatest stress concentration in the lining occurs along its inner boundary. Thus the tangential stress along the inner boundaries of the lining will be discussed. The normal stresses along the interface between the lining and surrounding rock will be discussed as well. In this section, we take Nb = 100 and consider four values of η : 0.2, 0.4, 0.6 and 0.8. The values of the other parameters remain unchanged. As η takes different values, the tangential stresses along the excavation boundary and the lining’s inner boundary are illustrated in Figs. 8 and 9, respectively.The following conclusions can be drawn from Figs. 8 and 9. The tangential stresses along the excavation boundary are compressive and increase with increasing η . The greater tangential stresses concentration occur as θ ¼60°, the foot of the arch or the top of the wall. The maximum values occur at θ ¼136°, the corner of the wall. The tangential stresses along the lining's inner boundary are also compressive but decrease with increasing η . The greatest tangential stress concentration occurs as θ ¼137°, the corner of the wall. From the calculation results, it also follows that the normal stresses along the rock-lining interface are generally compressive and decrease with increasing η . The maximum values occur at θ ¼141° smaller than the tangential stresses along the excavation boundary and the lining’s inner boundary.

Fig. 8. Values of σθ along the excavation boundary for different values of η .

189

Fig. 9. Values of σθ along the inner boundary of the lining for different values of η .

5.3. Comparative analysis of the tangential stresses along the inner boundary of the lining for two contact conditions From the results it follows that, the distribution of the tangential stress along the excavation boundary and the normal stress along the interface for the full-slip condition are similar to that for the perfect bond condition, respectively. While the tangential stress distributions along the lining's inner boundary are different for the two conditions. Thus, we only discuss the distributions of the tangential stresses along the lining's inner boundary. We take η ¼0.2 as an example. Figs. 10 and 11 illustrate the distributions for two contact conditions as the lateral pressure coefficient λ takes 0.5 and 1, respectively. Figs. 10 and 11 demonstrate that, the maximum value of the tangential stress for the full-slip condition is smaller than that for the perfect bond condition. Tensile stresses are more likely to occur under the perfect bond condition than under the full-slip condition. When λ ¼0.5, tensile stress occur at the centre of the floor for the perfect bond condition, with a maximum value of 0.17P, but it does not occur for the full-slip condition. When λ ¼1, no tensile stresses exist for both conditions. 5.4. Comparison with the numerical results using ANSYS software To simulate an infinite domain problem, the ANSYS model has

Fig. 10. Values of σθ along the inner boundary of the lining ( λ = 0.5).

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Acknowledgement The study was supported by the National Natural Science Foundation of China (Grant no. 11172101).

Appendix A The coefficients of Equation Set (21) are as follows: E1(Nb+ Nd+ 1) = − c2R0−4

E1(Nb+ Nd+ k ) = − (k − 1)ck − 1R0−2k − kck + 1R0−2k − 2, k = 2, … , Ne E1(2Nb+ Nd+ 1) = c0 E1(2Nb+ 2Nd+ 1) = E1(2Nb+ 3Nd+ k ) = E1(3Nb+ 3Nd+ 3) = E1(3Nb+ 3Nd+ 4) = Fig. 11. Values of σθ along the inner boundary of the lining ( λ = 1).

in-plane dimensions of 120 m  120 m. The four-node quadrilateral element PLANE42 is adopted. The lining includes 2691 nodes and 2332 elements, and the surrounding rock region includes 34,538 nodes and 34,168 elements in total. To reduce the calculation error, the grid cells closer to the hole boundary are smaller than those farther from the boundary. In order to simulate the full-slip interface, the contact elements TARGE169 and CONTA171 are set along the outer lining boundary and the surrounding rock excavation boundary, respectively. The contact mode is set as “No separation (always)”, and the contact friction coefficient is set as zero. We compared the analytical results of all cases with the numerical results. They are in good agreement. Fig. 12 illustrates the stress distributions as η ¼0.2. It should be mentioned that the tunnel cross-section after mapping was used in the analytical and numerical analyses.

E(j + 1)(Nb+ Nd+ k ) = − kck + j + 1R0−2k − j − 2, k = 1, … , j − 1 E(j + 1)(Nb+ Nd+ j ) = R0−j − jc2j + 1R0−3j − 2 E(j + 1)(Nb+ Nd+ j + 1) = − (j + 1)c2j + 2R 0−3j − 4(j ≤ Nb − 1) E(j + 1)(N + N + k ) = − (k − j − 1)ck − j − 1R0−2k + j − kck + j + 1R0−2k − j − 2 , b d

E(j + 1)(2Nb+ Nd+ j + 1) = (j + 1)c0R0j E(j + 1)(2Nb+ Nd+ j + 2) = (j + 2)R0j + 2

E(j + 1)(2Nb+ 3Nd+ j ) = R0j E(j + 1)(2Nb+ 3Nd+ k ) = − (k − j − 1)ck − j − 1R0j , k = j + 2, … , Nh

E(Nb+ 2)(Nb+ Nd+ k ) = − 2kckR0−2k − 1, k = 1… , Nb E(Nb+ 2)(2Nb+ Nd+ 1) = 2R0

E(Nb+ 2)(2Nb+ 2Nd+ 1) = R0−1 E(Nb+ 2)(2Nb+ 3Nd+ k ) = − kckR0−1, k = 1… , Nh E(Nb+ j + 1)(Nb+ Nd+ k ) = − (k + j − 1)ck + j − 1R0−2k − j , k = 1, … , j − 3

E(Nb+ j + 1)(Nb+ Nd+ j − 2) = − (2j − 3)c2j − 3R0−3j + 4 − (j − 2)R0−j + 2 E(Nb+ j + 1)(Nb+ Nd+ j − 1) = − (2j − 2)c2j − 2R0−3j + 2 − (j − 1)c0R0−j(j ≤ Nb + 1) b+ j + 1)(Nb+ Nd + k )

The stress and displacement fields for a lined non-circular tunnel excavated in great depth are derived. The effect of an already occurred displacement before support installation is considered. The solution is derived based on the conformal transformation method of the complex variable method. We use the power series method and provide the stress and displacement distributions for an inverted U-shaped tunnel. The accuracy of the solutions depends on the number of terms in the power series and on the complexity of the geometry of the tunnel. The more terms the series has, the more accurate the solutions are, especially for sections with sharp angles such as rectangular openings. According to the computer simulation software ANSYS, for an inverted U-shaped tunnel, a power series with 100 terms can provide good agreement between the analytical and numerical results. Even with a larger number of terms, one can easily implement the governing equations for the stresses and displacements into codes in FORTRAN, Matlab, Maple, etc., and the results can be obtained in a quick and accurate manner. In addition, for most support materials with low-tensile strengths, such as concrete, the maximum value of the tangential stress for the full-slip condition is smaller than that for the perfect bond condition which indicates that the carrying capacity of the lining can be increased if sliding materials are installed between the lining and the surrounding rock mass.

k = j + 2, … , Ne

E(j + 1)(2Nb+ Nd+ k ) = kcj − k + 1R02k − j − 2, k = 1, … , j

E(N

6. Conclusions

2R02 − (k − 1)ck − 1, k = 2, … , Nh 1 1

= − (k + j − 1)ck + j − 1R0−2k − j − kck − j + 1R0−2k − 2 + j ,

k = j, … , Ne

E(Nb+ j + 1)(2Nb+ Nd+ k ) = − (j − k − 1)cj − k − 1R02k − j , k = 1, … , j − 2

E(Nb+ j + 1)(2Nb+ Nd+ j ) = R0j E(Nb+ j + 1)(2Nb+ 2Nd+ k ) = − (j − k − 1)cj − k − 1R0−j , k = 1, … , j − 2

E(Nb+ j + 1)(2Nb+ 2Nd+ j ) = R0−j E(Nb+ j + 1)(2Nb+ 3Nd+ k ) = − (k + j − 1)ck + j − 1R0−j , k = j, … , Nh E(Nb+ j + 1)(3Nb+ 3Nd+ 3) = − (j − 1)cj − 1R0−j E(Nb+ j + 1)(3Nb+ 3Nd+ 4) = − (j − 1)cj − 1R0−j

Appendix B The coefficients of Equation Set (22) are as follows: D11 = − c2 D1k = − (k − 1)ck − 1 − kck + 1, k = 2, … , Nb D1(Nb+ Nd+ 1) = c2 D1(Nb+ Nd+ k ) = (k − 1)ck − 1 + kck + 1, k = 2, … , Ne D1(2Nb+ Nd+ 1) = − c0 D1(2Nb+ Nd+ 2) = − 2 D1(2Nb+ 3Nd+ k ) = (k − 1)ck − 1, k = 2, … , Nh D1(3Nb+ 3Nd+ 1) = 1 D1(3Nb+ 3Nd+ 2) = 1 D1(3Nb+ 3Nd+ 3) = − 1 D1(3Nb+ 3Nd+ 4) = − 1 D(j + 1)k = − kck + j + 1, k = 1, … , j − 1 D(j + 1)j = 1 − jc2j + 1

A.-z. Lu et al. / International Journal of Rock Mechanics & Mining Sciences 79 (2015) 183–192

A

B

C

D

191

Fig. 12. Comparison between the numerical and analytical results (η ¼ 0.2). (A) Tangential stress along the excavation boundary. (B) Tangential stress along the outer boundary of the lining. (C) Normal stress along the outer boundary of the lining. (D) Tangential stress along the inner boundary of the lining.

D(j + 1)(j + 1) = − (j + 1)c2j + 2(j ≤ Nb − 1) D(j + 1)k = − (k − j − 1)ck − j − 1 − kck + j + 1, k = j + 2, … , Nb D(j + 1)(Nb+ Nd+ k ) = kck + j + 1, k = 1, … , j − 1 D(j + 1)(Nb+ Nd+ j ) = − 1 + jc2j + 1 D(j + 1)(Nb+ Nd+ j + 1) = (j + 1)c2j + 2(j ≤ Nb − 1) D(j + 1)(Nb+ Nd+ k ) = (k − j − 1)ck − j − 1 + kck + j + 1, k = j + 2, … , Ne D(j + 1)(2Nb+ Nd+ k ) = − kcj − k + 1, k = 1, … , j D(j + 1)(2Nb+ Nd+ j + 1) = − (j + 1)c0 D(j + 1)(2Nb+ Nd+ j + 2) = − (j + 2) D(j + 1)(2Nb+ 3Nd+ j ) = − 1 D(j + 1)(2Nb+ 3Nd+ k ) = (k − j − 1)ck − j − 1, k = j + 2, … , Nh D(Nb+ 2)k = − 2kck , k = 1… , Nb D(Nb+ 2)(Nb+ 1) = 1 D(Nb+ 2)(Nb+ Nd+ k ) = 2kck , k = 1… , Ne D(Nb+ 2)(2Nb+ Nd+ 1) = − 2 D(Nb+ 2)(2Nb+ 2Nd+ 1) = − 1 D(Nb+ 2)(2Nb+ 3Nd+ k ) = kck , k = 1… , Nh D(Nb+ j + 1)k = − (k + j − 1)ck + j − 1, k = 1, … , j − 3 D(Nb+ j + 1)(j − 2) = − (2j − 3)c2j − 3 − (j − 2) D(Nb+ j + 1)(j − 1) = − (2j − 2)c2j − 2 − (j − 1)c0(j ≤ Nb + 1) D(Nb+ j + 1)k = − (k + j − 1)ck + j − 1 − kck − j + 1, k = j, … , Nb D(Nb+ j + 1)(Nb+ k ) = − (j − k − 1)cj − k − 1, k = 1, … , j − 2 D(Nb+ j + 1)(Nb+ j ) = 1 D(Nb+ j + 1)(Nb+ Nd+ k ) = (k + j − 1)ck + j − 1, k = 1, … , j − 3 D(Nb+ j + 1)(Nb+ Nd+ j − 2) = (2j − 3)c2j − 3 + (j − 2) D(Nb+ j + 1)(Nb+ Nd+ j − 1) = (2j − 2)c2j − 2 + (j − 1)c0(j ≤ Nb + 1) D(Nb+ j + 1)(Nb+ Nd+ k ) = (k + j − 1)ck + j − 1 + kck − j + 1, k = j, … , Ne D(Nb+ j + 1)(2Nb+ Nd+ k ) = (j − k − 1)cj − k − 1, k = 1, … , j − 2

D(Nb+ j + 1)(2Nb+ Nd+ j ) = − 1 D(Nb+ j + 1)(2Nb+ 2Nd+ k ) = (j − k − 1)cj − k − 1, k = 1, … , j − 2 D(Nb+ j + 1)(2Nb+ 2Nd+ j ) = − 1 D(Nb+ j + 1)(2Nb+ 3Nd+ k ) = (k + j − 1)ck + j − 1, k = j, … , Nh D(Nb+ j + 1)(3Nb+ 3Nd+ 1) = − (j − 1)cj − 1 D(Nb+ j + 1)(3Nb+ 3Nd+ 2) = − (j − 1)cj − 1 D(Nb+ j + 1)(3Nb+ 3Nd+ 3) = (j − 1)cj − 1 D(Nb+ j + 1)(3Nb+ 3Nd+ 4) = (j − 1)cj − 1

Appendix C The values of wk and w′k in Eq. (24) are as follows: wn − 3 = 2c1cn wn − 4 = 2c1cn − 1 + 6c2cn j+1 j−1 wn − j − 3 = ∑k = 1 k(k + 1)ckcn − j + k − 1 + ∑k = 1 kckwn − j + k − 2 , j = 2, … , n − 4 n−2

n−3

w′1 = ∑k = 1 k(k + 1)ckck + 2 + ∑k = 2 (k − 1)ck − 1wk n−1

n−3

w′2 = ∑k = 1 k(k + 1)ckck + 1 + ∑k = 2 kckwk w′3 = c1w′1 +

n ∑k = 1 k(k

n−3

+ 1)ckck + ∑k = 2 (k + 1)ck + 1wk

w′4 = c1w′2 + 2c2w′1 + 2c1c0 +

n ∑k = 2 k(k

+ 1)ckck − 1

n−3 ∑k = 1 (k

+ + 2)ck + 2wk w′n + 4 − j = (n − j )(n − j + 1)cn − j + (n − j + 1)(n − j + 2)cn − j + 1c0+ n+2−j

∑k = 1

n

kckw′n − k + 3 − j + ∑k = n − j + 2 k(k + 1)ckcj − n + k − 1

n−3 ∑k = 1 (k

+ + n + 2 − j )ck + n + 2 − jwk j = n − 1, … , 2 n w′n + 3 = (n − 1)ncn − 1 + n(n + 1)cnc0 + ∑k = 1 kckw′n − k + 2 n w′n + 4 = n(n + 1)cn + ∑k = 1 kckw′n − k + 3 n

w′n + 4 + j = ∑k = 1 kckw′n − k + 3 − j , j = 1, … , ∞

,

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Appendix D The coefficients of Equation Set (25) are as follows: Fjk = k(k + 1)cj + k + kwk + j − 1 − kw′ j + 2 − k , k = 1, … , j − 2 Fj(j − 1) = j(j − 1)c2j − 1 − j(j − 1) + (j − 1)w2j − 2 − (j − 1)w′3 (j ≥ 2) Fjj = j(j + 1)c2j − j(j + 1)c0 + jw2j − 1 − jw′2 (j ≤ Nb )

C(j + 1)(2Nb+ 3Nd+ k ) = C(j + 1)(3Nb+ 3Nd+ 1) = C(j + 1)(3Nb+ 3Nd+ 2) = C(j + 1)(3Nb+ 3Nd+ 3) = C(j + 1)(3Nb+ 3Nd+ 4) = n

Fj(j + 1) = (j + 1)(j + 2)c2j + 1 − (j + 1)(j + 2)c1 + (j + 1)w2j − (j + 1)w′1 (j ≤ Nb − 1) Fjk = k(k + 1)cj + k − k(k + 1)ck − j + kwk + j − 1 − kwk − j − 1, k = j + 2, … , Nb

Fj(Nb+ k ) = − k(j − k )cj − k , k = 1, … , j − 1 Fj(Nb+ j + 1) = j + 1

I3(k − j )ck − j + I3(k + j )ck + j , k = j + 1, … , Nh κ1jcj − jcj − I4jcj I3jcj n−2

H1 = − I2 ∑k = 1 kckak − I1 ∑k = 1 kckuk + I1v1 H2 =

n − I2 ∑k = 2 (k − 1)ck − 1ak n−2 ∑k = 2 (k − 1)ck − 1uk

n

− I2 ∑k = 1 (k + 1)ck + 1ak−I1

n−2

−I1 ∑k = 1 (k + 1)ck + 1uk +I1u0 − I1c1u0 + I1v2 n

n

Hj + 1 = − I2 ∑k = j + 1 (k − j )ck − jak − I2 ∑k = 1 (k + j )ck + jak n−2

− I1 ∑k = j + 1 (k − j )ck − juk

Appendix E

n−2

The coefficients and values of Hj in Equation Set (27) are as follows: C1k = kck(κ1 − 1), k = 1, … , Nb C1(Nb+ 1) = 1 C1(Nb+ Nd+ k ) = I3kck(1 − κ2), k = 1, … , Ne C1(2Nb+ Nd+ 1) = I3(κ2 − 1) C1(2Nb+ 2Nd+ 1) = − I3 C1(2Nb+ 3Nd+ k ) = I3kck , k = 1, … , Nh C21 = 2κ1c2 − c2 − c0 C2k = κ1(k − 1)ck − 1 + κ1(k + 1)ck + 1 − kck + 1 − kck − 1, k = 2, … , Nb C2(Nb+ 2) = 1 C2(Nb+ Nd+ 1) = I3c2(1 − 2κ2) + I3c0 C2(N

b+ Nd

+ k ) = − I4(k − 1)ck − 1 − I4(k + 1)ck + 1 + I3kck + 1 + I3kck − 1,

,

j−1

−I1 ∑k = 1 (k + j )ck + 1uk − I1 ∑k = j + 1 (j − k )cj − kvk + I2aj − 1 − I1jcju0 + I1uj − 1 + I1vj + 1

j = 2, …, n − 3 Hj + 1 = − I2 ∑kn= j + 1 (k − j )ck − jak − I2 ∑kn= 1 (k + j )ck + jak− −2 j−1 I1 ∑kn= 1 (k + j )ck + juk − I1 ∑k = 1 (j − k )cj − kvk + I2aj − 1 − I1jcju0

, j = n − 2, … , n − 1

n−1

Hn + 1 = − I1 ∑k = 1 (n − k )cn − kvk + I2an − 1 − I1ncnu0 n−1

Hn + 2 = − I1 ∑k = 1 (n + 1 − k )cn + 1 − kvk + I2an Hj + 1 = −

n I1 ∑k = 1 (j

− k )cj − kvk , j = n + 2, …, 2n

References

k = 2, … , Ne

C2(2Nb+ Nd+ 1) = − I3c0 C2(2Nb+ Nd+ 2) = I3(κ2 − 2) C2(2Nb+ 2Nd+ 2) = − I3 C2(2Nb+ 3Nd+ 1) = 2c2I3 C2(2Nb+ 3Nd+ k ) = I3(k − 1)ck − 1 + I3(k + 1)ck + 1, k = 2, … , Nb C2(3Nb+ 3Nd+ 1) = − κ1(1 − c1) C2(3Nb+ 3Nd+ 2) = 1 − c1 C2(3Nb+ 3Nd+ 3) = I4(1 − c1) C2(3Nb+ 3Nd+ 4) = − I3(1 − c1) C(j + 1)k = κ1(k + j )ck + j − kck + j , k = 1, … , j − 2 C(j + 1)(j − 1) = − κ1 + κ1(2j − 1)c2j − 1 − (j − 1)c2j − 1 − (j − 1) C(j + 1)j = κ1(2j )c2j − jc2j − c0j(j ≤ Nb) C(j + 1)k = κ1(k − j )ck − j + κ1(k + j )ck + j − kck + j − kck − j , k = j + 1, … , Nb C(j + 1)(Nb+ k ) = − (j − k )cj − k , k = 1, … , j − 1 C(j + 1)(Nb+ j + 1) = 1 C(j + 1)(Nb+ Nd+ k ) = − I4(k + j )ck + j + I3kck + j , k = 1, … , j − 2 C(j + 1)(Nb+ Nd+ j − 1) = I4 − I4(2j − 1)c2j − 1 + I3(j − 1) + I3(j − 1)c2j − 1 C(j + 1)(Nb+ Nd+ j ) = − I4(2j )c2j + I3jc2j + I3jc0 (j ≤ Nb) C(j + 1)(Nb+ Nd+ k ) = − I4(k − j )ck − j − I4(k + j )ck + j + I3kck + j + I3kck − j , k = j + 1, … , Nb C(j + 1)(2Nb+ Nd+ k ) = − I4(j − k )cj − k − I3kcj − k , k = 1, … , j − 1 C(j + 1)(2Nb+ Nd+ j ) = − I3jc0 C(j + 1)(2Nb+ Nd+ j + 1) = I4 − I3(j + 1) C(j + 1)(2Nb+ 2Nd+ k ) = I3(j − k )cj − k , k = 1, … , j − 1 C(j + 1)(2Nb+ 2Nd+ j + 1) = − I3 C(j + 1)(2Nb+ 3Nd+ k ) = I3(k + j )ck + j , k = 1, … , j − 2 C(j + 1)(2Nb+ 3Nd+ j − 1) = − I3 + I3(2j − 1)c2j − 1 C(j + 1)(2Nb+ 3Nd+ j ) = I3(2j )c2j(j ≤ Nb)

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