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Closed-form elastic solution for irregular frozen wall of inclined shaft considering the interaction with ground
International Journal of Rock Mechanics and Mining Sciences 100 (2017) 62–72
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International Journal of Rock Mechanics and Mining Sciences journal homepage: www.elsevier.com/locate/ijrmms
Closed-form elastic solution for irregular frozen wall of inclined shaft considering the interaction with ground
T
⁎
Renshu Yanga,b, Qianxing Wanga, , Liyun Yanga,b a b
School of Mechanics and Civil Engineering, China University of Mining and Technology (Beijing), Beijing, China State Key Laboratory for Geomechanics and Deep Underground Engineering, Beijing, China
A R T I C L E I N F O
A B S T R A C T
Keywords: Irregular frozen wall Interaction Non-uniform stresses Stresses relaxation Complex variable method
An analytical solution is derived that provides a closed-form formulation for stresses and displacements of a deep inclined shaft with irregular frozen wall liner in an infinite elastic medium, subjected to a non-uniform stresses. This solution is based on: (i) the conformal mapping of a doubly connected domain of prescribed shape onto a circular ring by an appropriate numerical optimization scheme which is achieved by MATLAB program, (j) the complex variable method. For a particular case, the distributions of hoop stress and radial displacement inside frozen wall are presented with analytical solution. The accuracy of stress functions is verified by comparison of different number of coefficients in stress functions and it show that the number of coefficients k≥15 is advised. The validity of analytical solution is verified by a comparison between its result and that obtained from the finite element program ANSYS. Noting that the shaft excavation can be regarded as relaxation of initial ground stresses around shaft excavation boundary, this paper presents the mechanical problem as two separate problems: (i) initial stresses and displacements model where stresses keep constant and displacements equal zero around frozen wall and ground, (j) stresses relaxation model with relaxed normal and shear stresses acted on excavated boundary. The solution is worked out by the principle of superposition in elastic medium. The Young's modulus ratio (β) of the frozen wall to the ground reflects the influence of ground to frozen wall, and the influence decreases with the growth of β. The proposed solution can provide analytical basis for frozen wall design and be used as a quick-solver for back-analysis of in situ stresses and displacements.
1. Introduction Artificial ground freezing (AGF) is a special construction method, which is used to freeze the soft or water-bearing ground into an enclosed structure, to increase the strength and stability of ground soil or rocks, and to make them impervious to water seepage.1 AGF has been widely used in mine construction, urban underground tunnel, and deep foundation excavation and so on. Tunnel and shaft usually have a circular cross section because of its structural advantages due to the large surrounding pressures. Analytical solutions for excavation of circular holes in elastic and homogeneous materials subjected to uniform stresses or non-uniform stresses are discussed in textbooks of elastic theory and rock mechanics due to their importance as basic tools for introducing the subject of underground tunnel excavations.2,3 Einstein and Schwartz4 perhaps at the first time worked out the solution for excavation and support problem of deep circular lined tunnel in an infinite elastic medium subjected to nonuniform stresses. However, they ignored the effect of support delaying. As supplementary, with consideration of the delaying effect of support ⁎
and different contact conditions on the interface between ground and lining, Carranza-Torres et al.5 found an analytical elastic solution for deep circular lined tunnel. Mason and Stacey6 considered three contact conditions on the interface between ground and support, and then analyzed the problem of concrete spraying support in deep circular tunnel subjected to a uniform shear stress at infinity. In most cases of tunnel and inclined shaft excavation, non-uniform stresses are acted on models. As an effective method, complex variable method can be used to solve the problem of circular or irregular tunnels or shaft subjected to uniform or non-uniform stresses. Poulos and Davis7 analyzed the tunnels with and without lining in his work. Pender8 and Sagaseta9 presented the stress-displacement solution for circular lined tunnel without regard to gravity as well. Li and Wang10,11 achieved the solutions for deep circular lined tunnel subjected to uniform internal pressures, and stress release of excavation to support and water seepage were taken into consideration. Based on these solutions, Yang and Wang12 obtained the solutions for circular shaft frozen wall and ground subjected to non-uniform stresses, and they considered the stresses relaxation model of shaft excavation.
Corresponding author. E-mail address: [email protected] (Q. Wang).
http://dx.doi.org/10.1016/j.ijrmms.2017.10.008 Received 27 September 2016; Received in revised form 7 May 2017; Accepted 16 October 2017 Available online 05 November 2017 1365-1609/ © 2017 Elsevier Ltd. All rights reserved.
International Journal of Rock Mechanics and Mining Sciences 100 (2017) 62–72
R. Yang et al.
Although the circular cross section is the first choice for inclined shaft excavation, the horseshoe-like section (a section with semi-circular arch roof and straight side wall and inverted arch floor) is the most common section of inclined shaft limited to the excavation condition and complex ground condition. It is assumed that an irregular boundary in the z-plane can be mapped conformally to a circle in the ζplane.13,14 The exact analytical solution for circular lined tunnel and simple irregular tunnel, like lined oval tunnel, could be worked out by complex variable theory, while only closed-form solution could be obtained for relatively complex tunnel. Based on this theory, Gercek15 presented the solution for stresses around tunnel with conventional shapes for the first time. However, Gercek did not compute the displacement solution for the ground, and did not consider the support either. Exadaktylos et al.16,17 presented the closed-form solutions for stresses and displacements around semi-circular and notched circular tunnels. Exadaktylos also did not take support into consideration. Huo and Bobet18 analyzed the regularities of stresses distribution around deep lined tunnel in earthquake region subjected to shear stresses. Clearly, all published works present solutions for circular supported tunnel or irregular tunnel without support. Different with them, frozen wall is a kind of doubly connected domain, which plays a significant role in the construction of shaft or tunnel, however there is no published work presenting the analytical solution for deep lined irregular tunnel or shaft subjected to non-uniform stresses (frozen wall is a kind of temporary liner). Now because of the irregularity of inclined shaft frozen wall, the design of inclined shaft frozen wall is based on method of circular shaft and empirical method, there is no theoretical solution for irregular frozen wall. Unfortunately, many inclined shafts were flooded because of weak frozen wall in western China. Therefore addressing the deep irregular inclined frozen shaft, this paper presents a closed-form solution of stresses and displacements for irregular frozen wall and ground with consideration of the interaction between frozen wall and ground.
This paper is conducted based on three assumptions presented as follows: (1) Frozen wall and ground are regarded as elastic and homogeneous materials. (2) Frozen wall and ground are assumed to be fully contacted, that is to say, the radial stress and shear stress are continuous at the interface between frozen wall and ground, as well as the radial and tangential displacements. (3) The variation of body forces are ignored, as well as the frost heave, i.e., field quantities (stresses and displacements) remain unchanged after freezing. 2.1. Stresses relaxation model The excavation of a deep inclined frozen shaft can be regarded as stresses relaxation around excavation boundary in an infinite elastic medium. Considering the stress path of shaft excavation, the problem of this paper can be divided into two new problems, stresses relaxation model (see Fig. 1a) and initial model, and the final solution is worked out by the principle of superposition in the elasticity. stresses relaxation model shows the excavation problem of an irregular inclined frozen shaft in the elastic material subjected to normal stress σrF0 and shear stress τrθF0 at the inner boundary, and normal displacements ux=0 and uy=0 at the far-boundary(r→∞). Initial model shows the initial stresses and displacements of frozen wall and ground where stresses keep constant and ux=0 and uy=0. Fig. 1a also can be separated by two parts, mechanical model for frozen wall as shown in Fig. 1b and mechanical model for ground as shown in Fig. 1c. Fig. 1b shows the frozen wall problem with normal stress and shear stress at inner and outer boundaries. Fig. 1c shows the ground problem with normal stress and shear stress at inner boundary, and normal displacements ux=0 and uy=0 at the far-boundary (r→∞). In Fig. 1, σrF0 and τrθF0 are radial and shear stresses on inner boundary of frozen wall, respectively. σrF1 and τrθF1 are radial and shear stresses on the interface of frozen wall and ground, respectively. In this paper, complex variable method is used to work out the solution for irregular frozen wall and ground in an infinite elastic material. In the complex variable method,19–21 the solution is expressed in terms of two analytical functions φ(z) and ψ(z), the stresses are related to these functions by the following equations
2. Problem statement This paper refers to the excavation problem of an irregular inclined frozen shaft in an infinite elastic medium under plane strain condition, subjected to non-uniform stresses, the cross section of inclined frozen shaft is shown in Fig. 1a, which consists of frozen wall and ground. In Fig. 1a, r0 and r1 are inner and outer radii of semi-circular arch of frozen wall, respectively, h0 and h1 are inner and outer heights of straight side wall, respectively, rd0 and rd1 are inner and outer radii of inverted arch, respectively.
⎧ σy + σx = 4 Re φ′ (z ) ⎨ σy − σx + 2iτxy = 2[z φ′ ′ (z ) + ψ′ (z )] ⎩
(1)
Fig. 1. Mechanical model for (a) stresses relaxation considering the interaction of frozen wall and ground with relaxation stresses on inner boundary of frozen wall in an infinite plane which can be separated into two parts: (b) frozen wall model and (c) ground model.
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and the conversion of stress functions in the z-plane and ζ-plane are presented as follows19:
where σx and σy are horizontal and vertical stresses at any point, respectively. The displacements are related to these functions by
(u x + iu y ) =
1 κφ (z ) − zφ′ (z ) − ψ (z ) 2G
φ (ζ ) = φ1 (z ) = φ1 [ω (ζ )],
= ψ1 [ω (ζ )],}
(2)
Φ (ζ ) = φ1′ (z ) = φ′ (ζ )/ ω′ (ζ ), Ψ (ζ ) = ψ1′ (z )
where G is the shear modulus of the elastic material, and κ is related to Poisson's ratio µ by κ= (3–4µ) for plane strain condition and κ= (3-µ)/ (1+µ) for plane stress condition. In this paper, plane strain condition is assumed. There are displacements or surface tractions prescribed along the boundaries. For displacements, the quantity (ux+iuy) is prescribed by Eq. (2). The surface tractions are related to the complex stress functions φ(z) and ψ(z) by
[φ (z ) + zφ′ (z ) + ψ (z )] = i
∫ (fx + ify ) ds
ψ (ζ ) = ψ1 (z )
= ψ′ (ζ )/ ω′ (ζ ),} ′
Φ′ (ζ ) = φ1′ (z )⋅ω′ (ζ ) (10) Stress and displacement variables in polar coordinates are related to stress functions and they are presented as follows:
⎧ σr + σθ = σx + σy = 4 Re[Φ (ζ )],
(3)
⎨ σθ − σr + 2iτrθ = ⎩
where fx and fy are horizontal and vertical boundary surface forces, respectively.
ur + iuθ =
2ζ 2 [ω (ζ )Φ′ (ζ ) + ω′ (ζ ) Ψ (ζ )] r 2ω′ (ζ )
ζ ω′ (ζ ) (u x + iu y ) r ω′ (ζ )
(11)
(12)
2.2. Boundary conditions 2.4. Scaling of variables
The boundary condition of inner frozen wall is expressed in terms of the integral of the surface relaxation forces, integrated along the boundary, which is described by
[φ1 (z ) + zφ1′ (z ) + ψ1 (z )]
z1
=i
∫ (fx + ify ) ds
In this paper, load and deformation quantities are considered as dimensionless by normalizing them with respect to properly selected variables. The unscaled quantities will be denoted by the use of the tilde symbol (~) on top of the variable. The radius of inner boundary r͠0 is used to scale the radius and deformation r͠i , u͠ i and u͠ θ respectively, i.e.,
(4)
where φ1(z) and ψ1(z) are two analytical functions of frozen wall in the z-plane, z1 is a point on the inner boundary of frozen wall. Along the interface between frozen wall and ground (z=z2), the radial and shear stresses, the radial and tangential displacements are continuous, so the stress boundary condition at z=z2 is written as
[φ1 (z ) + zφ1′ (z ) + ψ1 (z )]
z = z2
= [φ2 (z ) + zφ2′ (z ) + ψ2 (z )]
ur =
where φ2(z) and ψ2(z) are two analytical functions of ground in the zplane, z2 is a point on the interface of frozen wall and ground. The displacement boundary condition at z=z2 is written as
Γ [κ1 φ1 (z ) − zφ1′ (z ) − ψ1 (z )]
z = z2
= κ2 φ2 (z ) − zφ2′ (z ) − ψ2 (z ) z = z
2
σr =
(6)
3
(7)
As for plane strain condition, the parameters Г, κ1 and κ2 are given by
Γ=
E2 (1 + μ1) , κ1 = 3 − 4μ1 , κ2 = 3 − 4μ 2 E1 (1 + μ 2 )
(8)
where E1 and E2 are Young's moduli of the frozen wall and ground, and µ1 and µ2 are Poisson's ratios of the frozen wall and ground.
In order to work out these complex functions in terms of the known boundary conditions, mapping equation z=ω(ζ) is used to map the region in the z-plane with z = x + iy = ρe iϕ onto the region in the ζplane with ζ=ξ+iη=reiθ. A circular or irregular shape in the z-plane can be mapped conformally onto a circle or circular ring in the ζ-plane ,19 and the solution for circular shape could be worked out easily. Xu19 presented that the angle ϕ of the normal direction of a point in the z-plane, is related to the mapping function ω(ζ) by
ζ ω′ (ζ ) r ω′ (ζ )
(14)
∼ and The Poisson's ratio of frozen wall and ground, the quantities μ 1 ∼ are dimensionless, and they are scaled by μ 2 ∼,μ =μ ∼ μ1 = μ (17) 1 2 2
2.3. Conformal mapping
e−iϕ =
τ͠ F σ͠ F σ͠ r σ͠ τ͠ , σθ = θ , τrθ = rθ , σrF = r , τrθF = rθ p͠ p͠ p͠ p͠ p͠
Similarly, the horizontal and vertical stresses, acted on the frozen ∼ ∼ wall, the quantities fx and fy are scaled as follows: ∼ ∼ fy f fx = x , fy = ͠p p͠ (15) ∼ The Young's moduli of frozen wall and ground, the quantities E1 and ∼ ∼ E2 , the shear moduli of frozen wall and ground, the quantities G1 and ∼ G2 , respectively, are scaled as follows: ∼ ∼ ∼ ∼ E E G G E1 = 1 , E2 = 2 , G1 = 1 , G2 = 2 p͠ p͠ p͠ p͠ (16)
The last boundary condition is that the infinite boundary z=z3 is fixed, i.e., the radial and tangential displacements are both equal to zero. This gives
κ2 φ2 (z ) − zφ2′ (z ) − ψ2 (z ) z = z = 0
(13)
where the subscript ii=0 represents inner boundary of frozen wall, ii=1 represents interface between frozen wall and ground, ii=2 represents infinite boundary of ground. The vertical stress in far-field p͠ is used to scale normal and shear stresses, σ͠ r , σ͠ θ and τ͠ rθ in frozen wall and ground, respectively, i.e.,
(5)
z = z2
u͠ r u͠ r͠ r͠ , uθ = θ , r = , rii = ii (ii = 0, 1, 2) r͠0 r͠0 r͠0 r͠0
3. The mapping representation of frozen wall The references20,21 presented closed-form solutions for mapping a complex shape hole onto a circle. While different to conventional tunnels, the frozen wall is a kind of temporary support, and it is a doubly connected domain in a finite plane. The mapping of simply connected domain problem is not suitable for doubly connected domain any more. Considering that, Lv20 raised a solution for mapping the irregular doubly connected domain onto a circular ring, this method takes both the inner and outer boundaries into account, and keep the mapping
(9) 64
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Fig. 2. Conformal mapping of an irregular frozen wall shape to a circular ring: (a) the frozen wall shape before conformal mapping in the z-plane; (b) the frozen wall shape after conformal mapping in the ζ-plane.
n
precision at a relatively high level for inner and outer boundaries. The frozen wall region in the z-plane can be mapped conformally onto a circular ring in the ζ-plane, bounded by the circles |ζ|=r0 < 1 and |ζ|=r1=1, as shown in Fig. 2. The circle |ζ|=r0 corresponds to the inner boundary of frozen wall, and the circle |ζ|=r1 corresponds to the outer boundary of frozen wall. The appropriate conformal transformation can be described by Laurent series expansion as follows20:
rBj = R [cos (βBj − αBj ) +
∑ Ck ζ −k ) k=0
Eq. (18) gives the conformal transformation function, so the problem turns to solve R and Ck (k=0, 1, 2…n), these parameters can be solved by optimization method which is stated as follows: In the z-plane, x-axis is regarded as the symmetric axis for frozen wall, as shown in Figs. 2a, and (m+1) points are selected along inner and outer boundaries, respectively. It is assumed that the mapped point A0′ is coincident with A0, and this gives
(18)
R=
where R is a positive real number, which reflects the size of the ring, Ck are a series of parameters, which reflect the accuracy of transformation, k=0,1,2,…,n. It is assumed that any point Aj (rAj, αAj) lying on the inner boundary of frozen wall in the z-plane is conformally mapped onto the circle |ζ|=r0 in the ζ-plane corresponding to A′j (r0, βAj), and any point Bj (rBj, αBj) lying on the outer boundary of frozen wall in the z-plane is conformally mapped onto the circle |ζ|=r1 in the ζ-plane corresponding to B′j (r1, βBj). Then in terms of Eq. (18), the conformal transformation of inner boundary of frozen wall is given by
r1 n r0 + ∑k = 0 Ck r −k
m
f=
∑ Ck r0−k e−ikβAj)
+
k=0
(20)
The imaginary part of Eq. (19) is given by n
sin(βAj − αAj ) −
∑ Ck r0−k sin(kβAj + αAj) = 0 k=0
(21)
and the real part of Eq. (19) is given by n
rAj = R [cos (βAj − αAj ) +
∑ Ck r0−k cos(kβAj + αAj)] k=0
k=0
⎩
⎣
ai ≤ x i ≤ bi , (i = 1, 2, ... ,m)
2
⎫
k=0
⎦⎭
(27)
The main idea of this method is to collect m points as the initial compound, and it is better that m ≥ 2n. Compute the objective function value of each point and compare them one by one, remove the bad point which provides the maximum value for the objective function f (X), and replaces it by a new point which can improve the objective
n
∑ Ck r1−k sin(kβBj + αBj) = 0
n
⎧
∑ ⎨rBj* − R ⎡⎢cos (βBj − αBj) + ∑ Ck r1−k cos(kβBj + αBj) ⎤⎥ ⎬
gi (X ) ≥ 0, (i = 1, 2, ... ,m)
(22)
Similarly, the imaginary part and the real parts of Eq. (20) are given by Eqs. (23) and (24), respectively.
sin(βBj − αBj ) −
k=0
where f can reach minimum value when m > n. When the boundary shape of frozen wall can be mapped precisely by Eq. (18), the Ck series, which give the minimum value of Eq. (26), are accurate solution for Eq. (18), and f=0. Otherwise, when the boundary shape of frozen wall cannot be mapped precisely by Eq. (18), the Ck series are closed-form solution for Eq. (18), and f > 0. The complex optimum method is applied in this paper to solve the mapping function addressing the problem stated as follows .22–24 Work out the design value X={x1, ×2, …, xn}T which can provide the minimum value of the target function f(X) and meet the constraints presented by
n
∑ Ck r1−k e−ikβBj)
Ck ⎤⎫ cos(kβAj + αAj ) ⎥ ⎬ r0k ⎦⎭
(26)
and the outer boundary mapping is given by
rBj e iαBj = R (r1 e iβBj +
⎣
⎩
j=0
(19)
2
n
⎧
∑ ⎨rAj* − R ⎡⎢cos (βAj − αAj) + ∑ j=0
n k=0
(25)
According to Eqs. (22) and (24), when k=0,1,2,…,n, and n is a limited positive real number, the left side of (22) and (24) is not strictly equal to the right side, respectively. While the optimization method can be used to solve R and Ck series (k=0, 1, 2,…, n), the objective function is given by
m
rAj e iαAj = R (r0 e iβAj +
(24)
k=0
n
z = ω (ζ ) = R (ζ +
∑ Ck r1−k cos(kβBj + αBj)]
(23) 65
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expressed by
function and meet the constraints, and then make objective function f (X) close to the optimum solution gradually. In this paper, MATLAB software is used to accomplish the procedure and work out the optimal solution. Chen21 and Xu19 indicated that Ck should meet the relationship n ∑k = 0 Ck < 1, while Lv20 improved this condition by
F (z ) = i
kCk < 1
(fx + ify ) ds = iσx Im(zP − z A) + σy Re (zP − z A)
(36)
where zA is the start point of integration with angle 0 on the inner boundary of frozen wall, and zP is the ending point of integration which is an arbitrary point on the inner boundary of frozen wall, as shown in Fig. 2(a). The integration must obey the anti-clockwise direction. The integrated function F(z) in Eq. (36) can be expressed by Fourier series expansion21
(28)
k=0
zP
A
n
∑
∫z
4. Solution for irregular frozen wall and ground ∞
F (z ) =
As presented in Section 2, in elastic medium, the solution for irregular frozen wall and ground is divided into two part, solution for initial stresses and displacements problem and solution for stresses relaxation problem
k=0
(37)
k=0 1
2π
1
2π
where Ak = 2π ∫0 F (z ) e−ikθdθ , Bk = 2π ∫0 F (z ) e ikθdθ In the ζ-plane, after conformal mapping, the boundary conditions (4)–(7) are expressed by
4.1. Initial stresses and displacements of frozen wall and ground
φ1 (σ1) +
Based on the third assumption, before excavation, the ground is undisturbed, so the initial stresses of frozen wall and ground are given by 0 σx0 = −λp , σy0 = −p , τxy =0
∞
∑ Ak e ikθ + ∑ Bk e−ikθ
φ1 (σ2) +
(29)
ω (σ1) φ ′ (σ1) + ψ1 (σ1) = ω′ (σ1) 1
∞
∞
∑ Ak e ikθ + ∑ Bk e−ikθ k=0
k=0
ω (σ2) ω (σ2) φ ′ (σ2) + ψ1 (σ2) = φ2 (σ2) + φ ′ (σ2) + ψ2 (σ2) ω′ (σ2) 1 ω′ (σ2) 2
in Cartesian coordinates, and
(39)
1+λ λ−1 p− p cos 2ϕ σr0 = − 2 2
(30)
1+λ λ−1 p+ p cos 2ϕ 2 2
(31)
σθ0 = −
τrθ0 =
λ−1 p sin 2ϕ 2
Γ [κ1 φ1 (σ2) −
κ2 φ2 (σ3) −
(32)
ζ ω′ (ζ ) ζ ω′ (ζ ) )) ))orϕ = −arcsin(Im( r ω′ (ζ ) r ω′ (ζ )
∞
= L−n ζ −n + L−(n − 1) ζ −n + 1 + ⋯+L−1 ζ −1 +
∑ Lk ζ k k=0
(33)
(42)
It can be obtained that L−1=C1 when n=1; L−1=C1 and L−2=C2 when n=2; and when n≥3,
Complex stress functions in the ζ-plane can be considered as functions of ζ, the functions of frozen wall are given by ∞
(34)
and the functions of ground are given by
⎧ L−n = Cn, ⎪ L−(n − 1) = Cn − 1, ⎨ Cj − 1 − k j−2 ⎪ L−(n − j + 1) = ∑k = 1 (j − 1 − k ) r 2(j − k ) L−(n − k + 1) + Cn − j + 1 ⎩
(43)
−4 −2n ⎧ L0 = C0 + C1 r L−2 +⋯+(n − 1) Cn − 1 r L−n , ⎪ L1 = 1 + C1 r −4L−1 + 2C2 r −6L−2 +⋯+nCn − 1 r −2(n + 1) L−n , ⎨ ⎪ Li = ∑ik−=11 kCk r −2(k + 1) Li − k + 1 + ∑nk = i kCk r −2(k + 1) L−(k − i + 1) ⎩
(44)
where j=3,4,…,n, and i=2,3,…,n. After substituting the ω (ζ )/ ω′ (ζ ) with boundary values, Eqs. (38)–(41) can be simplified with equations without denominator. The infinite algebraic equations containing unsolved parameters are obtained by setting the coefficients of eikθ at the left side equal to the coefficients of eikθ at the right side where k=-n,…,−1,0,1,…,n. The infinite algebraic equations are given by
∞
−k ⎧ φ2 (ζ ) = ∑k = 0 fk ζ , ⎨ ψ2 (ζ ) = ∑∞ h ζ −k k=0 k ⎩
(41)
ω (ζ ) ζ + C0 + C1 ζ −1 + C2 ζ −2+⋯+Cn ζ −n = 1 − C1 r −4ζ 2 − 2C2 r −6ζ 3−⋯−nCn r −2nζ n + 1 ω′ (ζ )
4.2. Solution for stresses relaxation model
∞
ω (σ3) φ ′ (σ3) − ψ2 (σ3) = 0 ω′ (σ3) 2
where σ1 and σ2 represent points on the inner and outer boundaries of frozen wall, and σ3 represents a point on the infinite boundary of ground in the ζ-plane. In Eqs. (38)–(41), the coefficient ω (ζ )/ ω′ (ζ ) can be expanded by
Therefore, the initial stresses of frozen wall and ground can be worked out by Eqs. (29)–(32).
k −k ⎧ φ1 (ζ ) = ∑k = 1 ak ζ + ∑k = 1 bk ζ , ∞ ∞ ⎨ ψ1 (ζ ) = ∑k = 0 ck ζ k + ∑k = 0 dk ζ −k ⎩
ω (σ2) ω (σ2) φ ′ (σ2) − ψ1 (σ2)] = κ2 φ2 (σ2) − φ ′ (σ2) − ψ2 (σ2) ω′ (σ2) 1 ω′ (σ2) 2 (40)
in polar coordinates. And the vertical and horizontal displacements u x0 and u y0 are both equal to 0.19 0 In Eqs. (29)–(32), σx0 , σy0 and τxy are the initial horizontal, vertical 0 and shear stresses, respectively; σr , σθ0 and τrθ0 are initial radial, hoop and shear stresses, respectively, and the tensile stress is ordered as positive; λ is the lateral pressure coefficient; p is the vertical stress added on the model; ϕ is the angle of normal direction of a point in the z-plane, the relation of ϕ and mapping function ω(ζ) is given by Eq. (9), so ϕ is obtained from Eq. (9) by
ϕ = arccos(Re(
(38)
(35)
The parameters ak, bk, ck, dk, fk and hk in Eqs. (34) and (35)are all real because of the symmetry along x-axis. The integral of surface tractions at the right side of Eq. (4) can be
66
International Journal of Rock Mechanics and Mining Sciences 100 (2017) 62–72
R. Yang et al. ∞
∞
2(k − 1) − ∑k = 1 L−(k + n + 1) kbk r0−2(k + 1) ] + cn r0−2n} r0−n r = r ⎧ {bn + [ ∑k = 1 kLk − n − 1 ak r0 0 ⎪ ⎪ = Bn , ⎪ ∞ ∞ −2(k + 1) + ∑k = 1 Lk + m − 1 kak r02(k − 1) ] + dm r0−2m} r0n r = r ⎪ {am + [− ∑k = 1 kL−k − m − 1 bk r0 0 ⎪ ⎪ = Am , ⎪ b ∞ c ⎪ {b + [ ∑∞ kL 2(k − 1) − ∑k = 1 L−(k + n + 1) k 2(kk+ 1) ] + 2nn } k − n − 1 ak r1 k=1 ⎪ n r1 r1 r = r1 ⎪ ⎪ fk n ⎪ = fn − ∑k = 1 L−(k + n + 1) k 2(k + 1) , r1 ⎪ r = r1 ⎪ ⎪ b ∞ ∞ d 2(k − 1) − ∑k = 1 kL−k − m − 1 2(kk+ 1) ] + 2mm } ⎪ {am + [ ∑k = 1 Lk + m − 1 kak r1 r1 r1 r = r1 ⎪ ⎪ ⎪ fk ∞ hm , = 2m − ∑k = 1 kL−k − m − 1 2(k + 1) r1 r1 ⎨ r = r1 ⎪ ⎪ ∞ L−(k + n + 1) kbk ∞ c 2(k − 1) ] − 2nn } − ∑k = 1 ⎪ Γ {κ1 bn − [ ∑k = 1 kLk − n − 1 ak r1 r1 r 12(k + 1) ⎪ r = r1 ⎪ ⎪ = κ f + ∑n L−(k + n + 1) kfk , 2 n k=1 ⎪ r 12(k + 1) r = r1 ⎪ ⎪ b ∞ kL ∞ ⎪ Γ {κ1 am + [ ∑k = 1 −k2(−km+−1)1 k − ∑k = 1 Lk + m − 1 kak r12(k − 1) ] − d2mm } r1 r1 ⎪ r = r1 ⎪ ⎪ ∞ kL−k − m − 1 fk hm , ⎪ = − r12m + ∑k = 1 r 2(k + 1) 1 r = r1 ⎪ n ⎪ −2(k + 1) = 0, ⎪ κ2 fn + ∑k = 1 L−(k + n + 1) kfk r2 r=r2 ⎪ ∞ −2(k + 1) −2m kL f r h r − = 0(m = 1, 2, ⋯; n = 0, 1, 2, ⋯) ∑ ⎪ k = 1 −k − m − 1 k 2 m 2 r=r2 ⎩
Fig. 3. The original shape and mapping shape of frozen wall.
function. The mapping solution for frozen wall after scaled is presented by Eq. (47). The radius of inner boundary r0=0.33951 which is calculated by Eq. (47).
z = ω (ζ ) = 3.07296(ζ − 0.02009 − 0.00136ζ −1 + 0.00208ζ −2 − 0.00049ζ −3 + 0.00006ζ −4 )
(47)
(45) Fig. 3 presents the conformal mapped shape of frozen wall with Eq. (47). The solid line represents the original shape of frozen wall, and the dashed line represents the mapped shape of frozen wall. It can be concluded that the mapping shape and the original shape of frozen wall are seen to agree well, especially the semi-circular arch section of frozen wall. The mapping boundaries of inner and outer frozen wall are basically in coincidence with the original boundaries. There are certain difference located on side wall and invert section outside frozen wall, because the mapping is based on the assumption that the point A0 at original inner boundary of frozen wall is in coincidence with A0′ at mapped inner boundary. Nevertheless, the conformal mapping based on Eq. (47) could perform well. In order to assess the solution accuracy for stresses and displacements of frozen wall, three number of coefficients ak, bk, ck, dk, fk and hk are discussed, i.e., k=10, 15 and 20. Fig. 4 gives the comparison of stresses and displacements along inner frozen wall considering different coefficients number, and it shows that the coefficients number of stress functions does have influence on the solution of frozen wall. Fig. 4a shows that the radial stress varies evidently at the region where angel θ=90°~180° when k=10, and the peak stress reaches ± 0.3 (dimensionless). The radial stress changes little when k=15 or 20, especially when k=20. According to the theoretical solution, radial stress of inner boundary is zero when inner boundary of frozen wall is free of any constraint of force or displacement. Fig. 4b shows that coefficients number has relatively less influence on hoop stress and radial displacement of inner boundary of frozen wall. However the hoop compression stress grows up at the region angel θ=170°~180°, which violates the variation law of hoop stress. Therefore k≥15 should be considered for obtaining stress and displacement solutions accurately. In this paper k=20 is adopted, that is to say, totally 123 unknown coefficients are needed to be solved. Fig. 5 shows the distributions of radial displacement and hoop stress within frozen wall, which are obtained with analytical solution, the radial displacement is presented at the left side of frozen wall and the hoop stress is presented at the right side because of the symmetry of frozen wall. The outward displacement and tensile stress are ordered as positive. Fig. 5 shows that the displacement at arch roof and floor is bigger than the value at side wall of frozen wall, and the displacement
which contain four parts corresponding to four boundary conditions and are deduced by Eqs. (38)–(41). Note that the Lk and L-k series in Eq. (45) are related to radius r in the ζ-plane. Eq. (45) is a linear equation group with infinite equations, which contain the unsolved coefficients ak, bk, ck, dk, fk and hk, so the linear equation group can be simplified as (46)
MX = N
where M is a n×n matrix of coefficients ak, bk, ck, dk, fk and hk, X is a n×1 vector of coefficients ak, bk, ck, dk, fk and hk, and X={a0, a1,…, aii, b1,…, bii, c0, c1,…, cii, d1,…, dii, f0, f1,…, fii, h1,…, hii}T, N is a known n×1vector, and N={B1,…,Bii,A0,A1,…Aii, 0,…0}T with n=6ii+3 and ii is an assumed number on the premise of accuracy. 5. Example application and comparison of analytical and numerical solutions 5.1. Example application An inclined shaft excavation is taken as an example, the cross section of frozen shaft can be regarded as a horseshoe, as shown in Fig. 1a. Table 1 shows the dimension parameters of frozen wall and ground. The ∼ Young's moduli of frozen wall and ground are E1=300 MPa and ∼ E2 =100 MPa, respectively, and the Poisson's ratios of frozen wall and ∼ =0.24 and μ ∼ =0.35, respectively. The volume-weight of ground are μ 1 2 frozen/unfrozen soil γ͠ =20 kN/m3, and the lateral pressure coefficient λ=0.65. The parameters as shown in Table 1 are all explained in Fig. 1 a. In Eq. (18), it is assumed that the parameter n equals 4 (that is C0~C4), and on the basis of complex optimum method, the program MATLAB is used to get the closed-form solution for conformal mapping Table 1 Dimension parameters of tunnel and frozen wall. Parameters
r͠ 0 /m
r͠ 1 /m
h͠ 0 /m
h͠ 1/m
r͠ d0 /m
r͠ d1 /m
Values
3.5
10.5
2.6
5.8
7
14
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Fig. 4. The solutions of (a) radial stresses and (b) hoop stresses and radial displacements of frozen wall along the inner boundary with coefficients number k=10, 15 and 20.
Fig. 6. Numerical simulation model for example problem.
whole model to show the region of frozen wall clearly. The far boundaries in numerical model locate at a distance of 30 times the radius of the tunnel semi-circular arch. The left boundary of numerical model is set to be symmetric. The bottom boundary is set with normal displacement equaling 0. Horizontal pressure, which equals lateral pressure coefficient timing the weight of overlying soil, is acted on the right boundary, and vertical pressure, which equals the weight of overlying soil, is acted on the top boundary. Gravity is ignored in this model as with analytical model. The horizontal and vertical pressures are given by
Fig. 5. Distribution of radial displacement (left) and hoop stress (right) of frozen wall with analytical solution.
at floor is greater than the value at arch roof when the lateral pressure coefficient λ is less than 1. The stress concentrates on the side wall and corner place of frozen wall, that is to say, the side wall and corner place are the first to yield. 5.2. Comparison of analytical and numerical solutions with ANSYS program
pv = γh, ph = λγh
A corresponding elastic numerical model under plane strain condition was established by the ANSYS program to verify the validity of analytical solution for stresses and displacements in frozen wall and ground. The numerical model, as shown in Fig. 6, was established based on the parameters stated in Table 1. Only half of the problem has been modeled because of the symmetry of the problem. The ANSYS model considers three-stage sequence corresponding to the deduction of the analytical solution in this paper, i.e., on a first stage the boundary conditions are acted and the initial equilibrium state is computed, and on a second stage the frozen wall is formed, and on a third stage the tunnel is excavated and the equilibrium state is again computed. Fig. 6 represents the stage when the tunnel is excavated, so the soil and elements in tunnel are removed. Fig. 6 only presents part of the
(48)
where pv and ph represent the vertical and horizontal pressures respectively; h is the depth. Take p=1 for example, the stress function coefficients of frozen wall and ground can be solved by substituting the design parameters of tunnel and frozen wall as well as parameters of mapping function into Eq. (45) and then stresses and displacements for frozen wall and ground are worked out by substituting these coefficients into Eqs. (11) and (12). The load and deformation quantities in this chapter are considered dimensionless. The diagrams of Fig. 7 represent the analytical and numerical results for frozen wall along inner and outer boundaries after excavation. Fig. 7a represents the distribution of hoop stress along inner and outer boundaries for frozen wall as a function of θ. Fig. 7b represents the distribution of radial displacement along inner and outer boundaries for 68
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Fig. 7. Distribution of (a) hoop stresses and (b) radial displacements for frozen wall along the inner and outer boundaries as obtained with analytical and numerical solutions.
frozen wall as a function of θ. The analytical and numerical results for frozen wall quantities after excavation, as represented by solid/dashed lines and squares/circles in Fig. 7, are seen to agree well. The hoop stress around outer boundary of frozen wall changes slightly, the maximum stress difference is close to 0.2. The difference existing at the side wall and invert sections are caused by conformal mapping, as shown in Fig. 3. The diagrams of Fig. 8 represent the analytical and numerical results for frozen wall and ground along certain radial paths after excavation. Fig. 8a represents the distribution of hoop and radial stresses for frozen wall and ground as a function of radial distance (a radial scanline with respect to the spring line of the frozen wall at angle θ=0°). Fig. 8b represents the distribution of hoop and radial stresses for frozen wall and ground as a function of radial distance at angle θ=90°. Fig. 8c represents the distribution of hoop and radial stresses for frozen wall and ground as a function of radial distance at angle θ=180°. Fig. 8d represents the distribution of radial displacements for frozen wall and ground as a function of radial distance at angles θ=0°, θ=90° and θ=180°. The analytical and numerical results for frozen wall and ground quantities after excavation, as represented by solid/dashed lines and squares/circles/triangles in Fig. 8, are seen to agree wall. The diagrams in Figs. 7 and 8 present that the absolute hoop stress
of point D is smaller than the value of point A, and the radial displacement of point D is greater than the value of point A, which means point D is easier to break than point A. 6. Results and discussion The load and deformation quantities in this chapter are considered dimensionless. In this chapter, the law of stress and displacement distributions on frozen wall are analyzed with respect to depth, lateral pressure coefficient, thickness of frozen wall and the ratio of Young's moduli (the ratio of frozen wall Young's modulus to ground Young's modulus) 6.1. Influence of depth The diagrams in Fig. 9 represent the relationship between the increase of scaled hoop stress and the increase of depth which is presented with scaled pressure. Under the same parameters of frozen wall, the hoop stress increase with the growth of pressure. The angle θ changes from 0° to 180° means from point A to point D. It can be concluded from Fig. 9a that pressure has great effect on hoop stress at inner side section (from point B to point C) and inner semi-circular arch Fig. 8. Distribution of stresses at (a) angle θ=0°; (b) angle θ=90°; (c) angle θ=180° and (d) displacements at angles θ=0°, θ=90° and θ=180° for frozen wall and ground as obtained with analytical and numerical solutions.
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Fig. 9. Distribution of hoop stresses for frozen wall along (a) the inner boundary and (b) the outer boundary under different scaled pressure.
Fig. 10. Distribution of radial displacements for frozen wall along (a) the inner boundary and (b) the outer boundary under different scaled pressure.
Fig. 11. The variations of (a) hoop stresses and (b) radial displacements of different key points on inner boundary of frozen wall with lateral pressure coefficient.
Fig. 12. The variations of (a) hoop stresses and (b) radial displacements of different key points on inner boundary of frozen wall with frozen wall thickness.
increase of scaled radial displacement and the increase of scaled pressure. Under the same design values of frozen wall, the radial displacement along inner and outer boundaries of frozen wall increase with the growth of pressure. The diagrams in Fig. 10 present that pressure has great effect on radial displacement at arch and invert arch sections, the radial displacement changes little at the side section of
section (from point A to point B), but it has little effect on hoop stress in the center region of inner invert section (region close to point D). Different with the inner boundary of frozen wall, the changes of pressure have similar effect on all points along outer boundary of frozen wall, as shown in Fig. 9b. The diagrams in Fig. 10 represent the relationship between the 70
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Fig. 13. The variation of hoop stress and radial displacement of (a) points A and D and (b) point B and C on inner boundary of frozen wall with Young's modulus ratio β.
point A and point D. The radial displacement of point D is greater than the value of point A. However, the radial displacement difference between point A and point D decreases gradually with the increase of ratio β. The difference is 5.91 when ratio β equals1, and the difference decreases to 0.37 when ratio β grows to 19. Fig. 13b presents that different with point A and point D, the hoop stresses of point B and point C increase with the growth of ratio β, however the radial displacements of point B and point C decrease with the growth of ratio β. The radial displacement of point B is almost equal to the value of point C. Fig. 13a and b present that the hoop stress and radial displacement vary obviously when β < 9, and the variation slow down when β > 9, especially after β > 13
frozen wall. The difference of displacement between point A and point D changes little with the increase of depth. It can be concluded from Figs. 9 and 10 that the hoop stress and radial displacement increase linearly with the increase of depth. 6.2. Influence of lateral pressure coefficient The diagrams in Fig. 11 represent the influence of lateral pressure coefficient λ, which increases from 0.1 to 2.8, on stress and displacement at key points on the inner and outer boundaries of frozen wall. The lateral pressure coefficient λ has a significant influence on hoop stress and radial displacement along inner and outer boundaries of frozen wall, hoop stress and radial displacement change linearly with the lateral pressure coefficient λ. Fig. 11a presents the linear increase of hoop compressive stress of point A and point D, and linear decrease of hoop stress of point B and point C. Point A and point D are under tension when λ < 0.4, but point C is under tension when λ > 2.8, and it can be concluded that all points along inner boundary are under compression when 0.4 < λ < 2.8. Fig. 11b presents the linear increase of radial displacement of point B and point C, and linear decrease of radial displacement of point A and point D. Point B deforms outwards when λ < 0.4, and point A deforms outwards when λ > 2.4, that is to say all points deforms inwards when 0.4 < λ < 2.4. As the tensile strength of frozen soil or rock is much less than the compressive strength, the tensile region of frozen wall should call more attention.
7. Conclusions This paper presents a closed-form elastic solution for stresses and displacements along irregular frozen wall and ground in an infinite elastic medium under plane strain condition, subjected to non-uniform stresses. The problem considers the interaction of frozen wall and ground on the basis of excavation model. The solution can be used to analyze the distribution of stresses and displacements of irregular frozen wall, and provide analytical foundation for safety tunnel or shaft freezing design and construction. It is a valuable method for understanding how certain parameters affect the stresses and displacements and figuring out key parameters which could significantly affect the stresses and displacements. Numerical results have been presented with regard to analytical model. By comparison of analytical and numerical results, it is demonstrated that the analytical solution can be effectively used not only for studying the stresses and displacements around irregular freezing shaft but also for studying the stresses and displacements around any lined irregular tunnel or shaft construction subjected to uniform or nonuniform stresses. Based on the mapping method for doubly connected domain problem presented in,20 the irregular frozen wall shape (doubly connected domain) was mapped conformally by using compound optimization design method, and the solution also can be easily used for other problems with irregular support or tunnel excavation boundary. The stresses and displacements vary linearly with the increase of pressure and lateral pressure coefficient λ under elasticity. Lateral pressure coefficient λ could reverse the stresses and displacements state, for example the stress state changes from tension to compression and the displacement state changes from positive to negative, so λ has great influence on frozen wall design and tunnel excavation. Stresses and displacements perform differently with the variation of frozen wall thickness, hoop stress of side wall region (from point B to point C) are sensitive to wall thickness while radial displacement of semi-circular arch and invert arch sections are sensitive to wall thickness. To some extent, the ratio β represents the constraint effect of ground to frozen wall, the greater the value of β, the smaller the effect of ground to
6.3. Influence of wall thickness The diagrams in Fig. 12 represent the influence of wall thickness on stresses and displacements at key points on the inner and outer boundaries of frozen wall. Fig. 12a presents that the increase of wall thickness can decrease the hoop stress for frozen wall, especially for point B and point C. Fig. 12b presents that the increase of wall thickness can decrease the radial displacement for frozen wall, especially for point A and point D. Different with the change of hoop stress, the radial displacement of point B and point C change slightly with the increase of wall thickness. Under this condition, the radial displacement of point B is approximately equal to the value of point C. with the increase of wall thickness, the effect of wall thickness to stress and displacement decrease gradually. 6.4. Influence of Young's modulus ratio The ratio of frozen wall Young's modulus E1 to ground Young's modulus E2 is marked as β. The diagrams in Fig. 13 represent the variation of hoop stresses and radial displacements of key points on the inner and outer boundaries of frozen wall. It can be concluded that the ratio β has significant influence on hoop stress and radial displacement. Fig. 13a presents that the hoop stresses of point A and point D decrease with the growth of ratio β, as well as the radial displacements of 71
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9. Sagaseta C. Analysis of undrained soil deformation due to ground loss. Geotechnology. 1988;38(4):647–649. 10. Li SC, Wang MB. Elastic analysis of stress –displacement field for a lined circular tunnel at great depth due to ground loads and internal pressure. Tunn Undergr Sp Technol. 2008;23:609–617. 11. Li SC, Wang MB. An elastic stress-displacement solution for a lined tunnel at great depth. Int J Rock Mech Min Sci. 2008;45(4):486–494. 12. Yang RS, Wang QX. Elastic analysis and design of circular horizontal frozen wall based on interaction between frozen wall and surrounding rock. J China Coal Soc. 2016;41(5):1069–1077. 13. England AH. Complex Variable Methods in Elasticity. Mineola, New York: Dover Publications Inc; 2003. 14. Muskhelishvili NI. Some Basic Problems of the Mathematical Theory of Elasticity. Groningen, Holland: Noordhoff International Publishing; 1953. 15. Gerçek H. An elastic solution for stresses around tunnels with conventional shapes. Int J Rock Mech Min Sci. 1997;34:96. 16. Exadaktylos GE, Stavropoulou MC. A closed-form elastic solution for stresses and displacements around tunnels. Int J Rock Mech Min Sci. 2002;39(7):905–916. 17. Exadaktylos GE, Liolios PA, Stavropoulou MC. A semi-analytical elastic stress–displacement solution for notched circular openings in rocks. Int J Solids Struct. 2003;40(5):1165–1187. 18. Huo H, Bobet A, Fernandez G, et al. Analytical solution for deep rectangular structures subjected to far-field shear stresses. Tunn Undergr Sp Technol. 2006;21(6):613–625. 19. Xu ZL. Elasticity. Beijing: Higher Education Press; 2006. 20. Lv AZ. Complex Variable Method in Mechanical Analysis of Underground Tunnel. Beijing: Science Press; 2007. 21. Chen ZY. Analytical Method in the Mechanical Analysis of Surrounding Rock. Beijing: China Coal Industry Publishing House; 1994. 22. Wan YQ, Liang GR. Assembly Programs of Optimum Computing Method. Beijing: China Workers Press; 1983. 23. Cao WH, Guo Z. Optimization Method and the Realization of MATLAB. Beijing: Chemical Industry Press; 2005. 24. Gong C, Wang ZL. Optimization Computing of MATLAB. Beijing: Publishing House of Electronics Industry; 2005.
frozen wall, so it is essential to take ground into consideration when analyzing the stresses and displacements of frozen wall. Acknowledgments This research was supported by a National Natural Science Foundation of China (No. 51274203) and a National Key Research and Development Program of China (No. 2016YFC0600903), which are gratefully acknowledged. The authors would like to acknowledge the contribution of Prof. Weihao Yang from the State Key Laboratory for Geomechanics and Deep Underground Engineering at China University of Mining and Technology, who provided valuable comments for improving this work. References 1. Andersland OB, Ladanyi B. Frozen Ground Engineering. Hoboken, New Jersey: John Wiley & Sons Inc; 2004. 2. Goodman RE. Introduction to Rock Mechanics. New York: John Wiley & Sons; 1980. 3. Jaeger JC, Cook NG, Zimmerman R. Fundamentals of Rock Mechanics. Malden, Mass: Blackwell publishing Ltd; 2007. 4. Einstein HH, Schwartz CW. Simplified analysis for tunnel supports. J Geotech Geoenviron Eng. 1979;104(4):499–518. 5. Carranza-Torres C, Rysdahl B, Kasim M. On the elastic analysis of a circular lined tunnel considering the delayed installation of the support. Int J Rock Mech Min Sci. 2013;61:57–85. 6. Mason DP, Stacey TR. Support to rock excavations provided by sprayed liners. Int J Rock Mech Min Sci. 2008;45(5):773–788. 7. Poulos HG, Davis EH. Elastic Solutions for Soil and Rock Mechanics. New York: John Wiley & Sons; 1974. 8. Pender MJ. Elastic solutions for a deep circular tunnel. Geotech. 1980;30(2):216–222.
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International Journal of Rock Mechanics and Mining Sciences journal homepage: www.elsevier.com/locate/ijrmms
Closed-form elastic solution for irregular frozen wall of inclined shaft considering the interaction with ground
T
⁎
Renshu Yanga,b, Qianxing Wanga, , Liyun Yanga,b a b
School of Mechanics and Civil Engineering, China University of Mining and Technology (Beijing), Beijing, China State Key Laboratory for Geomechanics and Deep Underground Engineering, Beijing, China
A R T I C L E I N F O
A B S T R A C T
Keywords: Irregular frozen wall Interaction Non-uniform stresses Stresses relaxation Complex variable method
An analytical solution is derived that provides a closed-form formulation for stresses and displacements of a deep inclined shaft with irregular frozen wall liner in an infinite elastic medium, subjected to a non-uniform stresses. This solution is based on: (i) the conformal mapping of a doubly connected domain of prescribed shape onto a circular ring by an appropriate numerical optimization scheme which is achieved by MATLAB program, (j) the complex variable method. For a particular case, the distributions of hoop stress and radial displacement inside frozen wall are presented with analytical solution. The accuracy of stress functions is verified by comparison of different number of coefficients in stress functions and it show that the number of coefficients k≥15 is advised. The validity of analytical solution is verified by a comparison between its result and that obtained from the finite element program ANSYS. Noting that the shaft excavation can be regarded as relaxation of initial ground stresses around shaft excavation boundary, this paper presents the mechanical problem as two separate problems: (i) initial stresses and displacements model where stresses keep constant and displacements equal zero around frozen wall and ground, (j) stresses relaxation model with relaxed normal and shear stresses acted on excavated boundary. The solution is worked out by the principle of superposition in elastic medium. The Young's modulus ratio (β) of the frozen wall to the ground reflects the influence of ground to frozen wall, and the influence decreases with the growth of β. The proposed solution can provide analytical basis for frozen wall design and be used as a quick-solver for back-analysis of in situ stresses and displacements.
1. Introduction Artificial ground freezing (AGF) is a special construction method, which is used to freeze the soft or water-bearing ground into an enclosed structure, to increase the strength and stability of ground soil or rocks, and to make them impervious to water seepage.1 AGF has been widely used in mine construction, urban underground tunnel, and deep foundation excavation and so on. Tunnel and shaft usually have a circular cross section because of its structural advantages due to the large surrounding pressures. Analytical solutions for excavation of circular holes in elastic and homogeneous materials subjected to uniform stresses or non-uniform stresses are discussed in textbooks of elastic theory and rock mechanics due to their importance as basic tools for introducing the subject of underground tunnel excavations.2,3 Einstein and Schwartz4 perhaps at the first time worked out the solution for excavation and support problem of deep circular lined tunnel in an infinite elastic medium subjected to nonuniform stresses. However, they ignored the effect of support delaying. As supplementary, with consideration of the delaying effect of support ⁎
and different contact conditions on the interface between ground and lining, Carranza-Torres et al.5 found an analytical elastic solution for deep circular lined tunnel. Mason and Stacey6 considered three contact conditions on the interface between ground and support, and then analyzed the problem of concrete spraying support in deep circular tunnel subjected to a uniform shear stress at infinity. In most cases of tunnel and inclined shaft excavation, non-uniform stresses are acted on models. As an effective method, complex variable method can be used to solve the problem of circular or irregular tunnels or shaft subjected to uniform or non-uniform stresses. Poulos and Davis7 analyzed the tunnels with and without lining in his work. Pender8 and Sagaseta9 presented the stress-displacement solution for circular lined tunnel without regard to gravity as well. Li and Wang10,11 achieved the solutions for deep circular lined tunnel subjected to uniform internal pressures, and stress release of excavation to support and water seepage were taken into consideration. Based on these solutions, Yang and Wang12 obtained the solutions for circular shaft frozen wall and ground subjected to non-uniform stresses, and they considered the stresses relaxation model of shaft excavation.
Corresponding author. E-mail address: [email protected] (Q. Wang).
http://dx.doi.org/10.1016/j.ijrmms.2017.10.008 Received 27 September 2016; Received in revised form 7 May 2017; Accepted 16 October 2017 Available online 05 November 2017 1365-1609/ © 2017 Elsevier Ltd. All rights reserved.
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Although the circular cross section is the first choice for inclined shaft excavation, the horseshoe-like section (a section with semi-circular arch roof and straight side wall and inverted arch floor) is the most common section of inclined shaft limited to the excavation condition and complex ground condition. It is assumed that an irregular boundary in the z-plane can be mapped conformally to a circle in the ζplane.13,14 The exact analytical solution for circular lined tunnel and simple irregular tunnel, like lined oval tunnel, could be worked out by complex variable theory, while only closed-form solution could be obtained for relatively complex tunnel. Based on this theory, Gercek15 presented the solution for stresses around tunnel with conventional shapes for the first time. However, Gercek did not compute the displacement solution for the ground, and did not consider the support either. Exadaktylos et al.16,17 presented the closed-form solutions for stresses and displacements around semi-circular and notched circular tunnels. Exadaktylos also did not take support into consideration. Huo and Bobet18 analyzed the regularities of stresses distribution around deep lined tunnel in earthquake region subjected to shear stresses. Clearly, all published works present solutions for circular supported tunnel or irregular tunnel without support. Different with them, frozen wall is a kind of doubly connected domain, which plays a significant role in the construction of shaft or tunnel, however there is no published work presenting the analytical solution for deep lined irregular tunnel or shaft subjected to non-uniform stresses (frozen wall is a kind of temporary liner). Now because of the irregularity of inclined shaft frozen wall, the design of inclined shaft frozen wall is based on method of circular shaft and empirical method, there is no theoretical solution for irregular frozen wall. Unfortunately, many inclined shafts were flooded because of weak frozen wall in western China. Therefore addressing the deep irregular inclined frozen shaft, this paper presents a closed-form solution of stresses and displacements for irregular frozen wall and ground with consideration of the interaction between frozen wall and ground.
This paper is conducted based on three assumptions presented as follows: (1) Frozen wall and ground are regarded as elastic and homogeneous materials. (2) Frozen wall and ground are assumed to be fully contacted, that is to say, the radial stress and shear stress are continuous at the interface between frozen wall and ground, as well as the radial and tangential displacements. (3) The variation of body forces are ignored, as well as the frost heave, i.e., field quantities (stresses and displacements) remain unchanged after freezing. 2.1. Stresses relaxation model The excavation of a deep inclined frozen shaft can be regarded as stresses relaxation around excavation boundary in an infinite elastic medium. Considering the stress path of shaft excavation, the problem of this paper can be divided into two new problems, stresses relaxation model (see Fig. 1a) and initial model, and the final solution is worked out by the principle of superposition in the elasticity. stresses relaxation model shows the excavation problem of an irregular inclined frozen shaft in the elastic material subjected to normal stress σrF0 and shear stress τrθF0 at the inner boundary, and normal displacements ux=0 and uy=0 at the far-boundary(r→∞). Initial model shows the initial stresses and displacements of frozen wall and ground where stresses keep constant and ux=0 and uy=0. Fig. 1a also can be separated by two parts, mechanical model for frozen wall as shown in Fig. 1b and mechanical model for ground as shown in Fig. 1c. Fig. 1b shows the frozen wall problem with normal stress and shear stress at inner and outer boundaries. Fig. 1c shows the ground problem with normal stress and shear stress at inner boundary, and normal displacements ux=0 and uy=0 at the far-boundary (r→∞). In Fig. 1, σrF0 and τrθF0 are radial and shear stresses on inner boundary of frozen wall, respectively. σrF1 and τrθF1 are radial and shear stresses on the interface of frozen wall and ground, respectively. In this paper, complex variable method is used to work out the solution for irregular frozen wall and ground in an infinite elastic material. In the complex variable method,19–21 the solution is expressed in terms of two analytical functions φ(z) and ψ(z), the stresses are related to these functions by the following equations
2. Problem statement This paper refers to the excavation problem of an irregular inclined frozen shaft in an infinite elastic medium under plane strain condition, subjected to non-uniform stresses, the cross section of inclined frozen shaft is shown in Fig. 1a, which consists of frozen wall and ground. In Fig. 1a, r0 and r1 are inner and outer radii of semi-circular arch of frozen wall, respectively, h0 and h1 are inner and outer heights of straight side wall, respectively, rd0 and rd1 are inner and outer radii of inverted arch, respectively.
⎧ σy + σx = 4 Re φ′ (z ) ⎨ σy − σx + 2iτxy = 2[z φ′ ′ (z ) + ψ′ (z )] ⎩
(1)
Fig. 1. Mechanical model for (a) stresses relaxation considering the interaction of frozen wall and ground with relaxation stresses on inner boundary of frozen wall in an infinite plane which can be separated into two parts: (b) frozen wall model and (c) ground model.
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and the conversion of stress functions in the z-plane and ζ-plane are presented as follows19:
where σx and σy are horizontal and vertical stresses at any point, respectively. The displacements are related to these functions by
(u x + iu y ) =
1 κφ (z ) − zφ′ (z ) − ψ (z ) 2G
φ (ζ ) = φ1 (z ) = φ1 [ω (ζ )],
= ψ1 [ω (ζ )],}
(2)
Φ (ζ ) = φ1′ (z ) = φ′ (ζ )/ ω′ (ζ ), Ψ (ζ ) = ψ1′ (z )
where G is the shear modulus of the elastic material, and κ is related to Poisson's ratio µ by κ= (3–4µ) for plane strain condition and κ= (3-µ)/ (1+µ) for plane stress condition. In this paper, plane strain condition is assumed. There are displacements or surface tractions prescribed along the boundaries. For displacements, the quantity (ux+iuy) is prescribed by Eq. (2). The surface tractions are related to the complex stress functions φ(z) and ψ(z) by
[φ (z ) + zφ′ (z ) + ψ (z )] = i
∫ (fx + ify ) ds
ψ (ζ ) = ψ1 (z )
= ψ′ (ζ )/ ω′ (ζ ),} ′
Φ′ (ζ ) = φ1′ (z )⋅ω′ (ζ ) (10) Stress and displacement variables in polar coordinates are related to stress functions and they are presented as follows:
⎧ σr + σθ = σx + σy = 4 Re[Φ (ζ )],
(3)
⎨ σθ − σr + 2iτrθ = ⎩
where fx and fy are horizontal and vertical boundary surface forces, respectively.
ur + iuθ =
2ζ 2 [ω (ζ )Φ′ (ζ ) + ω′ (ζ ) Ψ (ζ )] r 2ω′ (ζ )
ζ ω′ (ζ ) (u x + iu y ) r ω′ (ζ )
(11)
(12)
2.2. Boundary conditions 2.4. Scaling of variables
The boundary condition of inner frozen wall is expressed in terms of the integral of the surface relaxation forces, integrated along the boundary, which is described by
[φ1 (z ) + zφ1′ (z ) + ψ1 (z )]
z1
=i
∫ (fx + ify ) ds
In this paper, load and deformation quantities are considered as dimensionless by normalizing them with respect to properly selected variables. The unscaled quantities will be denoted by the use of the tilde symbol (~) on top of the variable. The radius of inner boundary r͠0 is used to scale the radius and deformation r͠i , u͠ i and u͠ θ respectively, i.e.,
(4)
where φ1(z) and ψ1(z) are two analytical functions of frozen wall in the z-plane, z1 is a point on the inner boundary of frozen wall. Along the interface between frozen wall and ground (z=z2), the radial and shear stresses, the radial and tangential displacements are continuous, so the stress boundary condition at z=z2 is written as
[φ1 (z ) + zφ1′ (z ) + ψ1 (z )]
z = z2
= [φ2 (z ) + zφ2′ (z ) + ψ2 (z )]
ur =
where φ2(z) and ψ2(z) are two analytical functions of ground in the zplane, z2 is a point on the interface of frozen wall and ground. The displacement boundary condition at z=z2 is written as
Γ [κ1 φ1 (z ) − zφ1′ (z ) − ψ1 (z )]
z = z2
= κ2 φ2 (z ) − zφ2′ (z ) − ψ2 (z ) z = z
2
σr =
(6)
3
(7)
As for plane strain condition, the parameters Г, κ1 and κ2 are given by
Γ=
E2 (1 + μ1) , κ1 = 3 − 4μ1 , κ2 = 3 − 4μ 2 E1 (1 + μ 2 )
(8)
where E1 and E2 are Young's moduli of the frozen wall and ground, and µ1 and µ2 are Poisson's ratios of the frozen wall and ground.
In order to work out these complex functions in terms of the known boundary conditions, mapping equation z=ω(ζ) is used to map the region in the z-plane with z = x + iy = ρe iϕ onto the region in the ζplane with ζ=ξ+iη=reiθ. A circular or irregular shape in the z-plane can be mapped conformally onto a circle or circular ring in the ζ-plane ,19 and the solution for circular shape could be worked out easily. Xu19 presented that the angle ϕ of the normal direction of a point in the z-plane, is related to the mapping function ω(ζ) by
ζ ω′ (ζ ) r ω′ (ζ )
(14)
∼ and The Poisson's ratio of frozen wall and ground, the quantities μ 1 ∼ are dimensionless, and they are scaled by μ 2 ∼,μ =μ ∼ μ1 = μ (17) 1 2 2
2.3. Conformal mapping
e−iϕ =
τ͠ F σ͠ F σ͠ r σ͠ τ͠ , σθ = θ , τrθ = rθ , σrF = r , τrθF = rθ p͠ p͠ p͠ p͠ p͠
Similarly, the horizontal and vertical stresses, acted on the frozen ∼ ∼ wall, the quantities fx and fy are scaled as follows: ∼ ∼ fy f fx = x , fy = ͠p p͠ (15) ∼ The Young's moduli of frozen wall and ground, the quantities E1 and ∼ ∼ E2 , the shear moduli of frozen wall and ground, the quantities G1 and ∼ G2 , respectively, are scaled as follows: ∼ ∼ ∼ ∼ E E G G E1 = 1 , E2 = 2 , G1 = 1 , G2 = 2 p͠ p͠ p͠ p͠ (16)
The last boundary condition is that the infinite boundary z=z3 is fixed, i.e., the radial and tangential displacements are both equal to zero. This gives
κ2 φ2 (z ) − zφ2′ (z ) − ψ2 (z ) z = z = 0
(13)
where the subscript ii=0 represents inner boundary of frozen wall, ii=1 represents interface between frozen wall and ground, ii=2 represents infinite boundary of ground. The vertical stress in far-field p͠ is used to scale normal and shear stresses, σ͠ r , σ͠ θ and τ͠ rθ in frozen wall and ground, respectively, i.e.,
(5)
z = z2
u͠ r u͠ r͠ r͠ , uθ = θ , r = , rii = ii (ii = 0, 1, 2) r͠0 r͠0 r͠0 r͠0
3. The mapping representation of frozen wall The references20,21 presented closed-form solutions for mapping a complex shape hole onto a circle. While different to conventional tunnels, the frozen wall is a kind of temporary support, and it is a doubly connected domain in a finite plane. The mapping of simply connected domain problem is not suitable for doubly connected domain any more. Considering that, Lv20 raised a solution for mapping the irregular doubly connected domain onto a circular ring, this method takes both the inner and outer boundaries into account, and keep the mapping
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Fig. 2. Conformal mapping of an irregular frozen wall shape to a circular ring: (a) the frozen wall shape before conformal mapping in the z-plane; (b) the frozen wall shape after conformal mapping in the ζ-plane.
n
precision at a relatively high level for inner and outer boundaries. The frozen wall region in the z-plane can be mapped conformally onto a circular ring in the ζ-plane, bounded by the circles |ζ|=r0 < 1 and |ζ|=r1=1, as shown in Fig. 2. The circle |ζ|=r0 corresponds to the inner boundary of frozen wall, and the circle |ζ|=r1 corresponds to the outer boundary of frozen wall. The appropriate conformal transformation can be described by Laurent series expansion as follows20:
rBj = R [cos (βBj − αBj ) +
∑ Ck ζ −k ) k=0
Eq. (18) gives the conformal transformation function, so the problem turns to solve R and Ck (k=0, 1, 2…n), these parameters can be solved by optimization method which is stated as follows: In the z-plane, x-axis is regarded as the symmetric axis for frozen wall, as shown in Figs. 2a, and (m+1) points are selected along inner and outer boundaries, respectively. It is assumed that the mapped point A0′ is coincident with A0, and this gives
(18)
R=
where R is a positive real number, which reflects the size of the ring, Ck are a series of parameters, which reflect the accuracy of transformation, k=0,1,2,…,n. It is assumed that any point Aj (rAj, αAj) lying on the inner boundary of frozen wall in the z-plane is conformally mapped onto the circle |ζ|=r0 in the ζ-plane corresponding to A′j (r0, βAj), and any point Bj (rBj, αBj) lying on the outer boundary of frozen wall in the z-plane is conformally mapped onto the circle |ζ|=r1 in the ζ-plane corresponding to B′j (r1, βBj). Then in terms of Eq. (18), the conformal transformation of inner boundary of frozen wall is given by
r1 n r0 + ∑k = 0 Ck r −k
m
f=
∑ Ck r0−k e−ikβAj)
+
k=0
(20)
The imaginary part of Eq. (19) is given by n
sin(βAj − αAj ) −
∑ Ck r0−k sin(kβAj + αAj) = 0 k=0
(21)
and the real part of Eq. (19) is given by n
rAj = R [cos (βAj − αAj ) +
∑ Ck r0−k cos(kβAj + αAj)] k=0
k=0
⎩
⎣
ai ≤ x i ≤ bi , (i = 1, 2, ... ,m)
2
⎫
k=0
⎦⎭
(27)
The main idea of this method is to collect m points as the initial compound, and it is better that m ≥ 2n. Compute the objective function value of each point and compare them one by one, remove the bad point which provides the maximum value for the objective function f (X), and replaces it by a new point which can improve the objective
n
∑ Ck r1−k sin(kβBj + αBj) = 0
n
⎧
∑ ⎨rBj* − R ⎡⎢cos (βBj − αBj) + ∑ Ck r1−k cos(kβBj + αBj) ⎤⎥ ⎬
gi (X ) ≥ 0, (i = 1, 2, ... ,m)
(22)
Similarly, the imaginary part and the real parts of Eq. (20) are given by Eqs. (23) and (24), respectively.
sin(βBj − αBj ) −
k=0
where f can reach minimum value when m > n. When the boundary shape of frozen wall can be mapped precisely by Eq. (18), the Ck series, which give the minimum value of Eq. (26), are accurate solution for Eq. (18), and f=0. Otherwise, when the boundary shape of frozen wall cannot be mapped precisely by Eq. (18), the Ck series are closed-form solution for Eq. (18), and f > 0. The complex optimum method is applied in this paper to solve the mapping function addressing the problem stated as follows .22–24 Work out the design value X={x1, ×2, …, xn}T which can provide the minimum value of the target function f(X) and meet the constraints presented by
n
∑ Ck r1−k e−ikβBj)
Ck ⎤⎫ cos(kβAj + αAj ) ⎥ ⎬ r0k ⎦⎭
(26)
and the outer boundary mapping is given by
rBj e iαBj = R (r1 e iβBj +
⎣
⎩
j=0
(19)
2
n
⎧
∑ ⎨rAj* − R ⎡⎢cos (βAj − αAj) + ∑ j=0
n k=0
(25)
According to Eqs. (22) and (24), when k=0,1,2,…,n, and n is a limited positive real number, the left side of (22) and (24) is not strictly equal to the right side, respectively. While the optimization method can be used to solve R and Ck series (k=0, 1, 2,…, n), the objective function is given by
m
rAj e iαAj = R (r0 e iβAj +
(24)
k=0
n
z = ω (ζ ) = R (ζ +
∑ Ck r1−k cos(kβBj + αBj)]
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expressed by
function and meet the constraints, and then make objective function f (X) close to the optimum solution gradually. In this paper, MATLAB software is used to accomplish the procedure and work out the optimal solution. Chen21 and Xu19 indicated that Ck should meet the relationship n ∑k = 0 Ck < 1, while Lv20 improved this condition by
F (z ) = i
kCk < 1
(fx + ify ) ds = iσx Im(zP − z A) + σy Re (zP − z A)
(36)
where zA is the start point of integration with angle 0 on the inner boundary of frozen wall, and zP is the ending point of integration which is an arbitrary point on the inner boundary of frozen wall, as shown in Fig. 2(a). The integration must obey the anti-clockwise direction. The integrated function F(z) in Eq. (36) can be expressed by Fourier series expansion21
(28)
k=0
zP
A
n
∑
∫z
4. Solution for irregular frozen wall and ground ∞
F (z ) =
As presented in Section 2, in elastic medium, the solution for irregular frozen wall and ground is divided into two part, solution for initial stresses and displacements problem and solution for stresses relaxation problem
k=0
(37)
k=0 1
2π
1
2π
where Ak = 2π ∫0 F (z ) e−ikθdθ , Bk = 2π ∫0 F (z ) e ikθdθ In the ζ-plane, after conformal mapping, the boundary conditions (4)–(7) are expressed by
4.1. Initial stresses and displacements of frozen wall and ground
φ1 (σ1) +
Based on the third assumption, before excavation, the ground is undisturbed, so the initial stresses of frozen wall and ground are given by 0 σx0 = −λp , σy0 = −p , τxy =0
∞
∑ Ak e ikθ + ∑ Bk e−ikθ
φ1 (σ2) +
(29)
ω (σ1) φ ′ (σ1) + ψ1 (σ1) = ω′ (σ1) 1
∞
∞
∑ Ak e ikθ + ∑ Bk e−ikθ k=0
k=0
ω (σ2) ω (σ2) φ ′ (σ2) + ψ1 (σ2) = φ2 (σ2) + φ ′ (σ2) + ψ2 (σ2) ω′ (σ2) 1 ω′ (σ2) 2
in Cartesian coordinates, and
(39)
1+λ λ−1 p− p cos 2ϕ σr0 = − 2 2
(30)
1+λ λ−1 p+ p cos 2ϕ 2 2
(31)
σθ0 = −
τrθ0 =
λ−1 p sin 2ϕ 2
Γ [κ1 φ1 (σ2) −
κ2 φ2 (σ3) −
(32)
ζ ω′ (ζ ) ζ ω′ (ζ ) )) ))orϕ = −arcsin(Im( r ω′ (ζ ) r ω′ (ζ )
∞
= L−n ζ −n + L−(n − 1) ζ −n + 1 + ⋯+L−1 ζ −1 +
∑ Lk ζ k k=0
(33)
(42)
It can be obtained that L−1=C1 when n=1; L−1=C1 and L−2=C2 when n=2; and when n≥3,
Complex stress functions in the ζ-plane can be considered as functions of ζ, the functions of frozen wall are given by ∞
(34)
and the functions of ground are given by
⎧ L−n = Cn, ⎪ L−(n − 1) = Cn − 1, ⎨ Cj − 1 − k j−2 ⎪ L−(n − j + 1) = ∑k = 1 (j − 1 − k ) r 2(j − k ) L−(n − k + 1) + Cn − j + 1 ⎩
(43)
−4 −2n ⎧ L0 = C0 + C1 r L−2 +⋯+(n − 1) Cn − 1 r L−n , ⎪ L1 = 1 + C1 r −4L−1 + 2C2 r −6L−2 +⋯+nCn − 1 r −2(n + 1) L−n , ⎨ ⎪ Li = ∑ik−=11 kCk r −2(k + 1) Li − k + 1 + ∑nk = i kCk r −2(k + 1) L−(k − i + 1) ⎩
(44)
where j=3,4,…,n, and i=2,3,…,n. After substituting the ω (ζ )/ ω′ (ζ ) with boundary values, Eqs. (38)–(41) can be simplified with equations without denominator. The infinite algebraic equations containing unsolved parameters are obtained by setting the coefficients of eikθ at the left side equal to the coefficients of eikθ at the right side where k=-n,…,−1,0,1,…,n. The infinite algebraic equations are given by
∞
−k ⎧ φ2 (ζ ) = ∑k = 0 fk ζ , ⎨ ψ2 (ζ ) = ∑∞ h ζ −k k=0 k ⎩
(41)
ω (ζ ) ζ + C0 + C1 ζ −1 + C2 ζ −2+⋯+Cn ζ −n = 1 − C1 r −4ζ 2 − 2C2 r −6ζ 3−⋯−nCn r −2nζ n + 1 ω′ (ζ )
4.2. Solution for stresses relaxation model
∞
ω (σ3) φ ′ (σ3) − ψ2 (σ3) = 0 ω′ (σ3) 2
where σ1 and σ2 represent points on the inner and outer boundaries of frozen wall, and σ3 represents a point on the infinite boundary of ground in the ζ-plane. In Eqs. (38)–(41), the coefficient ω (ζ )/ ω′ (ζ ) can be expanded by
Therefore, the initial stresses of frozen wall and ground can be worked out by Eqs. (29)–(32).
k −k ⎧ φ1 (ζ ) = ∑k = 1 ak ζ + ∑k = 1 bk ζ , ∞ ∞ ⎨ ψ1 (ζ ) = ∑k = 0 ck ζ k + ∑k = 0 dk ζ −k ⎩
ω (σ2) ω (σ2) φ ′ (σ2) − ψ1 (σ2)] = κ2 φ2 (σ2) − φ ′ (σ2) − ψ2 (σ2) ω′ (σ2) 1 ω′ (σ2) 2 (40)
in polar coordinates. And the vertical and horizontal displacements u x0 and u y0 are both equal to 0.19 0 In Eqs. (29)–(32), σx0 , σy0 and τxy are the initial horizontal, vertical 0 and shear stresses, respectively; σr , σθ0 and τrθ0 are initial radial, hoop and shear stresses, respectively, and the tensile stress is ordered as positive; λ is the lateral pressure coefficient; p is the vertical stress added on the model; ϕ is the angle of normal direction of a point in the z-plane, the relation of ϕ and mapping function ω(ζ) is given by Eq. (9), so ϕ is obtained from Eq. (9) by
ϕ = arccos(Re(
(38)
(35)
The parameters ak, bk, ck, dk, fk and hk in Eqs. (34) and (35)are all real because of the symmetry along x-axis. The integral of surface tractions at the right side of Eq. (4) can be
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∞
2(k − 1) − ∑k = 1 L−(k + n + 1) kbk r0−2(k + 1) ] + cn r0−2n} r0−n r = r ⎧ {bn + [ ∑k = 1 kLk − n − 1 ak r0 0 ⎪ ⎪ = Bn , ⎪ ∞ ∞ −2(k + 1) + ∑k = 1 Lk + m − 1 kak r02(k − 1) ] + dm r0−2m} r0n r = r ⎪ {am + [− ∑k = 1 kL−k − m − 1 bk r0 0 ⎪ ⎪ = Am , ⎪ b ∞ c ⎪ {b + [ ∑∞ kL 2(k − 1) − ∑k = 1 L−(k + n + 1) k 2(kk+ 1) ] + 2nn } k − n − 1 ak r1 k=1 ⎪ n r1 r1 r = r1 ⎪ ⎪ fk n ⎪ = fn − ∑k = 1 L−(k + n + 1) k 2(k + 1) , r1 ⎪ r = r1 ⎪ ⎪ b ∞ ∞ d 2(k − 1) − ∑k = 1 kL−k − m − 1 2(kk+ 1) ] + 2mm } ⎪ {am + [ ∑k = 1 Lk + m − 1 kak r1 r1 r1 r = r1 ⎪ ⎪ ⎪ fk ∞ hm , = 2m − ∑k = 1 kL−k − m − 1 2(k + 1) r1 r1 ⎨ r = r1 ⎪ ⎪ ∞ L−(k + n + 1) kbk ∞ c 2(k − 1) ] − 2nn } − ∑k = 1 ⎪ Γ {κ1 bn − [ ∑k = 1 kLk − n − 1 ak r1 r1 r 12(k + 1) ⎪ r = r1 ⎪ ⎪ = κ f + ∑n L−(k + n + 1) kfk , 2 n k=1 ⎪ r 12(k + 1) r = r1 ⎪ ⎪ b ∞ kL ∞ ⎪ Γ {κ1 am + [ ∑k = 1 −k2(−km+−1)1 k − ∑k = 1 Lk + m − 1 kak r12(k − 1) ] − d2mm } r1 r1 ⎪ r = r1 ⎪ ⎪ ∞ kL−k − m − 1 fk hm , ⎪ = − r12m + ∑k = 1 r 2(k + 1) 1 r = r1 ⎪ n ⎪ −2(k + 1) = 0, ⎪ κ2 fn + ∑k = 1 L−(k + n + 1) kfk r2 r=r2 ⎪ ∞ −2(k + 1) −2m kL f r h r − = 0(m = 1, 2, ⋯; n = 0, 1, 2, ⋯) ∑ ⎪ k = 1 −k − m − 1 k 2 m 2 r=r2 ⎩
Fig. 3. The original shape and mapping shape of frozen wall.
function. The mapping solution for frozen wall after scaled is presented by Eq. (47). The radius of inner boundary r0=0.33951 which is calculated by Eq. (47).
z = ω (ζ ) = 3.07296(ζ − 0.02009 − 0.00136ζ −1 + 0.00208ζ −2 − 0.00049ζ −3 + 0.00006ζ −4 )
(47)
(45) Fig. 3 presents the conformal mapped shape of frozen wall with Eq. (47). The solid line represents the original shape of frozen wall, and the dashed line represents the mapped shape of frozen wall. It can be concluded that the mapping shape and the original shape of frozen wall are seen to agree well, especially the semi-circular arch section of frozen wall. The mapping boundaries of inner and outer frozen wall are basically in coincidence with the original boundaries. There are certain difference located on side wall and invert section outside frozen wall, because the mapping is based on the assumption that the point A0 at original inner boundary of frozen wall is in coincidence with A0′ at mapped inner boundary. Nevertheless, the conformal mapping based on Eq. (47) could perform well. In order to assess the solution accuracy for stresses and displacements of frozen wall, three number of coefficients ak, bk, ck, dk, fk and hk are discussed, i.e., k=10, 15 and 20. Fig. 4 gives the comparison of stresses and displacements along inner frozen wall considering different coefficients number, and it shows that the coefficients number of stress functions does have influence on the solution of frozen wall. Fig. 4a shows that the radial stress varies evidently at the region where angel θ=90°~180° when k=10, and the peak stress reaches ± 0.3 (dimensionless). The radial stress changes little when k=15 or 20, especially when k=20. According to the theoretical solution, radial stress of inner boundary is zero when inner boundary of frozen wall is free of any constraint of force or displacement. Fig. 4b shows that coefficients number has relatively less influence on hoop stress and radial displacement of inner boundary of frozen wall. However the hoop compression stress grows up at the region angel θ=170°~180°, which violates the variation law of hoop stress. Therefore k≥15 should be considered for obtaining stress and displacement solutions accurately. In this paper k=20 is adopted, that is to say, totally 123 unknown coefficients are needed to be solved. Fig. 5 shows the distributions of radial displacement and hoop stress within frozen wall, which are obtained with analytical solution, the radial displacement is presented at the left side of frozen wall and the hoop stress is presented at the right side because of the symmetry of frozen wall. The outward displacement and tensile stress are ordered as positive. Fig. 5 shows that the displacement at arch roof and floor is bigger than the value at side wall of frozen wall, and the displacement
which contain four parts corresponding to four boundary conditions and are deduced by Eqs. (38)–(41). Note that the Lk and L-k series in Eq. (45) are related to radius r in the ζ-plane. Eq. (45) is a linear equation group with infinite equations, which contain the unsolved coefficients ak, bk, ck, dk, fk and hk, so the linear equation group can be simplified as (46)
MX = N
where M is a n×n matrix of coefficients ak, bk, ck, dk, fk and hk, X is a n×1 vector of coefficients ak, bk, ck, dk, fk and hk, and X={a0, a1,…, aii, b1,…, bii, c0, c1,…, cii, d1,…, dii, f0, f1,…, fii, h1,…, hii}T, N is a known n×1vector, and N={B1,…,Bii,A0,A1,…Aii, 0,…0}T with n=6ii+3 and ii is an assumed number on the premise of accuracy. 5. Example application and comparison of analytical and numerical solutions 5.1. Example application An inclined shaft excavation is taken as an example, the cross section of frozen shaft can be regarded as a horseshoe, as shown in Fig. 1a. Table 1 shows the dimension parameters of frozen wall and ground. The ∼ Young's moduli of frozen wall and ground are E1=300 MPa and ∼ E2 =100 MPa, respectively, and the Poisson's ratios of frozen wall and ∼ =0.24 and μ ∼ =0.35, respectively. The volume-weight of ground are μ 1 2 frozen/unfrozen soil γ͠ =20 kN/m3, and the lateral pressure coefficient λ=0.65. The parameters as shown in Table 1 are all explained in Fig. 1 a. In Eq. (18), it is assumed that the parameter n equals 4 (that is C0~C4), and on the basis of complex optimum method, the program MATLAB is used to get the closed-form solution for conformal mapping Table 1 Dimension parameters of tunnel and frozen wall. Parameters
r͠ 0 /m
r͠ 1 /m
h͠ 0 /m
h͠ 1/m
r͠ d0 /m
r͠ d1 /m
Values
3.5
10.5
2.6
5.8
7
14
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Fig. 4. The solutions of (a) radial stresses and (b) hoop stresses and radial displacements of frozen wall along the inner boundary with coefficients number k=10, 15 and 20.
Fig. 6. Numerical simulation model for example problem.
whole model to show the region of frozen wall clearly. The far boundaries in numerical model locate at a distance of 30 times the radius of the tunnel semi-circular arch. The left boundary of numerical model is set to be symmetric. The bottom boundary is set with normal displacement equaling 0. Horizontal pressure, which equals lateral pressure coefficient timing the weight of overlying soil, is acted on the right boundary, and vertical pressure, which equals the weight of overlying soil, is acted on the top boundary. Gravity is ignored in this model as with analytical model. The horizontal and vertical pressures are given by
Fig. 5. Distribution of radial displacement (left) and hoop stress (right) of frozen wall with analytical solution.
at floor is greater than the value at arch roof when the lateral pressure coefficient λ is less than 1. The stress concentrates on the side wall and corner place of frozen wall, that is to say, the side wall and corner place are the first to yield. 5.2. Comparison of analytical and numerical solutions with ANSYS program
pv = γh, ph = λγh
A corresponding elastic numerical model under plane strain condition was established by the ANSYS program to verify the validity of analytical solution for stresses and displacements in frozen wall and ground. The numerical model, as shown in Fig. 6, was established based on the parameters stated in Table 1. Only half of the problem has been modeled because of the symmetry of the problem. The ANSYS model considers three-stage sequence corresponding to the deduction of the analytical solution in this paper, i.e., on a first stage the boundary conditions are acted and the initial equilibrium state is computed, and on a second stage the frozen wall is formed, and on a third stage the tunnel is excavated and the equilibrium state is again computed. Fig. 6 represents the stage when the tunnel is excavated, so the soil and elements in tunnel are removed. Fig. 6 only presents part of the
(48)
where pv and ph represent the vertical and horizontal pressures respectively; h is the depth. Take p=1 for example, the stress function coefficients of frozen wall and ground can be solved by substituting the design parameters of tunnel and frozen wall as well as parameters of mapping function into Eq. (45) and then stresses and displacements for frozen wall and ground are worked out by substituting these coefficients into Eqs. (11) and (12). The load and deformation quantities in this chapter are considered dimensionless. The diagrams of Fig. 7 represent the analytical and numerical results for frozen wall along inner and outer boundaries after excavation. Fig. 7a represents the distribution of hoop stress along inner and outer boundaries for frozen wall as a function of θ. Fig. 7b represents the distribution of radial displacement along inner and outer boundaries for 68
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Fig. 7. Distribution of (a) hoop stresses and (b) radial displacements for frozen wall along the inner and outer boundaries as obtained with analytical and numerical solutions.
frozen wall as a function of θ. The analytical and numerical results for frozen wall quantities after excavation, as represented by solid/dashed lines and squares/circles in Fig. 7, are seen to agree well. The hoop stress around outer boundary of frozen wall changes slightly, the maximum stress difference is close to 0.2. The difference existing at the side wall and invert sections are caused by conformal mapping, as shown in Fig. 3. The diagrams of Fig. 8 represent the analytical and numerical results for frozen wall and ground along certain radial paths after excavation. Fig. 8a represents the distribution of hoop and radial stresses for frozen wall and ground as a function of radial distance (a radial scanline with respect to the spring line of the frozen wall at angle θ=0°). Fig. 8b represents the distribution of hoop and radial stresses for frozen wall and ground as a function of radial distance at angle θ=90°. Fig. 8c represents the distribution of hoop and radial stresses for frozen wall and ground as a function of radial distance at angle θ=180°. Fig. 8d represents the distribution of radial displacements for frozen wall and ground as a function of radial distance at angles θ=0°, θ=90° and θ=180°. The analytical and numerical results for frozen wall and ground quantities after excavation, as represented by solid/dashed lines and squares/circles/triangles in Fig. 8, are seen to agree wall. The diagrams in Figs. 7 and 8 present that the absolute hoop stress
of point D is smaller than the value of point A, and the radial displacement of point D is greater than the value of point A, which means point D is easier to break than point A. 6. Results and discussion The load and deformation quantities in this chapter are considered dimensionless. In this chapter, the law of stress and displacement distributions on frozen wall are analyzed with respect to depth, lateral pressure coefficient, thickness of frozen wall and the ratio of Young's moduli (the ratio of frozen wall Young's modulus to ground Young's modulus) 6.1. Influence of depth The diagrams in Fig. 9 represent the relationship between the increase of scaled hoop stress and the increase of depth which is presented with scaled pressure. Under the same parameters of frozen wall, the hoop stress increase with the growth of pressure. The angle θ changes from 0° to 180° means from point A to point D. It can be concluded from Fig. 9a that pressure has great effect on hoop stress at inner side section (from point B to point C) and inner semi-circular arch Fig. 8. Distribution of stresses at (a) angle θ=0°; (b) angle θ=90°; (c) angle θ=180° and (d) displacements at angles θ=0°, θ=90° and θ=180° for frozen wall and ground as obtained with analytical and numerical solutions.
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Fig. 9. Distribution of hoop stresses for frozen wall along (a) the inner boundary and (b) the outer boundary under different scaled pressure.
Fig. 10. Distribution of radial displacements for frozen wall along (a) the inner boundary and (b) the outer boundary under different scaled pressure.
Fig. 11. The variations of (a) hoop stresses and (b) radial displacements of different key points on inner boundary of frozen wall with lateral pressure coefficient.
Fig. 12. The variations of (a) hoop stresses and (b) radial displacements of different key points on inner boundary of frozen wall with frozen wall thickness.
increase of scaled radial displacement and the increase of scaled pressure. Under the same design values of frozen wall, the radial displacement along inner and outer boundaries of frozen wall increase with the growth of pressure. The diagrams in Fig. 10 present that pressure has great effect on radial displacement at arch and invert arch sections, the radial displacement changes little at the side section of
section (from point A to point B), but it has little effect on hoop stress in the center region of inner invert section (region close to point D). Different with the inner boundary of frozen wall, the changes of pressure have similar effect on all points along outer boundary of frozen wall, as shown in Fig. 9b. The diagrams in Fig. 10 represent the relationship between the 70
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Fig. 13. The variation of hoop stress and radial displacement of (a) points A and D and (b) point B and C on inner boundary of frozen wall with Young's modulus ratio β.
point A and point D. The radial displacement of point D is greater than the value of point A. However, the radial displacement difference between point A and point D decreases gradually with the increase of ratio β. The difference is 5.91 when ratio β equals1, and the difference decreases to 0.37 when ratio β grows to 19. Fig. 13b presents that different with point A and point D, the hoop stresses of point B and point C increase with the growth of ratio β, however the radial displacements of point B and point C decrease with the growth of ratio β. The radial displacement of point B is almost equal to the value of point C. Fig. 13a and b present that the hoop stress and radial displacement vary obviously when β < 9, and the variation slow down when β > 9, especially after β > 13
frozen wall. The difference of displacement between point A and point D changes little with the increase of depth. It can be concluded from Figs. 9 and 10 that the hoop stress and radial displacement increase linearly with the increase of depth. 6.2. Influence of lateral pressure coefficient The diagrams in Fig. 11 represent the influence of lateral pressure coefficient λ, which increases from 0.1 to 2.8, on stress and displacement at key points on the inner and outer boundaries of frozen wall. The lateral pressure coefficient λ has a significant influence on hoop stress and radial displacement along inner and outer boundaries of frozen wall, hoop stress and radial displacement change linearly with the lateral pressure coefficient λ. Fig. 11a presents the linear increase of hoop compressive stress of point A and point D, and linear decrease of hoop stress of point B and point C. Point A and point D are under tension when λ < 0.4, but point C is under tension when λ > 2.8, and it can be concluded that all points along inner boundary are under compression when 0.4 < λ < 2.8. Fig. 11b presents the linear increase of radial displacement of point B and point C, and linear decrease of radial displacement of point A and point D. Point B deforms outwards when λ < 0.4, and point A deforms outwards when λ > 2.4, that is to say all points deforms inwards when 0.4 < λ < 2.4. As the tensile strength of frozen soil or rock is much less than the compressive strength, the tensile region of frozen wall should call more attention.
7. Conclusions This paper presents a closed-form elastic solution for stresses and displacements along irregular frozen wall and ground in an infinite elastic medium under plane strain condition, subjected to non-uniform stresses. The problem considers the interaction of frozen wall and ground on the basis of excavation model. The solution can be used to analyze the distribution of stresses and displacements of irregular frozen wall, and provide analytical foundation for safety tunnel or shaft freezing design and construction. It is a valuable method for understanding how certain parameters affect the stresses and displacements and figuring out key parameters which could significantly affect the stresses and displacements. Numerical results have been presented with regard to analytical model. By comparison of analytical and numerical results, it is demonstrated that the analytical solution can be effectively used not only for studying the stresses and displacements around irregular freezing shaft but also for studying the stresses and displacements around any lined irregular tunnel or shaft construction subjected to uniform or nonuniform stresses. Based on the mapping method for doubly connected domain problem presented in,20 the irregular frozen wall shape (doubly connected domain) was mapped conformally by using compound optimization design method, and the solution also can be easily used for other problems with irregular support or tunnel excavation boundary. The stresses and displacements vary linearly with the increase of pressure and lateral pressure coefficient λ under elasticity. Lateral pressure coefficient λ could reverse the stresses and displacements state, for example the stress state changes from tension to compression and the displacement state changes from positive to negative, so λ has great influence on frozen wall design and tunnel excavation. Stresses and displacements perform differently with the variation of frozen wall thickness, hoop stress of side wall region (from point B to point C) are sensitive to wall thickness while radial displacement of semi-circular arch and invert arch sections are sensitive to wall thickness. To some extent, the ratio β represents the constraint effect of ground to frozen wall, the greater the value of β, the smaller the effect of ground to
6.3. Influence of wall thickness The diagrams in Fig. 12 represent the influence of wall thickness on stresses and displacements at key points on the inner and outer boundaries of frozen wall. Fig. 12a presents that the increase of wall thickness can decrease the hoop stress for frozen wall, especially for point B and point C. Fig. 12b presents that the increase of wall thickness can decrease the radial displacement for frozen wall, especially for point A and point D. Different with the change of hoop stress, the radial displacement of point B and point C change slightly with the increase of wall thickness. Under this condition, the radial displacement of point B is approximately equal to the value of point C. with the increase of wall thickness, the effect of wall thickness to stress and displacement decrease gradually. 6.4. Influence of Young's modulus ratio The ratio of frozen wall Young's modulus E1 to ground Young's modulus E2 is marked as β. The diagrams in Fig. 13 represent the variation of hoop stresses and radial displacements of key points on the inner and outer boundaries of frozen wall. It can be concluded that the ratio β has significant influence on hoop stress and radial displacement. Fig. 13a presents that the hoop stresses of point A and point D decrease with the growth of ratio β, as well as the radial displacements of 71
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