# The Longitudinal Deformation Profile of a Rock Tunnel: An Elastic Analysis

Authors: Yonghong Wang; Wen Du; Guohui Zhang; Yang Song
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

The longitudinal deformation profile (LDP) is the profile of wall displacement versus the distance from the tunnel face. To develop LDP equations, numerical methods and in situ experiments have been used to obtain the deformation of a tunnel in three-dimensional space. However, extant approaches are inadequate in terms of explaining the mechanical relation between the wall displacement and the conditions of a tunnel (e.g., properties of rock). In this paper, an analytical approach is proposed to develop a new LDP equation. First, on the basis of the axisymmetric elastic model of a tunnel, a closed-form solution of wall displacement is derived. Then, a new LDP equation is presented according to the solution developed above; the coefficientβ, defined as the ratio of the effective range of the “face effect” to the radius of the tunnel, is proposed for the first time. Finally, a case study is proposed to validate the practicability of this equation.


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## Body

## 1. Introduction

The distribution of wall displacement around a tunnel face is one of the most important topics in civil and mining engineering. According to the convergence-confinement method, the tunnel face itself carries a significant portion of the load of its surroundings, and this “face effect” decreases naturally along the tunnel axis [1]. Consequently, tunnel wall displacements increase gradually as the distance from the face increases. The profile of wall displacement versus the distance from the face is defined as the longitudinal deformation profile (LDP), as shown in Figure 1.Figure 1
(a) A circular opening excavated in rock mass. (b) A form of the longitudinal deformation profile.
(a)(b)In practice, the wall displacements are usually normalized with respect to the maximum wall displacement at the far end of the tunnel in order to eliminate the effects of elastic properties (E and μ) and tunnel radius (r).The mathematical difficulties associated with calculating the wall displacements in three-dimensional space have motivated researchers to the utilization of numerical methods and in situ experiments. LEE [2] proposed a dimensionless equation for a circular opening excavated under hydrostatic stress conditions as follows:(1)u∗=0.5⋅1−tanh13−x∗2,where u∗ = u/umax, in which u is the wall displacement at a specified x and umax is the ultimate radial displacement far away from the tunnel face, and x∗ = x/r0, in which x is the distance from the face and r0 is the radius of the tunnel.Carranza-Torres and Fairhurst [1] proposed another empirical equation suggested by E. Hoek as follows:(2)u∗=1+exp−x∗1.10−1.7.Based on linear elastic numerical models, Unlu and Gercek [3] developed equations considering Poisson’s ratio as follows:(3)u∗=u0∗+Aa1−expBax∗ahead of the face,x<0,u∗=u0∗+Ab1−BbBb+x∗2behind the face,x>0,u0∗=0.22μ+0.19at the face,x=0,Aa=−0.22μ−0.19,Ba=0.73μ+0.81,Ab=−0.22μ+0.81,Bb=0.39μ+0.65,where u0∗ is the accumulated displacement release rate at the face and μ is Poisson’s ratio.According to numerical models of elastic-perfect-plastic (EPP) rock masses, Vlachopoulos and Diederichs [4] proposed a series of exponential functions as follows:(4)u∗=u0∗expx∗ahead of the face,x<0,u∗=1−1−u0∗exp−1.5x∗R∗behind the face,x>0,u0∗=13exp−0.15R∗at the face,x=0,where R∗ = Rp/r0, in which Rp is the plastic radius of the tunnel and r0 is the radius of the tunnel.Based on numerical models of EPP rock masses, Basarir et al. [5] related to the radial displacement of a tunnel to the rock mass rating (RMR) was as follows:(5)u=a⋅r0⋅RMRbcx/2r0ahead of the face,x<0,u=a⋅r0⋅RMRbx2r0cbehind the face,x>0,where a, b, and c are coefficients assessed by depth, RMR is the rock mass rating, and r0 is the radius of the tunnel.Shen et al. [6] performed an in situ test as part of the Dagangshan Hydropower Station project in China, and the authors suggested an equation as follows:(6)s=s02tanhax2r0−c+1,where s0 is the ultimate wall displacement observed at the far end of the tunnel, a and c are the best-fit coefficients given the measured data, and r0 is the radius of the tunnel.Alejano et al. [7] developed a simplified approximate equation of the plastic radius for a tunnel excavated in strain-softening (SS) rock masses as follows:(7)Rss∗=REPP∗+qu/pi∗0.0244⋅GSI−0.53⋅1−m/50−1/0.0018⋅GSI+0.0783⋅1+Kp,where Rss∗is the plastic radius for a tunnel excavated in SS material, REPP∗ is the plastic radius for a tunnel excavated in EPP material, qu is the unconfined compressive strength, Kp is the passive Earth pressure, pi is the internal pressure, GSI is the geological strength index, and m is the Hoek–Brown parameter of intact rock.Li et al. [8] performed another in situ test in the laboratory of Beishan Exploration Tunnel in China, and the authors recommended the following equation as the analytical equation of LDP on the basis of (2):(8)u∗=1+exp−x∗H1H2,where H1 and H2 are the best-fit coefficients given the measured data.The foregoing equations, which are acquired using numerical methods and in situ experiments, only reflect the quantitative relations between wall displacement and distance from the face. As a matter of fact, the wall displacement is independent from the distance from the face but strictly depends on the properties of rock and the radius of the tunnel [5, 7]. To reveal the core principle of wall displacement, it is essential to develop an analytical solution of wall displacement which considers those two aspects.In this paper, a closed-form solution of wall displacement is developed based on an axisymmetric elastic model for the rock behind the face; then, an analytical equation of LDP is presented according to the solution developed above; finally, an in situ test is introduced to confirm the practicability of the proposed LDP equation.


## 2. Model Conditions

### 2.1. Geometry

In equations (3)–(5), the LDPs ahead of the face (i.e., x < 0) and the LDPs behind the face (i.e., x > 0) are expressed separately. In equations (1), (2), (6), and (8), only the LDPs behind the face are proposed. The LDP ahead of the face reflects the wall deformation that has already been released before any excavation.It is not necessary to determine the exact LDP equation ahead of the face because only the LDP behind the face is considered in the calculation of actual wall displacement when the supports are installed [9, 10]. Therefore, this paper focuses on wall displacement behind the face, and the model for the rock behind the face is presented as follows.For a circular opening excavated in rock, the rock behind the face is considered to be a hollow cylinder located in a cylindrical coordinate system with limited length and radius, as shown in Figure2. The length of the model is L, the inner radius is r1, and the outer radius is r2. The rock is considered to be elastic homogeneous material, Poisson’s ratio is μ, and elastic modulus is E.Figure 2
Axisymmetric model for the rock behind the face.


### 2.2. Boundary Conditions

Carranza-Torres and Fairhurst [1] pointed out that the load carried by the face decreases as the tunnel advances, which also means that the load carried by the surrounding rock increases along the tunnel. It is assumed that the load carried by the surrounding rock is expressed by a nonlinear monotonic increasing function, for instance, the sine function.Figure3 illustrates the hypothetical load carried by the surrounding rock. The load applied on the inner circular boundary (i.e., r = r1) is expressed as pin = p1·sin(z·π/2L), where z is the distance from the face, L is the length of the model (see Figure 2), and p1 is the maximum pressure applied on the inner circular boundary at the end of the model (i.e., z = L). The load applied on the outer circular boundary (i.e., r = r2) is expressed as pout = p2·sin(z·π/2L), where p2 is the maximum pressure applied on the outer circular boundary at the end of the model.Figure 3
Hypothetical load carried by the surrounding rock.


## 2.1. Geometry

In equations (3)–(5), the LDPs ahead of the face (i.e., x < 0) and the LDPs behind the face (i.e., x > 0) are expressed separately. In equations (1), (2), (6), and (8), only the LDPs behind the face are proposed. The LDP ahead of the face reflects the wall deformation that has already been released before any excavation.It is not necessary to determine the exact LDP equation ahead of the face because only the LDP behind the face is considered in the calculation of actual wall displacement when the supports are installed [9, 10]. Therefore, this paper focuses on wall displacement behind the face, and the model for the rock behind the face is presented as follows.For a circular opening excavated in rock, the rock behind the face is considered to be a hollow cylinder located in a cylindrical coordinate system with limited length and radius, as shown in Figure2. The length of the model is L, the inner radius is r1, and the outer radius is r2. The rock is considered to be elastic homogeneous material, Poisson’s ratio is μ, and elastic modulus is E.Figure 2
Axisymmetric model for the rock behind the face.


## 2.2. Boundary Conditions

Carranza-Torres and Fairhurst [1] pointed out that the load carried by the face decreases as the tunnel advances, which also means that the load carried by the surrounding rock increases along the tunnel. It is assumed that the load carried by the surrounding rock is expressed by a nonlinear monotonic increasing function, for instance, the sine function.Figure3 illustrates the hypothetical load carried by the surrounding rock. The load applied on the inner circular boundary (i.e., r = r1) is expressed as pin = p1·sin(z·π/2L), where z is the distance from the face, L is the length of the model (see Figure 2), and p1 is the maximum pressure applied on the inner circular boundary at the end of the model (i.e., z = L). The load applied on the outer circular boundary (i.e., r = r2) is expressed as pout = p2·sin(z·π/2L), where p2 is the maximum pressure applied on the outer circular boundary at the end of the model.Figure 3
Hypothetical load carried by the surrounding rock.


## 3. Solutions to the Proposed Model

### 3.1. The Solution Satisfies the Harmonic Equation

For the axisymmetric problem, a stress function satisfying the biharmonic equation is required, which is(9)∇2∇2φ=∂2∂r2+1r∂∂r+∂2∂z2∂2∂r2+1r∂∂r+∂2∂z2φ=0.At the beginning, it is more rational to find a solution that satisfies the harmonic equation:(10)∇2φ=∂2∂r2+1r∂∂r+∂2∂z2φ=0.It is assumed that the function takes the separated form:(11)φ=grcosλz,where λ = π/2L. Substituting equation (11) into equation (10), and note that gr depends only on r, then(12)∇2φ=∂2∂r2+1r∂∂r+∂2∂z2grcosλz=∂2gr∂r2+1r∂gr∂r⋅cosλz−λ2⋅gr⋅cosλz=d2dr2+1rddr−λ2grcosλz=0.Definingf(ρ) = gr, where ρ = λr,λ = π/2L, the differential components of gr can be written in terms of f(ρ) as(13)dgrdr=dfρdρ⋅dρdr=dfρdρ⋅λ,d2grdr2=λ⋅ddfρ/dρdρ⋅dρdr=λ2⋅d2fρdρ2.Thus, equation (12) can be expressed as(14)∇2φ=λ2⋅d2fρdρ2+λ⋅λρ⋅dfρdρ−λ2fρ⋅cosλz=λ2⋅d2fρdρ2+1ρdfρdρ−fρ⋅cosλz=0.To find the solution to equation (14), a more general expression called Bessel equation with imaginary arguments is introduced as(15)d2fρdρ2+1ρdfρdρ−1+m2ρ2fρ=0.When the integral orderm of equation (15) is 0, the correspondence answer is the solution to equation (10). The solution to equation (15) is known as(16)fρ=c1Imρ+c2Kmρ,Imρ=∑k=0∞1k!m+k!ρ2m+2k,Kmρ=π2⋅i⋅eπi/2m⋅Hm1ρ,where Im functions are m-order Bessel functions with imaginary arguments and Km functions are m-order Hankel functions with imaginary arguments.Therefore, the solution to equation (10) can be written as(17)φ=a0I0λr+b0K0λrcosλz,or(18)φ=a0I0ρ+b0K0ρcosλz,where a0 and b0 are coefficients determined by boundary conditions, ρ = λr, λ = π/2L, and L is the length of the model.


### 3.2. Stress Function Satisfies the Biharmonic Equation

The stress function satisfying the biharmonic equation is developed on the basis of equation (18). The stress function is further extended as(19)φ=a0I0ρ+b0K0ρ+a1ρI1ρ+b1ρK1ρcosλz,where a1ρI1(ρ) is linearly independent from a0I0(ρ) and b1ρK1(ρ) is linearly independent from b0K0(ρ), in which ρ = λr, λ = π/2L, and L is the length of the model.Substituting equation (19) into equation (9), the biharmonic equation is found to be(20)∇2∇2φ=∇2∇2a0I0ρ+b0K0ρcosλz+∇2∇2a1ρI1ρ+b1ρK1ρcosλz=∇2∇2a1ρI1ρ+b1ρK1ρcosλz,while(21)∇2∇2ρI1ρcosλz=∇2∂2∂r2+1r∂∂r+∂2∂z2ρI1ρcosλz=∇2d2dr2+1rddr−λ2ρI1ρ⋅cosλz,(22)∇2ρK1ρcosλz=∇2∂2∂r2+1r∂∂r+∂2∂z2ρK1ρcosλz=∇2d2dr2+1rddr−λ2ρK1ρ⋅cosλz.For theIη and Kη functions with general order η, the following differential cyclic relations are given:(23)ddρρηIηρ=ρη⋅Iη−1ρ,ddρρ−ηIηρ=ρ−η⋅Iη+1ρ,(24)ddρρηKηρ=−ρη⋅Kη−1ρ,ddρρ−ηKηρ=−ρ−η⋅Kη+1ρ.According to equation (23), the differential relations between I1 and I0 are(25)ddρρI1ρ=I1ρ+ρ⋅ddρI1ρ=ρ⋅I0ρ,ddρI0ρ=I1ρ.Equation (25) can be written explicitly as(26)ddρI1ρ=I0ρ−I1ρρ,ddρI0ρ=I1ρ.Equation (26) is derived further by differential operator d/dr as(27)ddrI1λr=λI0λr−I1λrr,ddrI0λr=λI1λr.The first-order and second-order differential equations ofλrI1(λr) are expressed as(28)ddrλrI1λr=λI1λr+λrλI0λr−I1λrr=λ2rI0λr,d2dr2λrI1λr=ddrλ2rI0λr=λ2I0λr+λrI1λr.Substituting equation (28) into equation (21) gives(29)∇2∇2ρI1ρcosλz=∇2d2λrI1λrdr2+1rdλrI1λrdr−λ3rI1λr⋅cosλz=∇2λ2I0λr+λ3rI1λr+λ2I0λr−λ3rI1λr⋅cosλz=∇22λ2I0λrcosλz.Similarly, the first-order and second-order differential equations ofλrK1(λr) can be found as(30)ddrλrK1λr=λK1λr+λr−λK0λr−K1λrr=−λ2rK0λr,d2dr2λrK1λr=ddr−λ2rK0λr=−λ2K0λr−λrK1λr.Substituting equation (30) into equation (22) gives(31)∇2ρK1ρcosλz=∇2d2λrK1λrdr2+1rdλrK1λrdr−λ3rK1λr⋅cosλz=∇2−λ2K0λr+λ3rK1λr−λ2K0λr−λ3rK1λr⋅cosλz=∇2−2λ2K0λrcosλz.Finally, equation (20) can be simplified as(32)∇2∇2φ=∇2∇2a1λrI1λr+b1λrK1λrcosλz=∇22λ2a1I0λrcosλz−2λ2b1K0λrcosλz=2λ2⋅∇2a1I0λrcosλz−b1K0λrcosλz.Recalling that the solution satisfies the harmonic equation, as suggested in equations (18), (32) equals zero. Equation (19) is proved to be the stress function satisfying the biharmonic equation.


### 3.3. Radical Displacements

The radical displacement (behind the tunnel face) in terms of the stress function is solved by(33)ur=−1+μE∂2∂r∂zφ.Substituting equation (19) into equation (33), the radical displacement is expressed explicitly as(34)ur=−1+μE∂2∂r∂za0I0ρ+ρ⋅a1I1ρ+b0K0ρ+ρ⋅b1K1ρcosλz=1+μE⋅λsinλz⋅ddra0I0λr+a1λrI1λr+b0K0ρ+ρ⋅b1K1ρ=1+μEsinλz⋅a0λ2I1λr+a1λ3rI0λr−b0λ2K1λr−b1λ3rK0λr=1+μEsinλz⋅a0A0+a1A1+b0B0+b1B1,where A0 = λ2I1, A1 = λ3rI0, B0 = −λ2K1, and B1 = −λ3rK0; I0 and I1 are the Bessel functions with imaginary arguments; K0 and K1 are the Hankel functions with imaginary arguments; and a0, a1, b0, and b1 are the coefficients determined by boundary conditions.Identically, the stresses can also be solved by the stress function (Appendix A).


## 3.1. The Solution Satisfies the Harmonic Equation

For the axisymmetric problem, a stress function satisfying the biharmonic equation is required, which is(9)∇2∇2φ=∂2∂r2+1r∂∂r+∂2∂z2∂2∂r2+1r∂∂r+∂2∂z2φ=0.At the beginning, it is more rational to find a solution that satisfies the harmonic equation:(10)∇2φ=∂2∂r2+1r∂∂r+∂2∂z2φ=0.It is assumed that the function takes the separated form:(11)φ=grcosλz,where λ = π/2L. Substituting equation (11) into equation (10), and note that gr depends only on r, then(12)∇2φ=∂2∂r2+1r∂∂r+∂2∂z2grcosλz=∂2gr∂r2+1r∂gr∂r⋅cosλz−λ2⋅gr⋅cosλz=d2dr2+1rddr−λ2grcosλz=0.Definingf(ρ) = gr, where ρ = λr,λ = π/2L, the differential components of gr can be written in terms of f(ρ) as(13)dgrdr=dfρdρ⋅dρdr=dfρdρ⋅λ,d2grdr2=λ⋅ddfρ/dρdρ⋅dρdr=λ2⋅d2fρdρ2.Thus, equation (12) can be expressed as(14)∇2φ=λ2⋅d2fρdρ2+λ⋅λρ⋅dfρdρ−λ2fρ⋅cosλz=λ2⋅d2fρdρ2+1ρdfρdρ−fρ⋅cosλz=0.To find the solution to equation (14), a more general expression called Bessel equation with imaginary arguments is introduced as(15)d2fρdρ2+1ρdfρdρ−1+m2ρ2fρ=0.When the integral orderm of equation (15) is 0, the correspondence answer is the solution to equation (10). The solution to equation (15) is known as(16)fρ=c1Imρ+c2Kmρ,Imρ=∑k=0∞1k!m+k!ρ2m+2k,Kmρ=π2⋅i⋅eπi/2m⋅Hm1ρ,where Im functions are m-order Bessel functions with imaginary arguments and Km functions are m-order Hankel functions with imaginary arguments.Therefore, the solution to equation (10) can be written as(17)φ=a0I0λr+b0K0λrcosλz,or(18)φ=a0I0ρ+b0K0ρcosλz,where a0 and b0 are coefficients determined by boundary conditions, ρ = λr, λ = π/2L, and L is the length of the model.


## 3.2. Stress Function Satisfies the Biharmonic Equation

The stress function satisfying the biharmonic equation is developed on the basis of equation (18). The stress function is further extended as(19)φ=a0I0ρ+b0K0ρ+a1ρI1ρ+b1ρK1ρcosλz,where a1ρI1(ρ) is linearly independent from a0I0(ρ) and b1ρK1(ρ) is linearly independent from b0K0(ρ), in which ρ = λr, λ = π/2L, and L is the length of the model.Substituting equation (19) into equation (9), the biharmonic equation is found to be(20)∇2∇2φ=∇2∇2a0I0ρ+b0K0ρcosλz+∇2∇2a1ρI1ρ+b1ρK1ρcosλz=∇2∇2a1ρI1ρ+b1ρK1ρcosλz,while(21)∇2∇2ρI1ρcosλz=∇2∂2∂r2+1r∂∂r+∂2∂z2ρI1ρcosλz=∇2d2dr2+1rddr−λ2ρI1ρ⋅cosλz,(22)∇2ρK1ρcosλz=∇2∂2∂r2+1r∂∂r+∂2∂z2ρK1ρcosλz=∇2d2dr2+1rddr−λ2ρK1ρ⋅cosλz.For theIη and Kη functions with general order η, the following differential cyclic relations are given:(23)ddρρηIηρ=ρη⋅Iη−1ρ,ddρρ−ηIηρ=ρ−η⋅Iη+1ρ,(24)ddρρηKηρ=−ρη⋅Kη−1ρ,ddρρ−ηKηρ=−ρ−η⋅Kη+1ρ.According to equation (23), the differential relations between I1 and I0 are(25)ddρρI1ρ=I1ρ+ρ⋅ddρI1ρ=ρ⋅I0ρ,ddρI0ρ=I1ρ.Equation (25) can be written explicitly as(26)ddρI1ρ=I0ρ−I1ρρ,ddρI0ρ=I1ρ.Equation (26) is derived further by differential operator d/dr as(27)ddrI1λr=λI0λr−I1λrr,ddrI0λr=λI1λr.The first-order and second-order differential equations ofλrI1(λr) are expressed as(28)ddrλrI1λr=λI1λr+λrλI0λr−I1λrr=λ2rI0λr,d2dr2λrI1λr=ddrλ2rI0λr=λ2I0λr+λrI1λr.Substituting equation (28) into equation (21) gives(29)∇2∇2ρI1ρcosλz=∇2d2λrI1λrdr2+1rdλrI1λrdr−λ3rI1λr⋅cosλz=∇2λ2I0λr+λ3rI1λr+λ2I0λr−λ3rI1λr⋅cosλz=∇22λ2I0λrcosλz.Similarly, the first-order and second-order differential equations ofλrK1(λr) can be found as(30)ddrλrK1λr=λK1λr+λr−λK0λr−K1λrr=−λ2rK0λr,d2dr2λrK1λr=ddr−λ2rK0λr=−λ2K0λr−λrK1λr.Substituting equation (30) into equation (22) gives(31)∇2ρK1ρcosλz=∇2d2λrK1λrdr2+1rdλrK1λrdr−λ3rK1λr⋅cosλz=∇2−λ2K0λr+λ3rK1λr−λ2K0λr−λ3rK1λr⋅cosλz=∇2−2λ2K0λrcosλz.Finally, equation (20) can be simplified as(32)∇2∇2φ=∇2∇2a1λrI1λr+b1λrK1λrcosλz=∇22λ2a1I0λrcosλz−2λ2b1K0λrcosλz=2λ2⋅∇2a1I0λrcosλz−b1K0λrcosλz.Recalling that the solution satisfies the harmonic equation, as suggested in equations (18), (32) equals zero. Equation (19) is proved to be the stress function satisfying the biharmonic equation.


## 3.3. Radical Displacements

The radical displacement (behind the tunnel face) in terms of the stress function is solved by(33)ur=−1+μE∂2∂r∂zφ.Substituting equation (19) into equation (33), the radical displacement is expressed explicitly as(34)ur=−1+μE∂2∂r∂za0I0ρ+ρ⋅a1I1ρ+b0K0ρ+ρ⋅b1K1ρcosλz=1+μE⋅λsinλz⋅ddra0I0λr+a1λrI1λr+b0K0ρ+ρ⋅b1K1ρ=1+μEsinλz⋅a0λ2I1λr+a1λ3rI0λr−b0λ2K1λr−b1λ3rK0λr=1+μEsinλz⋅a0A0+a1A1+b0B0+b1B1,where A0 = λ2I1, A1 = λ3rI0, B0 = −λ2K1, and B1 = −λ3rK0; I0 and I1 are the Bessel functions with imaginary arguments; K0 and K1 are the Hankel functions with imaginary arguments; and a0, a1, b0, and b1 are the coefficients determined by boundary conditions.Identically, the stresses can also be solved by the stress function (Appendix A).


## 4. Case Study

### 4.1. An LDP Equation Based on the Proposed Displacement Solution

As a closed-form solution of displacement, the maximum displacement without considering the accumulated displacement ahead of the face can be solved from equation (34) as(35)umax=−1+μE⋅a0A0+a1A1+b0B0+b1B1.Without considering the accumulated displacement release rate ahead of the face, an LDP equation can be proposed as(36)ubehind∗=uumax=sinπz2L,where z is the distance from the face and L is the effective range of the “face effect.”At the face (z = 0), the accumulated displacement release rate is estimated by equation (3) in the following form:(37)u0∗=0.22μ+0.19.Considering the accumulated displacement release rate at the face, a total LDP equation can be proposed as(38)u∗=u0∗+1−u0∗⋅ubehind∗=u0∗+1−u0∗⋅sinπz2L=u0∗+1−u0∗⋅sinπ2β⋅zr1=u0∗+1−u0∗⋅sinπ2β⋅z∗,where u0∗is the displacement release rate at the face which can be estimated by equation (37), z∗ = z/r1, z is the distance from the face, r1 is the radius of the tunnel, β = L/r1, and L is the effective range of the “face effect.”The coefficientβ in equation (38) reflects the range of the “face effect”; this viewpoint has not hitherto been proposed in the literature. In future research, β could be related to the characteristics of rock mass (e.g., quality of rock mass and buried depth) as an empirical variant.


### 4.2. Practicability of the Proposed LDP Equation

The Beishan Exploration Tunnel is an underground laboratory in tonalite rock for the storage of radioactive waste in the Northwest of China. In this section, the in situ test performed by Li et al. [8] in the Beishan Exploration Tunnel is introduced to confirm the practicability of equation (38). The deformation of rock in this experiment was believed to be elastic.As shown in Figure4, a horseshoe-shaped tunnel (neighbor tunnel) had been excavated before the test. Three boreholes (B1, B2, and B3) were drilled from the neighbor tunnel to the sidewall of the test tunnel. In each borehole, three measuring points were installed to record rock deformation. The height and width of the test tunnel are 2.8 m and 3.0 m, respectively. According to Su et al. [11], the equivalent radius of the test tunnel is calculated by(39)req=h+w4=2.8+3.04=1.45m.Figure 4
The (a) plan view and (b) elevation view of the test performed by Li et al. [8].
(a)(b)The measuring points in borehole 1 and borehole 2 were installed immediately after the boreholes were created, but the measuring points in borehole 3 were installed until the advancing face of the test tunnel had reached borehole 3. Therefore, only the wall displacements recorded by the first measuring point in borehole 1 (B1-1) and the first measuring point in borehole 2 (B2-1) were supposed to be the complete deformation of rock.Figure5 shows the LDP data presented by Li et al. [8], and the effective range of the “face effect” is 10 times the equivalent radius. Poisson’s ratio of rock, which is not reported by Li et al. [8], is estimated as 0.30 (the typical Poisson’s ratio of rock). By substituting μ = 0.30 and β = 10.0 into equation (38), the LDP equation is expressed as(40)u∗=0.256+0.744⋅sinπ20⋅x∗.Figure 5
Comparisons between the data reported by Li et al. [8] and the predicted LDPs.The LDPs predicted by equations (1)–(3) and (40) are also illustrated in Figure 5. Equations (1)–(3) clearly overestimate the displacement release rate (u∗) for a short distance from the face (e.g., x∗<3), and equation (40) is reasonably consistent with the LDP data presented by Li et al. [8].


## 4.1. An LDP Equation Based on the Proposed Displacement Solution

As a closed-form solution of displacement, the maximum displacement without considering the accumulated displacement ahead of the face can be solved from equation (34) as(35)umax=−1+μE⋅a0A0+a1A1+b0B0+b1B1.Without considering the accumulated displacement release rate ahead of the face, an LDP equation can be proposed as(36)ubehind∗=uumax=sinπz2L,where z is the distance from the face and L is the effective range of the “face effect.”At the face (z = 0), the accumulated displacement release rate is estimated by equation (3) in the following form:(37)u0∗=0.22μ+0.19.Considering the accumulated displacement release rate at the face, a total LDP equation can be proposed as(38)u∗=u0∗+1−u0∗⋅ubehind∗=u0∗+1−u0∗⋅sinπz2L=u0∗+1−u0∗⋅sinπ2β⋅zr1=u0∗+1−u0∗⋅sinπ2β⋅z∗,where u0∗is the displacement release rate at the face which can be estimated by equation (37), z∗ = z/r1, z is the distance from the face, r1 is the radius of the tunnel, β = L/r1, and L is the effective range of the “face effect.”The coefficientβ in equation (38) reflects the range of the “face effect”; this viewpoint has not hitherto been proposed in the literature. In future research, β could be related to the characteristics of rock mass (e.g., quality of rock mass and buried depth) as an empirical variant.


## 4.2. Practicability of the Proposed LDP Equation

The Beishan Exploration Tunnel is an underground laboratory in tonalite rock for the storage of radioactive waste in the Northwest of China. In this section, the in situ test performed by Li et al. [8] in the Beishan Exploration Tunnel is introduced to confirm the practicability of equation (38). The deformation of rock in this experiment was believed to be elastic.As shown in Figure4, a horseshoe-shaped tunnel (neighbor tunnel) had been excavated before the test. Three boreholes (B1, B2, and B3) were drilled from the neighbor tunnel to the sidewall of the test tunnel. In each borehole, three measuring points were installed to record rock deformation. The height and width of the test tunnel are 2.8 m and 3.0 m, respectively. According to Su et al. [11], the equivalent radius of the test tunnel is calculated by(39)req=h+w4=2.8+3.04=1.45m.Figure 4
The (a) plan view and (b) elevation view of the test performed by Li et al. [8].
(a)(b)The measuring points in borehole 1 and borehole 2 were installed immediately after the boreholes were created, but the measuring points in borehole 3 were installed until the advancing face of the test tunnel had reached borehole 3. Therefore, only the wall displacements recorded by the first measuring point in borehole 1 (B1-1) and the first measuring point in borehole 2 (B2-1) were supposed to be the complete deformation of rock.Figure5 shows the LDP data presented by Li et al. [8], and the effective range of the “face effect” is 10 times the equivalent radius. Poisson’s ratio of rock, which is not reported by Li et al. [8], is estimated as 0.30 (the typical Poisson’s ratio of rock). By substituting μ = 0.30 and β = 10.0 into equation (38), the LDP equation is expressed as(40)u∗=0.256+0.744⋅sinπ20⋅x∗.Figure 5
Comparisons between the data reported by Li et al. [8] and the predicted LDPs.The LDPs predicted by equations (1)–(3) and (40) are also illustrated in Figure 5. Equations (1)–(3) clearly overestimate the displacement release rate (u∗) for a short distance from the face (e.g., x∗<3), and equation (40) is reasonably consistent with the LDP data presented by Li et al. [8].


## 5. Conclusions

This paper proposes a closed-form solution of wall displacement on the basis of an axisymmetric elastic model in a cylindrical coordinate system. It was hypothesized that the load carried by the surrounding rock mass takes the form of a sine function.Considering the stress function in the separated form written as equation (11), the closed-form solution of wall displacement to the proposed model is derived. According to this closed-form solution, an analytical LDP equation (equation (38)) is presented, where the coefficient β, defined as the ratio of the effective range of the “face effect” to the radius of the tunnel, is proposed for the first time.Finally, the in situ test performed by Li et al. [8] is introduced; the LDP predicted by equation (38) is found to be consistent with the data presented in this test, and it is acceptable to estimate the LDP by the analytical equation proposed in this paper.



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*Source: 6684035-2021-04-05.xml*