# Optimization of Measuring Points Layout around a Tunnel in the Transversely Isotropic Rock Mass

Authors: Zhizeng Zhang; Yongtao Li; Weili Yang; Shunchuan Wu; Xiaoli Liu; Bing Zhang
Journal: Shock and Vibration
 (2021)

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## Abstract

The arrangement of measuring points has a great impact on the uniqueness and accuracy of the back analysis of displacement. To explore the arrangement method of measuring points in the anisotropic rock mass, the principle of maximum displacement is applied to study the optimization of measuring points layout around a tunnel in the transversely isotropic rock mass. Firstly, the criterion of maximum displacement is deduced. Secondly, the variations of displacement with angle and radius are analyzed, respectively, and two key lateral pressure coefficients are obtained. Thirdly, some principles of measuring points arrangement are summarized according to the criterion of maximum displacement. Finally, an example is given to prove the correctness of these principles. These principles can be used to guide the arrangement of displacement measuring points in the transversely isotropic rock mass, especially suitable for shaft excavation in horizontal sedimentary rock.


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## Body

## 1. Introduction

The arrangement of measuring points has a great impact on the uniqueness and accuracy of the back analysis of displacement [1]. Therefore, many researchers have studied the optimal arrangement of displacement measuring points. Kernevez et al. [2] found that it is not that the bigger the number of measuring points the better; instead, it was found that the spatial location of the measuring points is more important. Cividini et al. [3] and Venclik [4] discussed the issue of the number of measuring points and demonstrated that the number, quality, and spatial location of the measuring points should be considered comprehensively. Sun et al. [5] studied that the effect of increasing the number of measuring points on the back analysis results was not significant. The measuring point arrangement is also affected by other factors, such as sensitivity and the size of the measurement value. Zhou et al. [6] studied the in situ monitoring of auxiliary tunnel in deep mine. However, there is no unified standard to guide the layout of measuring points.Currently, there are three principles that have formed theoretical systems: the principle of maximum displacement [7], the principle of maximum sensitivity [8], and the principle of minimum variance [9]. Xiang et al. [10, 11] proposed a criterion for optimizing the layout of measuring points considering the comprehensive influence of many factors and applied this criterion to the back analysis of material parameters in the field of geotechnical engineering and achieved good results. Su et al. [12] discussed the application of the least-square method in the displacement back analysis and used the three-dimensional finite element method in combination with the least-square method to obtain parameters that are more in line with the actual engineering. The application of the optimized arrangement of measuring points in underground engineering has also been studied [13–17].The rock mass was considered to be isotropic in the above studies. However, with the development of rock mechanics theory and experimental research, it is gradually recognized that anisotropy is an important property of rock mass. However, the measuring points layout around a tunnel in the anisotropic rock mass was rarely studied.Based on the criterion of maximum displacement, this paper attempts to make a preliminary study on the optimal arrangement of measuring points in the back analysis of transversely isotropic rock masses and find some principles of displacement measuring points arrangement. Transversely isotropic rock masses mean that the rock mass has the same material constants in any direction parallel to the layer, while the material parameters in the vertical layers are different.


## 2. The Criterion of Maximum Displacement

Any measurement has an error, and it exists in lots of scientific experiments. Therefore, the error is usually used to characterize the accuracy of the measured data. There are four main sources of error. The first case is the device error, which is mainly caused by instrument accuracy. The second case is the environmental error, which is caused by the inconsistency of various environmental factors and the required standard status. The third case is human error, which is caused by the measurer himself. The fourth case is the method error, which is caused by the approximation of function types.This article assumes that measurement errors only come from device errors.x∗ is set to be the exact displacement of a measuring point and x is set to be the measured value. Therefore, x∗−x is called the absolute error of x and x∗−x/x∗ is called the relative error. Because the exact value is often unknown in practice, x∗−x/x is often used as the relative error of x.If the reading accuracy of the instrument isδ, then the absolute error function satisfies(1)x∗−x≤δ,where δ is the absolute error bound ofx and δ/x is the relative error bound of x.If there are multiple displacement measuring points around the tunnel, the average relative error bound can be expressed as(2)e=1n∑i=1nδxi=δn∑i=1n1xi,where xi is the measured value of the i-th measuring point; n is the total number of measuring points. The expression of e∗ is set to(3)e∗=1n∑i=1n1xi.Becauseδ is a fixed value, so the value of e can be directly evaluated by the value of e∗without having to consider the specific value of δ. If the measuring points are arranged in the area where the absolute value of the displacement is larger, then the smaller relative errors will be obtained.Set the number of displacement measuring lines on the cross section of the tunnel ask, the number of measuring points on each line as l, and the displacement measurement value of each measuring point as uij. The formula of the average value of the relative error for the i-th measuring line can be written as(4)ei=δn∑j=1l1uij,or(5)ei∗=1n∑j=1l1uij.As can be seen from equation (5), the direction with the smallest ei∗ is the most suitable, and the magnitude of the relative error can be effectively reduced with the increased value of uij.Based on the above analysis, the criterion for the optimal arrangement of displacement measuring points can be simply summarized as that the place with a larger absolute displacement value is more suitable to arrange measuring points.


## 3. Displacement Analytical Solution

Figure1 presents an idealized model for a circular tunnel in the transversely isotropic rock mass.Figure 1
Model for a circular tunnel in transverse isotropy.Throughout this paper, the model with the following assumptions is used. The rock is continuous and linearly elastic, and the displacement and strain are tiny. The tunnel is deep. The cross section is circular and parallel to the transverse isotropic surface. The initial stress field in the rock mass is regarded as a normal stressp which acts in the vertical direction and a normal stress q which acts in the horizontal direction. The tunnel length is much bigger than its cross-sectional dimensions; thus, it can be simplified as a plane strain problem.Zhang deduced the displacement analytical solution as follows [18]:(6)ur=p+q2a2r1+μE+p−q2a4r31+μE−4a2r1E−μ′2E′cos2θ,where ur is the radial displacement, E and μ are, respectively, the elastic modulus and Poisson’s ratio in the isotropic plane, E′ and μ′ are corresponding elastic constants in the direction perpendicular to the isotropic plane, a is the radius of the tunnel, and r and θ are the polar diameter and angle in polar coordinates.Introduce the side pressure coefficientλ=q/p and equation (6) can be rewritten as(7)ur=p21+λa2r1+μE+p21+λ⋅a4r31+μE−4a2r1E−μ′2E′cos2θ,where ur>0 means that the displacement direction points to the excavation surface and ur<0 means that the displacement direction is far from the excavation surface.There exist the following restrictions between the parameters of transverse isotropy:(8)1E<1+μE<21E−μ′2E′.Sincea≤r, it is easy to get(9)0<a4r3<2a2r.Considering formulas (8) and (9) synthetically, the following formulas could be obtained:(10)a4r31+μE−4a2r1E−μ′2E′<0,p21+λa2r1+μE>0.These two formulas can be used to judge the monotonicity of the function in the following derivation.


## 4. Displacement Distribution Law of a Circular Tunnel in the Transversely Isotropic Rock Mass

### 4.1. Variation of Displacement with Angle

According to the value ofλ, the distribution of displacement with angle is discussed in three cases.(1)
λ=1In this case, equation (7) becomes(11)ur=p21+λa2r1+μE.It can be seen that the displacement has nothing to do withθ, and the displacement is equal on a circle centered on the origin.(2)
0≤λ<1In this case,ur is a subtraction function of cos2θ. In the area satisfying the relation cos2θ=−1, ur gets the maximum value.(12)urmax=p21+λa2r1+μE−p21−λ⋅a4r31+μE−4a2r1E−μ′2E′.In the area satisfying the relationcos2θ=1, ur gets the minimum value.(13)urmin=p21+λa2r1+μE+p21−λ⋅a4r31+μE−4a2r1E−μ′2E′.(3)
λ>1In this case,uris an increasing function of cos2θ. In the area satisfying the relation cos2θ=1, ur gets the maximum value.(14)urmax=p21+λa2r1+μE+p21−λ⋅a4r31+μE−4a2r1E−μ′2E′.In the area satisfying the relationcos2θ=−1, ur gets the minimum value.(15)urmin=p21+λa2r1+μE−p21−λ⋅a4r31+μE−4a2r1E−μ′2E′.Based on the above analysis, it can be seen thaturmax always appears in the direction of the larger initial ground stress, and urmin always appears in the direction of the smaller initial ground stress.Sinceurmax is always positive and the relation urmax≥urminis always satisfied, the relation urmax=urmax can be obtained. However, urmin is not always equal to urmin, and the value of urmin is affected by the value of urmin. If urmin≥0, then urmin=urmin. If urmin<0, then urmin=0, and urmin appears in the area of ur=0, where cos2θ can be expressed as(16)cos2θ=−1+λ1+μ/E1−λa2/r21+μ/E−41/E−μ′2/E′.


### 4.2. Variation of Displacement with Radius

Equation (7) can be rewritten as(17)ur=p2a2r1+λ1+μE−41−λ1E−μ′2E′cos2θ+p21−λa4r31+μEcos2θ.If1−λcos2θ≠0, then equation (17) can be rewritten as(18)ur=p2a2r1+λ1−λcos2θ1+μE−41E−μ′2E′+a2r21+μE1−λcos2θ.Deviation ofr yields(19)durdr=−p2a2r21+λ1−λcos2θ1+μE−41E−μ′2E′+3a2r21+μE⋅1−λcos2θ.Let equation (19) be equal to zero, so the extreme point can be obtained as follows:(20)re=341/E−μ′2/E′/1+μ/E−1+λ/1−λcos2θa.The following relation can be derived from formula (8):(21)2<41/E−μ′2/E′1+μ/E<4.In addition, it is easy to prove the following formulas:(22)1+λ1−λ≥1,1+λ1−λcos2θ≥1.According to the different combinations of parameters, the distribution of displacement with radius is discussed in two cases.


#### 4.2.1. Case 1:1−λcos2θ≤1+λ/41/E−μ′2/E′−3/41+μ/E1+μ/E

In this case,urmax always appears on the wall of the tunnel. There are four situations to discuss.(1)
1−λcos2θ<0It is easy to deduce(23)41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ>3.The relationre<a can be obtained by substituting formula (23) into formula (20), and it is easy to judge that re is the maximum point according to formula (19). Because of ur|r=a>0 and ur|r=∞=0, ur monotonically decreases from ur|r=a to 0 in the area where r≥a. Correspondingly, urmaxappears on the tunnel wall, where r=a, and urmin appears at infinity.(2)
1−λcos2θ=0Equation (16) can be simplified as(24)ur=p2a2r1+λ1+μE.This case has the same conclusion as the case of1−λcos2θ<0.(3)
1+λ/41/E−μ′2/E′1+μ/E≥1−λcos2θ>0The following formula can be derived:(25)41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ≤0.In this case, formula (20) is meaningless, which shows that there are no local extreme points. Substituting formula (25) into formula (19) yields dur/dr<0. This case has the same conclusion as that of the case of 1−λcos2θ<0.(4)
1+λ/41/E−μ′2/E′−3/41+μ/E1+μ/E≥1−λcos2θ>1+λ/41/E−μ′2/E′1+μ/EThe following formula can be derived:(26)34≥41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ>0.The relationre>a can be obtained by substituting formula (26) into formula (20) and it is easy to judge that re is the minimum point according to formula (19). According to formula (18), there are(27)ur|r=a=p2a1+λ1−λcos2θ1+μE−41E−μ′2E′+1+μE⋅1−λcos2θ>0,ur|r=re=−39pa41E−μ′2E′−1+λ1−λcos2θ1+μE3/2⋅1+μE−1/21−λcos2θ<0,and ur|r=∞=0. In the area of r≥a, ur monotonically decreases from ur|r=a to ur|r=re and then monotonically increases to 0. Because of ur|r=re/ur|r=a≤1, urmax appears on the tunnel wall. urmin appears not only at infinity but also in the area indicated by the following equation: (28)r=a41/E−μ′2/E′/1+μ/E−1+λ/1−λcos2θ.


#### 4.2.2. Case 2:1−λcos2θ>1+λ/41/E−μ′2/E′−3/41+μ/E1+μ/E

In this case,urmax always appears at re. There are two situations to discuss.(1)
1+λ/41/E−ν′2/E′−1+ν/E1+ν/E>1−λcos2θ>1+λ/41/E−ν′2/E′−3/41+ν/E1+ν/EThe following formula can be deduced:(29)1>41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ>34.The relationre>a can be obtained by substituting formula (29) into formula (20), and it is easy to judge that re is the minimum point according to formula (19). Because of ur|r=∞=0, ur monotonically decreases to ur|r=re and then monotonically increases to 0 in the area where r≥a. Due to ur|r=re/ur|r=a>1, urmaxappears at re. urmin appears not only at infinity but also in the area indicated by equation (28).(2)
1−λcos2θ≥1+λ/41/E−ν′2/E′−1+μ/E1+μ/EThe following formula can be deduced:(30)3>41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ≥1.The equationre>a can be obtained by substituting formula (30) into formula (20), and it is easy to judge that re is the minimum point according to formula (19). According to formula (18), there are (31)ur|r=a=p2a1+λ1−λcos2θ1+μE−41E−μ′2E′+1+μE⋅1−λcos2θ≤0,ur|r=re=−39pa41E−μ′2E′−1+λ1−λcos2θ1+μE3/2⋅1+μE−1/21−λcos2θ<0.Because ofur|r=∞=0, urmonotonically decreases to ur|r=a and then monotonically increases to 0 in the area of r≥a. Correspondingly, urmaxappears at re, and urmin appears at infinity.


#### 4.2.3. The Influence of Lateral Pressure Coefficient on the Variation of Displacement with Radius

According to the above analysis, it is obvious that the variation of the displacement with radius caused by different combinations of parameters is different. The effect of the lateral pressure coefficientλ on the variation of displacement with radius is further analyzed below.We have the following formula:(32)1−λcos2θ=1+λ41/E−μ′2/E′−3/41+μ/E1+μE.Then, there is(33)cos2θ=1+λ1+μ/E41/E−μ′2/E′−3/41+μ/E1−λ.Let equation (33) be equal to –1 and 1, respectively; λ1 and λ2 can be obtained:(34)λ1=41/E−μ′2/E′+1/41+μ/E41/E−μ′2/E′−7/41+μ/E,(35)λ2=41/E−μ′2/E′−7/41+μ/E41/E−μ′2/E′+1/41+μ/E.It is easy to acquire the following two relations:λ1>1>λ2>0 and λ1λ2=1. Here, λ1 and λ2 are referred to as the critical pressure coefficients of displacement. The discussion is divided into three cases.(1)
Whenλ1≥λ≥λ2, all surrounding rock areas meet the conditions of formula (36), which is in full compliance with Section 4.2.1. So, the value of urmax is on the tunnel wall in all surrounding rock areas.(36)1−λcos2θ≤1+λ41/E−μ′2/E′−3/41+μ/E1+μE.(2)
Whenλ>λ1, in the area indicated by formula (37), it is easy to deduce formula (38), which conforms to Section 4.2.1. So, the value of urmax is on the tunnel wall. In addition, this area decreases with the increase of λ.(37)1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ≤cos2θ≤1,(38)1−λcos2θ≤1+λ41/E−μ′2/E′−3/41+μ/E1+μE.However, in the area indicated by formula (39), it is easy to deduce formula (40), which conforms to Section 4.2.2. So, the value of urmax is at re. In addition, this area increases with the increase of λ.(39)−1≤cos2θ<1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ,(40)1−λcos2θ>1+λ41/E−μ′2/E′−3/41+μ/E1+μE.(3)
When0≤λ<λ2, in the area indicated by formula (41), it is easy to deduce formula (42), which accords with Section 4.2.1. So, the value of urmax is on the tunnel wall. In addition, this area increases with the increase of λ.(41)−1≤cos2θ≤1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ,(42)1−λcos2θ≤1+λ41/E−μ′2/E′−3/41+μ/E1+μE.However, in the area indicated by formula (43), it is easy to deduce formula (44), which accords with Section 4.2.2. So, the value of urmaxis at re. In addition, this area decreases with the increase of λ.(43)1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ<cos2θ≤1,(44)1−λcos2θ>1+λ41/E−μ′2/E′−3/41+μ/E1+μE.Based on the above analysis, it is known thaturmax does not always appear on the tunnel wall; it may also appear at re under certain conditions.


## 4.1. Variation of Displacement with Angle

According to the value ofλ, the distribution of displacement with angle is discussed in three cases.(1)
λ=1In this case, equation (7) becomes(11)ur=p21+λa2r1+μE.It can be seen that the displacement has nothing to do withθ, and the displacement is equal on a circle centered on the origin.(2)
0≤λ<1In this case,ur is a subtraction function of cos2θ. In the area satisfying the relation cos2θ=−1, ur gets the maximum value.(12)urmax=p21+λa2r1+μE−p21−λ⋅a4r31+μE−4a2r1E−μ′2E′.In the area satisfying the relationcos2θ=1, ur gets the minimum value.(13)urmin=p21+λa2r1+μE+p21−λ⋅a4r31+μE−4a2r1E−μ′2E′.(3)
λ>1In this case,uris an increasing function of cos2θ. In the area satisfying the relation cos2θ=1, ur gets the maximum value.(14)urmax=p21+λa2r1+μE+p21−λ⋅a4r31+μE−4a2r1E−μ′2E′.In the area satisfying the relationcos2θ=−1, ur gets the minimum value.(15)urmin=p21+λa2r1+μE−p21−λ⋅a4r31+μE−4a2r1E−μ′2E′.Based on the above analysis, it can be seen thaturmax always appears in the direction of the larger initial ground stress, and urmin always appears in the direction of the smaller initial ground stress.Sinceurmax is always positive and the relation urmax≥urminis always satisfied, the relation urmax=urmax can be obtained. However, urmin is not always equal to urmin, and the value of urmin is affected by the value of urmin. If urmin≥0, then urmin=urmin. If urmin<0, then urmin=0, and urmin appears in the area of ur=0, where cos2θ can be expressed as(16)cos2θ=−1+λ1+μ/E1−λa2/r21+μ/E−41/E−μ′2/E′.


## 4.2. Variation of Displacement with Radius

Equation (7) can be rewritten as(17)ur=p2a2r1+λ1+μE−41−λ1E−μ′2E′cos2θ+p21−λa4r31+μEcos2θ.If1−λcos2θ≠0, then equation (17) can be rewritten as(18)ur=p2a2r1+λ1−λcos2θ1+μE−41E−μ′2E′+a2r21+μE1−λcos2θ.Deviation ofr yields(19)durdr=−p2a2r21+λ1−λcos2θ1+μE−41E−μ′2E′+3a2r21+μE⋅1−λcos2θ.Let equation (19) be equal to zero, so the extreme point can be obtained as follows:(20)re=341/E−μ′2/E′/1+μ/E−1+λ/1−λcos2θa.The following relation can be derived from formula (8):(21)2<41/E−μ′2/E′1+μ/E<4.In addition, it is easy to prove the following formulas:(22)1+λ1−λ≥1,1+λ1−λcos2θ≥1.According to the different combinations of parameters, the distribution of displacement with radius is discussed in two cases.


### 4.2.1. Case 1:1−λcos2θ≤1+λ/41/E−μ′2/E′−3/41+μ/E1+μ/E

In this case,urmax always appears on the wall of the tunnel. There are four situations to discuss.(1)
1−λcos2θ<0It is easy to deduce(23)41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ>3.The relationre<a can be obtained by substituting formula (23) into formula (20), and it is easy to judge that re is the maximum point according to formula (19). Because of ur|r=a>0 and ur|r=∞=0, ur monotonically decreases from ur|r=a to 0 in the area where r≥a. Correspondingly, urmaxappears on the tunnel wall, where r=a, and urmin appears at infinity.(2)
1−λcos2θ=0Equation (16) can be simplified as(24)ur=p2a2r1+λ1+μE.This case has the same conclusion as the case of1−λcos2θ<0.(3)
1+λ/41/E−μ′2/E′1+μ/E≥1−λcos2θ>0The following formula can be derived:(25)41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ≤0.In this case, formula (20) is meaningless, which shows that there are no local extreme points. Substituting formula (25) into formula (19) yields dur/dr<0. This case has the same conclusion as that of the case of 1−λcos2θ<0.(4)
1+λ/41/E−μ′2/E′−3/41+μ/E1+μ/E≥1−λcos2θ>1+λ/41/E−μ′2/E′1+μ/EThe following formula can be derived:(26)34≥41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ>0.The relationre>a can be obtained by substituting formula (26) into formula (20) and it is easy to judge that re is the minimum point according to formula (19). According to formula (18), there are(27)ur|r=a=p2a1+λ1−λcos2θ1+μE−41E−μ′2E′+1+μE⋅1−λcos2θ>0,ur|r=re=−39pa41E−μ′2E′−1+λ1−λcos2θ1+μE3/2⋅1+μE−1/21−λcos2θ<0,and ur|r=∞=0. In the area of r≥a, ur monotonically decreases from ur|r=a to ur|r=re and then monotonically increases to 0. Because of ur|r=re/ur|r=a≤1, urmax appears on the tunnel wall. urmin appears not only at infinity but also in the area indicated by the following equation: (28)r=a41/E−μ′2/E′/1+μ/E−1+λ/1−λcos2θ.


### 4.2.2. Case 2:1−λcos2θ>1+λ/41/E−μ′2/E′−3/41+μ/E1+μ/E

In this case,urmax always appears at re. There are two situations to discuss.(1)
1+λ/41/E−ν′2/E′−1+ν/E1+ν/E>1−λcos2θ>1+λ/41/E−ν′2/E′−3/41+ν/E1+ν/EThe following formula can be deduced:(29)1>41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ>34.The relationre>a can be obtained by substituting formula (29) into formula (20), and it is easy to judge that re is the minimum point according to formula (19). Because of ur|r=∞=0, ur monotonically decreases to ur|r=re and then monotonically increases to 0 in the area where r≥a. Due to ur|r=re/ur|r=a>1, urmaxappears at re. urmin appears not only at infinity but also in the area indicated by equation (28).(2)
1−λcos2θ≥1+λ/41/E−ν′2/E′−1+μ/E1+μ/EThe following formula can be deduced:(30)3>41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ≥1.The equationre>a can be obtained by substituting formula (30) into formula (20), and it is easy to judge that re is the minimum point according to formula (19). According to formula (18), there are (31)ur|r=a=p2a1+λ1−λcos2θ1+μE−41E−μ′2E′+1+μE⋅1−λcos2θ≤0,ur|r=re=−39pa41E−μ′2E′−1+λ1−λcos2θ1+μE3/2⋅1+μE−1/21−λcos2θ<0.Because ofur|r=∞=0, urmonotonically decreases to ur|r=a and then monotonically increases to 0 in the area of r≥a. Correspondingly, urmaxappears at re, and urmin appears at infinity.


### 4.2.3. The Influence of Lateral Pressure Coefficient on the Variation of Displacement with Radius

According to the above analysis, it is obvious that the variation of the displacement with radius caused by different combinations of parameters is different. The effect of the lateral pressure coefficientλ on the variation of displacement with radius is further analyzed below.We have the following formula:(32)1−λcos2θ=1+λ41/E−μ′2/E′−3/41+μ/E1+μE.Then, there is(33)cos2θ=1+λ1+μ/E41/E−μ′2/E′−3/41+μ/E1−λ.Let equation (33) be equal to –1 and 1, respectively; λ1 and λ2 can be obtained:(34)λ1=41/E−μ′2/E′+1/41+μ/E41/E−μ′2/E′−7/41+μ/E,(35)λ2=41/E−μ′2/E′−7/41+μ/E41/E−μ′2/E′+1/41+μ/E.It is easy to acquire the following two relations:λ1>1>λ2>0 and λ1λ2=1. Here, λ1 and λ2 are referred to as the critical pressure coefficients of displacement. The discussion is divided into three cases.(1)
Whenλ1≥λ≥λ2, all surrounding rock areas meet the conditions of formula (36), which is in full compliance with Section 4.2.1. So, the value of urmax is on the tunnel wall in all surrounding rock areas.(36)1−λcos2θ≤1+λ41/E−μ′2/E′−3/41+μ/E1+μE.(2)
Whenλ>λ1, in the area indicated by formula (37), it is easy to deduce formula (38), which conforms to Section 4.2.1. So, the value of urmax is on the tunnel wall. In addition, this area decreases with the increase of λ.(37)1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ≤cos2θ≤1,(38)1−λcos2θ≤1+λ41/E−μ′2/E′−3/41+μ/E1+μE.However, in the area indicated by formula (39), it is easy to deduce formula (40), which conforms to Section 4.2.2. So, the value of urmax is at re. In addition, this area increases with the increase of λ.(39)−1≤cos2θ<1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ,(40)1−λcos2θ>1+λ41/E−μ′2/E′−3/41+μ/E1+μE.(3)
When0≤λ<λ2, in the area indicated by formula (41), it is easy to deduce formula (42), which accords with Section 4.2.1. So, the value of urmax is on the tunnel wall. In addition, this area increases with the increase of λ.(41)−1≤cos2θ≤1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ,(42)1−λcos2θ≤1+λ41/E−μ′2/E′−3/41+μ/E1+μE.However, in the area indicated by formula (43), it is easy to deduce formula (44), which accords with Section 4.2.2. So, the value of urmaxis at re. In addition, this area decreases with the increase of λ.(43)1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ<cos2θ≤1,(44)1−λcos2θ>1+λ41/E−μ′2/E′−3/41+μ/E1+μE.Based on the above analysis, it is known thaturmax does not always appear on the tunnel wall; it may also appear at re under certain conditions.


## 4.2.1. Case 1:1−λcos2θ≤1+λ/41/E−μ′2/E′−3/41+μ/E1+μ/E

In this case,urmax always appears on the wall of the tunnel. There are four situations to discuss.(1)
1−λcos2θ<0It is easy to deduce(23)41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ>3.The relationre<a can be obtained by substituting formula (23) into formula (20), and it is easy to judge that re is the maximum point according to formula (19). Because of ur|r=a>0 and ur|r=∞=0, ur monotonically decreases from ur|r=a to 0 in the area where r≥a. Correspondingly, urmaxappears on the tunnel wall, where r=a, and urmin appears at infinity.(2)
1−λcos2θ=0Equation (16) can be simplified as(24)ur=p2a2r1+λ1+μE.This case has the same conclusion as the case of1−λcos2θ<0.(3)
1+λ/41/E−μ′2/E′1+μ/E≥1−λcos2θ>0The following formula can be derived:(25)41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ≤0.In this case, formula (20) is meaningless, which shows that there are no local extreme points. Substituting formula (25) into formula (19) yields dur/dr<0. This case has the same conclusion as that of the case of 1−λcos2θ<0.(4)
1+λ/41/E−μ′2/E′−3/41+μ/E1+μ/E≥1−λcos2θ>1+λ/41/E−μ′2/E′1+μ/EThe following formula can be derived:(26)34≥41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ>0.The relationre>a can be obtained by substituting formula (26) into formula (20) and it is easy to judge that re is the minimum point according to formula (19). According to formula (18), there are(27)ur|r=a=p2a1+λ1−λcos2θ1+μE−41E−μ′2E′+1+μE⋅1−λcos2θ>0,ur|r=re=−39pa41E−μ′2E′−1+λ1−λcos2θ1+μE3/2⋅1+μE−1/21−λcos2θ<0,and ur|r=∞=0. In the area of r≥a, ur monotonically decreases from ur|r=a to ur|r=re and then monotonically increases to 0. Because of ur|r=re/ur|r=a≤1, urmax appears on the tunnel wall. urmin appears not only at infinity but also in the area indicated by the following equation: (28)r=a41/E−μ′2/E′/1+μ/E−1+λ/1−λcos2θ.


## 4.2.2. Case 2:1−λcos2θ>1+λ/41/E−μ′2/E′−3/41+μ/E1+μ/E

In this case,urmax always appears at re. There are two situations to discuss.(1)
1+λ/41/E−ν′2/E′−1+ν/E1+ν/E>1−λcos2θ>1+λ/41/E−ν′2/E′−3/41+ν/E1+ν/EThe following formula can be deduced:(29)1>41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ>34.The relationre>a can be obtained by substituting formula (29) into formula (20), and it is easy to judge that re is the minimum point according to formula (19). Because of ur|r=∞=0, ur monotonically decreases to ur|r=re and then monotonically increases to 0 in the area where r≥a. Due to ur|r=re/ur|r=a>1, urmaxappears at re. urmin appears not only at infinity but also in the area indicated by equation (28).(2)
1−λcos2θ≥1+λ/41/E−ν′2/E′−1+μ/E1+μ/EThe following formula can be deduced:(30)3>41/E−μ′2/E′1+μ/E−1+λ1−λcos2θ≥1.The equationre>a can be obtained by substituting formula (30) into formula (20), and it is easy to judge that re is the minimum point according to formula (19). According to formula (18), there are (31)ur|r=a=p2a1+λ1−λcos2θ1+μE−41E−μ′2E′+1+μE⋅1−λcos2θ≤0,ur|r=re=−39pa41E−μ′2E′−1+λ1−λcos2θ1+μE3/2⋅1+μE−1/21−λcos2θ<0.Because ofur|r=∞=0, urmonotonically decreases to ur|r=a and then monotonically increases to 0 in the area of r≥a. Correspondingly, urmaxappears at re, and urmin appears at infinity.


## 4.2.3. The Influence of Lateral Pressure Coefficient on the Variation of Displacement with Radius

According to the above analysis, it is obvious that the variation of the displacement with radius caused by different combinations of parameters is different. The effect of the lateral pressure coefficientλ on the variation of displacement with radius is further analyzed below.We have the following formula:(32)1−λcos2θ=1+λ41/E−μ′2/E′−3/41+μ/E1+μE.Then, there is(33)cos2θ=1+λ1+μ/E41/E−μ′2/E′−3/41+μ/E1−λ.Let equation (33) be equal to –1 and 1, respectively; λ1 and λ2 can be obtained:(34)λ1=41/E−μ′2/E′+1/41+μ/E41/E−μ′2/E′−7/41+μ/E,(35)λ2=41/E−μ′2/E′−7/41+μ/E41/E−μ′2/E′+1/41+μ/E.It is easy to acquire the following two relations:λ1>1>λ2>0 and λ1λ2=1. Here, λ1 and λ2 are referred to as the critical pressure coefficients of displacement. The discussion is divided into three cases.(1)
Whenλ1≥λ≥λ2, all surrounding rock areas meet the conditions of formula (36), which is in full compliance with Section 4.2.1. So, the value of urmax is on the tunnel wall in all surrounding rock areas.(36)1−λcos2θ≤1+λ41/E−μ′2/E′−3/41+μ/E1+μE.(2)
Whenλ>λ1, in the area indicated by formula (37), it is easy to deduce formula (38), which conforms to Section 4.2.1. So, the value of urmax is on the tunnel wall. In addition, this area decreases with the increase of λ.(37)1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ≤cos2θ≤1,(38)1−λcos2θ≤1+λ41/E−μ′2/E′−3/41+μ/E1+μE.However, in the area indicated by formula (39), it is easy to deduce formula (40), which conforms to Section 4.2.2. So, the value of urmax is at re. In addition, this area increases with the increase of λ.(39)−1≤cos2θ<1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ,(40)1−λcos2θ>1+λ41/E−μ′2/E′−3/41+μ/E1+μE.(3)
When0≤λ<λ2, in the area indicated by formula (41), it is easy to deduce formula (42), which accords with Section 4.2.1. So, the value of urmax is on the tunnel wall. In addition, this area increases with the increase of λ.(41)−1≤cos2θ≤1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ,(42)1−λcos2θ≤1+λ41/E−μ′2/E′−3/41+μ/E1+μE.However, in the area indicated by formula (43), it is easy to deduce formula (44), which accords with Section 4.2.2. So, the value of urmaxis at re. In addition, this area decreases with the increase of λ.(43)1+μ/E41/E−μ′2/E′−3/41+μ/E1+λ1−λ<cos2θ≤1,(44)1−λcos2θ>1+λ41/E−μ′2/E′−3/41+μ/E1+μE.Based on the above analysis, it is known thaturmax does not always appear on the tunnel wall; it may also appear at re under certain conditions.


## 5. The Principles of Measuring Points Arrangement for a Circular Tunnel in the Transversely Isotropic Rock Mass

According to the law of displacement distribution analyzed inSection 4, some principles of measuring point arrangement for a circular tunnel in the transversely isotropic rock mass can be obtained. (1)
The optimal direction of the displacement measuring line is the direction of the larger initial ground stress.(2)
If the displacement of the measuring points in the direction of the smaller initial ground stress is greater than or equal to zero, then this direction is the least suitable for measuring line layout. Conversely, if the displacement in the direction of the smaller initial ground stress is smaller than zero, the least suitable measuring line can be given by equation (16).(3)
Whenλ1≥λ≥λ2, the most suitable measuring points are on the tunnel wall. When λ>λ1or 0≤λ<λ2, in one part of the areas, the most suitable measuring points are on the tunnel wall, while in the other part of the areas, the most suitable measuring points are at re.Figure2 shows the flow chart for determining the optimal measuring line and measuring points.Figure 2
Flow chart for determining the optimal measuring line and measuring points.In the displacement back analysis, if the initial ground stress is known, the measuring points can be arranged according to the research results in this section. If the ground stress is unknown, some additional measuring points can be arranged on a certain section of the tunnel firstly, and the initial ground stress can be identified based on the displacement of these measuring points.


## 6. Example of Measuring Points Layout

The parameters of a circular tunnel are, respectively,a=2m, p=10MPa, E=1GPa, E′=0.8GPa, μ=0.25, and μ′=0.3.Due to the symmetry, only 1/4 of the surrounding rock area is used to study the measuring points layout, as shown in Figure3. Seven radial displacement measuring lines are arranged on the measurement cross section. The angle between each measuring line is 15 degrees, and each line is 9 m in length. There are ten evenly arranged measuring points with a spacing of 1 m on each line, and each measuring point is numbered 1 ∼ 10 according to the direction away from the tunnel wall.Figure 3
Measuring lines and points layout of a circular tunnel.By substituting the relevant parameters of this circular tunnel into equations (34) and (35), the critical pressure coefficients of displacement are obtained as λ1=2.83 and λ2=0.35. The layout of measuring points is discussed when λ is equal to 0, 0.5, 1, and 5, respectively, as follows.


### 6.1.λ=0

λ=0 represents the situation of 0≤λ<λ2. The radial displacement of each measuring point is calculated according to equation (7), and then the average relative error of each measuring line is calculated according to equation (5). The results are listed in Table 1.Table 1
The radial displacements and average relative errors whenλ = 0.
LineRadial displacement of each measuring pointei∗ (cm−1)N1 (cm)N2 (cm)N3 (cm)N4 (cm)N5 (cm)N6 (cm)N7 (cm)N8 (cm)N9 (cm)N10 (cm)L1−1.05−1.16−0.99−0.84−0.72−0.63−0.56−0.50−0.45−0.411.55L2−0.74−0.90−0.78−0.66−0.57−0.50−0.44−0.39−0.36−0.331.97L30.10−0.16−0.18−0.17−0.15−0.14−0.12−0.11−0.10−0.097.96L41.250.830.630.500.420.360.310.280.250.232.60L52.401.831.431.170.990.850.750.670.600.551.10L63.242.562.031.661.401.211.060.950.860.780.78L73.552.832.241.841.551.341.181.050.950.870.70The following conclusions can be obtained from Table1:(1)
The displacement increases as the angle increases from 0° to 90° for the measuring points with the same depth on different measuring lines. The measuring point with the maximum displacement (also the maximum displacement absolute value) appears in the L7 line, which is parallel to the larger initial ground stress. The measuring point with the minimum displacement always appears in the L1 line, which is parallel to the smaller initial ground stress. The measuring point with the minimum displacement absolute value appears in the L3 line. These verify the variation of displacement with an angle when0≤λ<1 in Section 4.1.(2)
In the L4, L5, L6, and L7 lines, the measuring points with the maximum displacement absolute value appear on the cave wall, while in the L1, L2, and L3 lines, they do not appear on the cave wall. According to the deduction inSection 4.2, when −1≤cos2θ≤0.4784 (corresponding to 30.71°≤θ≤90°), the measuring point with the maximum displacement absolute value is on the cave wall; when 0.4784<cos2θ≤1 (corresponding to 0°≤θ<30.71°), the measuring point is at re. These verify the variation of displacement with radius when 0≤λ<λ2 in Section 4.2.(3)
It can be seen fromei∗ that the order of the layout of each measuring line is L7 > L6 > L5 > L1 > L2 > L4 > L3. This is by the principles of measuring points arrangement for a circular tunnel in the transversely isotropic rock mass in Section 5.


### 6.2.λ=0.5

λ=0.5 represents the situation of λ1≥λ≥λ2. The radial displacements and average relative errors are calculated and listed in Table 2.Table 2
The radial displacements and average relative errors whenλ = 0.5.
LineRadial displacement of each measuring pointei∗ (cm−1)N1 (cm)N2 (cm)N3 (cm)N4 (cm)N5 (cm)N6 (cm)N7 (cm)N8 (cm)N9 (cm)N10 (cm)L10.730.250.130.080.060.040.030.030.020.0221.53L20.880.390.240.170.130.110.090.080.070.068.30L31.300.750.530.420.340.290.250.220.200.183.17L41.881.250.940.750.630.540.470.420.380.341.73L52.451.751.341.090.910.780.690.610.550.501.19L62.872.111.641.331.120.960.840.750.680.620.97L73.032.251.751.421.191.030.900.800.730.660.91The following conclusions can be obtained from Table2:(1)
The displacement increases as the angle increases from 0° to 90° for the measuring points with the same depth on different measuring lines. The measuring point with the maximum displacement (also the maximum displacement absolute value) appears in the L7 line, which is parallel to the larger initial ground stress. The measuring point with the minimum displacement (also the minimum displacement absolute value) appears in the L1 line, which is parallel to the smaller initial ground stress. These verify the variation of displacement with an angle when0≤λ<1 in Section 4.1.(2)
The measuring points with the maximum displacement absolute value appear on the cave wall in all the measuring lines. This verifies the variation of displacement with radius whenλ1≥λ≥λ2 in Section 4.2.(3)
It can be seen fromei∗ that the order of the layout of each measuring line is L7 > L6 > L5 > L4 > L3 > L2 > L1. This is by the principles of measuring points arrangement for a circular tunnel in the transversely isotropic rock mass in Section 5.


### 6.3.λ=1

λ=1 corresponds to the axisymmetric problem, which is often encountered in engineering. λ=1 represents the situation of λ1≥λ≥λ2. The radial displacements and average relative errors are calculated and listed in Table 3.Table 3
The radial displacements and average relative errors whenλ = 1.
LineRadial displacement of each measuring pointei∗ (cm−1)N1 (cm)N2 (cm)N3 (cm)N4 (cm)N5 (cm)N6 (cm)N7 (cm)N8 (cm)N9 (cm)N10 (cm)L12.501.671.251.000.830.710.630.560.500.451.30L22.501.671.251.000.830.710.630.560.500.451.30L32.501.671.251.000.830.710.630.560.500.451.30L42.501.671.251.000.830.710.630.560.500.451.30L52.501.671.251.000.830.710.630.560.500.451.30L62.501.671.251.000.830.710.630.560.500.451.30L72.501.671.251.000.830.710.630.560.500.451.30The following conclusions can be obtained from Table3:(1)
The displacements are all positive and do not vary with an angle for the measuring points with the same depth on different measuring lines. This verifies the variation of displacement with an angle whenλ=1 in Section 4.1.(2)
The measuring points with the maximum displacement absolute value appear on the cave wall in all the measuring lines. This verifies the variation of displacement with radius whenλ1≥λ≥λ2 in Section 4.2.(3)
As can be seen fromei∗, the average relative errors of all measuring lines are equal, which demonstrates that there is no distinction between good and bad for different measuring lines.


### 6.4.λ=5

λ=5 represents the situation of λ>λ1. The radial displacements and average relative errors are calculated and listed in Table 4.Table 4
The radial displacements and average relative errors whenλ = 5.
LineRadial displacement of each measuring pointei∗ (cm−1)N1 (cm)N2 (cm)N3 (cm)N4 (cm)N5 (cm)N6 (cm)N7 (cm)N8 (cm)N9 (cm)N10 (cm)L116.7012.9910.238.367.056.085.354.774.303.920.15L215.4711.929.367.646.445.564.884.353.923.570.17L312.108.996.995.684.774.113.613.222.902.640.23L47.505.003.753.002.502.141.881.671.501.360.43L52.901.010.510.320.230.170.140.120.100.095.38L6−0.47−1.92−1.86−1.64−1.44−1.27−1.13−1.02−0.92−0.850.94L7−1.70−2.99−2.73−2.36−2.05−1.80−1.60−1.43−1.30−1.190.57The following conclusions can be obtained from Table4:(1)
The displacement decreases as the angle increases from 0° to 90° for the measuring points with the same depth on different measuring lines. The measuring point with the maximum displacement (also the maximum displacement absolute value) appears in the L1 line, which is parallel to the larger initial ground stress. The measuring point with the minimum displacement appears in the L7 line, which is parallel to the smaller initial ground stress. The measuring point with the minimum displacement absolute value appears in the L5 or L6 line. These verify the variation of displacement with an angle whenλ>1 in Section 4.1.(2)
In the L1, L2, L3, L4, and L5 lines, the measuring points with the maximum displacement absolute value appear on the cave wall, while in the L6 and L7 lines, they do not appear on the cave wall. According to the deduction inSection 4.2, when 0.7177<cos2θ≤1 (corresponding to 0°≤θ<67.93°), the measuring point with the maximum displacement absolute value is on the cave wall; when −1≤cos2θ≤0.7177 (corresponding to 67.93°≤θ≤90°), the measuring point is at re. These verify the variation of displacement with radius when 0≤λ<λ2 in Section 4.2.(3)
It can be seen fromei∗ that the order of the layout of each measuring line is L1 > L2 > L3 > L4 > L7 > L6 > L5. This is by the principles of measuring points arrangement for a circular tunnel in the transversely isotropic rock mass in Section 5.


## 6.1.λ=0

λ=0 represents the situation of 0≤λ<λ2. The radial displacement of each measuring point is calculated according to equation (7), and then the average relative error of each measuring line is calculated according to equation (5). The results are listed in Table 1.Table 1
The radial displacements and average relative errors whenλ = 0.
LineRadial displacement of each measuring pointei∗ (cm−1)N1 (cm)N2 (cm)N3 (cm)N4 (cm)N5 (cm)N6 (cm)N7 (cm)N8 (cm)N9 (cm)N10 (cm)L1−1.05−1.16−0.99−0.84−0.72−0.63−0.56−0.50−0.45−0.411.55L2−0.74−0.90−0.78−0.66−0.57−0.50−0.44−0.39−0.36−0.331.97L30.10−0.16−0.18−0.17−0.15−0.14−0.12−0.11−0.10−0.097.96L41.250.830.630.500.420.360.310.280.250.232.60L52.401.831.431.170.990.850.750.670.600.551.10L63.242.562.031.661.401.211.060.950.860.780.78L73.552.832.241.841.551.341.181.050.950.870.70The following conclusions can be obtained from Table1:(1)
The displacement increases as the angle increases from 0° to 90° for the measuring points with the same depth on different measuring lines. The measuring point with the maximum displacement (also the maximum displacement absolute value) appears in the L7 line, which is parallel to the larger initial ground stress. The measuring point with the minimum displacement always appears in the L1 line, which is parallel to the smaller initial ground stress. The measuring point with the minimum displacement absolute value appears in the L3 line. These verify the variation of displacement with an angle when0≤λ<1 in Section 4.1.(2)
In the L4, L5, L6, and L7 lines, the measuring points with the maximum displacement absolute value appear on the cave wall, while in the L1, L2, and L3 lines, they do not appear on the cave wall. According to the deduction inSection 4.2, when −1≤cos2θ≤0.4784 (corresponding to 30.71°≤θ≤90°), the measuring point with the maximum displacement absolute value is on the cave wall; when 0.4784<cos2θ≤1 (corresponding to 0°≤θ<30.71°), the measuring point is at re. These verify the variation of displacement with radius when 0≤λ<λ2 in Section 4.2.(3)
It can be seen fromei∗ that the order of the layout of each measuring line is L7 > L6 > L5 > L1 > L2 > L4 > L3. This is by the principles of measuring points arrangement for a circular tunnel in the transversely isotropic rock mass in Section 5.


## 6.2.λ=0.5

λ=0.5 represents the situation of λ1≥λ≥λ2. The radial displacements and average relative errors are calculated and listed in Table 2.Table 2
The radial displacements and average relative errors whenλ = 0.5.
LineRadial displacement of each measuring pointei∗ (cm−1)N1 (cm)N2 (cm)N3 (cm)N4 (cm)N5 (cm)N6 (cm)N7 (cm)N8 (cm)N9 (cm)N10 (cm)L10.730.250.130.080.060.040.030.030.020.0221.53L20.880.390.240.170.130.110.090.080.070.068.30L31.300.750.530.420.340.290.250.220.200.183.17L41.881.250.940.750.630.540.470.420.380.341.73L52.451.751.341.090.910.780.690.610.550.501.19L62.872.111.641.331.120.960.840.750.680.620.97L73.032.251.751.421.191.030.900.800.730.660.91The following conclusions can be obtained from Table2:(1)
The displacement increases as the angle increases from 0° to 90° for the measuring points with the same depth on different measuring lines. The measuring point with the maximum displacement (also the maximum displacement absolute value) appears in the L7 line, which is parallel to the larger initial ground stress. The measuring point with the minimum displacement (also the minimum displacement absolute value) appears in the L1 line, which is parallel to the smaller initial ground stress. These verify the variation of displacement with an angle when0≤λ<1 in Section 4.1.(2)
The measuring points with the maximum displacement absolute value appear on the cave wall in all the measuring lines. This verifies the variation of displacement with radius whenλ1≥λ≥λ2 in Section 4.2.(3)
It can be seen fromei∗ that the order of the layout of each measuring line is L7 > L6 > L5 > L4 > L3 > L2 > L1. This is by the principles of measuring points arrangement for a circular tunnel in the transversely isotropic rock mass in Section 5.


## 6.3.λ=1

λ=1 corresponds to the axisymmetric problem, which is often encountered in engineering. λ=1 represents the situation of λ1≥λ≥λ2. The radial displacements and average relative errors are calculated and listed in Table 3.Table 3
The radial displacements and average relative errors whenλ = 1.
LineRadial displacement of each measuring pointei∗ (cm−1)N1 (cm)N2 (cm)N3 (cm)N4 (cm)N5 (cm)N6 (cm)N7 (cm)N8 (cm)N9 (cm)N10 (cm)L12.501.671.251.000.830.710.630.560.500.451.30L22.501.671.251.000.830.710.630.560.500.451.30L32.501.671.251.000.830.710.630.560.500.451.30L42.501.671.251.000.830.710.630.560.500.451.30L52.501.671.251.000.830.710.630.560.500.451.30L62.501.671.251.000.830.710.630.560.500.451.30L72.501.671.251.000.830.710.630.560.500.451.30The following conclusions can be obtained from Table3:(1)
The displacements are all positive and do not vary with an angle for the measuring points with the same depth on different measuring lines. This verifies the variation of displacement with an angle whenλ=1 in Section 4.1.(2)
The measuring points with the maximum displacement absolute value appear on the cave wall in all the measuring lines. This verifies the variation of displacement with radius whenλ1≥λ≥λ2 in Section 4.2.(3)
As can be seen fromei∗, the average relative errors of all measuring lines are equal, which demonstrates that there is no distinction between good and bad for different measuring lines.


## 6.4.λ=5

λ=5 represents the situation of λ>λ1. The radial displacements and average relative errors are calculated and listed in Table 4.Table 4
The radial displacements and average relative errors whenλ = 5.
LineRadial displacement of each measuring pointei∗ (cm−1)N1 (cm)N2 (cm)N3 (cm)N4 (cm)N5 (cm)N6 (cm)N7 (cm)N8 (cm)N9 (cm)N10 (cm)L116.7012.9910.238.367.056.085.354.774.303.920.15L215.4711.929.367.646.445.564.884.353.923.570.17L312.108.996.995.684.774.113.613.222.902.640.23L47.505.003.753.002.502.141.881.671.501.360.43L52.901.010.510.320.230.170.140.120.100.095.38L6−0.47−1.92−1.86−1.64−1.44−1.27−1.13−1.02−0.92−0.850.94L7−1.70−2.99−2.73−2.36−2.05−1.80−1.60−1.43−1.30−1.190.57The following conclusions can be obtained from Table4:(1)
The displacement decreases as the angle increases from 0° to 90° for the measuring points with the same depth on different measuring lines. The measuring point with the maximum displacement (also the maximum displacement absolute value) appears in the L1 line, which is parallel to the larger initial ground stress. The measuring point with the minimum displacement appears in the L7 line, which is parallel to the smaller initial ground stress. The measuring point with the minimum displacement absolute value appears in the L5 or L6 line. These verify the variation of displacement with an angle whenλ>1 in Section 4.1.(2)
In the L1, L2, L3, L4, and L5 lines, the measuring points with the maximum displacement absolute value appear on the cave wall, while in the L6 and L7 lines, they do not appear on the cave wall. According to the deduction inSection 4.2, when 0.7177<cos2θ≤1 (corresponding to 0°≤θ<67.93°), the measuring point with the maximum displacement absolute value is on the cave wall; when −1≤cos2θ≤0.7177 (corresponding to 67.93°≤θ≤90°), the measuring point is at re. These verify the variation of displacement with radius when 0≤λ<λ2 in Section 4.2.(3)
It can be seen fromei∗ that the order of the layout of each measuring line is L1 > L2 > L3 > L4 > L7 > L6 > L5. This is by the principles of measuring points arrangement for a circular tunnel in the transversely isotropic rock mass in Section 5.


## 7. Conclusions

This paper studies the optimization of the location of measuring points layout around a circular tunnel in the transversely isotropic rock mass. Based on the principle of maximum displacement, the variation of displacement with angle and radius was analyzed in detail, some layout principles were obtained, and these principles were verified through an example. The main conclusions are as follows:(1)
The criterion for the optimal arrangement of displacement measuring points can be simply summarized as that the place with a larger absolute displacement value is more suitable to arrange measuring points.(2)
The two critical side pressure coefficientsλ1 and λ2 suitable for a circular tunnel in the transversely isotropic rock mass are derived. When λ1≥λ≥λ2, the most suitable measuring points are on the tunnel wall. When λ>λ1or 0≤λ<λ2, in one part of the areas, the most suitable measuring points are on the tunnel wall, while in the other part of the areas, the most suitable measuring points are at re.(3)
The optimal direction of the displacement measuring line is the direction of the larger initial ground stress.(4)
The area near the cave wall in the direction of the larger principal stress is the most suitable area to arrange measuring points.



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*Source: 6643166-2021-01-27.xml*