# Elastoplastic Analysis of Two-Layered Circular Lining Based on the Unified Strength Theory

Authors: Ning Zhang; Haomin Guo; Tahir Khan; Erhao Zhu; Shaoshuai Shi; Qingliang Wu; Yinliang Tian; Xiaokun Xie
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

Based on the Unified Strength Theory (UST), elastoplastic analysis of the two-layered circular lining is carried out. The stresses, displacements, and the elastic and plastic zones in both layers are discussed under different values of Young’s moduli of the inner and outer layers. The results reveal that, compared to the single-layered lining, the tangential stress distributions in the two-layered linings are more reasonable along the radial direction, which is beneficial to enhance the overall elastic and plastic ultimate bearing capacities. When considering the intermediate stress (i.e., the axial load), the elastic ultimate bearing capacity will be higher. However, the plastic ultimate bearing capacity remains unchanged. Moreover, a comparison between the Unified Strength Theory and Tresca Criterion is analyzed as well.


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## Body

## 1. Introduction

Recently, along with the construction of underground caverns in depth, the support technique needs to develop accordingly, to assess the stability of the structures and surrounding rock [1]. Thick-walled hollow cylinder is one of the widely used support structures. To optimize the structural design and make the best usage of construction materials, pressurized hollow cylinders are common cases in deep-buried circular caverns.For the elastic plane problem on circular structures, a lot of literature can be found [2]. The well-known Lame solution presents the stress and displacement fields for a single-layered hollow cylinder subjected to axisymmetric loads. Wu [3] solved the solutions of the multilayered case. For the nonaxisymmetric case, Luet al. [4], by using the semi-inverse method, presented the elastic plane stress solution of a lined vertical shaft in isotropic ground. Wu and Lu [5] solved the stress fields for the plane problem of a thick-walled cylinder, by utilizing the complex variable function method, and they found that there exists a tensile tangential stress.For the elastoplastic analysis on single-layered thick-walled hollow cylinder, there also have been many solutions on stress, displacement distributions, and elastic, plastic zones based on different strength theories [6–11]. Durban and Kube [12] investigated the problem of a thick-walled cylindrical tube subjected to internal pressure using a finite strain elastoplastic flow theory. Detournay and Fairhurst [13] describe a two-dimensional elastoplastic model of a long, cylindrical cavity in an infinite rock mass subject to nonhydrostatic far-field stress loading. To discuss the plastic ultimate bearing capacity, Gaoet al. [14] considered the size effect of strain-hardening level. And in the fields of elastic-brittle-plastic analysis, Leu [15] applied sequential limit analysis, dealing with the quasistatic problem of hardening and weakening behaviors. Lee and Pietruszczak [16] calculated the stresses and displacements fields of a circular tunnel excavated in strain-softening Mohr-Coulomb or generalized Hoek-Brown rock mass. Wang et al. [17, 18] and Zhanget al. [19, 20] presented a solution of a spherical cavity in brittle–plastic medium under hydrostatic in situ stress, in which the linear Mohr-Coulomb and nonlinear Hoek-Brown yield criteria were applied.For nonplane problems, Luet al. [21] presented the elastoplastic solutions for a circular tunnel with considering the axial stress, showing that the distributions of plastic zones are dependent on the axial and horizontal in situ stresses. Wanget al. [22] suggested an elastic-brittle-plastic constitutive model with a nonassociated flow rule.From the above mentioned literature, there are few results on elastoplastic analysis for the multilayered thick-walled hollow cylinder. We herein will focus on the elastoplastic analysis of a multilayered thick-walled hollow cylinder (see Figure1). Moreover, the ultimate bearing capacities are to be investigated as well. In Figure 1,Ei,μi, and fci (i=1 and 2) denote the elastic modulus, Poisson’s ratio, and compressive strength of the inner and outer layers, respectively. The axisymmetric external load on the outer boundary isp. Here, we assume that the tunnel is deep-buried and materials of both layers are homogeneous and isotropic and satisfy the ideal elastoplastic model. Moreover, the interface boundary of the two layers is a pure bond, no slide happening along the tangential or longitudinal direction.Figure 1
A two-layered thick-walled hollow cylinder(a) subjected to external axisymmetric pressure(b).
(a) (b)


## 2. Basic Equations and Stresses and Displacements in Elastic Zones

The cylindrical coordinate system is employed. Moreover, for simplicity, the characteristic of axial symmetry is considered. The strain-displacement relationships for an axisymmetric problem are(1)εr=dudr,(2)εθ=ur,where εr and εθ are the radial and tangential strains, respectively. Moreover,u is the radial displacement, andr is the radial distance from the center of opening.The differential equation of equilibrium for an axisymmetric problem is(3)dσrdr+σr-σθr=0.where σθ and σr are the tangential and radial stresses, respectively.The compatibility equation of strain can be given as(4)dεθdr+εθ-εrr=0.The generalized Hooke’s Law is given as follows:(5)εr=1Eσr-μσθ+σz,(6)εθ=1Eσθ-μσr+σz,(7)εz=1Eσz-μσr+σθ.where σz is the intermediate principal stress.The expressions of stress components in elastic state for an axisymmetric problem can be given according to the Lame solutions as(8)σri=Ai-Bir2,(9)σθi=Ai+Bir2.The subscripti=1 and 2, denoting the inner and outer layers, respectively (similarly hereinafter). Substituting Eq. (8) into Eq. (6) and considering Eq. (2), the radial displacement in elastic zones is obtained as(10)ui=1Eir1-μiAir2+Bi1+μi-μiσzir2.Note that the expressions of the stress and displacement components, (8) and (10), are suitable for both inner and outer layers, just with different values of the coefficientsAi andBi (i=1 and 2).The Unified Strength Theory is applied herein. This strength theory can take the intermediate principal stressσz into consideration. For the axisymmetric plane strain problem, there exist three non-zero principal stresses (i.e., the radial, circumferential, and axial stresses, σr, σθ, and σz). Compression is taken as positive. For the plane strain problem, the axial stress σz can be given as follows [23]:(11)σzi=miσθi+σri20≤mi≤1,wherem is the parameter reflecting the state of the intermediate principal stress. In elastic zones, mi=2μi. As the in-plane plastic strains increase, σz approaches the average of the minor and major principal stresses (i.e.,mi tends to 1). For simplicity, in plastic zones,m is assumed to be 1 [24]. Then, (11) can be expressed as(12)σzie=μiσθi+σriintheelasticzone,(13)σzip=σθi+σri2intheplasticzone.The parameterm takes 0 and 2μ denoting the plane stress and strain problems, respectively. The intermediate principal stress in the elastic zone can be given by substituting (8) and (9) into (11) as(14)σzie=miAi.And the radial displacement in elastic zones can be obtained by substituting (14) into (10) as(15)ui=1Eir1-μi-miμiAir2+Bi1+μi.


## 3. Elastoplastic Analysis

### 3.1. Elastic Analysis

As the axisymmetric external loadp is relatively small, two layers are in the elastic state. Based on the stress boundary conditions on the inner and outer boundaries, and the stress and displacement continuity conditions on the interface between two layers, we have(16)A1-B1R02=0,(17)A1-B1R12=A2-B2R12,(18)1-μ1-μ1m1A1R12+1+μ1B1E1R1=1-μ2-μ2m2A2R12+1+μ2B2E2R1,(19)A2-B2R22=p.Based on (16)–(19), the undetermined coefficientsAi,Bi (i=1, 2) can be solved, which are presented in the Appendix.


### 3.2. Expressions of Stress and Displacement in the Plastic Zone

The Unified Strength Theory was proposed based on an orthogonal octahedron of a twin shear element model [23, 25]. It is a twin-shear element model [26]. It can be expressed using the principal stresses as(20)F=1ασ1-11+bbσ2+σ3=fc,σ2≤σ1+ασ31+α,(21)F′=1α1+bσ1+bσ2-σ3=fc,σ2≥σ1+ασ31+α,whereσ1>σ2>σ3, and α denotes the ratio of the uniaxial tensile to the compressive strengths (i.e., α=ft/fc) and 0<α⩽1 for most engineering materials. In this work, we take α=(1-sinφ)/(1+sinφ). Moreover,ft andfc are the uniaxial tensile and compressive strengths, respectively;b reflects the effect of the intermediate principal stress. Theoretically,b can take any value. Actually, the Unified Strength Theory can represent different criteria as α andb take specific values [23]. For example, ifb=0 and α=1, it is the Tresca Criterion. Ifb=1/2 or α=1 and b=1/1+3, it is the linear approximations of the von Mises criterion. The Mohr-Coulomb criterion can be obtained asb=0 and 0<α<1. Ifb=1 and 0<α<1, it becomes the generalized twin shear stress criterion. Specifically, when α=1 andb=1, it becomes the twin shear stress criterion. Thus, it has been widely used due to the generalization since it was proposed [7, 9, 23, 27, 28].According to (11), the axial stressσz should be the intermediate principal stress. The three principal stresses satisfy the relationship,σθ>σz>σr; thus, we have(22)σ1=σθ,σ2=σz,σ3=σrAnd we also have(23)σz=mσθ+σr2≤σθ+σr2≤σθ+ασr1+α.According to (22) and (23), (20) can be satisfied. Then,(24)1ασθ-11+bbσz+σr=fc.Equation (24) can be expressed as the form of yield function as follows:(25)f=σθ-ηbσz+σr-ft,where η=α/1+b and ft=αfc.Combining (3), (11), and (25), we have(26)σri=Cirζi+ξi.where ζi=2-1+ηibimi+ηi/2-ηibimi, ξi=fti/1-ηibimi-ηi.Combining Eq. (11), substituting Eq. (26) into Eq. (24), we have(27)σθi=Φirζi+Ψi.where Φi=2+bimiηiCi/2-ηibimi and Ψi=2fti+2+bimiηiξi/2-ηibimi.The flow rule related with the Unified Strength Theory is employed:(28)dεθp=∂f∂σθdλ=dλ,(29)dεrp=∂f∂σrdλ=-ηdλ,(30)dεzp=∂f∂σzdλ=-ηbdλ.From (28)–(30), we have(31)εθp=λ,(32)εrp=-ηλ,(33)εzp=-ηbλ.According to (31) and (32), we have(34)ηεθp+εrp=0.When the material enters the plastic state, the elastic and plastic strains constitute the strain, which is(35)εr=εre+εrp=1Eσr-μσθ+σz+εrp,(36)εθ=εθe+εθp=1Eσθ-μσr+σz+εθp,(37)εz=εze+εzp=1Eσz-μσr+σθ+εzp,where σr and σθ are the radial and tangential stresses in the plastic zone (see (26) and (27)), respectively;σz is the axial stress (i.e., (11)).Combining (34)–(36), we have(38)ηεθ+εr=ηEσθ-μσr+σz+1Eσr-μσθ+σz.Substituting (11) into (38) and combining (26) and (27), the tangential strain in the plastic zone is derived as follows:(39)εri=-ηiεθi+CiFirζi+Giwhere(40)Fi=--2ηi+2μi+ηiμimi+μimiΦi+2Ciηiμi-1+μimiCiηi+12EiCi,Gi=--2ηi+2μi+ηiμimi+μimiΨi+2ξiηiμi-1+μimiξiηi+12Ei.Substituting (39) into (4), we have(41)εθi=CiFi1+ηi+ζirζi+Dir-1+ηi+Gi1+ηi.Combining (2) and (41), the radial displacement in the plastic zone can be obtained as(42)ui=CiFi1+ηi+ζirζi+1+Dir-ηi+Gi1+ηir.To obtain the stresses and displacements in plastic zones, the coefficientsAi,Bi,Ci andDi (i=1, 2) need to be calculated.


### 3.3. Elastoplastic Analysis

When a multilayered thick-walled hollow cylinder is subjected to hydrostatic pressure on the outer boundary, the plastic yielding will always initiate from the inner boundary of each layer. Moreover, the extending path of plastic yielding is dependent on the elastic modulusEi, material strengthσci, and dimensionsRi. Specifically, for a two-layered hollow lining, plastic yielding may initiate in three different ways (see Figure 2; rp and rpi are the dimensions of plastic zones). Hereby, we only discuss the solution of the most complicated situation (Figure 2(c)). The other two simpler cases can be discussed in a similar way.Figure 2
The three stress states in a two-layered lining under hydrostatic in situ stresses.
(a)
Plastic zone in inner layer (R0≤rp≤R1) (b)
Plastic zone in outer layer (R1≤rp≤R2) (c)
Plastic zones in both layers (R0≤rp1≤R1,R1≤rp2≤R2)The equations for the undetermined coefficients can be obtained according to the boundary and continuity conditions, which include the stress boundary conditions at the inner boundary of the inner layerr=R0, σr1p=0; the radial stress and displacement continuity conditions at three interfaces (i.e., at the elastic-plastic interface of the inner layerr=rp1, σr1p=σr1e and u1p=u1e; at the inter-boundaryr=R1, σr1e=σr2p and u1e=u2p; at the elastic-plastic interface of the outer layerr=rp2, σr2p=σr2e and u2p=u2e); and the stress boundary conditions at the outer boundary of the outer layerr=R2, σr2e=p. According to the conditions, we have(43)C1R0ζ1+ξ1=0,(44)C1rp1ζ1+ξ1=A1-B1rp12,(45)C1F11+η1+ζ1rp1ζ1+1+D1rp1-η1+G11+η1rp1=1-μ1-m1μ1A1rp12+1+μ1B1E1rp1,(46)A1-B1R12=C2R1ζ2+ξ2,(47)1-μ1-μ1m1A1R12+1+μ1B1E1R1=C2F21+η2+ζ2R1ζ2+1+D2R1-η2+G21+η2R1,(48)A2-B2rp22=C2rp2ζ2+ξ2,(49)C2F21+η2+ζ2rp2ζ2+1+D2rp2-η2+G21+η2rp2=1-μ2-μ2m2A2rp22+1+μ2B2E2rp2,(50)A2-B2R22=p.There are eight undetermined variablesAi,Bi,Ci,Di (i=1 and 2) and eight equations (i.e., (43)–(50)). To obtain the radii of the plastic zones (i.e.,rp1 andrp2), two more equations are required. From the extending path of plastic yielding, the stresses on the inner boundaries of the two elastic zones should also satisfy the Unified Strength Theory. Thus, we have(51)σθ1e-η1bσz1+σr1e-α1fc1=0,(52)σθ2e-η2bσz2+σr2e-α2fc2=0.Substituting (8) and (14) into (51) and (52), we have(53)A1+B1rp12-η1b1m1A1+A1-B1rp12=ft1,(54)A2+B2rp22-η2b2m2A2+A2-B2rp22=ft2.With the increase of the loadp, the extending path of plastic yielding can be obtained from (53) and (54).


## 3.1. Elastic Analysis

As the axisymmetric external loadp is relatively small, two layers are in the elastic state. Based on the stress boundary conditions on the inner and outer boundaries, and the stress and displacement continuity conditions on the interface between two layers, we have(16)A1-B1R02=0,(17)A1-B1R12=A2-B2R12,(18)1-μ1-μ1m1A1R12+1+μ1B1E1R1=1-μ2-μ2m2A2R12+1+μ2B2E2R1,(19)A2-B2R22=p.Based on (16)–(19), the undetermined coefficientsAi,Bi (i=1, 2) can be solved, which are presented in the Appendix.


## 3.2. Expressions of Stress and Displacement in the Plastic Zone

The Unified Strength Theory was proposed based on an orthogonal octahedron of a twin shear element model [23, 25]. It is a twin-shear element model [26]. It can be expressed using the principal stresses as(20)F=1ασ1-11+bbσ2+σ3=fc,σ2≤σ1+ασ31+α,(21)F′=1α1+bσ1+bσ2-σ3=fc,σ2≥σ1+ασ31+α,whereσ1>σ2>σ3, and α denotes the ratio of the uniaxial tensile to the compressive strengths (i.e., α=ft/fc) and 0<α⩽1 for most engineering materials. In this work, we take α=(1-sinφ)/(1+sinφ). Moreover,ft andfc are the uniaxial tensile and compressive strengths, respectively;b reflects the effect of the intermediate principal stress. Theoretically,b can take any value. Actually, the Unified Strength Theory can represent different criteria as α andb take specific values [23]. For example, ifb=0 and α=1, it is the Tresca Criterion. Ifb=1/2 or α=1 and b=1/1+3, it is the linear approximations of the von Mises criterion. The Mohr-Coulomb criterion can be obtained asb=0 and 0<α<1. Ifb=1 and 0<α<1, it becomes the generalized twin shear stress criterion. Specifically, when α=1 andb=1, it becomes the twin shear stress criterion. Thus, it has been widely used due to the generalization since it was proposed [7, 9, 23, 27, 28].According to (11), the axial stressσz should be the intermediate principal stress. The three principal stresses satisfy the relationship,σθ>σz>σr; thus, we have(22)σ1=σθ,σ2=σz,σ3=σrAnd we also have(23)σz=mσθ+σr2≤σθ+σr2≤σθ+ασr1+α.According to (22) and (23), (20) can be satisfied. Then,(24)1ασθ-11+bbσz+σr=fc.Equation (24) can be expressed as the form of yield function as follows:(25)f=σθ-ηbσz+σr-ft,where η=α/1+b and ft=αfc.Combining (3), (11), and (25), we have(26)σri=Cirζi+ξi.where ζi=2-1+ηibimi+ηi/2-ηibimi, ξi=fti/1-ηibimi-ηi.Combining Eq. (11), substituting Eq. (26) into Eq. (24), we have(27)σθi=Φirζi+Ψi.where Φi=2+bimiηiCi/2-ηibimi and Ψi=2fti+2+bimiηiξi/2-ηibimi.The flow rule related with the Unified Strength Theory is employed:(28)dεθp=∂f∂σθdλ=dλ,(29)dεrp=∂f∂σrdλ=-ηdλ,(30)dεzp=∂f∂σzdλ=-ηbdλ.From (28)–(30), we have(31)εθp=λ,(32)εrp=-ηλ,(33)εzp=-ηbλ.According to (31) and (32), we have(34)ηεθp+εrp=0.When the material enters the plastic state, the elastic and plastic strains constitute the strain, which is(35)εr=εre+εrp=1Eσr-μσθ+σz+εrp,(36)εθ=εθe+εθp=1Eσθ-μσr+σz+εθp,(37)εz=εze+εzp=1Eσz-μσr+σθ+εzp,where σr and σθ are the radial and tangential stresses in the plastic zone (see (26) and (27)), respectively;σz is the axial stress (i.e., (11)).Combining (34)–(36), we have(38)ηεθ+εr=ηEσθ-μσr+σz+1Eσr-μσθ+σz.Substituting (11) into (38) and combining (26) and (27), the tangential strain in the plastic zone is derived as follows:(39)εri=-ηiεθi+CiFirζi+Giwhere(40)Fi=--2ηi+2μi+ηiμimi+μimiΦi+2Ciηiμi-1+μimiCiηi+12EiCi,Gi=--2ηi+2μi+ηiμimi+μimiΨi+2ξiηiμi-1+μimiξiηi+12Ei.Substituting (39) into (4), we have(41)εθi=CiFi1+ηi+ζirζi+Dir-1+ηi+Gi1+ηi.Combining (2) and (41), the radial displacement in the plastic zone can be obtained as(42)ui=CiFi1+ηi+ζirζi+1+Dir-ηi+Gi1+ηir.To obtain the stresses and displacements in plastic zones, the coefficientsAi,Bi,Ci andDi (i=1, 2) need to be calculated.


## 3.3. Elastoplastic Analysis

When a multilayered thick-walled hollow cylinder is subjected to hydrostatic pressure on the outer boundary, the plastic yielding will always initiate from the inner boundary of each layer. Moreover, the extending path of plastic yielding is dependent on the elastic modulusEi, material strengthσci, and dimensionsRi. Specifically, for a two-layered hollow lining, plastic yielding may initiate in three different ways (see Figure 2; rp and rpi are the dimensions of plastic zones). Hereby, we only discuss the solution of the most complicated situation (Figure 2(c)). The other two simpler cases can be discussed in a similar way.Figure 2
The three stress states in a two-layered lining under hydrostatic in situ stresses.
(a)
Plastic zone in inner layer (R0≤rp≤R1) (b)
Plastic zone in outer layer (R1≤rp≤R2) (c)
Plastic zones in both layers (R0≤rp1≤R1,R1≤rp2≤R2)The equations for the undetermined coefficients can be obtained according to the boundary and continuity conditions, which include the stress boundary conditions at the inner boundary of the inner layerr=R0, σr1p=0; the radial stress and displacement continuity conditions at three interfaces (i.e., at the elastic-plastic interface of the inner layerr=rp1, σr1p=σr1e and u1p=u1e; at the inter-boundaryr=R1, σr1e=σr2p and u1e=u2p; at the elastic-plastic interface of the outer layerr=rp2, σr2p=σr2e and u2p=u2e); and the stress boundary conditions at the outer boundary of the outer layerr=R2, σr2e=p. According to the conditions, we have(43)C1R0ζ1+ξ1=0,(44)C1rp1ζ1+ξ1=A1-B1rp12,(45)C1F11+η1+ζ1rp1ζ1+1+D1rp1-η1+G11+η1rp1=1-μ1-m1μ1A1rp12+1+μ1B1E1rp1,(46)A1-B1R12=C2R1ζ2+ξ2,(47)1-μ1-μ1m1A1R12+1+μ1B1E1R1=C2F21+η2+ζ2R1ζ2+1+D2R1-η2+G21+η2R1,(48)A2-B2rp22=C2rp2ζ2+ξ2,(49)C2F21+η2+ζ2rp2ζ2+1+D2rp2-η2+G21+η2rp2=1-μ2-μ2m2A2rp22+1+μ2B2E2rp2,(50)A2-B2R22=p.There are eight undetermined variablesAi,Bi,Ci,Di (i=1 and 2) and eight equations (i.e., (43)–(50)). To obtain the radii of the plastic zones (i.e.,rp1 andrp2), two more equations are required. From the extending path of plastic yielding, the stresses on the inner boundaries of the two elastic zones should also satisfy the Unified Strength Theory. Thus, we have(51)σθ1e-η1bσz1+σr1e-α1fc1=0,(52)σθ2e-η2bσz2+σr2e-α2fc2=0.Substituting (8) and (14) into (51) and (52), we have(53)A1+B1rp12-η1b1m1A1+A1-B1rp12=ft1,(54)A2+B2rp22-η2b2m2A2+A2-B2rp22=ft2.With the increase of the loadp, the extending path of plastic yielding can be obtained from (53) and (54).


## 4. Results and Discussions

### 4.1. Material Properties

For comparing, the geometry and material properties (see Table1) are taken the same as in [29]. LetK denote the ratio of Young’s moduli; that is,K=E1/E2.Table 1
Parameters of examples.
Single-layered lining Two-layered lining (Case i) Two-layered lining (Case ii) Radii,R (m) R 0=0.3, R2=0.5 R 0=0.3, R1=0.4, R2=0.5 R 0=0.3, R1=0.4, R2=0.5 Young’s modulus,E (GPa) E 1=E2=41 (K=41/41) E 1=25, E2=41 (K=25/41) E 1=41, E2=25 (K=41/25) Poisson’s ratio,μ 0.24 0.24 0.24 UCS,fc (MPa) 50 f c 1 = f c 2 =50 f c 1 = f c 2 =50


### 4.2. Effect of K on Stresses, Displacements, and Plastic Zones

Under the parameters listed in Table1, the radial and tangential stresses are discussed. To compare the stresses and displacements distributions in different situations, we take the value of support pressurep=16 and 20 MPa as examples. In this section, the intermediate principal stress is considered, takingb1=b2=0.5. The ratioK=25/41 and 41/25, denoting two kinds of multilayered structures, andK=41/41 representing the traditional homogenous single-layered lining. The radial and tangential stresses in the radial direction under different confining pressures are illustrated in Figures 3 and 4. The distributions of the tangential stress σθ differ significantly for the three conditions, but the difference in the radial stress σr is relatively small.Figure 3
Stresses under confining pressurep=16 Mpa.Figure 4
Stresses under confining pressurep=20 Mpa.The tangential stress concentration is the main reason of the plastic yielding. Usually, the largest tangential stress concentration always occurs on the inner boundary of each layer. Besides, the stresses there are two-dimensional, so the inner boundary is the most yielding-prone zone. Asp=16 MPa (see Figure 3), the plastic yielding only can be found in the area of the inner boundary of the inner layer asK=41/25. For this case, the greatest tangential stress concentration factorσθ/p=3.538, while, for the other two casesK=41/41 and 25/41, at the same position, the tangential stress concentration factorσθ/p is 3.125 and 2.486, decreasing by 11.7% and 29.7% compared to the case ofK=41/25, respectively. Moreover, for the caseK=25/41, the greatest tangential stress concentration,σθ/p=3.078, occurs at the inner boundary of the outer layer where the radial stress is nonzero, so the stress there is three-dimensional. As known to all, the material under three-dimensional stress has higher strength than that under two-dimensional stress. Thus, for the three cases, the most reasonable stress distribution is in the lining ofK=25/41, and the worst is that ofK=41/25. As p=20 MPa (see Figure 4), the plastic yielding can be found in the cases ofK=41/25 and 41/41, with the thicknesses of 0.097 m and 0.028 m, respectively. However, the whole lining ofK=25/41 is still in the elastic state. Thus, according to the stress distributions and the plastic area, we can come to the fact that the structure ofK=25/41 is the optimal one in the three cases.Figure5 illustrates the extending path of the plastic yielding forK=25/41, 41/41, and 41/25. To illustrate the elastoplastic stages, the radial displacement at the inner boundary of the linings is plotted in Figure 6. With the increase ofp, for the case ofK=41/25, the plastic yielding first happens in the inner layer, initiating from the inner boundary asp=14.639 MPa (i.e., the elastic ultimate bearing capacity, point H in Figures 5 and 6) and finishing at the outer boundary asp=20.080 MPa (point I). After that, the structure turns into the elastic state (stage IJ) before plastic yielding happens in the outer layer asp=24.425 MPa (point J), and finally the outer layer totally turns into the plastic yielding state asp=26.399 MPa (i.e., the plastic ultimate bearing capacity point K).Figure 5
Extending path of the plastic zone.Figure 6
Radial displacements on the inner boundary.For the conventional single-layered structure (K=41/41), the plastic yielding is found asp=17.360 MPa (point E, the elastic ultimate bearing capacity) at the inner boundary and extends to the outer boundary of the whole lining asp=26.399 MPa (point G, the plastic ultimate bearing capacity). In contrast with the other two-layered structures, the single-layered lining is considered as a two-layered lining as well, just with the same material parameters in both layers. In this way, point F corresponds to the moment that the plastic yielding extends from the inner layer to the outer layer, of which the confining pressurep=24.425 MPa. Note that the plastic zone extending stage FG is exactly the same as the stage JK of the two-layered lining (K=41/25), but with different radial displacements (Figure 6).By contrast, for the other two-layered lining (K=25/41), the plastic yielding happens asp=20.560 MPa (point A) from the inner boundary of the outer layer, not from the inner layer. The plastic yielding cannot be found in the inner layer until the confining pressurep increases to 21.769 MPa, when the dimension of the plastic zone in the outer layer is 0.014 m, about 14% of the layer thickness. With the increase ofp, the plastic yielding extends in both layers simultaneously (stages B′C and BC′), then the outer layer first turns into the plastic state totally asp=25.925 MPa (point C), and finally the inner layer asp=26.399 MPa (point D, the plastic ultimate bearing capacity).The case ofK=25/41 has the greatest elastic ultimate bearing capacity 20.560 MPa, about 40.4% and 18.4% higher than the cases ofK=41/25 and 41/41, respectively. Moreover, the three cases have the same plastic ultimate bearing capacity 26.399 MPa. Thus, from the aspect of the elastic bearing capacity, the lining ofK=25/41 performs the best among the three structures. Actually, for a multilayered lining, the most expected condition is like this: the plastic yielding initiates and finishes in every layer simultaneously; in this way, the material strength can be most utilized. This expected condition can be realized by optimizing the ratioK, the radiusR0, and the thicknesses of each layer [29]. According to the inverse analysis, the general rule of the optimized result is that Young’s modulus needs to be increased monotonously from the inner to the outer layer [29].


### 4.3. Comparison with the Tresca Criterion

As mentioned in Section3.2, parameterb can reflect the effect of the intermediate principal stress on the plastic yielding. Usually, the intermediate principal stress is helpful in increasing the material strength. To analyze this issue, the variations of the elastic bearing capacity withb are plotted in Figure 7. The analytical results support the above point: with the increase in valueb, the elastic ultimate bearing capacity increases monotonously (see Figure 7). Specifically, as the value b increases from 0.0 to 1.0, the elastic ultimate bearing capacities of the three cases ofK=25/41, 41/41, and 41/25 increase by 6.5%, 13.3%, and 13.3%, respectively. In other words, when the intermediate stress is considered, the elastic ultimate bearing capacity will increase, which is in line with the expectations.Figure 7
Relationship betweenb and the elastic bearing capacity.To compare with the Tresca Criterion [29], takingb=0 and α=1, UST becomes the Tresca Criterion. The extending paths of the plastic zone and the radial displacement of the inner boundary based on the two criteria are plotted in Figures 8 and 9, respectively. Under a given confining pressure, the Tresca-based plastic zone is larger than the UST-based plastic zone. The UST-based elastic and plastic bearing capacities are greater than the corresponding Tresca-based bearing capacities. Specifically, for the single-layered lining (i.e.,K=41/41), the UST-based elastic bearing capacity is 17.360 MPa, 8.5% higher than the Tresca-based value. For the two-layered lining (i.e.,K=25/41), the corresponding increment is 4.1%, from 19.741 to 20.560 MPa. For a certain yielding criterion, the plastic bearing capacity keeps unchanged, which depends on the material strength. The UST-based plastic bearing capacities of the single-layered and two-layered structures are the same, 26.399 MPa, 3.4% higher than the Tresca-based plastic bearing capacity, 25.541 MPa. The two figures also show that, for any of the two criteria, the two-layered lining (K=25/41) has greater elastic bearing capacity than the single-layered lining. The test data of the two casesK=25/41 and 41/41 are plotted in Figure 6 as well [29], which verifies the conclusion. The radial displacements by tests are smaller than the theoretical results, which is related to the methods of the tests.Figure 8
Extending path of the plastic zone.Figure 9
Radial displacements on the inner boundary.


## 4.1. Material Properties

For comparing, the geometry and material properties (see Table1) are taken the same as in [29]. LetK denote the ratio of Young’s moduli; that is,K=E1/E2.Table 1
Parameters of examples.
Single-layered lining Two-layered lining (Case i) Two-layered lining (Case ii) Radii,R (m) R 0=0.3, R2=0.5 R 0=0.3, R1=0.4, R2=0.5 R 0=0.3, R1=0.4, R2=0.5 Young’s modulus,E (GPa) E 1=E2=41 (K=41/41) E 1=25, E2=41 (K=25/41) E 1=41, E2=25 (K=41/25) Poisson’s ratio,μ 0.24 0.24 0.24 UCS,fc (MPa) 50 f c 1 = f c 2 =50 f c 1 = f c 2 =50


## 4.2. Effect of K on Stresses, Displacements, and Plastic Zones

Under the parameters listed in Table1, the radial and tangential stresses are discussed. To compare the stresses and displacements distributions in different situations, we take the value of support pressurep=16 and 20 MPa as examples. In this section, the intermediate principal stress is considered, takingb1=b2=0.5. The ratioK=25/41 and 41/25, denoting two kinds of multilayered structures, andK=41/41 representing the traditional homogenous single-layered lining. The radial and tangential stresses in the radial direction under different confining pressures are illustrated in Figures 3 and 4. The distributions of the tangential stress σθ differ significantly for the three conditions, but the difference in the radial stress σr is relatively small.Figure 3
Stresses under confining pressurep=16 Mpa.Figure 4
Stresses under confining pressurep=20 Mpa.The tangential stress concentration is the main reason of the plastic yielding. Usually, the largest tangential stress concentration always occurs on the inner boundary of each layer. Besides, the stresses there are two-dimensional, so the inner boundary is the most yielding-prone zone. Asp=16 MPa (see Figure 3), the plastic yielding only can be found in the area of the inner boundary of the inner layer asK=41/25. For this case, the greatest tangential stress concentration factorσθ/p=3.538, while, for the other two casesK=41/41 and 25/41, at the same position, the tangential stress concentration factorσθ/p is 3.125 and 2.486, decreasing by 11.7% and 29.7% compared to the case ofK=41/25, respectively. Moreover, for the caseK=25/41, the greatest tangential stress concentration,σθ/p=3.078, occurs at the inner boundary of the outer layer where the radial stress is nonzero, so the stress there is three-dimensional. As known to all, the material under three-dimensional stress has higher strength than that under two-dimensional stress. Thus, for the three cases, the most reasonable stress distribution is in the lining ofK=25/41, and the worst is that ofK=41/25. As p=20 MPa (see Figure 4), the plastic yielding can be found in the cases ofK=41/25 and 41/41, with the thicknesses of 0.097 m and 0.028 m, respectively. However, the whole lining ofK=25/41 is still in the elastic state. Thus, according to the stress distributions and the plastic area, we can come to the fact that the structure ofK=25/41 is the optimal one in the three cases.Figure5 illustrates the extending path of the plastic yielding forK=25/41, 41/41, and 41/25. To illustrate the elastoplastic stages, the radial displacement at the inner boundary of the linings is plotted in Figure 6. With the increase ofp, for the case ofK=41/25, the plastic yielding first happens in the inner layer, initiating from the inner boundary asp=14.639 MPa (i.e., the elastic ultimate bearing capacity, point H in Figures 5 and 6) and finishing at the outer boundary asp=20.080 MPa (point I). After that, the structure turns into the elastic state (stage IJ) before plastic yielding happens in the outer layer asp=24.425 MPa (point J), and finally the outer layer totally turns into the plastic yielding state asp=26.399 MPa (i.e., the plastic ultimate bearing capacity point K).Figure 5
Extending path of the plastic zone.Figure 6
Radial displacements on the inner boundary.For the conventional single-layered structure (K=41/41), the plastic yielding is found asp=17.360 MPa (point E, the elastic ultimate bearing capacity) at the inner boundary and extends to the outer boundary of the whole lining asp=26.399 MPa (point G, the plastic ultimate bearing capacity). In contrast with the other two-layered structures, the single-layered lining is considered as a two-layered lining as well, just with the same material parameters in both layers. In this way, point F corresponds to the moment that the plastic yielding extends from the inner layer to the outer layer, of which the confining pressurep=24.425 MPa. Note that the plastic zone extending stage FG is exactly the same as the stage JK of the two-layered lining (K=41/25), but with different radial displacements (Figure 6).By contrast, for the other two-layered lining (K=25/41), the plastic yielding happens asp=20.560 MPa (point A) from the inner boundary of the outer layer, not from the inner layer. The plastic yielding cannot be found in the inner layer until the confining pressurep increases to 21.769 MPa, when the dimension of the plastic zone in the outer layer is 0.014 m, about 14% of the layer thickness. With the increase ofp, the plastic yielding extends in both layers simultaneously (stages B′C and BC′), then the outer layer first turns into the plastic state totally asp=25.925 MPa (point C), and finally the inner layer asp=26.399 MPa (point D, the plastic ultimate bearing capacity).The case ofK=25/41 has the greatest elastic ultimate bearing capacity 20.560 MPa, about 40.4% and 18.4% higher than the cases ofK=41/25 and 41/41, respectively. Moreover, the three cases have the same plastic ultimate bearing capacity 26.399 MPa. Thus, from the aspect of the elastic bearing capacity, the lining ofK=25/41 performs the best among the three structures. Actually, for a multilayered lining, the most expected condition is like this: the plastic yielding initiates and finishes in every layer simultaneously; in this way, the material strength can be most utilized. This expected condition can be realized by optimizing the ratioK, the radiusR0, and the thicknesses of each layer [29]. According to the inverse analysis, the general rule of the optimized result is that Young’s modulus needs to be increased monotonously from the inner to the outer layer [29].


## 4.3. Comparison with the Tresca Criterion

As mentioned in Section3.2, parameterb can reflect the effect of the intermediate principal stress on the plastic yielding. Usually, the intermediate principal stress is helpful in increasing the material strength. To analyze this issue, the variations of the elastic bearing capacity withb are plotted in Figure 7. The analytical results support the above point: with the increase in valueb, the elastic ultimate bearing capacity increases monotonously (see Figure 7). Specifically, as the value b increases from 0.0 to 1.0, the elastic ultimate bearing capacities of the three cases ofK=25/41, 41/41, and 41/25 increase by 6.5%, 13.3%, and 13.3%, respectively. In other words, when the intermediate stress is considered, the elastic ultimate bearing capacity will increase, which is in line with the expectations.Figure 7
Relationship betweenb and the elastic bearing capacity.To compare with the Tresca Criterion [29], takingb=0 and α=1, UST becomes the Tresca Criterion. The extending paths of the plastic zone and the radial displacement of the inner boundary based on the two criteria are plotted in Figures 8 and 9, respectively. Under a given confining pressure, the Tresca-based plastic zone is larger than the UST-based plastic zone. The UST-based elastic and plastic bearing capacities are greater than the corresponding Tresca-based bearing capacities. Specifically, for the single-layered lining (i.e.,K=41/41), the UST-based elastic bearing capacity is 17.360 MPa, 8.5% higher than the Tresca-based value. For the two-layered lining (i.e.,K=25/41), the corresponding increment is 4.1%, from 19.741 to 20.560 MPa. For a certain yielding criterion, the plastic bearing capacity keeps unchanged, which depends on the material strength. The UST-based plastic bearing capacities of the single-layered and two-layered structures are the same, 26.399 MPa, 3.4% higher than the Tresca-based plastic bearing capacity, 25.541 MPa. The two figures also show that, for any of the two criteria, the two-layered lining (K=25/41) has greater elastic bearing capacity than the single-layered lining. The test data of the two casesK=25/41 and 41/41 are plotted in Figure 6 as well [29], which verifies the conclusion. The radial displacements by tests are smaller than the theoretical results, which is related to the methods of the tests.Figure 8
Extending path of the plastic zone.Figure 9
Radial displacements on the inner boundary.


## 5. Conclusion

The multilayered circular structures have been proved to have some advantages compared to the traditional single-layered structure. An elastoplastic analysis details such structure; a two-layered circular lining is conducted based on the Unified Strength Theory, which can take the intermediate principal stress into consideration. The purpose of the work is to discuss the bearing capacities of different kinds of structures. The results show that the distributions of stresses in the two-layered lining are more reasonable compared to the traditional single-layered lining. The two-layered Lining has greater elastic bearing capacity, but the plastic bearing capacity keeps unchanged for the same criterion. Considering the intermediate principal stress, the elastic bearing capacity increases correspondingly. Compared to the Tresca-based results, the UST-based elastic bearing capacities are greater, and the plastic zones are smaller, which is due to the intermediate principal stress. The analytical conclusions are also verified by the results of the model tests.



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*Source: 4949364-2018-10-02.xml*