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

1 Institute of Hydroelectric and Geotechnical Engineering, North China Electric Power University, Beijing 102206, China 2School of Qilu Transportation, Shandong University, Jinan 250061, China 3College of Engineering and Technology, Southwest University, Chongqing 400715, China 4State Key Laboratory of Mining Disaster Prevention and Control, Shandong University of Science and Technology, Qingdao 266590, China 5Guizhou Hongxin Changda Engineering Detection & Consultation Co. LTD., Guiyang 550008, China


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 deepburied 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, Lu et al. [4], by using the semiinverse 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 thickwalled hollow cylinder, there also have been many solutions on stress, displacement distributions, and elastic, plastic zones based on different strength theories [6][7][8][9][10][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, Gao et al. [14] considered the size effect of strain-hardening level.And in the fields of elasticbrittle-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 Zhang et 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, Lu et 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.Wang et 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 Figure 1).Moreover, the ultimate bearing capacities are to be investigated as well.In Figure 1, E i ,  i , and  c (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 is p.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.

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 where   and   are the radial and tangential strains, respectively.Moreover, u is the radial displacement, and r is the radial distance from the center of opening.The differential equation of equilibrium for an axisymmetric problem is where   and   are the tangential and radial stresses, respectively.
The compatibility equation of strain can be given as The generalized Hooke's Law is given as follows: where   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 The subscript i=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 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 coefficients A i and B i (i=1 and 2).
The Unified Strength Theory is applied herein.This strength theory can take the intermediate principal stress   into consideration.For the axisymmetric plane strain problem, there exist three non-zero principal stresses (i.e., the radial, circumferential, and axial stresses,   ,   , and   ).Compression is taken as positive.For the plane strain problem, the axial stress   can be given as follows [23]: where m is the parameter reflecting the state of the intermediate principal stress.In elastic zones,   = 2  .As the inplane plastic strains increase,   approaches the average of the minor and major principal stresses (i.e., m i tends to 1).For simplicity, in plastic zones, m is assumed to be 1 [24].Then, (11) can be expressed as The parameter m takes 0 and 2 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 p And the radial displacement in elastic zones can be obtained by substituting ( 14) into (10) as

Elastoplastic Analysis
. .Elastic Analysis.As the axisymmetric external load p 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 Based on ( 16)-( 19), the undetermined coefficients A i , B i (i=1, 2) can be solved, which are presented in the Appendix.

. . 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 where  1 >  2 >  3 , and  denotes the ratio of the uniaxial tensile to the compressive strengths (i.e., =f t /f c ) and 0<⩽1 for most engineering materials.In this work, we take =(1-sin)/(1+sin).Moreover, f t and f c 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  and b take specific values [23].For example, if b=0 and =1, it is the Tresca Criterion.If b=1/2 or =1 and b = 1/(1 + √ 3), it is the linear approximations of the von Mises criterion.The Mohr-Coulomb criterion can be obtained as b=0 and 0<<1.If b=1 and 0<<1, it becomes the generalized twin shear stress criterion.Specifically, when =1 and b=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 stress  z should be the intermediate principal stress.The three principal stresses satisfy the relationship,   >  z >  r ; thus, we have And we also have According to ( 22) and ( 23), ( 20) can be satisfied.Then, Equation ( 24) can be expressed as the form of yield function as follows: where  = /(1 + ) and   =   .Combining (3), (11), and ( 25), we have where Combining Eq. ( 11), substituting Eq. ( 26) into Eq.( 24), we have where The flow rule related with the Unified Strength Theory is employed: From ( 28)-(30), we have According to (31) and (32), we have When the material enters the plastic state, the elastic and plastic strains constitute the strain, which is where   and   are the radial and tangential stresses in the plastic zone (see (26) and ( 27)), respectively;  z is the axial stress (i.e., (11)).Combining (34)-(36), we have Substituting ( 11) into (38) and combining ( 26) and ( 27), the tangential strain in the plastic zone is derived as follows: where Substituting (39) into (4), we have Combining ( 2) and (41), the radial displacement in the plastic zone can be obtained as To obtain the stresses and displacements in plastic zones, the coefficients A i , B i , C i and D i (i=1, 2) need to be calculated.
. .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 modulus E i , material strength  ci , and dimensions R i .Specifically, for a two-layered hollow lining, plastic yielding may initiate Figure 2: The three stress states in a two-layered lining under hydrostatic in situ stresses.
in three different ways (see Figure 2;  p and  p 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.
The equations for the undetermined coefficients can be obtained according to the and continuity conditions, which include the stress boundary conditions at the inner boundary of the inner layer r=R 0 ,   1 = 0; the radial stress and displacement continuity conditions at three interfaces (i.e., at the elastic-plastic interface of the inner layer r=r p1 , There are eight undetermined variables A i , B i , C i , D i (i=1 and 2) and eight equations (i.e., (43)-( 50)).To obtain the radii of the plastic zones (i.e., r p1 and r p2 ), 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 Substituting ( 8) and ( 14) into (51) and (52), we have With the increase of the load p, the extending path of plastic yielding can be obtained from ( 53) and (54).

Results and Discussions
. .Material Properties.For comparing, the geometry and material properties (see Table 1) are taken the same as in [29].Let K denote the ratio of Young's moduli; is, K=E 1 /E 2 .

. . Effect of K on Stresses, Displacements, and Plastic Zones.
Under the parameters listed in Table 1, the radial and tangential stresses are discussed.To compare the stresses and displacements distributions in different situations, we take the value of support pressure p=16 and 20 MPa as examples.In this section, the intermediate principal stress is considered, taking b 1 =b 2 =0.5.The ratio K=25/41 and 41/25, denoting two kinds of multilayered structures, and K=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   differ significantly for the three conditions, but the difference in the radial stress   is relatively small.
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.As p=16 MPa (see Figure 3), the plastic yielding only can be found in the area of the inner boundary of the inner layer as K=41/25.For this case, the greatest tangential stress concentration factor   /p=3.538,while, for the other two cases K=41/41 and 25/41, at the same position, the tangential stress concentration factor   /p is 3.125 and 2.486, decreasing by 11.7% and 29.7% compared to the case of K=41/25, respectively.Moreover, for the case K=25/41, the greatest tangential stress concentration,   /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 of K=25/41, and the worst is that of K=41/25.As p=20 MPa (see Figure 4), the plastic yielding can be found in the cases of K=41/25 and 41/41, with the thicknesses of 0.097 m and 0.028 m, respectively.However, the whole lining of K=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 of K=25/41 is the optimal one in the three cases.
Figure 5 illustrates the extending path of the plastic yielding for K=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 of p, for the case of K=41/25, the plastic yielding first happens in the inner layer, initiating from the inner boundary as p=14.639MPa (i.e., the elastic ultimate bearing capacity, point H in Figures 5  and 6) and finishing at the outer boundary as p=20.080MPa (point I).After that, the structure turns into the elastic state (stage IJ) before plastic yielding happens in the outer layer as p=24.425MPa (point J), and finally the outer layer totally turns into the plastic yielding state as p=26.399MPa (i.e., the plastic ultimate bearing capacity point K).
For the conventional single-layered structure (K=41/41), the plastic yielding is found as p=17.360MPa (point E, the elastic ultimate bearing capacity) at the inner boundary and extends to the outer boundary of the whole lining as p=26.399MPa (point G, the plastic ultimate bearing capacity).In contrast with the other two-layered structures, the singlelayered 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 pressure p=24.425MPa.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 as p=20.560MPa (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 pressure p increases to 21.769 MPa, when the  The case of K=25/41 has the greatest elastic ultimate bearing capacity 20.560 MPa, about 40.4% and 18.4% higher than the cases of K=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 of K=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 ratio K, the radius R 0 , 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].
. .Comparison with the Tresca Criterion.As mentioned in Section 3.2, parameter b 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 with b are plotted in Figure 7.The analytical results support the above point: with the increase in value b, 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 of K=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.
To compare with the Tresca Criterion [29], taking b=0 and =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 Trescabased 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 Mathematical Problems in Engineering     capacity than the single-layered lining.The test data of the two cases K=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.

Conclusion
The multilayered circular structures have been proved to have some advantages compared to the traditional single-layered structure.An

𝑒 1 ;
at the inter-boundary r=R 1 ,   1 =   2 and   1 =   2 ; at the elastic-plastic interface of the outer layer r=r p2 ,   2 =   2 and   2 =   2 ); and the stress boundary conditions at the outer boundary of the outer layer r=R 2 ,   2 = .According to the conditions, we have

Figure 7 :Figure 8 :
Figure 7: Relationship between b and the elastic bearing capacity.

Figure 9 :
Figure 9: Radial displacements on the inner boundary.

Table 1 :
Parameters of examples.
Figure 6: Radial displacements on the inner boundary.dimension of the plastic zone in the outer layer is 0.014 m, about 14% of the layer thickness.With the increase of p, the plastic yielding extends in both layers simultaneously (stages B  C and BC  ), then the outer layer first turns into the plastic state totally as p=25.925MPa (point C), and finally the inner layer as p=26.399MPa (point D, the plastic ultimate bearing capacity).
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.