外文翻译-原文 相邻基坑开挖对地铁隧道变形影响的分析研究

An analytic study on the deflection of subway tunnel dueto adjacent excavation of foundation pitZelin Zhou1Shougen Chen1Peng Tu1Haisheng Zhang1Received: 5 March 2015/Revised: 21 October 2015/Accepted: 23 October 2015/Published online: 19 November 2015C211 The Author(s) 2015. This article is published with open access at SAbstract Predicting and estimating the response of sub-way tunnel to adjacent excavation of foundation pit is aresearch focus in the field of underground engineering.Based on the principle of two-stage method and incre-mental method, an analytic approach is suggested in thispaper to solve this problem in an accurate and rapid way,and the upheavals of tunnel due to adjacent excavation aresolved by analytic method. Besides, the presented methodis used in the practical engineering case of Shenzhen MetroLine 11 and verified by numerical simulation and in situmeasurement. Finally, a parametric analysis is performedto investigate the influence of different factors on tunnelsdeflection. Some useful conclusions have been drawn fromthe research as below: The deflection results of tunnelobtained from analytic method are nearly consistent withthe results getting from numerical analysis and measureddata, which verified the accuracy and rationality of pre-sented method. The excavation size has a significantimpact on both the displacement values and influencedrange of tunnel. However, the relative distance onlyimpacts the displacement values of tunnel, but not theinfluenced range of tunnel. It may provide certain referenceto analyze the deflection of subway tunnel influenced byadjacent excavation.Keywords Subway tunnel C1 Upheaval deflection C1Excavation of foundation pit C1 Two-Stage Method C1Parametric analysis1 IntroductionWith the rapid development of urban underground space,more and more excavations adjacent to underground spaceare constructed, in which the soil unloading due to adjacentexcavations will lead to an uplift of underlying tunnels. Asa life line of the city transport, the criterion of allowabledeflection of subway tunnels is very strict. According to thedesign code of building foundation in China, the uplift ofexisting subway tunnels cannot be bigger than 10.0 mm,while the deflection radius cannot be less than 15,000.0 m.A typical case to damage a tunnel in the Pachiao line due tonearby excavations in Taipei caused a big loss 1.Therefore, effectively predicting the tunnel deflection insuch cases is very important in order to reduce the risk andhas recently become a big concern in undergroundconstructions.Many cases of interaction behavior between excavationsand existing tunnels have been studied using numericalmodeling and analytical studies. For example, Dolezalova2 used a 2D numerical model to analyze the deformationof a tunnel underlying a deep-open excavation. Gao et al.3 investigated the influence of excavations on a nearbyroad tunnel using 3D FEM. Hu et al. 4 used FEM toinvestigate the deformation of subway tunnels due toadjacent pit excavations. The numerical modeling is pow-erful to deal with the pit excavation steps and can considernonlinear interactions between tunnels and surroundingsoil. However, the reliability of the modeling resultdepends greatly on the constitutive model and hypotheticmaterial parameters.In general, the analytical method allows a convenient andrapid approach to estimate the tunnel deflection for engi-neers. Ji et al. 5 presented a simple analytical method,which is called residual stress method (RSM), to analyze they 2C0L=2C24L=2; z H;C1: x C0B=2; y 2C0L=2C24L=2; z 20C24H;C2: x B=2; y 2C0L=2C24L=2; z 20C24H;C3: x 2C0B=2C24B=2; y C0L=2; z 20C24H;C4: x 2C0B=2C24B=2; y L=2; z 20C24H:9=;1Assumptions are made for the analysis as follows: (1)the soil is assumed to be a homogenous and elastic solidmedium; (2) the underlying tunnel is assumed to be anEulerBernoulli beam, and the deflection between the(a)o2134x()BLTunnelDFoundation pit excavationy()(b)z()x()H1 2O0T1BGround surfacex0,y,z0Existing tunnelPit-excavationHorizontal force ofsupport structuresSoil horizontal released loadSoil vertical released loadT2TnT1T2Tn Fig. 1 Sketch of analytical model. a Front view. b Plan view288 Z. Zhou et al.123J. Mod. Transport. (2015) 23(4):287297tunnel and surrounding soil is compatible; and (3) thetunnel is much longer comparing to the tunnel diameter,the influence of tunnel cross section is ignored.2.2 Equivalent released load on the tunnel dueto the excavationThe tunnel deflection is caused by the released load invertical direction as being discussed in Sect. 3. As men-tioned previously, the release load comes not only from thevertical unloading at the pit bottom, but also from thehorizontal unloading and contribution of support structureson the pit sidewalls.2.2.1 Equivalent released load caused by the verticalunloading at the pit bottomBefore excavation, there is a vertical stress distributing atthe pit bottom, which can be calculated as cH, where c isthe unit weight of soil. Obviously, the excavation loadacting on the pit bottom P in the opposite direction is equalto cH.By employing Mindlin solution 9, the vertical stressat the tunnel (i.e., the equivalent released load on thetunnel) should ber0zx0;y;z0ZZC0P8pt1C1z0C0 HR31C0z0C0 HR32C18C19t2C13z0z0 H2R52( t33z0C0 H3R51C03Hz0 H5z0C0 HR52“30Hz0z0 H3R72#)de dg; 2where r0zx0;y;z0 is the vertical stress at the tunnel withcoordinates of (x0, y, z0); t is the Poissons ratio of soil; t1,t2, and t3are constants given by t1= (1-2t)/(1-t),t2= (3-4t)/(1-t), and t3= 1/(1-t), respectively;variables R1and R2areR1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix C0e2y C0g2z C0 H2q;R2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix C0e2y C0g2z H2q;9=;3where e;g;H is the coordinates of points at the bottom ofthe pit.2.2.2 Equivalent released load caused by horizontalunloading and supports on the pit sidewallsThe horizontal unloading on the sidewalls of the pitdepends on the interaction of the soil with the retainingwalls and lateral braces. The incremental method can beadopted to analyze the support effect on the sidewalls ofthe pit, using the computational model as shown in Fig. 2.Before excavation as shown in Fig. 2a, the soil pressureat the retaining wall can be computed as K0cH, where K0isthe lateral pressure coefficient.The excavation of the pit is usually from top to bottomin several steps and the depth of each step is hi. Accordingto the theory of incremental method 10, 11, there will bethree parts of incremental loads acting on the retaining wallat each step including the incremental soil load, theincremental spring load, and the support prestressing load.(1) The incremental soil load Dqnwhich is induced bythe previous step excavation, can be expressed asDqncz C0 hnC01K0hnC01C20zC20hnchnC0 hnC01K0zC21hn;C264where the superscript n denotes the step n, h(n-1), andh(n)are the excavation depths at the previous step andcurrent step, respectively. The load Dqncan beconverted to a load vector Dqnusing one-dimen-sional finite element modeling.(2) The incremental spring load f(n), which is induced bythe current step excavation (equivalent to eliminationof the springs), can be expressed as,fn Kns1C1DdnC01; 5where Kns1is the stiffness matrix of eliminatedsprings at the current step, and DdnC01is the dis-placement of retaining wall at the previous step.(3) The support prestressing load Tnis induced by theprestressing force being applied to the lateral braces.The three parts of the incremental load above are appliedto the retaining wall, lateral braces, and remaining springs.Therefore, the finite element equation can be expressed asKns2 Knr KnC01bC1Ddn Dqn0 fn Tn; 6where Kns2, Knr, and KnC01bare the stiffness matrix ofremaining springs at current step, the stiffness matrix ofretaining wall at current step, and the stiffness matrix oflateral braces at previous step, respectively.Since Eq. (6) is so complicated to solve theoretically, acomputing program has been developed based on the finiteAn analytic study on the deflection of subway tunnel due to adjacent excavation of 289123J. Mod. Transport. (2015) 23(4):287297element theory in order to solve Eq. (6). Once the pit isexcavated up to the pit bottom, the final displacement ofthe retaining wall can be calculated asd XNsn1Ddn; 7where Nsis number of excavation steps.The horizontal released load of soil mass on the sidewallof the pit, Qs, can be calculated asQsXNsn1Dqn: 8Due to the interaction of the retaining wall and soil, thehorizontal support force provided by the retaining wall, Qr,can be calculated asQr KNsrC1d: 9Similarly, the horizontal support force provided by theith lateral brace, Qbi, can be calculated asQbi EiAidi; 10where EiAiis stiffness of the ith lateral brace, and diishorizontal displacement of the ith lateral brace.The vectors Qsand Qrcan be fitted as a continuousfunction of Qsand Qrusing the spline function fittingmethod. Based on the above, the horizontal released loadand horizontal support force of retaining wall, Qrs, can beexpressed asQrs Qs Qr: 11Based on the Mindlin solution, the vertical stress at thetunnel level induced by both horizontal released load andhorizontal support force of retaining wall on sidewall C192 ofthe pit, r1zsr, can be integrated asr1zsrx0;y;z0ZZC1Qrsx0C0e8pC2 t1C1C01R311R32C06nz0 HR52C18C19 3t2C1z0 H2R52(t33z0C0 H2R51C06n2R52C030nz0z0 H2R72“#)dg dn;12where n is the Z-coordinates of points on the pit sidewalls.Similarly, the vertical stress at the tunnel level inducedby horizontal force of lateral braces on the sidewall C192 ofthe pit, r1zb, can be expressed asZone of earth pressure at rest Zone of earth pressure at rest K0zK0z(a)(b)q(2)h(1)h(2)(2)s1K(2)s2K(2)rK()nf(2)T(c)q(1)h(1)(1)rK(1)s2K(1)TFig. 2 Computational model. a Before excavation. b Excavation ofthe first bench. c Excavation of the second bench290 Z. Zhou et al.123J. Mod. Transport. (2015) 23(4):287297r1zbx0;y;z0XNbi1Qbix0C0ei8pC2 t1C1C01R31i1R32iC06nz0 HR52iC18C19C26 3t2C1z0 H2R52it3C13z0C0 H2R51i“C06n2R52iC030nz0z0 H2R72i#); 13where Nbis number of lateral braces located at sidewall C192;Variables R1iand R2iareR1iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix C0ei2y C0gi2z C0ni2q;R2iffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix C0ei2y C0gi2z ni2q;9=;14where ei;gi;ni is the coordinates of lateral braces on thesidewall of the pit.Therefore, the equivalent released load on the tunnelaxis induced by the horizontal released load and horizontalsupport force (including retaining walls and lateral braces)on the sidewall C192, r1zx0;y;z0, can be superimposed asr1zx0;y;z0r1zsrx0;y;z0r1zbx0;y;z0: 15Similarly, the equivalent released load on the tunnel axisinduced by the horizontal released load and horizontalsupport force on the sidewalls , , and can be derivedas r2zx0;y;z0, r3zx0;y;z0, and r4zx0;y;z0;respectively.Based on the superposition principle, the equivalentreleased load at the tunnel level, rzx0;y;z0, can beobtained fromrzx0;y;z0X4i0rizx0;y;z0: 162.3 Tunnel uplift due to the pit excavationThe unloading due to the pit excavation causes upliftw(y) in the longitudinal direction of the existing tunnelbelow the pit, as shown in Fig. 3a.In order to estimate the response of the existing tunnel tothe equivalent released load, the tunnel is assumed to be acontinuous and slender beam on an elastic foundation. Toobtain a more accurate and rational result, an elasticPasternak foundation model is adopted to simulate theinteraction between the tunnel and surrounding soil asshown in Fig. 3b. The Pasternak foundation is improvedfrom the traditional Winkler foundation by adding a layerof shear units on the foundation 12. It can not only reflectthe elastic deflection of the tunnel, but also embody thecontinuity of the soil.2.3.1 Differential equations of equilibriumAs shown in Fig. 4, the external loads acting on the tunnelinclude two parts: one is the load q(y) coming from theequivalent released load, which can be expressed asqyD C1rzx0;y;z0; 17where D is the tunnel diameter.The other is the interaction load p(y) coming from theshear units and spring units in the Pasternak model, whichcan be expressed aspyGd2wydy2C0 Kwy; 18where G is the foundation shear modulus, K is the bulkmodulus, and w(y) is the uplift along the tunnel axis.To obtain the equilibrium differential equation subjectedto the external loads on the Pasternak foundation, thebalance mechanism of a beam unit can be described asshown in Fig. 5.The force balance condition of RY = 0 for the beam unitcan be written asQ C0Q dQqydy C0 pydy 0; 19where Q is the shearing force of Z-axis direction.Equation (19) can be simplified asdQdy qyC0py: 20oxPoints on the exiting tunnel longitudinal axisUpheaval deflection of exiting tunnel w(y)(a)q(y)Existing tunnelLayer of shear unitsLayer of spring units(b)Fig. 3 Uplift of the existing tunnel in longitudinal axis (a) andPasternaks foundation model (b)An analytic study on the deflection of subway tunnel due to adjacent excavation of 291123J. Mod. Transport. (2015) 23(4):287297The force balance condition of RM = 0 for the beamunit can be written asM C0M dMQ dQdy qydy22C0 pydy22 0;21where M is the bending moments of Y-axis direction.Given that the second trace can be omitted, Eq. (21)issimplified asQ dMdy: 22Taking the derivative of Eq. (22) to get an expression asdQdyd2Mdy2; 23according to Eqs. (20) and (23), a relationship can beobtained asd2Mdy2 qyC0py: 24By substituting Eqs. (17) and (18) into Eq. (24), weobtain the displacement balance equation as follows:d2Mdy2C0 Gd2wydy2 KwyDrzx0;y;z0: 25Based on the material mechanics theory, the relationshipbetween bending moment and displacement of thecontinuous beam can be expressed asd2Mdy2 EId4wydy4; 26where EI is the longitudinal bending stiffness of the tunnel.By substituting Eq. (26) into Eq. (25), the differentialequations of equilibrium for the uplift of the tunnel can beobtained asd4wydy4C0GDEId2wydy2DKEIwyqyEI: 272.3.2 The solution of the differential equationFrom Eq. (27), a fourth-order homogeneous differentialequation corresponding to Eq. (27) can be written asd4wydy4C0GDEId2wydy2DKEIwy0: 28The characteristic equation of Eq. (28)isr4C0GDEIr2KDEI 0: 29The general solutions of Eq. (29) depend on D whichcan be expressed asD GDEIC18C192C04KDEI: 30(1) When D0, the solutions of Eq. (29) arer1;2C6ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiGD ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiGD2C0 4kDEIq2EIvuutC6a; 31r3;4C6ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiGD C0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiGD2C0 4kDEIq2EIvuutC6b: 32The general solution of Eq. (28) can be written aswyA1eay A2eC0ay B1ebyC0 B1eby: 33(2) When D = 0, the solutions of Eq. (29) arer1;2ffiffiffiffiffiffiffiffiGD2EIr a; r3;4C0ffiffiffiffiffiffiffiffiGD2EIrC0a: 34The general solution of Eq. (28) can be written aswyA1 A2eayB1 B2eby: 35(3) When D0, the solutions of Eq. (29) arer1;2;3;4C6ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKD4EIrGD4EIsC6ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiKD4EIrC0GD4EIsC1iC6aC6b: 36The general solution of Eq. (28) can be written aszyoq(y)p(y)dyylFig. 4 External loads acting on the tunnelM+dMQMQ+dQq(y)dyp(y)dyFig. 5 The balance mechanism of a beam unit292 Z. Zhou et al.123J. Mod. Transport. (2015) 23(4):287297wyA1cosbyA2sinbyC138C1 eayB1cosbyB2sinbyC138C1 eC0ay: 37Assuming that there is a concentrated load acting on thetunnel, when y tends to be infinity, the value of w(y) willtend to be zero, so, the expression of w(y) can be simplifiedas,wyA1eC0ay A2eC0byD0;A1 A2yeC0ayD 0;A1cosbyA2sinbyC138eC0ayD0:8=;39According to the general solution Eq. (38) and boundarycondition Eq. (39), the uplift of the tunnel induced by aconcentrated load of qgdg can be derived asUsing the integration method, the uplift of the tunnelinduced by the pit excavation can be obtained fromEq. (40)aswyZ1C01dwydg: 413 A case study for validationThe approach described above is applied to an actualproject case, and the analytical result is compared with theresults obtained from numerical modeling and monitoringfor validating the approach.3.1 Background of the projectThe net horizontal distance between the double-hole sub-way tunnels is 12.0 m. The outside diameter of the subwaytunnel is 6.2 m and the thickness of tunnel lining is 0.35 m.The net vertical distance from the bottom of open-cut pit tothe crown of the tunnels is 11.5 m. The net horizontaldistance between the pit center and the right subway tunnelis 6.9 m.The soil enc