Mechanical Model of Tunnel Support Pipe Support

Zuo Zhuo1 Fu Helin3 Peng Wenxuan4 1,2 Changsha University of Science and technology,410007 3,4 School of Civil Engineering e of Central South University Changsha 41007 China e-mail: [email protected]

ABSTRACT In this paper, based on the principle analysis of tunnel lining structure function, starting from the angle of structural mechanics, the elastic foundation beam theory to build the tunnel support function of piperoof mechanical model, and the application of combined with the specific project. It provides a new approach for stress analysis of tunnel support.

KEYWORDS: pipe shed support mechanical model of foundation beam theory

INTRODUCTION

In tunnel support, the pipe roof is an extremely effective means, and the mechanical model is shown in Figure 1.

Figure 1: pipe roof beam on elastic foundation model

Based on this, the excavation process can be divided into two cases, shown in Table 1, and the elastic foundation beam model can be established.

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Table 1: Tunnel excavation and the corresponding elastic foundation beam model No. Force characteristic Mechanical schema Beam model of semi infinite foundation The excavation section of AB length a for himself, has excavation without support BC segment length is s, the working range of the front CD relaxation length was D, the tunnel excavation height is h. BC shed pipe under only the rock pressure above Q (x) A and pass it to the rear of the initial support and the surrounding rock; CD can withstand the rock pressure Q (x), but also by the elastic resistance of P (x) DE segment; surrounding rock pressure is negligible, only under the pipe roof and surrounding rock deformation caused by the ground reaction force P(x) With pipe roof and initial support of the connecting end of the B as the origin of coordinates, which has been excavated section of supporting AB segment length is a, excavation without support BC

segment length is s, CD is in front of the working range of relaxation length is D, h is the height of B tunnel excavation. The BC section pipe roof only withstands the surrounding rock pressure Q (x) and passes it to the rear primary support and the front wall rock; the CD not only withstands the surrounding rock pressure Q (x), but also is subjected to the elastic resistance of P (x).

BEAM MODEL OF ELASTIC FOUNDATION WITH

TWO PARAMETERS

As shown in Figure 2, a layer of shear layer existing in the spring unit in the Kerr foundation beam model, and the layer can only produce shear deformation and incompressible shear, so there is interaction between the spring element, can be in good agreement with the surrounding rock stress. The theory is referred to as the "PAP" theory, and the "pipe to pipe" model is analyzed. Vol. 24 [2019], Bund. 01 259

Figure 2: B model

The shear layer is assumed to be isotropic in the X and Y planes and its shear modulus Gx=Gy=Gp,then ∂w τ xz = Gpγ xz = Gp ∂x （1）

∂w τ yz = Gpγ yz = Gp ∂y （2）

The total shear force per unit length of shear layer is:

1 ∂w N = τ dz = G s ∫ xz p ∂x 0 （3）

1 ∂w N = τ dz = G y ∫ yz p ∂y 0 （4） The equilibrium condition of the force in the direction of Z is obtained :

∂ ∂N x N y + + q − r0 = 0 ∂x ∂y （5）

In the formula, r0=kw, the formula (3) (4) will be given to the people q(x, y) = kw(x, y) − G ∇2w(x, y) p （6） Here：q(x,y)—Subgrade reaction ,，KN/m2 ；k—Coefficient of subgrade reaction ，KN/m3 ；Gp —Characterization of transfer shear parameters ;KN/mm， w—Foundation settlement ,m ；When Gp=0,i.e Winker foundation model Vol. 24 [2019], Bund. 01 260

IDENTIFICATION OF CALCULATION MODEL FOR

TWO PARAMETER ELASTIC FOUNDATION BEAM

The method of distinguishing two parameter foundation beam is shown in table 2.

Table 2: beam pattern discrimination Concentrated load position Length of beam Calculation type of beam （to beam endless x） Infinite beam Both ends of the beam are satisfied L ≥ 2π / λ x ≥ π / λ Semi infinite beam Only a section from the beam is satisfied L ≥ 2π / λ x ≥ π / λ Finite length beam Both ends of the beam are satisfied π/（4λ）< L < 2π / λ x < π / λ ≤ π λ L /(4 ) Rigid beam No connection λ = 4 k noted： 4EI ，It is related to the relative stiffness of the beam and foundation .

GOVERNING DIFFERENTIAL EQUATIONS OF

BEAMS ON ELASTIC FOUNDATIONS

Based on the Pasternak elastic foundation beam, the force model of the pipe roof is established, and the force is shown in Figure 3.

Figure 3: beam pattern discrimination

From balance condition of FG ∑ Y = 0 then： dQ(X ) = b[ p(x) − q(x)] （4） dx To F section, from ∑ M = 0 then： (dx)2 (dx)2 M − (M + dM ) + (Q + dQ)dx + q(x) − p(x) = 0 （5） 2 2 Vol. 24 [2019], Bund. 01 261

Omitting two orders of differentiation: dM (x) Q(x) = dx （6）

Bring in (4): dQ（x） dM 2 (x) = = − 2 b[ p(x) q(x)] dx dx （7） From Bernoulli-Euler typical theory, the force、displacement and inner force have following relationship：

dω(x)  θ（x）= dx  2  dω (x)  = − M (x) EI 2  dx  dM (x) dω 3 (x) Q(x) = = −EI  3  dx dx  （8）

The differential equation of beam deflection is brought into (7) type dω 4 (x) dω 2 (x) − * + *ω = EI 4 Gpb 2 kb (x) bq(x) dx dx （9） b* = b[1+ (G / k)1/ 2 / b] Here：b*--as equivalent width of beam,，satisfied p ；b- equivalent width of beam；I-as the moment of inertia of foundation beam ;；E- as The elastic modulus of the beam on the elastic foundation ; Gp-as foundation shear modulus； ω (x) -as Deflection of foundation beam ;k-as Coefficient of subgrade reaction. On the basis of formula (7) and (8), the deflection curve of two cases can be obtained by analyzing the model of pipe roof. 1) the governing equations of type A (1) in the BC section, the surrounding rock pressure is q0, the foundation reaction force is p (x) =0, and the governing equation is:

dω 4 (x) = EI 4 bq0 dx （10） (2)CD section, the surrounding rock pressure is q0, the foundation reaction

dω 2 (x) = ω − p(x) k (x) Gp 2 dx ，The governing equations are:

dω 4 (x) dω 2 (x) EI − G b* + kb*ω(x) = bq （11） dx4 p dx2 0 Vol. 24 [2019], Bund. 01 262

（3）CD section, the surrounding rock pressure is q(x)=0,the foundation reaction

dω 2 (x) = ω − p(x) k (x) Gp 2 dx ，The governing equations are:

dω 4 (x) dω 2 (x) − * + *ω = EI 4 Gpb 2 kb (x) 0 dx dx （12） 2)the governing equations of type B （1）in the BC section, the surrounding rock pressure is q0, the foundation reaction force is p(x)=0，The governing equations are: dω 4 (x) EI = bq dx4 0 （13） (2)CD section, the surrounding rock pressure is q0, the foundation reaction

dω 2 (x) = ω − p(x) k (x) Gp 2 dx ，The governing equations are:

dω 4 (x) dω 2 (x) − * + *ω = EI 4 Gpb 2 kb (x) bq0 dx dx （14）

CONTROL DIFFERENTIAL EQUATIONS

1) A type （1）(1) the governing equation (10) of the DE section pipe roof is a homogeneous equation, that is, the deflection equation of the foundation beam ,，assum

kb* λ4 = 4EI ，then：

4 4 2 1 dω (x) Gpλ dω (x) − + λ4ω = 4 2 (x) 0 4 dx k dx （15） Its characteristic equation is:

4 1 Gpλ r 4 − r 2 + λ4 = 0 4 k （16）

G λ2 p <1 Normally，k is much bigger than Gp，and λ <1，so k ，So the characteristic equation has two pairs of plural root

α = λ[1+ (G λ2 / k)]1/ 2 , β = λ[1− (G λ2 / k)]1/ 2 ±α ± iβ ，Here p p ，The general solution of the equation (16) is: Vol. 24 [2019], Bund. 01 263

ω（x）= eαx (B cos βx + B sin βx) + e−αx (B cos βx + B sin βx) 3 1 2 3 4 （17） Here B1，B2，B3，B4 are the undetermined coefficients can be derived from the boundary conditions of the pipe roof. （2）The characteristic equation of the BC segment control equation (11) is

喷射混凝土 4 轴力N

内压力 p r = 0 ，There are 4 roots 1 ，the general solution of the equation is

P σ P σ ，When the load of pipe shed is uniformly distributed load

P σ

P σ q0, the special solution is ，the equation (11) is the general solution ：

P σ

P σ （18）

Here A1，A2，A3，A4 are are the undetermined coefficients . (3)It is obvious that the governing equation (13) of the DE section pipe roof is the homogeneous equation of the differential control equation () of the CD section, so that the general solution of the CD section control equation (12) can be obtained: ω（x）= eαx (B cos βx + B sin βx) + e−αx (B cos βx + B sin βx) +ω * 3 1 2 3 4 2 （19）

ω * In the formula, the particular solution 2 is related to the load q(x) and boundary conditions. When the load on the pipe roof is uniformly distributed load q0, the boundary condition can be obtained by continuity at the D point: ′ ′ ω = θ = θ = ω ω = ω 2 2 x=s+d 3 x=s+d 3 2 x=s+d 3 x=s+d ， x=s+d x=s+d ，take special solution ：

* q ω = 0 − b − − b − − 2 * [1 cosh (x s d)cos (x s d)] kb 6 DEFLECTION EQUATION OF PIPE SHED

1) B type (1) in the BC section, the control differential equation (14) is the same as the control equation (12) of the A type, so the general solution is (16). (2) the general solution of the same CD segment is (20). ω The B end boundary condition can be obtained by the initial displacement 0 and the initial

θ ω = ω θ = ω′ = 0 rotation angle 0 of the B: 1 x=0 0 、 1 x=0 1 x=0 ，then A4=ω0、A3=θ0；To Vol. 24 [2019], Bund. 01 264 point C，The boundary conditions can be obtained according to the continuity condition of the pipe roof

ω = ω ω′ = ω′ ω′′ = ω′′ ω′′′ = ω′′′ 1 x=s 2 x=s 、 1 x=s 2 x=s 、 1 x=s 2 x=s 、 1 x=s 2 x=s ；D section is free

ω′′ = ω′′ ω′′′ = ω′′′ part，its boundary condition 1 x=s+d 2 x=s+d 、 1 x=s+d 2 x=s+d 。By formula（16）、（20）and its boundary condition the following formula will be gained：

 0 0 0 1 0 0 0 0  A1  ω0     θ   0 0 0 0 1 0 0 0 A2   0  3 2 (21)  s s s 1 ϕ35 ϕ36 ϕ37 ϕ38 A3  ψ 3   2     3s 2s 1 0 ϕ45 ϕ46 ϕ47 ϕ48 A4  ψ 4    =    6s 2 0 0 ϕ ϕ ϕ ϕ  B ψ  55 56 57 58  1   5       6 0 0 0 ϕ65 ϕ66 ϕ67 ϕ68  B2 ψ 6      0 0 0 0 ϕ ϕ ϕ ϕ  B 0  75 76 77 78  3         0 0 0 0 ϕ85 ϕ86 ϕ87 ϕ88 B4   0  Here： αs ϕ45 = −e [− β sin(βs) +α cos(βs)] ϕ = −eαs cos(βs) 35 ϕ = −eαs [β cos(βs) +α sin(βs)] ϕ = − αs β 46 36 e sin( s) −αs −αs ϕ47 = −e [− β sin(βs) −α cos(βs)] ϕ37 = −e cos(βs) ϕ = −e−αs [β cos(βs) −α sin(βs)] ϕ = −e−αs sin(βs) 48 38 αs 2 2 ϕ55 = −e [− 2αβ sin(βs) + (α − β )cos(βs)] αs 2 2 ϕ56 = −e [2αβ cos(βs) + (α − β )sin(βs)] ϕ = −αs β 3 − α 2β β + α 3 − αβ 2 β −αs 2 2 68 e [( 3 )cos( s) ( 3 )sin( s)] ϕ57 = −e [2αβ sin(βs) + (α − β )cos(βs)] α (d +s) 2 2 ϕ75 = −e {− 2αβ sin[β (d + s)]+ (α − β )cos[β (d + s)]} −αs 2 2 ϕ = −e [− 2αβ cos(βs) + (α − β )sin(βs)] α (d +s) 2 2 58 ϕ76 = −e {2αβ cos[β (d + s)]+ (α − β )sin[β (d + s)]} αs 3 2 3 2 ϕ = − −α (d +s) αβ β + + α 2 − β 2 β + ϕ65 = −e [(β − 3α β )sin(βs) + (α − 3αβ )cos(βs)] 77 e {2 sin[ (d s)] ( )cos[ (d s)]} −α + αs 3 2 3 2 (d s) 2 2 ϕ78 = −e {− 2αβ cos[β (d + s)]+ (α − β )sin[β (d + s)]} ϕ66 = e [(β − 3α β )cos(βs) − (α − 3αβ )sin(βs)] α (d +s) 3 2 3 2 −αs 3 2 3 2 ϕ85 = −e {(β − 3α β )sin[β (d + s)]+ (α − 3αβ )cos[β (d + s)]} ϕ67 = −e [(β − 3α β )sin(βs) − (α − 3αβ )cos(βs)] α (d +s) 3 2 3 2 ϕ86 = e {(β − 3α β )cos[β (d + s)]− (α − 3αβ )sin[β (d + s)]} −α (d +s) 3 2 3 2 ϕ87 = −e {(β − 3α β )sin[β (d + s)]− (α − 3αβ )cos[β (d + s)]} −α (d +s) 3 2 3 2 ϕ88 = e {(β − 3α β )cos[β (d + s)]+ (α − 3αβ )sin[β (d + s)]} q q ψ = − 0 s4 + 0 [1− cos(bd)cosh(bd)] 3 24EI kb* q q ψ = − 0 s3 + b 0 [cos(bd)sinh(bd) − cosh(bd)sin(bd)] 4 6EI kb* q q ψ = − 0 s2 + 2b 2 0 sin(bd)sinh(bd) 5 2EI kb* q q ψ = − 0 s − 2b 3 0 [cosh(bd)sin(bd) − cos(bd)sinh(bd)] 6 EI kb* The coefficients can be obtained by the equations (12) A1、A2、A3、A4、B1、B2、B3、B4，Taking into (16) and (20), the deflection equation ω (x) l of each section of B type pipe shed can be obtained. 2）Type A For A, when the face is far away from the front end of pipe roof, the pipe roof is regarded ω = 0 as semi-infinite elastic foundation beam, the boundary conditions for CE 3 x→∞ ，

θ = 0 ω 3 x→∞ ，then B1=B2=0；An elastic fixed end with a certain initial displacement 0 and Vol. 24 [2019], Bund. 01 265

′ θ = ω = 0 θ ω1 = ω0 1 x=0 1 an initial rotation angle 0 at the B end is inherent x=0 ， x=0 ，i.e.

A = ω A = θ 4 0 、 3 0 ；In point C,The pipe roof satisfies the continuity condition and the inherent boundary condition

′ ′ （3） （3） ω = ω ω = ω 1 2 ω ″ = ω ″ ω1 = ω2 1 x=s 2 x=s 、 x=s x=s 、 1 x=s 2 x=s 、 x=s x=s 。 By formula (16) ~ (20) and the boundary conditions mentioned above, The following equations can be obtained:

 0 0 0 1 0 0  A1  ω0     θ  （ ）  0 0 1 0 0 0 A2   0  22  s3 s2 s 1 ϕ ϕ A  ψ   35 36  3  =  3  2 ϕ ϕ ψ 3s 2s 1 0 45 46 A4   4   6s 2 0 0 ϕ ϕ B  ψ   55 56  3   5   6 0 0 0 ϕ65 ϕ66 B4  ψ 6  −αs ϕ35 = −e cos(bs)  −αs  ϕ36 = −e sin(bs)  −αs  ϕ45 = −e [−b sin(bs) −α cos(bs)] −αs  ϕ46 = −e [b cos(bs) −α sin(bs)]  Here −α ϕ = −e s[2αb sin(bs) + (α 2 − b 2 )cos(bs)]  55  −αs 2 2 ϕ56 = −e [−2αb cos(bs) + (α − b )sin(bs)]  −αs 3 2 3 2  ϕ65 = −e [(b − 3α b )sin(bs) − (α − 3αb )cos(bs)]  ϕ = −αs b 3 − α 2b b − α 3 − αb 2 b  66 e [( 3 )cos( s) ( 3 )sin( s)]  q q ψ = − 0 s4 + 0 [1− cos(bd)cosh(bd)]  3 24EI kb*  q q  ψ 0 3 b 0 b b b b 4 = − s + * [cos( d)sinh( d) − cosh( d)sin( d)] 6EI kb  q q ψ = − 0 s2 + 2b 2 0 sin(bd)sinh(bd)  5 2EI kb*  q0 3 q0  ψ 6 = − s − 2b [cosh(bd)sin(bd) + cos(bd)sinh(bd)] EI kb*  The coefficients can be obtained by the equations (21) A1、A2、A3、A4、B3、B4 and B1=B2=0，Bring all coefficients into (16), (19), and (20) forms, and obtain deflection equations in the case of A type.

PARAMETER ANALYSIS OF PIPE TUNNEL IN SLOPE

TUNNEL

Between the Liangjiayuanzi tunnel located in the Huaihua of Hunan Province, Xi Ling Xi Xiang Lin Chang to two Chahe, for double tunnel about separation, two cavern spacing in portal section of about 7 ~ 9m, 10 ~ 12m in the body of the cave. The left line tunnel starts at ZK75+470, and finally ZK76+124. The elevation at the bottom of the tunnel is 379.31 to 365.33m, and the total length is 654.00m. The right line tunnel starts at K75+440, and finally K76+124. The elevation of the tunnel bottom is 377.69 to 368.23m, the length is 684.00m, Vol. 24 [2019], Bund. 01 266 and the overall direction of the tunnel is 162~168 degrees. The maximum buried depth is about 102.82m (K75+940), and about 2% of the tunnel has a buried depth of more than 100m, and the opening is mainly a slope accumulation.

The deflection curve of pipe roof can be obtained by taking the above parameters into deflection curve equation by using MATLAB program.

1) the influence of the tunnel depth on the deflection of the pipe shed

On the basis of the above conditions, the buried depth of H, take 100m, 40m, 20m, 15m, 10m, respectively, using the matlab program to draw the deflection curve, as shown in Figure 4.

2) the influence of the excavation step length on the deflection of the pipe shed

In the case of other conditions unchanged, the deflection curves of b=0.3m, excavation steps s=2m, 1.5m 1 and m are shown in Figure 5.

隧道开挖方向（m） 隧道掘进方向（m） 0 1 2 3 4 5 6 7 8 9 10 -5 0 1 2 3 4 5 6 7 8 9 10 -5

0 0

5 5 H=10m s=2m

10 H=15m s=1.5m 10

s=1m

管棚挠度（ 15 mm） 管棚挠度（ 15 mm）

20 20 H=20m，40m ,100m

25 25 Figure 4: Effect of buried depth of tunnel on the Figure 5: Grid spacing on the deflection on pipe roof deflection of pipe roof

3)Influence of pipe shed distance on deflection of pipe shed On the basis of the above geological conditions, the deflection curves of the pipe roof can be obtained by taking the distance between the pipe shed, b=1.0m, 0.75m, 0.5m, 0.4m and 0.3m, as shown in Figure 6. 4) the effect of pipe diameter on deflection of pipe shed Under the same initial conditions, the pipe shed diameter is 108mm, 152mm, 180mm and 200mm respectively, and the deflection curve of the pipe roof can be obtained as shown in Figure 7. Vol. 24 [2019], Bund. 01 267

隧道掘进方向（m） 隧道掘进方向（m） 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 -5 -2

0 0 隧道掌子面 5 2 10 隧道掌子面 15 4 b=0.3m Φ200mm 20 6 b=0.4m Φ180mm

25 管棚挠度（ mm） b=0.5m 8 管棚挠度（ mm） 30 Φ152mm b=0.75m 10 35 Φ108mm 40 12 b=1.0m 45 Figure 6: pipe shed spacing effect on the Figure 7: pipe shed diameter influence on deflection of pipe roof deflection of pipe roof

CONCLUSION

Through the above analysis of the factors influencing the deflection of the pipe shed, we can draw the following conclusions:

(1) Proper excavation step can ensure the construction progress and ensure the safety of construction. The excavation step has a more obvious influence on the deflection of the pipe shed. The larger the excavation length is, the greater the deflection of the pipe roof is, the higher the deflection of the surrounding rock. Therefore, we can not pursue the pace step by step for the sake of progress, but ignore the necessary safety.

(2) Adopting reasonable caliber is both economical and safe. Under certain distance, the diameter of pipe increases, but the disturbance of the pipe becomes smaller, but the influence is not obvious when it increases to a certain extent. Therefore, reasonable diameter must be determined. In order to avoid the pipe shed diameter is too large, the optimization effect of support is not obvious, and cause economic waste;

(3) Take the spacing between pipes can be guaranteed reasonable, economical and safe, the maximum deflection of pipe roof decreases as the distance decreases, when the spacing between pipes is reduced to a certain extent, the deflection of pipe roof is small, support optimization effect is not obvious, therefore, can not simply rely on reducing pipe shed pitch to control pipe shed deflection.

ACKNOWLEDGEMENT This paper was sponsored by Project supported by China Natural Foundation (No.51538009, No.51578550) and Project of Chang Science Bureau, here thank them. Vol. 24 [2019], Bund. 01 268

REFERENCES [1] XING Xue-sheng, FU He-lin. Theoretical Solution of Bearing Capacity of Shallow Circular Broken Rock-mass Foundation According to Limit Equilibrium Analysis EJGE, 2016,21(2):

[2] XIONG Yi,FU He-lin.. Oversize Hall Type Karst Scheme Comparison in Tunnel EJGE, 2016,21(3):

[3] ZHANG Jia-bing,FU He-lin, HUANG Zhen. Prediction for Shield Tunneling-Induced Displacement Based on Elastic Complex Variable Theory[ EJGE, 2015,20(18:

[4] HUANG Zhen,FU He-lin,ZHANG Jia-bing.. Relationship between Wave Velocity and Strength of Compacted Expansive Soil, and Application of Acoustic Technology EGJE, 2016,21(6)

© 2019 ejge

Editor’s note. This paper may be referred to, in other articles, as: Zuo Zhuo, Fu Helin, and Peng Wenxuan: “ Mechanical Model of Tunnel Support Pipe Support” Electronic Journal of Geotechnical Engineering, 2019 (24.01), pp 257-268. Available at ejge.com.