Tunneling in soft soil

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(1)Tunneling in soft soil A thesis submitted for the Degree of Civil Engineering. By Andrés Ricardo Barrero Bernal. Advisor Prof. Dr.-Ing. Arcesio Lizcano. Universidad de Los Andes Faculty of Engineering Department of Civil and Environmental Engineering 2011.

(2) Table of content 1 Tunnel transversal section stability 1.1 Terzaghi’s method . . . . . . . . . 1.2 Mandel . . . . . . . . . . . . . . . . 1.3 Bierbäumer’s method . . . . . . . . 1.4 Caquot method . . . . . . . . . . . 1.5 Stability of cylindrical cavities . . . 1.6 Circular tunnels in cohesive soils . . 1.7 Cylindrical cavity with support . .. . . . . . . .. 1 1 5 6 9 15 17 17. 2 Tunnel excavation face stability 2.1 Solution for own weight support . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Stability of the excavation face according to bound theorem . . . . . . . . . 2.3 Proctor and White method . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 21 21 22 24. 3 Ground surface settlement 3.1 Lame solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Peck solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 27 27 29. . . . . . . .. 1. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . ..

(3) List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14. Terzaghi loading body . . . . . . . . . . . . . . . Terzaghi force equilibrium . . . . . . . . . . . . . Tunnel crown stress values for different depths . . Tunnel crown stress values for different diameters Mandel loading body . . . . . . . . . . . . . . . . Mandel loading body H/R>1.14 . . . . . . . . . . Bierbäumier loading body . . . . . . . . . . . . . Bierbäumier force equilibrium . . . . . . . . . . . Bierbäumier pressure distribution . . . . . . . . . Pressure distribution for different diameters . . . Caquot loading body . . . . . . . . . . . . . . . . Caquot force equilibrium . . . . . . . . . . . . . . Cross section of a circular tunnel . . . . . . . . . Stress of the cross section . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. 1 2 4 4 5 5 6 6 9 10 10 11 15 16. 2.1 2.2 2.3 2.4 2.5 2.6. Excavation face (picture took from D.Kolymbas(2005:P.329)) . . . . . . . Lower bound hemisphere (picture took from D.Kolymbas(2005:P.331)) . . . Proctor and White sliding blocks (picture took from F.Acosta (2007:P.27)) Left sliding block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Center sliding block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Right sliding block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .. 21 22 24 24 25 25. 3.1 3.2. Lame elastic space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Gauss distribution (picture took from F.Acosta (2007:P.39)) . . . . . . . . .. 27 31. 2. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . ..

(4) Table index 1.1 1.2. Values for equation 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Values for b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3 3. 2.1. Risk factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 26. 3.1. Values for a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 31. 3.

(5) Abstract Tunnel design has three main problems. Tunnel rock pressure for its self-support, excavation stability and possible settlement in the surface level. It’s important for the engineer to calculate the value of these three variables in order to guarantee the safety of the tunnel and the structures within the tunnel path. Several methods have been developed according different assumptions made to describe the real life problem in equations, such as analytical methods and empirical methods are going to de be discussed in this thesis. The document is divided in three parts; Tunnel transversal section stability, tunnel excavation face stability and ground surface settlement..

(6) Chapter 1 Tunnel transversal section stability In this chapter, several methods of the tunnel transversal section stability are going to be review. The objective is to consider the same assumptions made by the author and solve each method step by step in order to present a clear solution. Is to remark that some steps are omitted or not shown, consult the bibliography for full detail analysis.. 1.1. Terzaghi’s method. Figure 1.1 shows the Terzaghi’s surface failure. In this Figure B is the width and C is the distance among the surface level and the tunnel upper vault, which is the active zone of failure. If we take a differential height between the surface level and the tunnel upper vault, its possible to find wall pressure at the tunnel crown.. Figure 1.1: Terzaghi loading body. Assumptions:. 1.

(7) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. • The rigid zone has the same width of the tunnel. • Vertical surface failure. • Mohr-Coulomb criteria τ = c + σh tan ϕ Applying force equilibrium for a horizontal section of dz width, within the surface level and the tunnel upper vault.. Figure 1.2: Terzaghi force equilibrium. ΣFy = 0 = b(σv + dσv − σv ) − dW + 2cdz + 2kσv tan(ϕ) σh σh − ΣFx = 0 = dz dz dσv =bdzγ − 2(cdz + kσv tan ϕ) ( ) cdz + kσv tan(ϕ) dσv =γ − 2 dz b Solution:. dy kσv tan(ϕ) c dσv + P(x) y =Q(x) ⇒ +2 =γ−2 dx dz b b 2K tan(ϕ) 2K tan(ϕ) R dz z b b I(x) =e =e ) 2K tan(ϕ) ( 2K tan(ϕ) ) Z ( z z c b b σv = dz e γ−2 e b 0 2.

(8) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. σv =. (. bγ − 2c 2K tan(ϕ). −2K tan(ϕ) ) z b 1−e. )(. ICIV 201020 02. (1.1). Analyzing equation 1.1, b, γ, c, K, are constants assuming an homogeneous and isotropic soil. We can say, the only variable that changes within distance C is depth (z). As a consequence, equation 1.1 describes the tunnel crown pressure in terms of depth (σv ) → f (z). Taking for example values as given in Table 1.1 for equation 1.1, and plotting from z = [0, · · · , 100], we get the figure 1.3. As shown, there is an asymptotic value as depth increases. It’s obvious to say that for bigger diameters the curve should open at a mayor depth because the soil column acting upon the tunnel gets wider. In order to prove the previous sentence, three different values of b where chosen and plot (b1 < b2 < b3 ) the result is shown in figure 1.4 (see bi values in table 1.2). As b gets bigger the asymptotic line gets farther, on the other hand tunnel cost is proportional to tunnel diameter, so tunnel cost increases. Now, if at the ground surface the soil is loaded with a load q per unit area, then the boundary conditions at z=0 would be (σ(z=0) = q). This modify the equation 1.1, c[kPa] 12. γ[kN/m3 ] 21. ϕ[Degrees] 28. b[m] 4. Table 1.1: Values for equation 1.1. b1 [m] 4. b2 [m] 5. b3 [m] 6. Table 1.2: Values for b. σv =. (. bγ − 2c 2K tan(ϕ). −2K tan(ϕ) ) −2K tan(ϕ) z z b b 1−e + qe. )(. (1.2). Where σv [kPa] is the pressure at the tunnel crown, b[m] is the tunnel diameter, γ[kN/m3 ] is the soil unit weight, c[kPa] is soil cohesion, ϕ is the angle of internal friction, K is the earth pressure n at rest coefficient. Analyzing the previous equation it’s easy to say that at a o bγ−2c σv (z = ∞) = 2K tan(ϕ) . And the equation is only valid for a homogeneous soil, if there exist another soil with different γ. It’s necessary to calculate the overburden stress due to the soil above the new layer. Is to remark the value of K,√that can be expressed as K ≈ 1 − senϕ for normally consolidated clays or K ≈ (1 − senϕ) · OCR. The analysis was made without considering a water table, in such case the analysis would have to consider the pore pressure ∆U . 3.

(9) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. Figure 1.3: Tunnel crown stress values for different depths. Figure 1.4: Tunnel crown stress values for different diameters. 4. ICIV 201020 02.

(10) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. 1.2. ICIV 201020 02. Mandel. As the previous method, these consider rigid block sliding along vertical columns above the √ H tunnel. For r < 2. Pt =γ(H − r) − 2Cln. nH o r. + 0, 154. (1.3). Figure 1.5: Mandel loading body. For. H r. >. √. 2. Pt =γ(H − r) − C. rn. H2 o −1 r2. Figure 1.6: Mandel loading body H/R>1.14. 5. (1.4).

(11) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. 1.3. ICIV 201020 02. Bierbäumer’s method. Biermäumers theory was developed during the construction of the Alpine tunnels. The theory states that a tunnel is acted upon by the load of a soil mass bounded by the parabola height of h = αH as seen in figure 1.7. The method consist in finding α such that h = αH. So. Figure 1.7: Bierbäumier loading body. one approach was to assume that upon the tunnel excavation, the soil material tends to slide ϕ along the surface failure inclined at 45 + as figure 1.8 shows. In order to find the value 2. Figure 1.8: Bierbäumier force equilibrium. of P [kPa], which is the pressure over the tunnel crown. A force equilibrium in the vertical 6.

(12) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. direction has to be made:. ΣFy = 0 = P = γHB − 2τ. (1.5). 2τ =2CH + 2E tan{ϕ}. (1.6). Where τ is equal to:. E is equal to the lateral active earth pressure:. 1 E = H 2 γKa 2. (1.7). Ka = tan2 {45o − ϕ/2}. (1.8). If we substitute equation 1.7 in 1.8 and replace in equation 1.6, we get:. 1 2τ =2CH + 2 H 2 γ tan2 {45o − ϕ/2} tan{ϕ} 2. (1.9). Now replacing equation 1.9 in 1.6, we get the load acting above the tunnel upper vault:. P =γHB − 2CH − H 2 γ tan2 {45o − ϕ/2} tan{ϕ}. (1.10). But the pressure above the tunnel crown is:. p= 7. P B. (1.11).

(13) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. Where B = b + 2b tan{45o − ϕ/2}. So the equation changes (considering the cohesion to be zero, C = 0).. (. H tan2 {45o − ϕ/2} tan{ϕ} p =γH 1 − b + 2b tan{45o − ϕ/2}. ). (1.12). Otherwise if C 6= 0 the equation 1.12 would be:. (. H tan2 {45o − ϕ/2} tan{ϕ} p =γH 1 − b + 2b tan{45o − ϕ/2}. ). −. 2CH b + 2b tan{45o − ϕ/2}. (1.13). So its easy to see that the variable multiplying the soil unit weight and the tunnel crown depth is:. α=1−. Hγ tan2 {45o − ϕ/2} tan{ϕ} b + 2b tan{45o − ϕ/2}. (1.14). The equation 1.14 has two limit values, for very small overburden depths (H → 0) α = 1, so the final equation would be:. p = hγ. (1.15). On the other hand, whenever H → ∞|α = −∞, leading to:. p = −∞. (1.16). In order to understand how p changes as depth increases, the figure 1.9 was plot using the equation 1.13, taking the same values of table 1.1. As terzaghi’s method, the pressure increases until it reaches H/b ≈ 5. From this point the pressure in Bierbäumier method decreases an tends to −∞, it’s erroneous to think that for deep depths the pressure tunnel is 8.

(14) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. negative so it’s necessary to define possible boundaries in Bierbäumier method. As we can see in figure 1.9 the parabola intersects twice in the Y-axis, telling us that there are two depths (Z1 , Z2 ) where the pressure is equal to zero (p = 0). As a consequence the Bierbäumier method is valid within Z1 and Z2 values. Notice that the use of wall support for p(Z2 = 0 is useless, on the other hand p reaches a maximum value when H/b ≈ 5. Now if we increase tunnel diameter is to expect that tunnel pressure maximum value must increases according to H/b ≈ 5. The same values (Table 1.2) used in Terzaghi’s method where plot in figure 1.10. Is to remark that is assume a homogeneous and isotropic soil with no water table.. Figure 1.9: Bierbäumier pressure distribution. 1.4. Caquot method. Caquots method defines the surface failure taking into account the propagation of the plastic zone all around the circular tunnel. The figure 1.11 shows a representation of the plastic zone. In order to begin with the force equilibrium as shown in figure 1.12, several assumptions must be made: • The ground it’s in limit state and the rupture occurs during the tunnel drilling. In this way the soils tends to flow towards the tunnel center making concentric circles. 9.

(15) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. Figure 1.10: Pressure distribution for different diameters. Figure 1.11: Caquot loading body. 10. ICIV 201020 02.

(16) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. • The rupture takes place from the tunnel key until the isostatic circle tangent to the surface. • The soil is consider to be homogeneous and isotropic. In this way γ, ϕ remain constant. • The soil is consider to have an elastic-plastic behavior.. Figure 1.12: Caquot force equilibrium. If we apply a force equilibrium in the radius direction, it’s obtain:. ) dθ dθ ∂σr ∂σθ dθ)dr sin (r + dr)dθ + σθ dr sin + (σθ + ΣFr =σr rdθ − σr + ∂r 2 ∂θ 2 dθ − γrdrdθ cos θ + 2 (. Replacing.... . dθ dθ sin = 2 2 dθ cos θ + = cos θ θ dr2 dθ ⇒ 0 drdθ2 ⇒ 0 11.

(17) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. The expression reduces to:. ∂rσr ∂r Now taking into account de forces perpendicular to the radio direction: σθ = γr cos θ +. ΣF⊥R =σθ dr cos. . dθ 2. . (1.17). dθ ∂σθ dθ dr cos − σθ + + γrdrdθ cos θ ∂θ 2. Replacing.... dθ ⇒ 0 dθ cos =1 2 dθ sin θ + = sin θ 2 The expression reduces to:. ∂σθ = γr sin θ ∂θ Using an elastic-plastic model and the Mohr-Coulomb failure criterion.. (1.18). σθ =Kp σr + σc 1 + sin ϕ Kp = 1 − sin ϕ 2C cos ϕ σc = 1 − sin ϕ If we integrate equation 1.18:. Z. dσθ =. Z. γr sin θdθ. σθ = − γr cos θ + C 12. (1.19).

(18) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. From the border conditions we know: θ = 0, σθ (r, θ = 0) = σθ0 . So the constant C is equal to:. (1.20). C = σθ0 + γr Replacing the previous equation 1.20 in 1.19, we get:. (1.21). σθ = σθ0 + γr(1 − cos θ) Deduction for σr0 and σθ0 Replacing σθ = Kp σr + σc in the equation 1.17 for θ = 0, we obtain:. Kp σr0 + σc −. ∂rσr0 − γr = 0 ∂r. Re-ordering terms:. Kp ∂rσr0 − (rσro ) = σc − γr ∂r r Notice that has the same solution for a first degree equation:. R dy + P(x) y =Q(x) ⇒ ye P(x) dx = dx. Z. R. Q(x) e. P(x) dx. dx + cte. Solving the differential equation:. r(1−Kp ) σr0 =. γ σc r2−Kp + cte r(1−Kp ) − (1 − Kp ) 2 − Kp 13.

(19) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. From border conditions we know σr0 (r = H) = q, where q is the external load in the surface. Thus:. cte =H 1−Kp q −. γ σc H (1−Kp ) + H (2−Kp ) (1 − Kp ) (2 − Kp ). Finally we obtain σr0 and σtheta0. σr0 =. γH (Kp − 2). (. r r − H H. (Kp −1). ). +. σc (Kp − 1). (. r H. (Kp −1). ). −1. +q. r (Kp −1) H. (1.22). (1.23). σθ0 =Kp σr0 + σc Now if we find an expression for σp , it’s necessarily to solve:. ∂σr0 = Kp σr0 + σc + γr(1 − 2 cos θ) ∂r Replacing in the previous equation with σr0 (equation 1.22). Integrating and solving for cte = 0 using the border conditions we get an expression for σp .. σp =. Hγ Kp − 2. (. r r − H H. (Kp −1). ). +. σc (Kp − 1). (. r H. (Kp −1). ). −1. +q. (. r H. (Kp −1). ). + γr − γr cos θ. Which is equivalent to:. σr = σr0 + γr(1 − cos θ) 14. (1.24).

(20) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. Figure 1.13: Cross section of a circular tunnel. 1.5. Stability of cylindrical cavities. Consider a circular tunnel build in a homogeneous soil, subjected to a uniform σo at the top of the tunnel. The stress applied is consider to be in principle axes as the result of absence of any shear stress. Taking a cross section as in figure 1.14, we are going to analyze the equilibrium requirements for points a, b, c, d, which are inclined at an angle θ with respect to the vertical. Neglecting any shear stress acting along sides a − b and c − d, equilibrium equations can be stablished.. w ≈ γrdθdr. (1.25). γ is the unit weight of soil, r is the radius to the point a and c. Solving the forces radially, ≈ dθ , the same made in Caquot’s method. It follows that considering sin dθ 2 2. (σr + dσr )(r + dr)dθ − σr drdθ − σθ dr. dθ dθ − (σθ + dσθ )dr + γrdrdθ cos θ = 0 2 2 15.

(21) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. Figure 1.14: Stress of the cross section. σr (σr − σθ ) + + γ cos θ = 0 dr r. (1.26). Solving perpendicular to the radius,. σθ dr + γrdrdθ sin θ − (σθ + dσθ )dr = 0. σθ + γr sin θ = 0 dθ. (1.27). Equations 1.26 and 1.27 represent the equilibrium of an element near the tunnel. But it’s critic point (2006:P.660) is at the tunnel crown where θ = 0, in this point we can eliminate forces perpendicular to the radio. Reducing equation 1.26 to:. σr (σθ − σr ) = −γ dr r 16. (1.28).

(22) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. 1.6. ICIV 201020 02. Circular tunnels in cohesive soils. In short term, clays owe their strength to their undrained cohesion Cu . The failure criterion in passive mode (σθ > σr ) under undrained conditions is:. (σθ − σr ) = 2Cu Replacing in:. ∂σr σθ − σr = −γ ∂r r and integrating between σr = σa at r = a and σr = σ0 at r = R. It follows:. Z. σ0. dσr =. σa. Z r( a. ) 2Cu − γ dr r. R σa = σ0 + γ(R − a) − 2Cu ln a. 1.7. (1.29). Cylindrical cavity with support. Suppose a cavity within a homogeneous soil that has an elastic-plastic behavior. As a consequence the soil has a linear stress path in the elastic region, outside the elastic boundaries the soil response to a perfectly plastic model. It’s assume that the cavity has a uniform radial pressure [Pi ]. At the initial state we have σz = σr = σθ = σ0 . In the elastic zone according to Wood (2002:P.170). σe =. 2c cos ϕ + Kp σr 1 − sin ϕ. Assuming strain plane conditions between the plastic zone. 17. (1.30).

(23) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. Eθ =. u = σθ − ν(σr + σz ) − σ0 (1 − 2ν) r. (1.31). Er =. du = σr − ν(σθ + σz ) − σ0 (1 − 2ν) dr. (1.32). Ez = 0 = σz − ν(σθ + σr ) − σ0 (1 − 2ν). (1.33). Solving for σr in the previous equation, we get. σz = ν(σθ + σr ) + σ0 (1 − 2ν). (1.34). Replacing equation 1.34 in equations 1.31 and 1.33, we get two equations in function of σθ and σr .. Eθ = σθ − ν(σr + ν(σθ + σr ) + σ0 (1 − 2ν)) − σ0 (1 − 2ν). (1.35). Er = σr − ν(σθ + ν(σθ + σr ) + σ0 (1 − 2ν)) − σ0 (1 − 2ν) According to Wood, if we differentiate equation 1.35 with respect to r, we obtain. σθ − σr + (1 − ν)r. dσr dσθ − νr =0 dr dr. (1.36). It’s know that for an element to be in equilibrium the equation is. dσr σr − σθ =− dr r 18. (1.37).

(24) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. Therefore equations 1.36 and 1.37 leads us to. (1 − ν)rd(σθ + σr ) =0 dr σθ + σθ = 2σ0. (1.38) (1.39). Now analyzing the equations 1.30 and 1.36 for a point that exist in the elastic and the plastic regions. It’s possible to join the two equations in order to get σθ and σr .. 2c cos ϕ + Kp σr = 2σ0 − σr 1 − sin ϕ. (1.40). Using equations 1.40 and 1.38, we find a equation for σθ and σr in function of Kp , r, σ0 .. σr = σ0 −. σθ = σ0 +. (Kp − 1)σ0 +. 2c cos ϕ 1 − sin ϕ. (1.41). 2c cos ϕ 1 − sin ϕ. (1.42). 1 + Kp. (Kp − 1)σ0 +. 1 + Kp. In the plastic region, we know. σθ = Kp0 σr. (1.43). So replacing the previous equation 1.43 in the equation 1.37, we obtain the equilibrium for the plastic region.. dσr σr (1 − Kp0 ) =− dr r 19. (1.44).

(25) CHAPTER 1. TUNNEL TRANSVERSAL SECTION STABILITY. ICIV 201020 02. Solving for σr in equation 1.44:. Z r dσr dr (1 − Kp0 ) =− σr r σ0 r0 ln σr − ln σ0 = (Kp0 − 1) ln r − ln r0 r σr = (Kp0 − 1) ln ln σ0 r0 Z. σr. σr = σ0. . r r0. (1−Kp0 ). (1.45). If we get σr1 from equation 1.44 and having σr2 at r = ri from the previous equation 1.45. It’s possible to say that σr1 = σr2 .. Pi. . ri r0. (1−Kp0 ). =. 2σ0 − Kp0 1 + Kp0. (1.46). So the pressure applied inside the cavity must be equal to equation 1.46 to prevent possible tunnel collapse. On the other hand, it’s possible to design the radius of the tunnel by solving the equation 1.46 for ri given the pressure inside the cavity.. r i = r0. . 1 2σ0 − Kp0 Kp0 − 1 Pi (1 + Kp0 ). 20. (1.47).

(26) Chapter 2 Tunnel excavation face stability The following methods calculates the minimum excavation face support that prevents tunnel collapse.. 2.1. Solution for own weight support. Regarding the distribution of the vertical stress between the spherical zone (r = r0 ) and the ground surface, as seen in figure 2.1. The stress distribution is approximate with a quadratic parabola with the material strength fully mobilized at the tunnel crown. The necessary support pressure pc is given as D.Kolymbas did in (2005:P.330):. Figure 2.1: Excavation face (picture took from D.Kolymbas(2005:P.329)). 21.

(27) CHAPTER 2. TUNNEL EXCAVATION FACE STABILITY. ICIV 201020 02. c cos ϕ rf (1 − sin ϕ) pc = h 2h sin ϕ 1+ rf (1 − sin ϕ) γ−. (2.1). Is seen that for the unsupported excavation face is stable if:. 0≥γ−. c cos ϕ rf (1 − sin ϕ). c ≥ γrf. 2.2. 1 − sin ϕ cos ϕ. (2.2). Stability of the excavation face according to bound theorem. Figure 2.2: Lower bound hemisphere (picture took from D.Kolymbas(2005:P.331)). The lower-bound theorem supplies a simple but conservative solution of the necessary P at the excavation face to preserve the tunnel stability. An hemisphere is assumed as scene in figure 2.2. In the first case is assumed γ = 0 within a spherical radio (r = r0 + H) around the excavation face is assumed a limit condition (σθ + σc = 2c), that is fulfilled in the spherical zone. The equation of equilibrium in radial direction: 22.

(28) CHAPTER 2. TUNNEL EXCAVATION FACE STABILITY. ICIV 201020 02. dσr 2 + (σr − σθ ) = 0 dr r Solving the previous equation and using boundary conditions such as σr(r=ro ) = p. The equation gives:. p = σr − 4c ln. . r r0. . (2.3). Assuming a constant hydrostatic stress outside the boundaries of the spherical zone (σr = σθ = q). Analyzing p in the boundary of the two zones (r = r0 + H) requires σr = q, thus:. H p = q − 4c ln 1 + r0. (2.4). Now taking a second case γ > 0 and the hydrostatic stress as σii = γz for directions {x, y, z} the support pressure increases linearly with depth:. H p = γz + q − 4c ln 1 + r0. (2.5). For a infinite long strip circular tunnel the support pressure can be calculated similar to Caquot method:. H p = γz + q − 2c ln 1 + r0. (2.6). On the hand the upper bound theorem considers a different method to estimate the excavation support pressure. It’s assume two cylindrical rigid blocks made of rock sliding along a surface failure. As seen in D.Kolymbas(2005:P.332) Davis propose a stability ratio N :. N=. q − p + γ(h + r0 ) c 23. (2.7).

(29) CHAPTER 2. TUNNEL EXCAVATION FACE STABILITY. 2.3. ICIV 201020 02. Proctor and White method. In order to preserve tunnel excavation face stability, Proctor and White proposed a method to calculate the necessary pressure to retain the soil acting in the tunnel face. Such method consist in calculating the difference between the soil weight in points CEF HC and the shear stress acting along the vertical column CEF HC −AJA. The soil weight is calculated through. Figure 2.3: Proctor and White sliding blocks (picture took from F.Acosta (2007:P.27)). figures 2.4, 2.5,2.6. Soil weight for figures 2.4 and 2.6, are:. Figure 2.4: Left sliding block. 24.

(30) CHAPTER 2. TUNNEL EXCAVATION FACE STABILITY. ICIV 201020 02. Figure 2.5: Center sliding block. Figure 2.6: Right sliding block. π 2 H ·γ·L 4. (2.8). Pγ = H · B · γ · L. (2.9). Pγ = Soil weight for figure 2.5 is:. The total soil weight is calculated by the sum of equations 2.8 and 2.9.. Pγ = γ · L · H · B +. π 2 π H · γ · L + H2 · γ · L 4 4. n π o Pγ = γ · L H · B + H 2 2 25. (2.10). (2.11).

(31) CHAPTER 2. TUNNEL EXCAVATION FACE STABILITY. ICIV 201020 02. The resistance shear stress for a pure cohesive soil is:. n o R = cu · L π · H + B. (2.12). Finally the difference between Pγ − R, is the necessary for to support tunnel excavation face. It’s easy to establish that whenever if:. Pγ − R > 0. (2.13). It’s necessary the use external support to preserve the equilibrium. If:. Pγ − R ≡ 0. (2.14). External support is useless. In order to obtain the excavation face stress we must divide the force in the area where the load is applied. According to Figure 2.3, the area is equal to points CEF HC. So the necessary stress to maintain tunnel excavation face stability is:. Ps =. Pγ − R H2 · π B·H + 2. (2.15). According to the authors (2007:P.27), the following considerations must be taking into account handling with soil simple compression resistance. p − σT ≤ 0.5 σc p/σc < 0.5 0.5 < p/σc < 1.0 p/σc ≥ 1.0. Everytime p/σc is not satisfactory, a support pressure σT must act for possible front inestability Tunnel excavation face stability is reach in short time periods Tunnel excavation face support it’s necessary for resisting possible displacements Tunnel excavation face instability is presented Table 2.1: Risk factors. 26.

(32) Chapter 3 Ground surface settlement Ground surface settlement is important as excavation face stability or wall pressure. It’s impossible to predict the total settlement due to the non-linear soil stiffness. Being aware of the settlement due to the excavation of the tunnel. Analytical and empirical methods focus on having a limit value for the ground surface settlement in order to prevent damages in buildings or constructions near the tunnel boring. As a consequence tunnel design is limited.. 3.1. Lame solution. Considering a cylindrical cavity in a weightless elastic space, loaded by a hydrostatic stress σ∞ . The vertical component is δv of the displacement δ, from the figure 3.1 we can say:. Figure 3.1: Lame elastic space. 27.

(33) CHAPTER 3. GROUND SURFACE SETTLEMENT. sin α =. ICIV 201020 02. H r. (3.1). We know that any vector in R2 has X and Y components, so:. (3.2). δv = sin α · δ Replacing equation 3.1 in 3.2:. δv =. H δ r. (3.3). With r2 = x2 + H 2 , we obtain Lame’s solution (2005:P.340):. δv =. σ∞ p r02 H 2G x2 + H 2. (3.4). It’s easy to see from the figure 3.1 that the maximum settlement or displacement occurs when x = 0.. δv =. δv,max 2 x 1+ H. (3.5). But the problem with Lame’s solution is the non realistic distribution, when it’s compare with measurements made by J.H Atkinson and D.M. Potts. It’s been prove that ground settlements are described much better using a Gauss distribution as Peck propose. For real measurement comparisons, see [6].. δv = δv,max exp 28. . −x2 2a2. . (3.6).

(34) CHAPTER 3. GROUND SURFACE SETTLEMENT. ICIV 201020 02. Also it’s inconsistent when it uses a full space for a half-space problem. Surface settlements in the horizontal direction can be found according to figure 3.1.. (3.7). δh = cos α · δ Remember that:. x r. (3.8). x δh = · δ r. (3.9). cos α = Replacing equation 3.8 in 3.7, we get:. If we do a triangle similarity in figure 3.1, we can establish a relation between vertical and horizontal settlements.. δv = δh. 3.2. H x. (3.10). Peck solution. Peck (1969) propose a more realistic distribution of the measured settlement with his formula:. δv = δv,max exp. . x2 − 2 2a. . (3.11). The parameter a depends on the tunnel diameter D and H is depth of the tunnel axis. For clay soils:. 29.

(35) CHAPTER 3. GROUND SURFACE SETTLEMENT. ICIV 201020 02. a ≈ (0.4 · · · 0.8)H. (3.12). a ≈ (0.25 · · · 0.45)H. (3.13). For non-cohesive soils:. Different investigations were made in order to find diverse expressions for the parameter a. When the equation 3.11 is integrate by per current tunnel meter, we get the volume of settlement.. Vδ =. √. 2πaδv,max. (3.14). The loss volume is express as an amount percentage (V1 ) of the tunnel dug area, for a circular tunnel.. Vδ = V1. πD2 4. Combining the previous equations 3.11 and 3.14 we get:. δmax =. 0.31V1 D2 KZ0. It’s important to remark that during the tunnel collapse there is a change of form, but the volume remains constant. Table 3.1 shows different values of a according to different authors (2007:P.36). Also in figure 3.2 different curves of settlement were plot using equation 3.11, each curve represent a different value of a according each author of table 3.1. It’s important to say that it doesn’t matter what curve of figure 3.2 is chosen. The important factor is, how reliable the parameter a was stablished. 30.

(36) CHAPTER 3. GROUND SURFACE SETTLEMENT. ICIV 201020 02. Figure 3.2: Gauss distribution (picture took from F.Acosta (2007:P.39)). Author Peck(1969) Attewel and Farmer (1974). a a/D = (H/2D)n n = 0.8 − 1 a/D = (H/2D). Atkinson and Potes (1977). a = 0.25(H + D) (loose sand). Clough and Schmidt (1981). a = 0.25(1.5H + 0.5D) (dense sand and SC clays) a/D = (H/2D)0.8. O’Reilly and New (1982). a = 0.43H + 1.1m. Mair (1983). (For cohesive soils (3 ≤ H ≤ 34m)) a = 0.28H − 0.12m (For granular soils (6 ≤ H ≤ 10m)) a = 0.5H. Leach(1985). a = (0.57 + 0.45H) ± 1.01m Table 3.1: Values for a. 31. Bases Field observations Field observations in UK tunnels Field observations and physical experiments. Field observations in UK tunnels Field observations in UK tunnels. Field observations and centrifuge experiments For sites where consolidations effects are neglected.

(37) Bibliography [1] F Acosta. Diseño de túneles basado en viscohipoplasticidad para los suelos blandos de bogotá. Master’s thesis, Andes University, 2007. [2] F Azizi. Applied analyses in geotechnics. Taylor and Francis, 2005. [3] J.P. Bardet. Experimental Soil Mechanics. Upper Saddle River, New Jersey, prentice hall edition, 1997. [4] J Bowles. Foundations and Analyses Design. New York, mcgraw hill. 5 edition edition, 1996. [5] B.H.G Brady and E.T Brown. Rock mechanics for underground mining. Springer, United States of America, third edition, 2005. [6] L.A Bulla and J.C Chamorro. Modelación centrífuga de túneles poco profundos en arcillas de la sabana de bogotá. Master’s thesis, Andes University, 2003. [7] J Chakrabarty. Theory of plasticity. Elsevier, Oxford, UK, third edition, 2006. [8] D Kolymbas. Tunelling and Tunnel Mechanics: A Rational Approach to Tunnelling. Springer, 2005. [9] M Lai, D Rubin, and E Krempl. Continuum mechanics, 1993. Third Edition. Butterworth Heinemann. [10] U.S Army Corps of Engineers. Tunnels ans shafts in rock. Department of the Army, Washington, DC, May 2007. [11] A. Schofield and P. Wroth. Critical state soil mechanics. London, mc graw hill edition, 1968.. 32.

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