ANALYTIC SOLUTIONS OF ELASTIC TUNNELING PROBLEMS Phần 4 doc

Section 4.2 Solution for a Deforming Hole in a Half-Plane 25 The expansion of the left-hand side of (4.24) is not as trivial, and will be greatly simplified by multiplication with 1 − ασ . The right-hand side of the modified equation can then be expanded as follows. We write G  (ασ ) = 2µ(1 −ασ) h g m (ασ ) = ∞  k=−∞ A k σ k . (4.31) where we have assumed that the given displacements along the hole, multiplied by 2µ(1−ασ), can be expanded into a Fourier series. Such an expansion should be possible for all problems of practical significance. Note that the function G  (ασ ) has same definition as in [42]. The logarithmic terms in the right-hand side of (4.24) can be expanded with help of the formula given in (3.7): log 1 1 − ζ = ∞  n=1 ζ n n , | ζ | < 1, (4.32) where the principle branch of the logarithm has been chosen. The modified form of (4.25) resulting from the multiplication with 1 − ασ can be expanded as follows (using (4.32) and the series in(4.31)): (1 − ασ) h g m◦ (ασ ) = ∞  k=−∞ A k ◦ σ k (4.33) where A k ◦ =                                                                          A k + h F x + i h F y 2π  1 + (α 2 − 1)k k(k − 1)α k  k ≤−2, A −1 + α( h F x + i h F y ) 4π (2 − α 2 ) − ακ( h F x − i h F y ) 4π(1 + κ) k =−1, A 0 + C F − α 2 ( h F x + i h F y ) 2π − h F x − i h F y 4π(1 + κ) + κ( h F x − i h F y ) 4π(1 + κ) (1 − α 2 )k= 0, A 1 − αC F + ακ( h F x + i h F y ) 2π + h F x − i h F y 4πα(1 + κ) + α(κ − 1)( h F x − i h F y ) 4π(1 + κ) k = 1, A 2 − α 2 κ( h F x + i h F y ) 4π + h F x − i h F y 4π(1 + κ) k = 2, A k − κ( h F x + i h F y ) 2π  α k k(k − 1)  k ≥ 3. (4.34) 26 A Deforming Circular Tunnel Chapter 4 The expansion of the right-hand sides of both (4.21) and the modified form of (4.24) is complete. The left-hand sides of these equations can be expanded through substitution of (4.27) and (4.28), as mentioned before. The details of these expansions, which are straightforward, are omitted here. Determination of the Laurent Coefficients The unknown coefficients in the Laurent series (4.27) and (4.28) can be de- termined by comparing the coefficients of like powers of σ in the expanded boundary equations. The results of these calculations for the expansion of (4.21) are c 0 = B 0 ◦ − a 0 − 1 2 a 1 − 1 2 b 1 , (4.35) c k = B −k ◦ − b k − 1 2 (k + 1)a k+1 + 1 2 (k − 1)a k−1 ,k>0, (4.36) d k = B k ◦ − a k + 1 2 (k − 1)b k−1 − 1 2 (k + 1)b k+1 ,k>0. (4.37) These equations can be used to determine each of the coefficients c k and d k once a k and b k have been determined. Note that each of the coefficients c k and d k are dependant on three levels of coefficients (e.g. a k−1 , a k , and a k+1 ). Expressions for a k and b k are found by comparing the coefficients of like powers of σ in the expansion of the modified form of (4.24). This results in a complicated system of equations which are simplified considerably by eliminating c k and d k through use of (4.35) – (4.37). Equations for a k+1 and b k+1 are then obtained which are dependant on a single level of coefficients: (1 + κα 2k+2 )a k+1 + (1 − α 2 )(k + 1)b k+1 = α 2 (1 + κα 2k )a k + (1 − α 2 )kb k + α k+1 A k+1 ◦ + B k+1 ◦ − α 2 B k ◦ ,k≥ 0 (4.38) and α 2k (1 − α 2 )(k + 1)a k+1 − (κ + α 2k+2 )b k+1 = α 2k (1 − α 2 )ka k − (κ + α 2k )b k + α k A −k ◦ + α 2k B −k ◦ − α 2k+2 B −k−1 ◦ ,k≥ 0, (4.39) where we have multiplied through by α k in order to make the expressions numerically stable for large values of k. Analytic Solution of the System Equations (4.38) and (4.39) can be solved for a k and b k numerically, as suggested in [42]. We choose to solve the system analytically. After some algebraic manipulation, which will not be given in full here, the solution is given by a k+1 = k λ 11 a k + k λ 12 b k + k λ 13 , (4.40) b k+1 = k λ 21 a k + k λ 22 b k + k λ 23 , (4.41) Section 4.2 Solution for a Deforming Hole in a Half-Plane 27 where k λ 11 = κα 2 +[α 2 (κ 2 + α 2 ) + (1 − α 2 ) 2 (k + 1)k + κα 2k+4 ]α 2k κ +γ k α 2k , (4.42) k λ 12 = κα 2 − κ − (1 − α 2 )[(1 − α 2 )k + 1]α 2k κ +γ k α 2k , (4.43) k λ 13 = (κ +α 2k+2 )(B k+1 ◦ − α 2 B k ◦ ) +[α(κ + α 2k+2 )A k+1 ◦ ]α k κ +γ k α 2k + (1 − α 2 )(k + 1)(A −k ◦ + α k B −k ◦ − α k+2 B −k−1 ◦ )α k κ +γ k α 2k , (4.44) k λ 21 = (1 − α 2 )[α 2 − (1 − α 2 )k + κα 2k+2 ]α 2k κ +γ k α 2k , (4.45) k λ 22 = κ +[1 +κ 2 α 2 + (1 − α 2 ) 2 (k + 1)k + κα 2k+2 ]α 2k κ +γ k α 2k , (4.46) k λ 23 = (1 − α 2 )(k + 1)(α 2k+1 A k+1 ◦ + α k B k+1 ◦ − α k+2 B k ◦ )α k κ +γ k α 2k − (1 + κα 2k+2 )(A −k ◦ + α k B −k ◦ − α k+2 B −k−1 ◦ )α k κ +γ k α 2k , (4.47) with γ k = (1 + κ 2 )α 2 + (1 − α 2 ) 2 (k + 1) 2 + κα 2k+4 . (4.48) Here γ k and k λ mn ,n={3}, are real and k λ mn ,n={3}, is complex. Explicit Determination of the Coefficients By studying the expanded expressions for the first few coefficients, it becomes clear that explicit equations for a k and b k can be written as follows: a k = k β 011 a 0 + k β 012 a 0 + k−1  i=0 k β i13 k = 1, 2, 3, , (4.49) b k = k β 021 a 0 + k β 022 a 0 + k−1  i=0 k β i23 k = 1, 2, 3, , (4.50) where k β lmn =    l λ mn k = l + 1, k−1 λ mj k−1 β lj n k = l + 2,l+3,l+4, , (4.51) 28 A Deforming Circular Tunnel Chapter 4 with l = 0, 1, 2, , and where the summation convention (summation over like indices) has been adopted for the symbols λ and β, with m = 1, 2, n = 1, 2, 3, and j = 1, 2. Note that k β lmn ,n={3} is real and k β lmn ,n={3} is complex. Determination of the Remaining Unknown Coefficient It is apparent from these equations that the boundary conditions are not sufficient to determine the last unknown, a 0 , as noted in [42]. It can be seen from (4.49) – (4.50) and (4.35) – (4.37) that this unknown is incorporated in every coefficient of the Laurent series (4.27) and (4.28). The value of a 0 affects the convergence of the Laurent series; in fact, on the basis of (4.42) – (4.47) and (4.40) – (4.41), a k and b k will each approach a constant value in the limit that k →∞, the value of which is determined by the starting value of the coefficients, a 0 .An expression for a 0 can be determined by requiring that these constants tend to zero as k →∞, which is a necessary condition for the convergence of the Laurent series (4.27) and (4.28) along the outer boundary  s of the annulus. Enforcing the condition that a k → 0 and b k → 0ask →∞in the complex conjugate of (4.49) and in (4.50) gives 0 = ∞ β 012 a 0 + ∞ β 011 a 0 + ∞−1  i=0 ∞ β i13 , (4.52) 0 = ∞ β 021 a 0 + ∞ β 022 a 0 + ∞−1  i=0 ∞ β i23 , (4.53) where we have used the notation ∞ β lmn = lim k→∞ k β lmn , ∞−1 = lim k→∞ (k − 1). (4.54) Solving the system (4.52) – (4.53) for a 0 results in a 0 = ∞ β 012  ∞−1 i=0 ∞ β i23 − ∞ β 021  ∞−1 i=0 ∞ β i13 ∞ β 021 ∞ β 011 − ∞ β 022 ∞ β 012 . (4.55) We note that (4.55) gives the proper result for the limiting case that there are no resultant forces and no stresses or displacements given on either of the two boundaries L s and L h . In this case the expansion coefficients, B k ◦ and A k ◦ , are equal to zero for every value of k. By examining the terms of the form β ij 3 it is apparent that the leading factor in these terms will always be of the form λ j3 , causing them to be zero when all B k ◦ and A k ◦ are zero. The conclusion is that a 0 = 0 in this case, with the consequence that all of the coefficients vanish. This corresponds to the case that there are no stresses and no displacements in the entire half-plane, which is a direct consequence of unloaded condition along the boundaries. Section 4.3 Chapter Summary 29 Final Form of the Transformed Solution The final form of the solution in the z-plane is given by the potentials (4.1) and (4.2). They can be expressed in terms of ζ through use of (4.7), (4.17), (4.27), and (4.28). The result of the transformation is ϕ m (ζ ) =− h F x + i h F y 2π(1 + κ)  κ log(− i2a 1 − ζ ) + log(− i2aζ 1 − ζ )  + a 0 + ∞  k=1 a k ζ k + ∞  k=1 b k ζ −k , (4.56) and ψ m (ζ ) = h F x − i h F y 2π(1 + κ)  log(− i2a 1 − ζ ) + κ log(− i2aζ 1 − ζ )  + c 0 + ∞  k=1 c k ζ k + ∞  k=1 d k ζ −k , (4.57) where the coefficients a k , b k , c k , and d k are given by (4.55), (4.49) – (4.50), and (4.35) – (4.37), and where a is given by (4.8). These potentials, written in terms of the transformed parameter ζ , can be used to compute the stresses and displacements by conformally mapping the basic equations (2.1) - (2.5). Details of the implementation have been included in Appendix B. § 4.3 Chapter Summary An analytical solution for a deforming, buoyant tunnel in an elastic half-plane was derived in this chapter. The method used consists of first determining the general form of the potentials for a half-plane with a hole, when a resultant force on the hole due an excavation-related buoyancy effect is present. The potentials derived in Chapter 3 were used for this purpose, and it should be noted that they contain two logarithmic terms each, yielding unbounded displacements at infinity. The half-plane region, the potentials, and the boundary equations were then transformed to an annulus in order to facilitate an expansion in Laurent series. After the expansion of the potentials (excluding the logarithmic terms) and the boundary conditions was performed, the unknown Laurent coefficients were determined in an analytical limiting process. 30 A Deforming Circular Tunnel Chapter 4 Part II APPLICATIONS AND CONCLUSION 31 Chapter 5 BUOYANCY OF A RIGID TUNNEL Buoyancy effects arise after the excavation of a tunnel due to an unloading of the material under the tunnel. This can be understood mathematically by inte- grating the initial gravitational stresses around the circumference of the tunnel, which results in a force equal in magnitude to the weight of the excavated ma- terial (minus the weight of the tunnel lining). This force, which is referred to here as the buoyancy force, is represented in the solution of Chapter 4 by the resultant force h F x + i h F y acting on the tunnel. In the mathematical idealization of this problem all of the excavated material is removed instantaneously and simultaneously replaced by a tunnel lining of negligible, or zero, weight. The resultant force acts on the tunnel in the instant after excavation, and vanishes through a combination of a vertical movement of the tunnel and an equaliza- tion of the stresses in the ground through deformation. This process results in equilibrium of the tunnel and an unloading of the ground. The deformations and the final stresses acting on the tunnel lining are the focus of this chapter. The buoyancy effect was included byYu [45] in the solution for a completely rigid tunnel with a lining of negligible weight, excavated in an infinite plane. In this solution a vertically varying gravitational field is applied to an elastic plane, but the effects of the free surface are ignored. The buoyancy effect is also present in the solution for the excavation of an unlined (and thus stress-free) tunnel in an elastic half-plane by Mindlin [18]. In his solution, Mindlin only obtained equations for the stresses and did not determine the displacements. In addition to these solutions, Verruijt [44] determined the approximate stresses acting on a rigid, buoyant tunnel by assuming that the horizontal stresses directly above the tunnel, and the horizontal-vertical shear stresses in the entire field, are not affected by the excavation of the tunnel. When studying deep tunnels, a constant stress due to the overburden is often assumed, and the buoyancy effect is usually ignored (see for example [26, 8, 4]). For shallow tunnels, however, the approximation of a constant gravitational stress around the tunnel results in stresses along the nearby surface. To cir- cumvent this, a more realistic gravitational loading with a vertically varying 33 34 Buoyancy of a Rigid Tunnel Chapter 5 gradient may be applied, making it necessary to include buoyancy effects in order to ensure equilibrium of the tunnel. In this chapter, we treat the simple case of a shallow tunnel with a circular liner of negligible weight which is perfectly rigid and which is bonded to the surrounding soil. This will allow us to study the buoyancy effects in their simplest possible form. The solution can be superimposed on the solutions for other deformation modes in later chapters in order to include the stresses induced by gravity on those solutions as well. § 5.1 Solution of the Problem The solution of a rigid tunnel in a heavy half-plane is composed of two parts: the solution for the stresses induced by gravity in the half-plane and the solution for the stresses and displacements induced by the resultant force acting on the tunnel due to buoyancy. In geotechnical engineering it is common to write the solution for the stresses induced by gravity as σ xx = γy, (5.1) σ yy = K 0 γy, (5.2) σ xy = 0, (5.3) where γ is the unit weight of the material and K 0 is the (constant) coefficient of lateral earth pressure (K 0 is an independent parameter, determined by the geological history of the material). The displacements for this problem are deliberately left out, as we are only interested in deformations that occur after the material has been deposited [25]. The stresses and displacements induced by the resultant force acting on the tunnel can be obtained using the solution presented in section 4.2. There are two given components in this solution: the resultant force acting on the hole, and the boundary conditions along the tunnel. The resultant force is equal to the weight of the ground removed during excavation minus the weight of the tunnel lining. In this solution, we neglect the weight of the tunnel lining, and it follows that the resultant force on the hole in the solution of Chapter 4 can be written as h F x + i h F y = iγπr 2 , (5.4) where r (as in Chapter 4) denotes the radius of the tunnel. The boundary conditions for a rigid tunnel are that the displacements along the tunnel are constant. This insures that the tunnel does not deform, but allows a certain amount of rigid body motion due to the rebound of the ground underneath the tunnel. In this formulation of the problem the buoyancy force acting on the tunnel is known beforehand and is given by (5.4), but the amount of rigid [...]... result is given by (4. 56) and (4. 57) The results for this part of the solution can be normalized by setting the parameter P in Appendix B equal to γ h2 It follows from (B.13) that the expression for the resultant force (5 .4) becomes h h F x + i Fy r 2 = iπ , (5.6) P h which is only dependant on the radius to depth ratio of the tunnel Since the only component of the solution in Chapter 4 which is not dimensionless... g(z) = 0 in (4. 6) Consequently, the mapped h displacement function is also equal to zero: g (ασ ) = 0 It follows from (4. 31) m that h ∞ G (ασ ) = 2µ(1 − ασ )g (ασ ) = m Ak σ k = 0, (5.5) k=−∞ from which it then follows that Ak = 0 for all k The Fourier expansion given by (4. 34) can now be calculated by using these values for Ak and by using the value of the resultant force given by (5 .4) This completes... superposition of (5.7) – (5.9) § 5.2 Validation of the Solution The solution has been implemented in a computer program which can be used to generate plots of the normalized stresses and displacements These plots can be visually inspected to confirm that the stresses vanish along the surface (see Figure 5.1) and that the tunnel experiences only a rigid body motion (see Figure 5.2) x/ h -5 0 -1 -2 -3 -4 -5 -4 -3... γh Contours of the initial gravitational stresses generated from (5.7) – (5.9) will not depend on the value of h if the x and y coordinates are normalized by division by h, as will be done for the figures in this thesis In the following chapters the focus will be on plots of the incremental stresses due to the deformation modes that are being analyzed (in this chapter those are 36 Buoyancy of a Rigid... force, it follows from the discussion in Appendix B that dimensionless contours of the stresses and displacements can be constructed which will only depend on the parameters r/ h and ν The displacements will be given in terms of P /2µ = γ h2 /(2µ) and the stresses will be given in terms of P / h = γ h, as can be seen from (B. 24) – (B.26) The solution for the gravitational stresses can be made dimensionless... - y/ h 0 -1 -2 -3 -4 -5 Figure 5.1: Incremental stresses σyy (left-hand side) and σxy (right-hand side) for ν = 0 and r/ h = 0.7 The contours are given in increments of 0.035γ h In Figure 5.1 and in all the following plots of stress contours, the solid thin curves denote negative (compressive) stresses, the thick solid curves denote contours of zero stress, and the finely dotted curves...Solution of the Problem 35 Section 5.1 body motion of the tunnel is not Since an arbitrary rigid body motion can be added to the solution for the displacements [23], zero displacements along the tunnel boundary can be enforced without loss of generality (the actual boundary condition along the tunnel periphery is that the displacements... superposition of the stresses due to gravity, (5.7) – (5.9), as Figure 5.1 shows only the incremental stresses due to the excavation of the tunnel The initial stresses can be added to obtain the total stresses A rigid body motion has been superimposed on the displacements shown in Figure 5.2 such that the vertical displacements are zero at the two outermost points along the surface (x/ h = ±5) In equilibrium solutions. .. equilibrium solutions a rigid body motion is usually added such that the displacements are zero at infinity (see, for example, [42 ] and Chapter 6) In this case however, the displacements are infinite at infinity due to the resultant force acting on the tunnel, which makes the choice for the point of vertical constraint in the solution somewhat arbitrary ... 5.1) and that the tunnel experiences only a rigid body motion (see Figure 5.2) x/ h -5 0 -1 -2 -3 -4 -5 -4 -3 -2 -1 0 1 2 3 4 5 . 4 The expansion of the right-hand sides of both (4. 21) and the modified form of (4. 24) is complete. The left-hand sides of these equations can be expanded through substitution of (4. 27) and (4. 28),. on the basis of (4. 42) – (4. 47) and (4. 40) – (4. 41), a k and b k will each approach a constant value in the limit that k →∞, the value of which is determined by the starting value of the coefficients,. seen from (4. 49) – (4. 50) and (4. 35) – (4. 37) that this unknown is incorporated in every coefficient of the Laurent series (4. 27) and (4. 28). The value of a 0 affects the convergence of the Laurent