ANALYTIC SOLUTIONS OF ELASTIC TUNNELING PROBLEMS Phần 3 pot

Section 3.2 Complex Potentials for a Half-Plane with Holes 13 where the notation [] x b x a denotes the increase undergone by the expression inside the brackets along the integration path from x a to x b . The integral in (3.18) is path independent in portions of R beyond the expansion circle. Substituting (3.14) and (3.15) in (3.18) gives, for values of x a and x b outside a sufficiently large circle centered at the origin and containing all of the holes, i  x b x a (t x + it y ) ds = (ϒ + ϒ  ) ln | x a | | x b | + i( ϒ  − ϒ)π + ϕ 0 (x b ) + x b ϕ  0 (x b ) + ψ 0 (x b ) − ϕ 0 (x a ) − x a ϕ  0 (x a ) − ψ 0 (x a ). (3.19) Since (3.19) must remain finite when x a and x b approach infinite values inde- pendently, the coefficient of the real-valued logarithm must vanish. When x a and x b each approach infinite values simultaneously the integral must approach the sum of all external forces on the half-plane. It follows that we must have ϒ + ϒ  = 0 and (ϒ  − ϒ)π = T F x + i T F y , (3.20) where total resultant force T F x + i T F y = s F x + i s F y + h F x + i h F y (3.21) is given by the sum of the resultant forces acting on the surface ( s F x +i s F y ) and on the holes ( h F x +i h F y ). The corresponding values of ϒ and ϒ  agree with the coefficients of the logarithms obtained in [10] for the far-field behavior of the potentials in a semi-infinite plane with holes and vanishing stresses at infinity. Final Form of the Complex Potentials Using (3.20) and (3.16) to calculate γ and γ  and substituting the results in (3.8) and (3.9) allows us to obtain the general form of the potentials for a half-plane with holes, valid for all values of z in R. Equations (3.1) and (3.2) become ϕ(z) =−  s F x + i s F y 2π + κ( h F x + i h F y ) 2π(1 + κ)  log(z − z c ) − m  k=1 k F x + i k F y 2π(1 + κ) log(z − z k ) + ∞ σ xx 4 z + ϕ 0 (z), (3.22) 14 Multiple Holes in a Half-Plane Chapter 3 and ψ(z) =  s F x − i s F y 2π + h F x − i h F y 2π(1 + κ)  log(z − z c ) + m  k=1 κ( k F x − i k F y ) 2π(1 + κ) log(z − z k ) − ∞ σ xx 2 z + ψ 0 (z). (3.23) We note that in deriving these potentials we assumed that only a finite section of the surface was loaded. This assumption was used to expand the potential functions in Laurent series for large values of z so that we could examine their behavior there. This restriction is not a necessary condition for the potentials to take the forms (3.22) and (3.23), and was made in order to simplify the presentation. It has also been assumed that the forces on the holes are known beforehand. In problems consisting of equilibrium stresses on the holes, which occur often in practice, these forces will be zero. It should also be noted that the analysis presented here does not include concentrated forces acting on the surface of the half-plane. Such forces can be included by superimposing the corresponding components on the potentials (while being sure to disregard their contribution to s F x +i s F y in order to maintain equilibrium). § 3.3 Boundary Equations for a Half-Plane with Holes The unknown single-valued analytic functions ϕ 0 (z) and ψ 0 (z) in (3.22) and (3.23) can be determined from boundary conditions along the surface of the half- plane and along the holes. These boundary conditions can be given in terms of the stresses or in terms of the displacements (the case of mixed boundary conditions is not treated here). Stress Boundary Equations Boundary equations involving the stresses can be determined by writing the integral of the tractions (2.6) along boundary L j , j = s, 1, 2, 3, ,m: ϕ(z) + z ϕ  (z) + ψ(z)+ C j = j f(z) on L j , (3.24) where the loading function j f(z) = i  z z 0 (t x + it y ) ds (3.25) Section 3.3 Boundary Equations for a Half-Plane with Holes 15 is a known function on L j and where C j is a constant of integration, which is undetermined at this stage. Substitution of (3.22) and (3.23) in (3.24) results in ϕ 0 (z) + zϕ  0 (z) + ψ 0 (z) + C j = j f ◦ (z) on L j , (3.26) where j f ◦ (z) = j f(z)+  s F x + i s F y 2π + κ( h F x + i h F y ) 2π(1 + κ)  log(z − z c ) −  s F x + i s F y 2π + h F x + i h F y 2π(1 + κ)  log(z − z c ) + m  k=1 k F x + i k F y 2π(1 + κ)  log(z − z k ) − κlog(z − z k )  +  s F x − i s F y 2π + κ( h F x − i h F y ) 2π(1 + κ)  z z − z c + m  k=1 k F x − i k F y 2π(1 + κ) · z z − z k + ∞ σ xx 2 ( z − z). (3.27) The function j f ◦ (z) approaches a constant for infinite values of z along L s as a result of the calculations leading to the potentials (3.22) and (3.23). In ad- dition, it is single-valued along each contour L k , k = 1, 2, 3, ,msince the only multi-valued terms are the logarithms with singularities inside L k and the loading function, whose continuously increasing values cancel each other out for each circuit around L k . Both of the properties mentioned here are necessary since the left-hand side of (3.26) contains only single-valued functions which approach constants near infinity. It follows that (3.27) can be interpreted as an equation for new loading functions consisting of the original loading functions plus terms which remove the multi-valuedness inherent in the integral of the tractions when there is a resultant force acting on the boundary. Displacement Boundary Equations Boundary equations involving the displacements can be determined by writing (2.1) along boundary L j , j = s, 1, 2, 3, ,m: κϕ(z) − z ϕ  (z) − ψ(z) = 2µ j g(z) on L j , (3.28) 16 Multiple Holes in a Half-Plane Chapter 3 where the displacement function j g(z) is a known function on L j . Substitution of (3.22) and (3.23) in (3.28) results in κϕ 0 (z) − zϕ  0 (z) − ψ 0 (z) = j g ◦ (z) on L j , (3.29) where j g ◦ (z) = 2µ j g(z) + T F x + i T F y 2π (κ −1) log(z − z c ) +  s F x + i s F y 2π + h F x + i h F y 2π(1 + κ)  log[(z − z c )(z − z c )] + m  k=1 κ( k F x + i k F y ) 2π(1 + κ) log[(z − z k )(z − z k )] −  s F x − i s F y 2π + κ( h F x − i h F y ) 2π(1 + κ)  z z − z c − m  k=1 k F x − i k F y 2π(1 + κ) · z z − z k + ∞ σ xx 2  1 − κ 2 z − z  . (3.30) The function j g ◦ (z) is single-valued and continuous for complete circuits around the closed boundaries L k , k = 1, 2, 3, ,msince log[(z −z k )(z −z k )] is the logarithm of a real number. It should be noted that the displacement boundary condition along the surface L s must be formulated in such a way that the conditions specified in the second paragraph of section 3.2 are met (for a more detailed discussion the reader is referred to [23], §90 where the formulation of the surface displacements for the case of a half-plane without holes is treated). Due to the single-valued and continuous nature of j f ◦ (z) and j g ◦ (z), the bound- ary equations (3.26) and (3.29) for the functions ϕ 0 (z) and ψ 0 (z) are now in the same form as they are for ϕ(z) and ψ(z) in the case that there are no resultant forces on the holes or on the surface of the half-plane. Specific examples can be solved by a variety of techniques [23, 34] § 3.4 Chapter Summary The general representation of the complex potentials for a lower half-plane with holes has been derived. Each potential includes a single logarithmic term in the upper half-plane which cancels out the singularities in the integral of the surface tractions caused by a total resultant force on the holes and the surface Section 3.4 Chapter Summary 17 of the half-plane. In addition, it has been shown that only a constant stress parallel to the surface of the half-plane may act at infinity if the stresses are to remain bounded in the lower half plane. Finally, the boundary conditions were formulated such that the remaining unknown analytic functions can be solved for directly. It was also shown that the multi-valued and singular terms in these boundary conditions cancel each other out, allowing a complete solution of the problem. In the derivation it was assumed that only a finite amount of loading was present on the surface and that only a finite section of the surface was loaded. The latter assumption is not a necessary condition for the potentials to take the form derived in this chapter, but was made in order to simplify the presentation. In addition, it was assumed that no concentrated forces act on the surface of the half-plane. Such forces can be included by superposition of the corresponding potentials. The potentials and boundary conditions presented here will be used (albeit in a much simplified form) to solve a problem involving a resultant force acting on a hole near the surface of a half plane. This problem arises after the excavation of tunnels in a gravity loaded medium, when buoyancy forces arise. Chapter 4 A DEFORMING CIRCULAR TUNNEL The focus of this chapter is the derivation of a solution for a deforming circular tunnel (or hole) in an elastic half-plane, using the complex variable method. The deformations along the tunnel are prescribed in the form of a given function, and they may entail a resultant force acting on the tunnel. It is assumed that the surface of the half-plane is stress-free. Each complex potential in the solution consists of logarithmic terms plus an infinite series with analytically determined coefficients. Immediately after the excavation of a tunnel, buoyancy effects may arise due to the placement of a lining which is lighter than the excavated material. These excavation-related buoyancy effects need to be included in order to ensure equilibrium of the tunnel. They appear in any solution with a gravitational stress gradient, and are due to the non-uniform gravitational stresses acting on the tunnel in the instant just after excavation. The buoyancy effect is included in this solution by means of a resultant force incorporated in the complex potentials, as discussed and derived in Chapter 3. This resultant force vanishes as soon as the medium around the tunnel (or hole) is allowed to deform, yielding equilibrium of the tunnel. The buoyancy effect was included in the solution byYu [45] for an infinitely deep tunnel, and was also present in the solution for a stress-free tunnel in an elastic half-plane by Mindlin [18], the displacements of which were later de- termined by Verruijt and Booker [41]. Verruijt and Booker included buoyancy effects for Mindlin’s stress-free boundary condition by superimposing the solu- tion by Melan [17], a compensating Boussinesq solution, and a complex variable solution which was used to remove the tractions along the tunnel boundary. In this chapter, the related problem of a buoyant tunnel with given displacements will be solved directly and a full analytic solution is obtained. The numerical analysis used to determine the starting value for the recursive system of equa- tions in [42] and [41] is circumvented. The work in this chapter is also presented in the form of a paper in [36]. 18 Section 4.1 The Complex Potentials and Geometry of the Problem 19 § 4.1 The Complex Potentials and Geometry of the Problem The problem of a buoyant tunnel with given displacements along its boundary is posed in an elastic region R consisting of an infinite half-plane minus a hole. The stress-free upper surface of R is denoted by L s and the hole in R is bounded by a contour L h . The origin of the coordinate system is at a point on L s directly above the center of the hole. The radius of the hole is denoted by r and the depth of the center of the hole is denoted by h. The geometry of the problem is shown in Figure 4.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x y r h θ L s L h R Figure 4.1: Half-plane with a hole. The removal of material during the excavation of the tunnel causes a resultant force equal to the weight of the excavated material minus theweight of the tunnel lining to act on the tunnel. We incorporate this resultant buoyancy force in the solution by using the potentials (3.22) and (3.23) for a half-plane with holes derived in Chapter 3 and simplify them for the current situation. In this problem, the horizontal stress at infinity included in Chapter 3 is assumed to vanish ( ∞ σ xx = 0) and the surface is assumed to be stress-free ( s F x + i s F y = 0). In addition, we have only one hole (m = 1) and the total force on all holes ( h F x + i h F y ) is equal to the force on the single hole (corresponding to h F x + i h F y = 1 F x + i 1 F y ). It follows that the potentials (3.22) and (3.23) reduce to ϕ(z) =− h F x + i h F y 2π(1 + κ)  κ log(z − z c ) + log(z − z c )  + ϕ 0 (z), (4.1) ψ(z) = h F x − i h F y 2π(1 + κ)  log(z − z c ) + κ log(z − z c )  + ψ 0 (z), (4.2) 20 A Deforming Circular Tunnel Chapter 4 where, as in Chapter 3, the point z c is located arbitrarily within the hole, and the functions ϕ 0 (z) and ψ 0 (z) are single-valued and analytic in R, including the point at infinity. The function log(z −z c ) is the multi-valued natural logarithm (with its primary singularity located at z c ). We are interested in the case in which the surface of the half-plane is stress- free, and we apply the boundary equation (3.26) along L s (which can also be obtained by substituting (4.1) and (4.2) directly in (2.6)). For the current problem, this equation reduces to ϕ 0 (z) + zϕ  0 (z) + ψ 0 (z) = s f ◦ (z) on L s , (4.3) where s f ◦ (z) = h F x − i h F y 2π(1 + κ)  κz z − z c + z z − z c  . (4.4) The compactness of (4.4) is a result of the cancelation of the logarithmic terms in (3.26) along the surface, due to symmetry. Since the integration constant along one of the two boundaries can be chosen arbitrarily due to the fact that an arbitrary rigid body motion can be added to the solution [23, 34], we have assigned the constant in (4.3) the value of zero. The boundary equation for the displacements along the hole is found by applying (3.29) along L h (which can also be obtained by substituting (4.1) and (4.2) directly in (2.1)). The result is κϕ 0 (z) − zϕ  0 (z) − ψ 0 (z) = h g ◦ (z) on L h , (4.5) where h g ◦ (z) = 2µ h g(z) + h F x + i h F y 2π (κ −1) log(z − z c ) + h F x + i h F y 2π(1 + κ) log[(z − z c )(z − z c )] + κ( h F x + i h F y ) 2π(1 + κ) log[(z − z c )(z − z c )] − h F x − i h F y 2π(1 + κ)  κz z − z c + z z − z c  . (4.6) As in Chapter 3, the function h g(z) represents the given displacements along the boundary of the tunnel, L h , as a function of z. Note that the function given by (4.6) is single-valued and continuous along the boundary, as required by the single-valued terms on the left-hand side of (4.5). Section 4.2 Solution for a Deforming Hole in a Half-Plane 21 § 4.2 Solution for a Deforming Hole in a Half-Plane In order to solve the problem given by the potentials (4.1) and (4.2) and the boundary equations (4.3) and (4.5) we use a method of solution similar to the one used in [42] for a hole deforming under equilibrium stresses in a half-plane with a stress-free surface. The method consists of conformally mapping the half-plane with a hole to an annular region and expanding the potentials and boundary equations into Laurent series in the transformed region. The solution technique used here differs from the one in [42] in that we find an analytic expression for the unknown starting value of the recursive equations for the Laurent coefficients instead of using a numerical procedure. Conformal Mapping to an Annulus We conformally map (for an explanation of this technique see [6, 16]) the region R in the z-plane onto an annular region R in the transformed plane, which we refer to as the ζ -plane (see Figure 4.2). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ξ η σ 1 α ϑ  s  h R Figure 4.2: The annulus in the conformally mapped ζ-plane. The conformal mapping is given by z = ω(ζ) =−ia 1 + ζ 1 − ζ , (4.7) where a = h 1 − α 2 1 + α 2 with α = 1 r  h −  h 2 − r 2  . (4.8) The tunnel (hole) depth h and radius r are shown in Figure 4.1. This function maps the elastic material outside the hole and below the surface of the half-plane 22 A Deforming Circular Tunnel Chapter 4 onto an annulus [42]. The boundary along the surface of the half-plane, L s , corresponds to the outer boundary of the annulus,  s , along which | ζ | = 1. The boundary along the hole in the half-plane, L h , corresponds to the inner boundary of the annulus,  h , along which | ζ | = α. By virtue of the substitution z = ω(ζ) the functions ϕ 0 (z) and ψ 0 (z) can be written in terms of ζ . We use the following notation: ϕ 0 (z) = ϕ 0 (ω(ζ)) = ϕ 0 m (ζ ), (4.9) ψ 0 (z) = ψ 0 (ω(ζ)) = ψ 0 m (ζ ), (4.10) where the small m denotes the conformally mapped version of the corresponding function in the z-plane. The same notation is used for the complete potentials and the forcing functions: ϕ(z) = ϕ m (ζ ), ψ(z) = ψ m (ζ ), (4.11) h g(z) = h g m (ζ ), h g ◦ (z) = h g m◦ (ζ ), (4.12) and h f ◦ (z) = h f m◦ (ζ ). (4.13) The boundary equation (4.3) contains the derivative of ϕ 0 (z) with respect to z. We use the chain rule in order to determine ϕ  0 (z) in terms of ζ : ϕ  0 m (ζ ) = dϕ 0 m dζ = dϕ 0 dz · dz dζ = ϕ  0 (z)ω  (ζ ) → ϕ  0 (z) = ϕ  0 m (ζ ) ω  (ζ ) . (4.14) Differentiation of (4.7) with respect to ζ yields ω  (ζ ) =−ia  1 1 − ζ + 1 + ζ (1 − ζ) 2  =−ia 2 (1 − ζ) 2 , (4.15) with the result that the second term in (4.3) and (4.5) can be written as z ϕ  0 (z) = ω(ζ) ω  (ζ ) ϕ  0 m (ζ ) =− (1 + ζ)(1 − ζ) 2 2(1 − ζ) ϕ  0 m (ζ ). (4.16) Wechoose the pointz c in the potentials(4.1) and (4.2) such that it corresponds the point ζ c = 0 at the center of the annulus in the transformed region: z c =−ia 1 + ζ c 1 − ζ c =−ia. (4.17) [...]... (4.26) κ(Fx + iFy ) + ln α π(1 + κ) Expansion of the Potentials and Boundary Equations The next step in the solution of the problem is an expansion of the unknown functions ϕ0 (ζ ) and ψ0 (ζ ) into Laurent series in the transformed annular rem m gion R These expansions are valid up to and including the boundaries of the annulus since the functions are analytic in R up to and including the point at...    4π(1 + κ)    h  h Bk = Fx − iFy ◦   4π   h h    Fx − i F y     4π(1 + κ)    0 k ≤ −2, k = −1, k = 0, (4 .30 ) k = 1, k ≥ 2 This completes the expansion of the right-hand side of (4.21) The left-hand side can easily be expanded through substitution of (4.27) and (4.28) in (4.21) These details are omitted here ... Half-Plane 23 Section 4.2 Equations (4.7) and (4.17) can be used to transform the remaining expressions in (4.4) and (4.6) The results are z − zc = − i2aζ , 1−ζ z − zc = z − zc = − i2a , 1−ζ z − zc = z (1 + ζ )(1 − ζ ) , =− z − zc 2ζ (1 − ζ ) i2aζ 1−ζ i2a 1−ζ , (4.18) , (4.19) z (1 + ζ )(1 − ζ ) =− z − zc 2(1 − ζ ) (4.20) Transformation of the Boundary Equations The boundary equation along the surface of the... along the surface of the half-plane can be transformed to the ζ -plane by substituting (4.7) and (4.17) into (4 .3) , with the help of (4.16) and (4.18) – (4.20) Along the circle s we note that |ζ | = 1 and we write ζ = σ , where σ = exp(iϑ), with ϑ defined as shown in Figure 4.2 Then ζ = σ −1 and (4 .3) becomes s 1 ϕ0 (σ ) + (1 − σ −2 )ϕ0 (σ ) + ψ0 (σ ) = f (σ ) on s , (4.21) m m◦ m m 2 where h h h h Fx −... ) = a0 + m bk ζ −k , (4.27) dk ζ −k (4.28) k=1 k=1 and ∞ ψ0 (ζ ) = c0 + m ∞ ck ζ k + k=1 k=1 In order to solve for the coefficients of these Laurent series, we will also expand the boundary equations (4.21) and (4.24) into series Beginning with the boundary equation (4 .3) along the surface, we write the loading function (4.22) as s ∞ f (σ ) = m◦ where Bk σ k , ◦ (4.29) k=−∞  0    h h    κ(Fx... circle h , |ζ | = α, and we write ζ = ασ Then ζ = ασ −1 and the coefficient of ϕ0 (ζ ) in (4.16) becomes s m ω(ασ ) −ασ − (1 − 2α 2 ) + α(2 − α 2 )σ −1 − α 2 σ −2 2(1 − ασ ) = ω (ασ ) The result is that (4.5) can be written as ω(ασ ) h ϕ0 (ασ ) − ψ0 (ασ ) = g (ασ ) κϕ0 (ασ ) − m m m◦ m ω (ασ ) where h h, (4.24) h m◦ on (4. 23) m g (ασ ) = 2µg (ασ ) + CF h h Fx + i F y 1 log + 2π 1 − ασ −1 h h + κ log . z k + ∞ σ xx 2 ( z − z). (3. 27) The function j f ◦ (z) approaches a constant for infinite values of z along L s as a result of the calculations leading to the potentials (3. 22) and (3. 23) . In ad- dition,. holes or on the surface of the half-plane. Specific examples can be solved by a variety of techniques [ 23, 34 ] § 3. 4 Chapter Summary The general representation of the complex potentials for a lower. to maintain equilibrium). § 3. 3 Boundary Equations for a Half-Plane with Holes The unknown single-valued analytic functions ϕ 0 (z) and ψ 0 (z) in (3. 22) and (3. 23) can be determined from boundary