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 ha
Trang 1Section 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 = F T x + i F T y , (3.20)
where total resultant force
T
F x + i F T y = F s x + i F s y + F h x + i F h y (3.21)
is given by the sum of the resultant forces acting on the surface (
s
F x + i F s y ) and
on the holes ( F h x + i F h 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 F s y
κ(
h
F x + i F h y )
2π(1 + κ)
log(z − z c )
−
m
k =1
k
F x + i F k y
2 π(1 + κ) log(z − z k ) +
∞
σ xx
4 z + ϕ 0 (z), (3.22)
Trang 214 Multiple Holes in a Half-Plane Chapter 3
and
ψ (z) =
s
F x − i F s y
h
F x − i F h y
2π(1 + κ)
log(z − z c )
+
m
k =1
κ( F k x − i F k y )
2 π(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 F s 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) + zϕ (z) + ψ(z) + C j = f (z) j on L j , (3.24) where the loading function
j
f (z) = i
z
z (t x + it y ) ds (3.25)
Trang 3Section 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 = f j
◦ (z) on L j , (3.26)
where
j
f
◦ (z) = f (z) j +
s
F x + i F s y
κ(
h
F x + i F h y )
2π(1 + κ)
log(z − z c )
−
s
F x + i F s y
h
F x + i F h y
2π(1 + κ)
log(z − z c )
+
m
k =1
k
F x + i F k y
2 π(1 + κ)
log(z − z k ) − κlog(z − z k )
+
s
F x − i F s y
κ(
h
F x − i F h y )
2π(1 + κ)
z
z − z c
+
m
k =1
k
F x − i F k 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, , m since 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) − zϕ (z) − ψ(z) = 2µ g(z) j on L , (3.28)
Trang 416 Multiple Holes in a Half-Plane Chapter 3
where the displacement function g(z) j 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) = g j
◦ (z) on L j , (3.29)
where
j g
◦ (z) = 2µ g(z) j +
T
F x + i F T y
2 π (κ − 1) log(z − z c )
+
s
F x + i F s y
h
F x + i F h y
2π(1 + κ)
log[(z − z c )(z − z c ) ] +
m
k =1
κ( F k x + i F k y )
2 π(1 + κ) log[(z − z k )(z − z k ) ]
−
s
F x − i F s y
κ(
h
F x − i F h y )
2π(1 + κ)
z
z − z c
−
m
k =1
k
F x − i F k y
2π(1 + κ) ·
z
z − z k +
∞
σ xx
2
1 − κ
2 z − z
.
(3.30)
The function g j
◦ (z) is single-valued and continuous for complete circuits around
the closed boundaries L k , k = 1, 2, 3, , m since 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
Trang 5Section 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.
Trang 6Chapter 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 by Yu [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 solusolu-tion, 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
Trang 7Section 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 the weight 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 ( F s x +
i F s y = 0) In addition, we have only one hole (m = 1) and the total force on
all holes ( F h x + i F h y ) is equal to the force on the single hole (corresponding to
h
F x + i F h y = F 1 x + i F 1 y ) It follows that the potentials (3.22) and (3.23) reduce to
ϕ(z) = −
h
F x + i F h y
2π(1 + κ)
κ log(z − z c ) + log(z − z c ) + ϕ 0 (z), (4.1)
ψ (z) =
h
F x − i F h y
2 π(1 + κ)
log(z − z c ) + κ log(z − z c ) + ψ 0 (z), (4.2)
Trang 820 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) = f s
◦ (z) on L s , (4.3)
where
s f
◦ (z) =
h
F x − i F h y
2π(1 + κ)
κz
z − z c
z − z c
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) = g h
◦ (z) on L h , (4.5)
where
h g
◦ (z) = 2µ g(z) h +
h
F x + i F h y
2 π (κ − 1) log(z − z c )
+
h
F x + i F h y
2 π(1 + κ) log[(z − z c )(z − z c ) ] + κ(
h
F x + i F h y )
2π(1 + κ) log[(z − z c )(z − z c ) ]
−
h
F x − i F h y
2π(1 + κ)
κz
z − z c + z
z − z c
.
(4.6)
As in Chapter 3, the function g(z) h 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).
Trang 9Section 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).
.
. . . .
. . .
.
. .
ξ
η
σ
1
α
ϑ
R
Figure 4.2: The annulus in the conformally mapped ζ -plane.
The conformal mapping is given by
z = ω(ζ) = −ia 1 + ζ
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
Trang 1022 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
ψ 0 (z) = ψ 0 (ω(ζ )) = ψ 0
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
h g(z) = g h
m (ζ ), g h
◦ (z) = g h
m ◦ (ζ ), (4.12)
and
h f
◦ (z) = f h
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 m 0
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
(1 − ζ) 2 , (4.15) with the result that the second term in (4.3) and (4.5) can be written as
zϕ
0 (z) = ω(ζ )
ω (ζ ) ϕ m 0 (ζ ) = − (1 + ζ)(1 − ζ) 2
2(1 − ζ ) ϕ m 0 (ζ ). (4.16)
We choose the point z 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
... terms of ζ :ϕ
0
m (ζ ) = dϕ m 0
dζ = dϕ 0
dz... (ζ ) = −ia
1
1 − ζ +
1 + ζ (1 − ζ) 2
(1 − ζ) , (4.15)... h
m ◦ (ζ ), (4.12)
and
h f
◦ (z) = f h
The