ANALYTIC SOLUTIONS OF ELASTIC TUNNELING PROBLEMS Phần 9 pot

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Section A.2 Evaluation of the Coefﬁcients in the Potentials 85
(due to the fact that Cauchy’s Integral Theorem, which guarantees path inde-
pendence in a simply-connected region, is not necessarily satisﬁed for contours
around the hole). It follows that

L
∗
(z) dz = C
c
, (A.19)
where C
c
is some complex constant (the same reasoning applies here as was
used to show that the value B
c
in (A.13) is constant).
Proceeding as before (see the reasoning used to determine equation (A.17)),
the integral of
∗
(z) can be written as

z
z
0
∗
(z) dz =
C
c
2πi
log(z − z
c

) +
∗∗
ϕ(z), (A.20)
where
∗∗
ϕ(z) is a single-valued analytic function. Substituting (A.20) in (A.18)
and expanding gives
ϕ(z) = A
c
z log(z − z
c
) − A
c
z
c
log(z − z
c
) − A
c
(z − z
c
)
+
C
c
2πi
log(z − z
c
) +
∗∗

ϕ(z)+ C.
(A.21)
Combining logarithmic terms and incorporating all single-valued analytic terms
in a new analytic function,
∗
ϕ(z), results in
ϕ(z) = A
c
z log(z − z
c
) + γ
c
log(z − z
c
) +
∗
ϕ(z), (A.22)
where the constant γ
c
and the new single-valued analytic function
∗
ϕ(z) have
been introduced for convenience.
The multi-valued nature of the potential ψ(z) can be determined by noting
that (z) = ψ

(z) is single-valued (this is because the stresses on the left-hand
side of (2.5) and the function ϕ

(z) on the right-hand side of the same equation

are all single valued). With the same reasoning used to determine (A.17) and
(A.20), it can be shown that the integral of (z) can be written as
ψ(z) =

z
z
0
(z)dz = γ

c
log(z − z
c
) +
∗
ψ(z). (A.23)
where γ

c
is a complex constant and
∗
ψ(z) is a single-valued analytic function.
§A.2 Evaluation of the Coefﬁcients in the Potentials
All that remains for a full determination of the multi-valued nature of the poten-
tials in a region with a hole is to determine the unknown coefﬁcients in (A.22)
and (A.23). This is accomplished as follows.
In order to ensure that the displacements associated with the multi-valued
potentials given by (A.22) and (A.23) are single-valued, equation (2.1) must not
86 Multi-Valued Complex Potentials Appendix A
become multi-valued anywhere in the region. Substituting (A.22) and (A.23)
in (2.1) gives

2µ(u + iv) = κ

A
c
z log(z − z
c
) + γ
c
log(z − z
c
) +
∗
ϕ(z)

− z

A
c
log(z − z
c
) +
A
c
z
z −
z
c
+
γ
c

z −
z
c
+
∗
ϕ

(z)

−
γ

c
log(z − z
c
) −
∗
ψ(z).
(A.24)
Separating the multi-valued parts from the single-valued parts results in
2µ(u + iv) = (κA
c
z + κγ
c
) log(z − z
c
)
− (zA
c
+ γ


c
)log(z − z
c
) + f(z),
(A.25)
where
f(z) = κ
∗
ϕ(z) −
A
c
zz
z − z
c
−
z
γ
c
z − z
c
− z
∗
ϕ

(z) −
∗
ψ(z), (A.26)
which is single-valued. Denoting by r
zc

the distance from z to z
c
, denoting by
ϑ
zc
the argument of z − z
c
(see Figure A.3), and splitting the logarithms into
their real and imaginary parts yields
2µ(u + iv) = (κzA
c
+ κγ
c
)(ln r
zc
+ iϑ
zc
)
− (zA
c
+ γ

c
)(ln r
zc
− iϑ
zc
) + f(z).
(A.27)
After collecting terms, we have

2µ(u + iv) = (κzA
c
+ κγ
c
− zA
c
− γ

c
) ln r
zc
+ i(κzA
c
+ κγ
c
+ zA
c
+ γ

c
)ϑ
zc
+ f(z).
(A.28)
The function ϑ
zc
in this expression increases continuously along each circuit
around the point z
c
. The other functions in (A.28) are single-valued. It follows

that the coefﬁcient of ϑ
zc
must be set to zero in order to ensure single-valued
displacements:
A
c
(κ + 1)z + κγ
c
+ γ

c
= 0. (A.29)
As a result
A
c
= 0 (A.30)
and
κγ
c
+ γ

c
= 0. (A.31)
The coefﬁcients γ
c
and γ

c
can be related to the resultant force on the hole
through use of (2.6). We take an integration path consisting of a complete circuit

around L
h
in a clockwise direction (see Figure A.3). Note that this integration
path is chosen such that the region is to the left, as speciﬁed for (2.6). Denoting
by []
L
h
the increase undergone by the expression inside the brackets along the
Section A.2 Evaluation of the Coefﬁcients in the Potentials 87

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y
ϑ
zc
r
zc
z
c
L
h
Figure A.3: Integration path.
integration path, integrating along L
h
in (2.6) and substituting equations (A.22)
and (A.23) gives

γ
c
log(z − z
c
) +
∗
ϕ(z) +
z
γ
c
z −
z
c
+ z
∗

ϕ

(z) + γ

c
log(z − z
c
) +
∗
ψ(z)

L
h
= i

L
h
(t
x
+ it
y
) ds = i(
h
F
x
+ i
h
F
y
), (A.32)

where
h
F
x
+ i
h
F
y
is the resultant force on the hole, and where we have utilized
(A.30). Using the same notation as above, the logarithms can be split into their
real and imaginary parts, resulting in

i(γ
c
− γ

c
)ϑ
zc
+ (γ
c
+ γ

c
) ln r
zc
+ g(z)

L
h

= i(
h
F
x
+ i
h
F
y
), (A.33)
where
g(z) =
z
γ
c
z − z
c
+
∗
ϕ(z) + z
∗
ϕ

(z) +
∗
ψ(z). (A.34)
We note that ϑ
zc
decreases by an amount 2π upon making a complete circuit
around L
h

, and that ln r
zc
and g(z) each return to their original values upon
making a complete circuit around L
h
. The result is that
(γ
c
− γ

c
)(−2π) =
h
F
x
+ i
h
F
y
. (A.35)
Solving for γ
c
and γ

c
using (A.31) and (A.35), applying (A.30), and substituting
the results in (A.22) and (A.23) ﬁnally yields
ϕ(z) =−
h
F

x
+ i
h
F
y
2π(1 + κ)
log(z − z
c
) +
∗
ϕ(z), (A.36)
88 Multi-Valued Complex Potentials Appendix A
and
ψ(z) =
κ(
h
F
x
− i
h
F
y
)
2π(1 + κ)
log(z − z
c
) +
∗
ψ(z). (A.37)
§A.3 Extension to more than one Hole

The potentials ϕ(z) and ψ(z) for a ﬁnite plane with multiple holes can be
obtained by a superimposition of the potentials for a single hole (given by (A.36)
and (A.36)). This is due to the fact that the coefﬁcients of the logarithms (or
more precisely, the integrals (A.13), (A.19), and (A.23) and the resultant force,
given by the integral (A.32)) are not effected by the presence of other holes. It
follows that the potentials for a ﬁnite plane with m holes can be written as
ϕ(z) =−
m

k=1
k
F
x
+ i
k
F
y
2π(1 + κ)
log(z − z
k
) +
∗
ϕ
n
(z), (A.38)
ψ(z) =
m

k=1
κ(

k
F
x
− i
k
F
y
)
2π(1 + κ)
log(z − z
k
) +
∗
ψ
n
(z), (A.39)
where the values z
k
denote points within their respective holes. The functions
∗
ϕ
n
(z) and
∗
ψ
n
(z) denote new arbitrary single-valued analytic functions.
Appendix B
IMPLEMENTATION IN A COMPUTER PROGRAM
The focus of this appendix is the implementation of the results from Chap-

ter 4 in a computer program. In order to facilitate this implementation, the
expressions from Chapter 4 are put into a form that is capable of generating
dimensionless stresses and displacements. Using this form, we take the deriva-
tives of the potentials and use them to write out the equations for the stresses
and displacements explicitly.
§ B.1 Dimensionless Complex Potentials
In order to generate dimensionless expressions for the stresses and the displace-
ments, we divide the complex potentials by a constant P that has dimensions
of force per unit length. This will generate dimensionless complex potentials
on the basis of (2.1) – (2.5). The constant P will be deﬁned according to the
parameters appearing in the solution of particular problems.
We start this process for the results derived in Chapter 4 by dividing the
values of the coefﬁcients in the Laurent expansions (4.27) and (4.28) by P .We
make the following deﬁnitions:
p
k
=
a
k
P
,q
k
=
b
k
P
,r
k
=
c

k
P
,s
k
=
d
k
P
, (B.1)
where a
k
and b
k
are the coefﬁcients in the Laurent expansion (4.27), given by
(4.49), (4.50), and (4.55). The coefﬁcients c
k
and d
k
in the Laurent expan-
sion (4.28) are given by (4.35) – (4.37). Dividing the Laurent expansions (4.27)
and (4.28) by P results in
∗
ϕ
0
(ζ ) = p
0
+
∞

k=1

p
k
ζ
k
+
∞

k=1
q
k
ζ
−k
, (B.2)
∗
ψ
0
(ζ ) = r
0
+
∞

k=1
r
k
ζ
k
+
∞

k=1

s
k
ζ
−k
. (B.3)
89
90 Implementation in a Computer Program Appendix B
where
∗
ϕ
0
(ζ ) =
ϕ
0
m
(ζ )
P
,
∗
ψ
0
(ζ ) =
ψ
0
m
(ζ )
P
. (B.4)
Dividing (4.35) – (4.37) by P gives
r

0
=
B
0
◦
P
−
p
0
−
1
2
p
1
−
1
2
q
1
, (B.5)
r
k
=
B
−k
◦
P
−
q
k

−
1
2
(k + 1)p
k+1
+
1
2
(k − 1)p
k−1
,k>0, (B.6)
s
k
=
B
k
◦
P
−
p
k
+
1
2
(k − 1)q
k−1
−
1
2
(k + 1)q

k+1
,k>0. (B.7)
On the basis of (4.42) – (4.47) and (4.51), the only parameters in (4.49), (4.50),
and (4.55) that are not dimensionless are
k
β
i13
and
k
β
i13
. These parameters have
the dimensions of the Fourier coefﬁcients
B
k
◦
and A
k
◦
. Dividing equations (4.49)
and (4.50) through by P results in
p
k
=
k
β
011
p
0
+

k
β
012
p
0
+
k−1

i=0
k
β
i13
P
k = 1, 2, 3, , (B.8)
q
k
=
k
β
021
p
0
+
k
β
022
p
0
+
k−1


i=0
k
β
i23
P
k = 1, 2, 3, (B.9)
The expression for a
0
, equation (4.55), cannot be implemented in a computer
program unless the limit k →∞is replaced by an expression like k = k
I
, where
k
I
is some large integer, say 10,000. Dividing (4.55) by P and replacing the
symbol ∞ with k
I
results in
p
0
=
k
I
β
012
k
I
−1


i=0
k
I
β
i13
P
−
k
I
β
021
k
I
−1

i=0
k
I
β
i23
P
k
I
β
021
k
I
β
011
−

k
I
β
022
k
I
β
012
. (B.10)
We now have complete and dimensionless expressions for
∗
ϕ
0
(ζ ) and
∗
ψ
0
(ζ )
which can be programmed into a computer.
Finally, we convert the complete potentials given by (4.56) and (4.57) to
their dimensionless forms. This gives
∗
ϕ(ζ) =−
κ(
h∗
F
x
+ i
h∗
F

y
)
2π(1 + κ)
log(−
i2a
1 − ζ
) −
h∗
F
x
+ i
h∗
F
y
2π(1 + κ)
log(−
i2aζ
1 − ζ
)
+ p
0
+
∞

k=1
p
k
ζ
k
+

∞

k=1
q
k
ζ
−k
,
(B.11)
Section B.2 Dimensionless Coordinates 91
and
∗
ψ(ζ) =
h∗
F
x
− i
h∗
F
y
2π(1 + κ)
log(−
i2a
1 − ζ
) +
κ(
h∗
F
x
− i

h∗
F
y
)
2π(1 + κ)
log(−
i2aζ
1 − ζ
)
+ r
0
+
∞

k=1
r
k
ζ
k
+
∞

k=1
s
k
ζ
−k
,
(B.12)
where

h∗
F
x
+ i
h∗
F
y
=
h
F
x
+ i
h
F
y
P
,
∗
ϕ(ζ) =
ϕ
m
(ζ )
P
,
∗
ψ(ζ) =
ψ
m
(ζ )
P

. (B.13)
§ B.2 Dimensionless Coordinates
The coordinates in the z-plane can be written in dimensionless form by dividing
(4.7) by h, the depth of the tunnel. This results in
∗
z =
∗
ω(ζ) =−i
∗
a
1 + ζ
1 − ζ
, (B.14)
where
∗
z = z/h and
∗
ω(ζ) are the dimensionless coordinates and dimensionless
mapping function, respectively, and where we have deﬁned the parameter
∗
a =
1 − α
2
1 + α
2
. (B.15)
This parameter is simply the dimensionless form of (4.8). The dimensionless
coordinates
∗
z can be obtained by using (B.14) instead of (4.7) to map the

coordinates back from the ζ -plane.
§ B.3 Evaluation of the Stresses and Displacements
It is most convenient to calculate (2.1) and (2.4) – (2.5) in the ζ -plane and then
to relate the resulting values to the dimensionless coordinates
∗
z through use
of (B.14). In order to perform these calculations we will need expressions for
∗
ϕ

(z),
∗
ϕ

(z), and
∗
ψ

(z) in terms of ζ . The derivative of ϕ(z) with respect to z
in terms of ζ was derived in (4.14). Similar results can be obtained for
∗
ϕ

(z)
and
∗
ψ

(z). They are
∗

ϕ

(z) =
1
h
∗
W
1
(ζ )
∗
ϕ

(ζ ),
∗
ψ

(z) =
1
h
∗
W
1
(ζ )
∗
ψ

(ζ ), (B.16)
where we have deﬁned the (dimensionless) function
∗
W

1
(ζ ) =
h
ω

(ζ )
= i
(1 − ζ)
2
2
∗
a
. (B.17)
92 Implementation in a Computer Program Appendix B
The chain rule can be used twice to determine
∗
ϕ

(z) in terms of ζ :
∗
ϕ

(ζ ) =
d
dζ

∗
ϕ

(z)ω


(ζ )

=
d
dz
∗
ϕ

(z)·
dz
dζ
·ω

(ζ ) +
∗
ϕ

(z)ω

(ζ ). (B.18)
It follows that
∗
ϕ

(z) =
1
h
2
[

∗
W
1
(ζ )]
2
∗
ϕ

(ζ ) −
1
h
2
∗
W
3
(ζ )
∗
ϕ

(ζ ), (B.19)
where we have deﬁned the (dimensionless) function
∗
W
3
(ζ ) = h
2
ω

(ζ )
[ω


(ζ )]
3
=−
(1 − ζ)
3
2
∗
a
2
. (B.20)
The derivatives of (B.11) and (B.12) with respect to ζ are:
∗
ϕ

(ζ ) =−
κ(
h∗
F
x
+ i
h∗
F
y
)
2π(1 + κ)
·
1
1 − ζ
−

h∗
F
x
+ i
h∗
F
y
2π(1 + κ)
·
1
ζ(1 − ζ)
+
∞

k=1
kp
k
ζ
k−1
−
∞

k=1
kq
k
ζ
−k−1
, (B.21)
∗
ψ


(ζ ) =
h∗
F
x
− i
h∗
F
y
2π(1 + κ)
·
1
1 − ζ
+
κ(
h∗
F
x
− i
h∗
F
y
)
2π(1 + κ)
·
1
ζ(1 − ζ)
+
∞


k=1
kr
k
ζ
k−1
−
∞

k=1
ks
k
ζ
−k−1
, (B.22)
and
∗
ϕ

(ζ ) =−
κ(
h∗
F
x
+ i
h∗
F
y
)
2π(1 + κ)
·

1
(1 − ζ)
2
+
h∗
F
x
+ i
h∗
F
y
2π(1 + κ)
·
1 − 2ζ
[ζ(1 − ζ)]
2
+
∞

k=1
k(k − 1)p
k
ζ
k−2
+
∞

k=1
k(k + 1)q
k

ζ
−k−2
. (B.23)
Expressions for the displacements in terms of ζ can be obtained by dividing
(2.1) by P and substituting (4.7), (B.16), (B.19), and (B.21) – (B.23) in the
resulting expression. This gives
2µ
P
(u + iv) = κ
∗
ϕ(ζ) −
∗
ω(ζ)
∗
W
1
(ζ )
∗
ϕ

(ζ ) −
∗
ψ(ζ). (B.24)
Expressions for the stresses can be obtained by multiplying (2.4) and (2.5) by
h/P and substituting the same expressions in the result. This yields
h
P
(σ
xx
+ σ

yy
) = 2

∗
W
1
(ζ )
∗
ϕ

(ζ ) +
∗
W
1
(ζ )
∗
ϕ

(ζ )

, (B.25)
Section B.4 Comparison of Numerical and Analytical Solutions 93
and
h
P
(σ
yy
− σ
xx
+ 2iσ

xy
)
= 2

∗
ω(ζ)[
∗
W
1
(ζ )]
2
∗
ϕ

(ζ ) −
∗
ω(ζ)
∗
W
3
(ζ )
∗
ϕ

(ζ ) +
∗
W
1
(ζ )
∗

ψ

(ζ )

. (B.26)
The dimensionless stresses and displacements have been fully determined in
terms of the mapping parameter ζ . The dimensionless coordinates x/h and y/h
corresponding to the values calculated using (B.24) - (B.26) can be obtained by
mapping ζ to
∗
z with (B.14).
§ B.4 Comparison of Numerical and Analytical Solutions
A computer program implementing the equations in the preceding sections is
approximately 40% slower than code which determines p
0
, p
k
and q
k
by solving
the system of equations (4.38) - (4.39) numerically and examining the behavior
of p
k
for large values of k and different values of p
0
, as performed by Ver-
ruijt [42]. It should be noted that part of this difference is due to the fact that
in Verruijt’s implementation it is assumed that all the coefﬁcients are purely
imaginary (so that all calculations involve only real variables) whereas the cal-
culations in this implementation are fully complex (making it possible to include

a non-vertical resultant force acting on the tunnel). It may be noted that both
the numerical and the analytical computation of the coefﬁcients of the Lau-
rent series result in extremely small errors in the normalized displacements and
stresses. Even so, there are some small differences between the two techniques,
as can be seen in Table B.1, and as will be discussed below.
Table B.1: Comparison of solutions for the ground loss problem with ν = 0.3 and with
no resultant forces acting on the boundaries. The values have been normalized through
use of the parameter P = 2µu
g
, where u
g
is the maximum convergence of the tunnel.
Solution r/h Num. Terms Max. Error in u/u
g
Max. σ
surface
·h/(2µu
g
)
analytic 0.01 4 9.9 × 10
−10
2.4 ×10
−15
numeric 0.01 4 1.5 ×10
−8
2.4 ×10
−15
analytic 0.26 11 3.3 ×10
−9
4.9 ×10

−15
numeric 0.26 10 3.4 ×10
−8
2.2 × 10
−13
analytic 0.9 41 1.5 × 10
−7
1.6 ×10
−13
numeric 0.9 38 3.4 × 10
−8
1.2 ×10
−12
94 Implementation in a Computer Program Appendix B
Based on the ground loss problem with no resultant forces acting on the
boundaries, the fully analytical solution is approximately one order of mag-
nitude more accurate for common tunnel depths (r/h < 0.5) and about one
order of magnitude less accurate for extremely shallow tunnels (r/h > 0.8).
The surface stresses (σ
yy
and σ
xy
) in the new solution are on the average of
one order of magnitude more accurate for all values of r/h. As noted earlier,
however, both solutions are very accurate and the differences discussed here are
negligible in comparison to the small errors in the boundary conditions.
Appendix C
THE OVALIZATION BOUNDARY CONDITION
It is the purpose of this appendix to derive the transformed Fourier expansion
of the boundary conditions for the ovalization of a circular tunnel used in this

thesis. The transformation is made from the physical z-plane containing the
tunnel to the conformally mapped annulus in the ζ -plane (see Figure C.1) using
the mapping function (4.7) presented in Chapter 4. The results are written in the
form of the given displacement expansion G

(ασ ) introduced in (4.31). This
solution was ﬁrst published in [35].
§ C.1 Boundary Condition Along the Tunnel
The displacements for the ovalization of a tunnel in the z-plane are given in
terms of the polar coordinates r and θ (see the left side of Figure C.1) by
u
r
+ iu
θ
= u
o
cos 2θ, (C.1)
where u
r
and u
θ
are the radial and tangential displacements along the tunnel
boundary, and where u
o
is the maximum displacement of the tunnel boundary.

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r
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θ
L
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ξ
η
σ
1
α
ϑ
Figure C.1: Ovalization of a tunnel in the z-plane; conformal mapping the ζ -plane.
95
96 The Ovalization Boundary Condition Appendix C
It is our goal to transform (7.1) to the ζ -plane, and then to expand it in a
Fourier series in order to determine the function G

(ασ ) given in (4.31). We start
by writing (7.1) in terms of the horizontal and vertical displacements u
t
+ iv
t
along the tunnel periphery. For this we use the relation (see [23])
u
t
+ iv
t

= (u
r
+ iu
θ
)e
iθ
= u
o
cos 2θ cos θ + iu
o
cos 2θ sin θ. (C.2)
After substituting the expressions
cos θ =
x
r
, sin θ =
y +h
r
and x =
z +
z
2
,y=
z −
z
2i
(C.3)
into (C.2), expanding the resulting equation, and rearranging and collecting
terms we obtain
2r

3
u
o
·
h
g(z) = z
3
+ i3hz
2
− 4h
2
z + zz
2
+ ihz
2
− i2hzz + 2h
2
z − i2h
3
, (C.4)
where the function
h
g(z), deﬁned in Chapter 4, represents the given displace-
ments along the tunnel boundary:
h
g(z) = u
t
+ iv
t
on L

h
. (C.5)
In (C.4) the given displacements along the tunnel have been written completely
in terms of z and
z, allowing substitution of the conformal mapping (4.7) for z
in order to transform the given displacements to the ζ -plane.
§ C.2 Transformation of the Boundary Conditions
The given displacements
h
g(z) can now be transformed to the ζ -plane by sub-
stituting (4.7) for z in (C.4). We use the following notation, introduced in
Chapter 4: ζ = αe
iϑ
= ασ and ζ = ασ = ασ
−1
, along the inner boundary
of the annulus in the ζ -plane. After substituting (4.7) for z in (C.4), combining
fractions, expanding the numerator, and collecting terms we obtain
−i
2α
3
u
o
·
h
g(z) =
c
1
σ
3

+ c
2
σ
2
+ c
3
σ + c
4
+ c
5
σ
−1
+ c
6
σ
−2
(1 − ασ)
3
(1 − ασ
−1
)
2
, (C.6)
where
c
1
= α
3
(1 + α
4

), c
2
=−α
4
(5 + 4α
2
+ α
4
), c
3
= 4α
5
(4 + α
2
),
c
4
=−4α
4
(1 + 4α
2
), c
5
= α
3
(1 + 4α
2
+ 5α
4
), c

6
=−α
4
(1 + α
4
),
and where we have simpliﬁed the coefﬁcient on the left-hand side of (C.6) by
using the relation
1 + α
2
2h
=
α
r
, (C.7)

ANALYTIC SOLUTIONS OF ELASTIC TUNNELING PROBLEMS Phần 9 pot

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