An analytical solution for a deforming, buoyant tunnel in an elastic half-plane was derived in this chapter.. Chapter 5BUOYANCY OF A RIGID TUNNEL Buoyancy effects arise after the excavat
Trang 1Section 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 ) g h
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 ) g h
m◦(ασ )= ∞
k=−∞
A◦k σ k (4.33)
where
A◦k =
A k+
h
F x+ iF h y
2π
1+ (α2− 1)k k(k − 1)α k
k ≤ −2,
A−1+α(
h
F x+ iF h y )
4π (2− α
2)−ακ(
h
F x− iF h y )
4π(1 + κ) k = −1,
A0+ C F −α2(
h
F x+ iF h y )
h
F x− iF h y
4π(1 + κ) +κ(
h
F x− iF h y )
4π(1 + κ) (1− α2) k = 0,
A1− αC F+ακ(
h
F x+ iF h y )
h
F x− iF h y
4πα(1 + κ)
+α(κ − 1)(
h
F x− iF h y )
A2−α2κ(
h
F x+ iF h y )
h
F x− iF h y
A k−κ(
h
F x+ iF h y )
2π
α k k(k − 1)
k ≥ 3.
(4.34)
Trang 226 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
c0= B◦ 0− a0−1
2a1−1
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 kand
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+1and
b k+1are then obtained which are dependant on a single level of coefficients:
(1+ κα 2k+2)a
k+1+ (1 − α2)(k + 1)b k+1= α2(1+ κα 2k )a k
+ (1 − α2)kb k + α k+1A
k+1
◦ + B◦k+1− α2B◦k , k≥ 0 (4.38) and
α 2k (1− α2)(k + 1)a k+1− (κ + α 2k+2)b
k+1= α 2k (1− α2)ka k
− (κ + α 2k )b k + α k A◦−k + α 2k B◦−k − α 2k+2B
−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 knumerically, 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=λ k11a k+λ k12b k+λ k13, (4.40)
b k+1=λ k21a k+λ k22b k+λ k23, (4.41)
Trang 3Section 4.2 Solution for a Deforming Hole in a Half-Plane 27
where
k
λ11 =κα2+ [α2(κ2+ α2) + (1 − α2)2(k + 1)k + κα 2k+4]α 2k
k
λ12 =κα2− κ − (1 − α2) [(1 − α2)k + 1]α 2k
k
λ13 =(κ + α
2k+2)(B
k+1
◦ − α2B◦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+2B −k−1
◦ )α k
k
λ21 =(1− α2) [α2− (1 − α2)k + κα 2k+2]α 2k
k
λ22 =κ + [1 + κ2α2+ (1 − α2)2(k + 1)k + κα 2k+2]α 2k
k
λ23 =(1− α
2)(k + 1)(α 2k+1A k+1
◦ + α k B◦k+1− α k+2B k
◦ )α k
κ + γ k α 2k
−(1+ κα
2k+2)(A −k
◦ + α k B◦−k − α k+2B −k−1
◦ )α k
with
γ k = (1 + κ2)α2+ (1 − α2)2(k + 1)2+ κα 2k+4. (4.48)
Hereγ kandλ k mn , n are real andλ k 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 kcan be written as follows:
a k =β k011a0+β k012a0+
k−1
i=0
k
β i13 k = 1, 2, 3, , (4.49)
b k =β k021a0+β k022a0+
k−1
i=0
k
β i23 k = 1, 2, 3, , (4.50)
where
k
β lmn=
l
k−1
λ mj
k−1
β k = l + 2, l + 3, l + 4, ,
(4.51)
Trang 428 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 β k lmn , n β k 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, a0, 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 a0affects 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, a0 An
expression for a0can 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 boundarysof the annulus
Enforcing the condition that a k → 0 and b k → 0 as k → ∞ in the complex
conjugate of (4.49) and in (4.50) gives
0=∞β012a0+∞β011a0+∞−1
i=0
∞
0=∞β021a0+∞β022a0+
∞−1
i=0
∞
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 a0results in
a0=
∞
β012∞−1
i=0
∞
β i23−∞β021∞−1
i=0
∞
β i13
∞
β021∞
β011−∞β022∞
β012
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◦kandA◦k, are
equal to zero for every value of k By examining the terms of the form β ij3
it is apparent that the leading factor in these terms will always be of the form
λ j3, causing them to be zero when allB◦k and A◦k are zero The conclusion is
that a0= 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
Trang 5Section 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+ iF h y
2π(1 + κ)
κ log(− i2a
1− ζ ) + log(−
i2aζ
1− ζ ) + a0+∞
k=1
a k ζ k+∞
k=1
b k ζ −k ,
(4.56)
and
ψ
m (ζ )=
h
F x− iF h y
2π(1 + κ)
log(− i2a
1− ζ ) + κ log(−
i2aζ
1− ζ ) + c0+∞
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
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
Trang 630 A Deforming Circular Tunnel Chapter 4
Trang 7Part II
APPLICATIONS AND CONCLUSION
31
Trang 9Chapter 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 forceF h x+ iF h 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 deformaequaliza-tion 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 by Yu [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
Trang 1034 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
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
where γ is the unit weight of the material and K0is the (constant) coefficient
of lateral earth pressure (K0is 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+ iF h y = iγ πr2, (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
Trang 11Section 5.1 Solution of the Problem 35
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 are constant)
Zero displacements along the tunnel are specified in the solution by setting
the given boundary functiong(z) h = 0 in (4.6) Consequently, the mapped
displacement function is also equal to zero:g h
m (ασ )= 0 It follows from (4.31) that
G(ασ ) = 2µ(1 − ασ ) g h
m (ασ )=
∞
k=−∞
A k σ k = 0, (5.5)
from which it then follows that A k = 0 for all k The Fourier expansion given
by (4.34) can now be calculated by using these values for A k and by using the
value of the resultant force given by (5.4) This completes the solution for a
resultant force acting on a rigid tunnel The 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
F x+ iF h y
P = iπ r
h
2
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 in this
case is the expression for the resultant 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 in a
similar fashion Multiplying the stresses (5.1) – (5.3) through by the same factor
h/P = 1/(γ h) as used above yields
σ xx
γ h =y
σ yy
γ h = K0
y
σ xy
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