Mathematically, the boundary conditions along the tunnel are given in terms of the polar coordinates r and θ see Figure 6.1 by the following equation: where u r is the displacement in th
Trang 1Section 6.1 Solution of the Problem 49
and ψ The stresses and displacements due to buoyancy can be added later by
superimposing the solution for the buoyant, rigid tunnel discussed in Chapter 5,
and the gravitational stresses can be superimposed through use of (5.7) – (5.9).
Mathematically, the boundary conditions along the tunnel are given in terms
of the polar coordinates r and θ (see Figure 6.1) by the following equation:
where u r is the displacement in the radial direction, u θ is the displacement in
the tangential direction, and u g is the amount that the tunnel contracts in the
radial direction due to ground loss (see Figure 6.1).
.
.
. .
.
. . .
.
x y
r
h θ
u g
L h
Figure 6.1: The ground loss problem.
Equation (6.1) can be written in terms of the horizontal and vertical
displace-ments u t + iv t along the tunnel periphery, which are needed for the solution in
Chapter 4, by utilizing the following transformation (see [23]):
u t + iv t = (u r + iu θ )e iθ = u g ( cos θ + i sin θ). (6.2)
After substituting the expressions
cos θ = x
r , sin θ = y + h
2 , y = z − z
2i (6.3) into (6.2) we obtain
h
g(z) = u t + iv t = −u g
z + ih
where the function g(z) h , defined in Chapter 4, represents the given
displace-ments along the tunnel boundary in terms of the variable z The conformal
mapping (4.7) can now be substituted in (6.4), facilitating the transformation
Trang 2of the given displacements to the ζ -plane and the determination of the function
G (ασ ) given in (4.31).
Substituting (4.7) for z and utilizing the relation ζ = ασ which is valid along the tunnel boundary L h yields
h
g
m (ασ ) = −iu g
α − σ
Multiplication by 2µ(1 − ασ ) finally gives
The Fourier expansion of this expression is trivial and contains only the two terms (the other terms being equal to zero)
A 0 = −2iµu g α and A 1 = 2iµu g σ. (6.7) The expansion given by (4.34) can now be calculated by using these values for
A k and by setting the value of the resultant force equal to zero This completes the solution for the ground loss problem in a linearly elastic half-plane with a circular tunnel The result is given by (4.56) and (4.57).
The solution can be normalized by setting the parameter P in Appendix B equal to 2µu g It is clear from (6.7) that division by P = 2µu g will result
in dimensionless coefficients which will be independent of µ and u g The only components of the solution in Chapter 4 which are not dimensionless are the Fourier coefficients given by (6.7) and the resultant force, which in this case is equal to zero It then follows from the discussion in Appendix B that dimensionless expressions for stresses and displacements will be obtained
which will only depend on the parameters r/ h and ν The displacements will
be given in terms of P /2µ = u g and the values for the stresses will be given in
terms of P / h = 2µu g / h , as is evident from (B.24) – (B.26).
Examination of the final solution shows that the displacements tend towards
a constant value for large values of z After adding a rigid body motion to the
solution such that the displacements vanish at infinity, a certain rigid body dis-placement of the tunnel as a whole is obtained It is found that this disdis-placement
is a downward displacement, see Figure 6.3 through Figure 6.5.
§ 6.2 Validation of the Solution
The solution for the ground loss problem has been implemented in a computer program which can be used to generate plots of the normalized stresses and displacements The plots can be visually inspected to confirm that the stresses vanish along the surface (see Figure 6.2) and that the tunnel does indeed contract uniformly (see Figure 6.3).
It is noted that in Figure 6.2, as in all plots of stress contours in this thesis, the solid thin curves denote negative (compressive) stresses, the thick solid curves denote contours of zero stress, and the finely dotted curves denote positive
Trang 3Section 6.2 Validation of the Solution 51
-5
-4
-3
-2
-1
0
-5 -4 -3 -2 -1 0
x/ h
- y/ h
. .
.
. .
.
.
.
.
.
Figure 6.2: The stresses σ yy (left-hand side) and σ xy (right-hand side) for ν = 0 and r/ h = 0.7 The contour interval is 0.25µu g / h (tensile) stresses It is emphasized that in most cases such tensile stresses will be eliminated through superposition of the stresses due to gravity, (5.7) – (5.9) The plots in this chapter only show the incremental stresses due to ground loss The initial stresses should be added to obtain the total stresses -5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 -5 -4 -3 -2 -1 0 x/ h - y/ h
.
.
.
.
.
.
.
.
It is interesting to observe that the tunnel settles as a whole due to the uniform
contraction around the cavity This is clearly depicted in Figure 6.3, where the
invert of the tunnel has remained practically fixed, and the crown has settled by
approximately 2u
Trang 4The programmed solution yields maximum errors in the normalized stresses
σ yy h/ 2µu g and σ xy h/ 2µu g along the surface ranging from ±4×10 −13 for very
shallow tunnels (r/ h = 0.9) to ±2 × 10 −15 for deep tunnels (r/ h = 0.01) The maximum errors of the components of the normalized displacements u t /u g and v t /u g along the tunnel boundary range from ±7 × 10 −8 for very shallow
tunnels to ±8 × 10 −10 for deep tunnels These values are for ν = 0, but the
maximum errors for other values of ν appear to be within an order of magnitude
of the errors given here.
§ 6.3 Comparison to Sagaseta’s Solution
Sagaseta [28] presented an approximate solution for the ground loss problem
in a half-plane made up of an incompressible material His solution consists
of three parts: the first is a singular point-sink solution corresponding to the contraction of point in an infinite plane, the second is an image of the first (obtained by reflecting the first solution across the surface of the half-plane), and the third solution is a Boussinesq stress distribution added to the surface
to ensure that normal stresses along the surface of the half-plane are zero The complete solution is approximate in the sense that the addition of the third part affects the displacements at the singular points, and thus along the tunnel boundary.
The influence of the third part of the solution can be clearly seen in Figure 6.4
in which the exact solution is compared to Sagaseta’s solution It appears that
-3
-2
-1
0
-3 -2 -1
0
x/ h
- y/ h
.
.
. . .
. .
. .
. .
. .
. .
. .
. .
. . .
.
.
.
.
.
. . .
. .
. . . .
.
. .
. . .
.
. .
.
.
.
. .
.
. .
.
.
Figure 6.4: Comparison between the exact solution (left-hand side) and Sagaseta’s
Trang 5Section 6.4 Stresses around the Tunnel 53
the surface load added by the Boussinesq solution has the effect of adding
an additional compression the tunnel, as is to be expected This results in
a moderate over-prediction of the surface settlements when using Sagaseta’s
solution for very shallow tunnels.
For very deep tunnels the effect of the Boussinesq part of the approximate
solution should become negligible and the complete solution should agree with
the solution presented in this chapter This is indeed the case, as can be observed
in Figure 7.5, where a deeper tunnel with an r/ h ratio of 0.3 is shown Even
-2
-1
0
-2 -1
0
x/ h
- y/ h
.
.
.
.
. .
.
.
.
.
.
.
. . .
.
.
. .
.
.
. .
.
.
.
Figure 6.5: Comparison between the exact solution (left-hand side) and Sagaseta’s
though the results do not yet agree fully with each other, the differences have
become negligible in this case The difference vanishes to within machine
accuracy for very deep tunnels (r/ h<0.01).
§ 6.4 Stresses around the Tunnel
The isotropic stresses 1
2 (σ xx + σ yy ) and deviatoric stresses
1
4 (σ yy − σ xx ) 2 + σ 2
xy
are shown in Figure 6.6 It appears from Figure 6.6 that both the isotropic and
deviatoric stresses around the tunnel are strongly affected by the presence of
the stress-free surface The isotropic stresses increase rapidly along the upper
half of the tunnel, and the deviatoric stresses decrease rapidly, especially near
the crown of the tunnel.
It can also be seen from the contours of the isotropic and deviatoric stresses in
Figure 6.6 that any plastic effects that may occur would probably be concentrated
around the bottom half of the tunnel The deviatoric stresses are highest there
Trang 6-5 -4 -3 -2 -1 0 1 2 3 4 5
-5
-4
-3
-2
-1
0
-5 -4 -3 -2 -1 0
x/ h
- y/ h
. . ..
.
.
.
.
.
.
Figure 6.6: The isotropic stress (left-hand side) and deviatoric stress (right-hand side) for the case that ν = 0 and r/h = 0.7 The contour interval is 0.15µu g / h and the isotropic stresses are lowest there It should be taken into consideration, however, that these lower isotropic stresses may be offset by the higher isotropic stresses near the bottom of the tunnel resulting from gravity loading § 6.5 Stresses on the Tunnel The radial and shear stresses σ rr and σ rt acting on the tunnel lining due to a contraction of the tunnel cavity are shown for different values of Poisson’s ratio and different relative tunnel depths in Figure 6.7 and Figure 6.8 The
.
.
. . . .
.
. .
. . . . . . . . . . . . σ rr ν = 0.5 .ν = 0.25 ν = 0 r/ h = 0.7
.
.
.
. . . .
.
. . . . . . . . . .
σ rr
ν = 0.5
ν = 0.25
ν = 0
r/ h = 0.1
Figure 6.7: The radial stresses acting on the tunnel as a function of ν.
most noticeable feature of Figure 6.7 is that the radial shear stresses σrt are relatively small for shallow tunnels and vanish completely for deeper tunnels.
Trang 7Section 6.5 Stresses on the Tunnel 55
. . .
.
.
σ rt ν = 0.5 .ν = 0.25 ν = 0 r/ h = 0.7
..
. . .
. .
.
σ rt ν = 0.5 .ν = 0.25 ν = 0 r/ h = 0.1 Figure 6.8: The shear stresses acting on the tunnel as a function of ν. It would appear that any radial shear stress present in the solution is due to the anti-symmetry introduced by the presence of the stress-free surface It is also apparent from these plots that the radial and shear stresses acting on the tunnel lining are only moderately affected by the relative depth of the tunnel The most noticeable effect in this regard is a fairly substantial reduction in the radial stress along the top of the tunnel The tangential hoop stress σ t t is shown in Figure 6.9 for different values of Poisson’s ratio and different relative tunnel depths One of the most noticeable
.
.
. .
.
.. .
.. .
.. .
. .
.
. . . . . . .
.. .
.. .
. .
. σ t t ν = 0.5 .ν = 0.25 ν = 0 r/ h = 0.7
.
.. .
.. .
.. .
.. .
. .
. . .
. .
. . . . . . .
.. .
.. .
. . .
σ t t
ν = 0.5
ν = 0.25
ν = 0
r/ h = 0.1
Figure 6.9: The hoop stress around the tunnel as a function of ν.
features of Figure 6.9 is that the hoop stress is constant for ν = 0 This effect
is probably related to the effect seen in the case of the rigid buoyant tunnel,
where the hoop stress vanishes completely for ν = 0 The hoop stress can
only be generated directly by deformation of the tunnel lining, and indirectly
through Poisson’s ratio by the radial and radial shear stresses In the case of the
rigid buoyant tunnel the hoop stresses vanish for ν = 0 because the tunnel does
not deform and therefore does not directly generate a hoop stress In the case
of a pure contraction of the tunnel boundary, the uniform contraction directly
Trang 8generates a constant hoop stress, and for ν = 0 the stress remains constant as there is no additional indirect generation of hoop stress.
§ 6.6 Displacements along the Surface
The scaled settlements, denoted by v/u g , for the ground loss problem are shown
in Figure 6.10 for different values of Poisson’s ratio.
-4 -3 -2 -1 0
-4 -3 -2 -1 0
2
u g
v
x/ h
.
.
.
.
ν = 0.5 .ν = 0.25 ν = 0 Figure 6.10: Surface settlements as a function of ν for r/ h = 0.7. One of the more prominent features of these curves is that the settlements increase substantially (by more than a factor of 1.5 for the maximum settlement for ν = 0 versus ν = 0.5) for smaller values of ν This means that the volume of the settlement trough will be larger than the volume of ground loss around the tunnel, since these volumes are equal for incompressible ground with ν = 0.5. A similar, but even more amplified, behavior occurs for deeper tunnels, as can be seen in Figure 6.11 -5 -4 -3 -2 -1 0 1 2 3 4 5 -4 -3 -2 -1 0 -4 -3 -2 -1 0 10 u g v x/ h
.
.
. .
.
ν = 0.5
ν = 0.25
ν = 0
Trang 9Section 6.6 Displacements along the Surface 57
In Figure 6.11 the difference between the trough depths for different ν values
has increased (to a factor of 2.0 between the maximum trough depth for ν = 0
and ν = 0.5).
A possible interpretation of the increasing settlements for lower values of
Poisson’s ratio is that undrained soil, which is incompressible, will react to
ground loss around the tunnel according (nearly exactly) to the curves drawn
for ν = 0.5 As the soil drains, or consolidates, the soil may settle progressively,
according to the curves drawn for lower values of ν These latter curves will not
be as accurate as the curves for the undrained case, as linear elasticity becomes
a poorer approximation for soil behavior for ν values smaller than 0.5.
The scaled maximum trough depths v max /u g versus the tunnel radius to depth
ratios have been plotted in Figure 6.12 for different values of ν.
r/ h
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
v max
u g
ν = 0
ν = 0.25
ν = 0.5
Figure 6.12: Maximum settlement as a function of r/ h and ν.
It appears from Figure 6.12 that there is an approximately linear relationship
between the normalized maximum trough depth v max /u g and r/ h, for values of
r/ h < 0.5 If ground loss around the tunnel is expected to be the predominant
mode of deformation and the type of ground can be well characterized by
linearly elastic behavior then this relationship can be used as a rule of thumb
to determine the trough depth for tunnels with r/ h < 0.5 A rule of thumb for
Trang 10pure ground loss could be written as
v max
u g ≈ 3.5(1 − ν) r
h . for r/ h < 0.5 (6.8)
It should be emphasized that this rule of thumb is only valid for a uniform contraction of the tunnel cavity and for linearly elastic behavior Equation (6.8) has not been verified with any practical cases.
§ 6.7 Comparison to a Finite Element Calculation
The solution presented in this chapter has also been compared to a Finite Element Method (FEM) calculation, using the program Plaxis, version 8.0 [3] As was the case for the FEM comparison in the previous chapter, this comparison is not intended as a validation of the solution or as a validation of the program Plaxis, but is merely an illustration of the differences that may be encountered.
A linearly elastic material model has again been chosen for the ground in the FEM calculation in order to afford for a more direct comparison to the exact solution The elements chosen are 15-noded triangular elements The tunnel is a solid linearly elastic cylinder, also modeled with 15-noded triangular elements For the particular details of the FEM implementation in the program Plaxis, see [3].
The ground loss problem was calculated in Plaxis by determining the initial stresses for a weighted elastic medium containing a stiff, solid cylinder with the same weight as the surrounding material The staged construction module was then used to apply a volume loss of 2.5% to the stiff material inside the tunnel, resulting in a uniform contraction of the tunnel cavity The values of the parameters used are given in Table 6.1.
Table 6.1: Parameters used in the FEM calculation for the ground loss problem The
abbreviations grnd and tunl refer to parameters for the ground and tunnel, respectively.
In the table, the values of x max and y max define the range of the FEM mesh
in the horizontal and vertical directions, respectively Note that the relative mesh size is larger than commonly used in geotechnical calculations This has been done in order to minimize the effect of the boundaries on the stresses and displacements near the tunnel.