TECHNICAL NOTE

Surface settlements due to deformation of a tunnel in an elastic halfplane

A. VERRUIJT and J. R. BOOKER{

KEYWORDS: deformation; elasticity; stress analysis;tunnels.

INTRODUCTION

Deformations of a tunnel may result in settlementsof the ground surface. In engineering practice thesesurface settlements are often described by empiri-cal formulae, based upon eld observations, forinstance a normal (Gaussian) distribution curve(Peck, 1969; Attewell & Woodman, 1982). In thisnote an approximate analytical solution is presented,considering the soil as a linear elastic material.Although it is realized that this is a very poorschematization of the real behaviour of soils, anelastic solution may well serve to investigate someof the characteristics of the resulting elds of stressand strain, and it may also serve as a reference formore rened (numerical) computations.

Two basic deformation mechanisms of thetunnel are considered: a uniform radial displace-ment (representing, in rst approximation, theground loss that may occur during constructionof the tunnel), and an ovalization of the tunnel,see Fig. 1. Even though in practice the tunnellingprocess may be executed very carefully andappropriate engineering techniques may be appliedto minimize the deformations, for instance theinjection of grout into the soil surrounding thetunnel, it remains of interest to study the defor-mations and stresses caused by both ground lossand ovalization.

The method used is an extension of a methodsuggested by Sagaseta (1987) for the case ofground loss in an incompressible soil. The startingpoint is the solution for a singularity at a point ofan elastic half plane (at the axis of the tunnel). Byadding the image solution for a singularity at a

point located symmetrically above the soil surface,the shear stresses at the surface are made equal tozero. In order to balance the normal stresses at thesurface, a third solution is added, by solving aBoussinesq-type problem. The approximation inestablishing the solution is that, in balancing thenormal and tangential stresses on the surface, thepresence of the tunnel, with its prescribed dis-placements, is disregarded. Although an exactsolution might well be obtainable by the methodsoutlined by Mindlin (1939), using bipolar coordi-nates, such a solution is very complicated. Thepresent approximate solution will be seen to bevery simple and easy to apply.

The solution given in this technical note is ageneralization of Sagaseta's solution in that it givesthe solution for the case of ground loss not onlyfor the incompressible case (with Poisson's ratioequal to 05), but for arbitrary values of Poisson'sratio, and that it includes the effect of ovalization.The process of ovalization has been studied beforefor a tunnel in an innite elastic medium (MuirWood, 1975). Here the surface displacements for atunnel in a semi-innite medium are considered, aswell as the displacements and stresses throughoutthe half space.

THE SOLUTION

The (approximate) solution of the problem issupposed to consist of three parts. The rst two

Verruijt, A. & Booker, J. R. (1996). Geotechnique 46, No. 4, 753756

753

Manuscript received 4 July 1995; revised manuscriptaccepted 30 October 1995.Discussion on this technical note closes 3 March 1997;for further details see p. ii. Delft University of Technology{ University of Sydney

Fig. 1. Ground loss and ovalization of a tunnel

parts are singular solutions from the theory ofelasticity in the points x 0, z h and x 0,z h, (Fig. 2). In these solutions the displace-ments are of order O(1/r), so that they correspondto the removal of a nite volume, and the stressesare of order O(1/r2). They are well-known basicsolutions from the theory of elasticity (Timoshenko& Goodier, 1951). All other possible singularsolutions are of larger negative orders of r, andthus will be small compared to the present one ifthe radius R of the tunnel is small enough.

Taken together, the expressions for the displace-ment components ux and uz for the singular partsof the solution and their images are

ux R2 xr21 x

r22

R2 x(x

2 kz21)r41

x(x2 kz22)

r42

(1)

uz R2 z1r21 z2

r22

R2 z1(kx

2 z21)r41

z2(kx2 z22)r42

(2)

where and are parameters indicating the relativedisplacement of the tunnel surface, for the uniformradial displacement case () and the ovalizationcase () respectively. In these solutions z1 z h,z2 z + h, and r1 and r2 are the distances from thesingular point and its image (Fig. 2). It can easilybe veried that these solutions satisfy the basicdifferential equations from the theory of planestrain elasticity.

Because of the symmetry of these two solutions,the shear stress zx and the vertical displacementuz will be zero at the surface z 0. The normal

stresses zz induced by the two singular solutionsare equal, and add up to

z 0 : zz q(x) 4R2 x2 h2

(x2 h2)2

8mR2

m 1h2(3x2 h2)(x2 h2)3 (3)

where is the shear modulus of the elasticmaterial, and m is an auxiliary elastic constant,related to Poisson's ratio by

m 11 2 (4)

In order to satisfy the boundary condition that thenormal stress zz 0 at the surface z 0, a thirdsolution must be added, which must balance thestress distribution q(x). This third part of thesolution is the solution of a Boussinesq problem,for the half plane z . 0, with the boundaryconditions

z 0 : zz q(x), zx 0: (5)Problems of this type can be solved mostconveniently by using Fourier transforms (Sneddon,1951). It can be shown that, if the load function iseven, q(x) q(x), the solution of the problemwith the boundary conditions (5) can be written as

ux 1

0

C(1 mz) exp (z) sin (x) d (6)

uz 1

0

C(1 m mz) exp (z) cos (x) d(7)

where C is dened by the integral

C 1m

1

10

q(x) cos (x) dx (8)

In this case, with the function q(x) dened by (3)the function C is found to be

C 2R2

mexp (h) 2R

2

m 1h exp (h)(9)

Substitution of this result into equations (6) and (7)gives, after evaluation of the Fourier integrals, usingknown integrals from Bateman (1954) and someresults that can easily be derived from them,

ux 2R2x

m

1

r22 2mzz2

r42

4R

2xh

m 1z2

r42 mz(x

2 3z22)r62

(10)

r2

h

h

z

x

r1

Fig. 2. A singularity and its image

754 VERRUIJT AND BOOKER

uz 2R2

m

(m 1)z2r22

mz(x2 z22)r42

2R2h x

2 z22r42

mm 1

2zz2(3x2 z22)

r62

(11)

The complete solution of the problem is the sum ofthis solution and the expressions (1) and (2).

All stresses and strains can be directly derivedfrom the solution given above. The verticaldisplacement of the surface z 0, denoted as u0,is of particular interest. Taking z 0 in equation(11) gives

u0 2R2 m 1m

h

x2 h2 2R2 h(x

2 h2)(x2 h2)2

(12)

This is the total settlement of the surface, becausefor z 0 the two singular solutions in equation (2)cancel.

The shape of the two settlement functions isshown in Fig. 3. The solid line is the settlementcurve for the rst term, due to the ground loss.The dashed line is the settlement curve for thesecond term, due to the ovalization of the tunnel.It appears that the width of the settlement troughis considerably smaller for the ovalization casethan for the ground loss case. This may beone explanation for the rather narrow settlementtroughs usually observed in tunnelling practice.Another explanation may of course be that the soilbehaviour is non-linear, exhibiting plastic effects.

The factor (m + 1)/m in equation (12) can alsobe written as 2(1 ). For 05 this factor is 1.For that value of the rst term of the solutionhas been presented earlier by Sagaseta (1987).

The total area (A) of the settlement trough isfound by integrating equation (12) from 1 to+1. The result is

A m 1m

2R2 4(1 )R2 (13)For 05 this is precisely the ground loss, 2R2,a result also mentioned by Sagaseta (1987). It is

interesting to note that for other values of thetotal area below the settlement curve is larger thanthe ground loss. This initially somewhat surprisingresult must be attributed to the deformations atinnity. For 0 the total area is double that foran incompressible material, and thus Sagaseta'ssolution will in general lead to an underestimationof the settlement under drained conditions. It mayalso be concluded that it is of great practicalimportance to keep the ground loss as small aspossible, in order to prevent surface settlements, afact long recognized in tunnel construction.

The shape of the settlement curve is indepen-dent of , but its values increase if decreasesfrom 05 to 0. In a consolidating, porous, elasticmedium this would mean that the immediatesettlement is equal to the ground loss, with theconsolidation settlement leading to a gradual butapproximately conformal increase of the settle-ments. This will be the subject of a subsequentpublication.

It may also be interesting to note that theaverage value of the ovalization term in thesolution is zero. In the zone x , h there is asettlement, but in the zone x . h there is someheave, and the two of these appear to balance onaverage.

In order to assess the accuracy of the approxi-mation, the total displacements at the tunnel'sradial circumference may be determined. The rstterm of the solution represents the constant radialdisplacement due to the singularity in an innitemedium. The other two solutions disturb thisdisplacement, and the relative magnitude of thisdisturbance is a measure of the accuracy of thesolution. This can be evaluated by comparing theradial displacements at the tunnel face due to thesecond and the third parts of the solution withthose due to the rst part. It appears that the erroris less than 10%, provided that the radius of thetunnel is less than one half of its depth.

STRESSES

The strain components can easily be derived bydifferentiation from the expressions for the dis-placements, and then the stress components can beobtained with the aid of Hooke's law. The isotropicstress 0 12(xx + zz) is found to be, for instance,

0 4R2 z22 x2

r42

R2

1 z41 x4

r61 z

42 x4

r62

4hz2(z

22 3x2)r62

(14)

A graphical representation of the isotropic stress 0

5 50x/h

u0

Fig. 3. Surface settlements

SURFACE SETTLEMENTS DUE TO TUNNEL 755

is shown in Fig. 4. The left half of the gure showscontours of 0 for the ground loss problem, and theright half shows contours of 0 for the ovalizationproblem. The thick lines indicate the contours for0 0, and the characters + and indicate zonesof tension and compression respectively. The con-tour intervals are 02(R/h)2 and 02(R/h)2/(1 ). Although in Fig. 4 it has been assumedthat R/h 05, the contours are valid for any valueof R/h. It is interesting to note that in the case ofground loss the stresses around the tunnel are allcompressive, whereas in the case of ovalizationthere are zones of tension above and below thetunnel. The construction of contours of other stress

components is a relatively simple and straight-forward exercise.

CONCLUSIONS

It appears to be possible to derive approximateanalytic solutions for two important cases of tunneldeformation in an elastic soil: a uniform radialdisplacement due to ground loss, and an ovaliza-tion of the tunnel. Simple analytical expressionscan be given for the displacements at arbitrarypoints of the half plane, and for the stress com-ponents.

REFERENCESAttewell, P. B. & Woodman, J. P. (1982). Predicting the

dynamics of ground settlement and its derivatives bytunnelling in soil. Ground Engineering 15, 1322.

Bateman, H. (1954). Tables of Integral Transforms. NewYork: McGraw-Hill.

Mindlin, R. D. (1939). Stress distribution around atunnel. Proc. ASCE 65, 619642.

Muir Wood, A. M. (1975). The circular tunnel in elasticground. Geotechnique 25, 115127.

Peck, R. B. (1969). Deep excavations and tunneling insoft ground, state of the art report. 7th Int. Conf. onSoil Mech. and Fdn Eng., Mexico City, 225290.

Sagaseta, C. (1987). Analysis of undrained soil deforma-tion due to ground loss. Geotechnique 37, 301320.

Sneddon, I. N. (1951). Fourier Transforms. New York:McGraw-Hill.

Timoshenko, S. P. & Goodier, J. N. (1951). Theory ofElasticity, 2nd edn. New York: McGraw-Hill.Fig. 4. Contours of isotropic stress

756 VERRUIJT AND BOOKER

Surface settlements due to deformation of a tunnel in an elastic halfplane

A. Verruijt and J.R. Booker

Geotechnique 46 (1996) 753-756

Erratum

In equations (1) and (2) : k = /(1 ).

Surface settlements due to ground lossand ovalization of a tunnel

A. Verruijt and J.R. Booker

Geotechnique, 46, 753-756, 1996.

Details of the analysis, and an alternative

The problem

The paper gives an approximate analytic solution for a tunnel in a homogeneous elastic halfspace, caused by ground loss and ovalization. The two basic processes are illustrated infigure 1. This note presents some additional material and derivations.

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Figure 1: Ground loss and Ovalization of a tunnel.

Basic solution for ground loss

The first solution, describing the ground loss, is a purely radial displacement, decreasinginversely proportional to the distance r from the point x = 0, y = 0.

ur = R2

r, (1)

where ur is the displacement in radial direction, and is a parameter indicating the relativedisplacement at the circumference of a tunnel of radius R. The total ground loss (per unitlength) is 2piR2, and the relative ground loss, per unit volume of the tunnel, is 2. Thehorizontal and vertical components of the displacement are

ux = R2x

r2, (2)

uy = R2y

r2, (3)

It can easily be verified that these expressions satisfy the basic differential equations for ahomogeneous isotropic linear elastic material, deforming under plane strain conditions,

me

x+

2uxx2

+2uxy2

= 0, (4)

1

me

y+

2uyx2

+2uyy2

= 0, (5)

where e is the volume strain,

e =uxx

+uyy

, (6)

and the elastic parameter m is defined by

m =1

1 2 , (7)

where is Poissons ratio.In this case the volume strain appears to be zero,

e = 0. (8)

The stresses can be determined using Hookes law for the case of plane strain,

xx2

= 12(m 1) e +uxx

=

1 2 e +uxx

, (9)

yy2

= 12(m 1) e+uyy

=

1 2 e +uyy

, (10)

xy2

= 12

(uxy

+uyx

). (11)

This gives

xx2

=R2

r4(x2 y2), (12)

yy2

=R2

r4(y2 x2), (13)

xy2

=R2

r42xy. (14)

Basic solution for ovalization

The basic form of the second type of singularity, describing the ovalization, is

ux =R2x(x2 ky2)

r4, (15)

uy =R2y(kx2 y2)

r4, (16)

where is a measure of the degree of ovalization (the relative increase and decrease of theradius R in the two coordinate directions), and where k is an elastic parameter, defined as

k =

1 =

+ 2. (17)

Again it can be verified that this solution satisfies the basic differential equations (4) and (5).

2

In this case the volume strain is

e =uxx

+uyy

=R2

r6(k 1)(x4 y4) = R

2

r61 21 (x

4 y4). (18)

The stresses are, with equations (9), (10) and (11),

xx2

= R2

(1 )r6x2(x2 3y2), (19)

yy2

= R2

(1 )r6y2(3x2 y2), (20)

xy2

= R2

(1 )r62xy(x2 y2). (21)

Solution for two singularities

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y

h

h

r1

r2

Figure 2: A singularity and its image.

Superposition of two singularities, with ground loss and ovalization, at y = h and at y = h,see figure 2, leads to the displacement field

ux = R2{ xr21

+x

r22

}+ R2

{x(x2 ky21)r41

+x(x2 ky22)

r42

}, (22)

uy = R2{y1r21

+y2r22

}+ R2

{ y1(kx2 y21)r41

+y2(kx2 y22)

r42

}, (23)

where y1 = y h and y2 = y + h, and r1 and r2 are the distances from the singular pointand its image, see figure 2, so that

r21 = x2 + (y h)2, r22 = x2 + (y + h)2. (24)

The stresses due to the two singularities are

xx2

= R2{x2 y21

r41+

x2 y22r42

} R21

{x2(x2 3y21)r61

+x2(x2 3y22)

r62

}, (25)

yy2

= R2{x2 y21r41

+x2 y22

r42

} R21

{ y21(3x2 y21)r61

+y22(3x2 y22)

r62

}, (26)

3

xy2

= R2{2xy1

r41+

2xy2r42

} R21

{2xy1(x2 y21)r61

+2xy2(x2 y22)

r62

}. (27)

At the surface y = 0: y1 = h and y2 = h, so that it follows from equations (23) and (27)that on that surface uy = 0 and xy = 0. This is a consequence, of course, of the symmetryof the solution. The normal stresses yy induced by the two parts of the solution are equal,and add up to

y = 0 : yy = q(x) = 4R2 x2 h2

(x2 + h2)2 4R

2

1 h2(3x2 h2)(x2 + h2)3

, (28)

In order to satisfy the boundary condition that the normal stress yy = 0 at the surfacey = 0, a third solution must be added, which should balance the stress distribution q(x).

The third part of the solution

The third part of the solution is a problem for an elastic half plane with zero shear stressesat the surface y = 0, and normal stresses

y = 0 : yy = q(x) = 4R2 x2 h2

(x2 + h2)2+

4R2

1 h2(3x2 h2)(x2 + h2)3

, (29)

This is a Boussinesq problem, which can be solved using the Fourier transform method.

The Fourier transform method

The Fourier transform method for solving problems of elasticity for a half plane (Sneddon,1951) leads to the general type of solution

ux = 0

C(1my) exp(y) sin(x) d, (30)

uy = 0

C(1 +m +my) exp(y) cos(x) d, (31)

where, as before, m is defined by

m =1

1 2 . (32)It can easily be verified that these equations satisfy the basic differential equations (4) and(5).

It follows from equations (30) and (31) that

e =uxx

+uyy

= 2 0

C exp(y) cos(x) d, (33)

f =uxx

uyy

= 2m 0

C2y exp(y) cos(x) d, (34)

g =uxy

+uyx

= 2m 0

C2y exp(y) cos(x) d. (35)

Using these equations the stresses can be expressed asxx2

= 12me + 1

2f = m

0

C(1 y) exp(y) cos(x) d, (36)

yy2

= 12me 12f = m 0

C(1 + y) exp(y) cos(x) d, (37)

xy2

= 12g = m 0

C2y exp(y) cos(x) d. (38)

4

Ground loss

In the present problem the boundary condition for the stress yy is, with equation (29), andconsidering the ground loss case only,

y = 0 :yy2

= 2R2x2 h2

(x2 + h2)2. (39)

With equation (37) this leads to the condition 0

C exp(y) cos(x) d = 2R2

m

x2 h2(x2 + h2)2

. (40)

Using the inversion theorem for the Fourier cosine integral (Sneddon, 1951) it follows that

C =4R2

pim

0

x2 h2(x2 + h2)2

cos(x) dx. (41)

Using the Fourier transform (114) from the Appendix gives

C = 2R2

mexp(h). (42)

Substitution into the expressions (30) and (31) for the displacements now gives

ux = 2R2

m

0

(1my) exp[(y + h)] sin(x) d, (43)

uy =2R2

m

0

(1 +m+my) exp[(y + h)] cos(x) d. (44)

These integrals are all given in the Appendix. The final results are, using the definition ofm, see equation (32),

ux = 2R2{(1 2)x

r22 2xyy2

r42

}, (45)

uy = 2R2{2(1 )y2

r22 y(x

2 y22)r42

}, (46)

where y2 = y + h and r22 = x2 + (y + h)2.With equation (42) the expressions for the stresses become

xx2

= 2R2 0

(1 y) exp(y2) cos(x) d, (47)

yy2

= 2R2 0

(1 + y) exp(y2) cos(x) d, (48)

xy2

= 2R2 0

2y exp(y2) cos(x) d, (49)

where y2 = y + h.All the integrals in these expressions are also given in the Appendix. This gives

xx2

= 2R2{x2 y22

r42 2yy2(3x

2 y22)r62

}, (50)

yy2

= 2R2{x2 y22

r42+

2yy2(3x2 y22)r62

}, (51)

xy2

= 2R22xy(x2 3y22)

r62. (52)

5

Ovalization

For the ovalization case the boundary condition for the stress yy is, with equation (29) and = 0,

y = 0 :yy2

=2R2

1 h2(3x2 h2)(x2 + h2)3

, (53)

With equation (37) this leads to the condition 0

C exp(y) cos(x) d = 2R2 1 21

h2(3x2 h2)(x2 + h2)3

. (54)

Using the inversion theorem for the Fourier cosine integral (Sneddon, 1951) it follows that

C =4R2

pi

1 21

0

h2(3x2 h2)(x2 + h2)3

cos(x) dx. (55)

This integral is determined in the Appendix, see equation (116). In this case the result is

C = 2R2

m + 1h exp(h), (56)

in which the factor 2/(m+ 1) can also be written as (1 2)/(1 ).Substitution of (56) into the expressions (30) and (31) for the displacements now gives

ux = 2R2

m + 1

0

h(1my) exp[(y + h)] sin(x) d, (57)

uy =2R2

m + 1

0

h(1 +m+my) exp[(y + h)] cos(x) d. (58)

The Fourier integrals in these expressions are given in the Appendix. This gives

ux = 4R2xh

m+ 1

{y2r42

+my(x2 3y2)2

r62

}, (59)

uy = 2R2h{x2 y22

r42+

2mm + 1

yy2(3x2 y22)r62

}. (60)

With equation (42) the expressions for the stresses become

xx2

= 2R2m

m+ 1

0

2h(1 y) exp(y2) cos(x) d, (61)

yy2

= 2R2m

m + 1

0

2h(1 + y) exp(y2) cos(x) d, (62)

xy2

= 2R2m

m + 1

0

3hy exp(y2) cos(x) d, (63)

where y2 = y + h.All the integrals in these expressions are also given in the Appendix. This gives

xx2

=2R2h1

{y2(3x2 y22)r62

+3y(x4 6x2y22 + y42)

r82

}, (64)

yy2

=2R2h1

{y2(3x2 y22)r62

3y(x4 6x2y22 + y42)

r82

}, (65)

xy2

=24R2h1

xyy2(x2 y22)r82

. (66)

6

An alternative solution for ovalization

It has been observed by Strack (2002) that in the solution for the ovalization case presented inthe paper considered above, the tangential displacement ut does not vanish at the boundaryof the cavity, in contrast with his complex variable solution. For the purpose of comparisonan alternative may be used, in which ut = 0 along the cavity boundary. This alternative is

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x

y

r

Figure 3: Cartesian and polar coordinates

presented here.The general solution for ovalization of a circular cavity in an infinite elastic plane is, in

polar coordinates, see figure 3,

ur = {BR2(m + 1)mr

+DR4

3r3} cos 2, (67)

ut = {BR2

mr+

DR4

3r3} sin 2, (68)

where, as before, m = 1/(1 2).The condition that ut = 0 for r = R gives

D =3Bm

. (69)

The solution now reduces to

ur =BR2

mr{(m + 1) + R

2

r2} cos 2, (70)

ut =BR2

mr{1 + R

2

r2} sin 2. (71)

The relation between polar coordinates and cartesian coordinates is

ux = ur sin + ut cos =x

rur +

y

rut, (72)

uy = ur cos ut sin = yrur x

rut. (73)

7

With

BR

m= R

m + 2, (74)

one obtains

ux =R2x

(m + 2)r2{1my

2 x2r2

R2(3y2 x2)

r4

}, (75)

uy = R2y

(m + 2)r2{1 +m

y2 x2r2

+R2(y2 3x2)

r4

}. (76)

It can easily be seen that

x = R, y = 0 : ux = R, uy = 0, x = R, y = 0 : ux = R, uy = 0, (77)

x = 0, y = R : ux = 0, uy = R, x = 0, y = R : ux = 0, uy = R, (78)

which is in agreement with the assumed ovalization of the circle of radius R.It follows from equations (75) and (76) that

uxx

=R2

(m + 2)r2{y2 x2

r2mx

4 6x2y2 + y4r4

3R2(x4 6x2y2 + y4)

r6

}, (79)

uxy

=2R2xy(m + 2)r4

{1m3x

2 y2r2

6R2(x2 y2]r4

}, (80)

uyx

=2R2xy(m + 2)r4

{1mx

2 3y2r2

6R2(x2 y2)

r4

}. (81)

uyy

=R2

(m + 2)r2{y2 x2

r2+m

x4 6x2y2 + y4r4

+3R2(x4 6x2y2 + y4)

r6

}. (82)

It now follows that

e =uxx

+uyy

=2R2

(m+ 2)r2{y4 x4

r4

}(83)

f =uxx

uyy

= 2R2

(m + 2)r2{mx4 6x2y2 + y4

r4+

3R2(x4 6x2y2 + y4)r6

}(84)

g =uxy

+uyx

= 8R2

(m + 2)r2{mxy(x2 y2)

r4+

3R2xy(x2 y2)r6

}. (85)

Using these relations the stresses are found to be

xx2

= 12(me + f) = R2

(m + 2)r2{m2x2(x2 3y2)

r4+

3R2(x4 6x2y2 + y4r6

}, (86)

yy2

= 12(me f) = R2

(m + 2)r2{m2y2(3x2 y2)

r4 3R

2(x4 6x2y2 + y4r6

}, (87)

xy2

= 12g = 4R

2

(m + 2)r2{mxy(x2 y2)

r4+

3R2xy(x2 y2)r6

}. (88)

8

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y

h

h

r1

r2

Figure 4: A singularity and its image.

Solution for two singularities

Superposition of two of these ovalization singularities, at y = h and at y = h, see figure 4,leads to the displacement field

ux =R2x

m + 2

{ 1r21my

21 x2r41

R2(3y21 x2)

r61

+1r22my

22 x2r42

R2(3y22 x2)

r62

}, (89)

uy = R2

m + 2

{y1r21

+my1(y21 x2)

r41+

R2y1(y21 3x2)r61

+y2r22

+my2(y22 x2)

r42+

R2y2(y22 3x2)r62

}, (90)

where y1 = y h, y2 = y + h, r21 = x2 + y21 and r22 = x2 + y22 .The stress field for these two singularities is

xx2

= R2

m + 2

{m2x2(x2 3y21)

r61+

3R2(x4 6x2y21 + y41)r81

+m2x2(x2 3y22)

r62+

3R2(x4 6x2y22 + y42)r82

}, (91)

yy2

= R2

m + 2

{m2y21(3x2 y21)

r61 3R

2(x4 6x2y21 + y41)r81

+m2y22(3x2 y22)

r62 3R

2(x4 6x2y22 + y42)r82

}, (92)

xy2

= 4R2

m + 2

{mxy1(x2 y21)r61

+3R2xy1(x2 y21)

r81

+mxy2(x2 y22)

r62+

3R2xy2(x2 y22)r82

}. (93)

At the free surface y = 0, so that y1 = h, y2 = h and r21 = r22 = x2 + h2. It then followsthat the vertical displacement uy and the shear stress xy on that surface vanish,

9

y = 0 : uy = 0, (94)

y = 0 : xy = 0. (95)

This is a consequence of the symmetry of the problem, of course.The vertical normal stress on the surface y = 0 is found to be

y = 0 : yy = 2f(x), (96)

where now

f(x) =2R2

m + 2

{m2h2(h2 3x2)(h2 + x2)3

+3R2(x4 6x2h2 + h4)

(h2 + x2)4}. (97)

This can also be written as

f(x) =2R2

m + 2

{6m h

2

(h2 + x2)2+ 8m

h4

(h2 + x2)3

+3R2

(h2 + x2)2 24R

2h2

(h2 + x2)3+

24R2h4

(h2 + x2)4}. (98)

Written in this form the numerators no longer contain terms dependent upon x. This makesthe Fourier integrals somewhat easier to determine.

The third part of the solution

The third part of the solution is a problem for an elastic half plane with zero shear stressesat the surface y = 0, and normal stresses

y = 0 : yy = 2f(x). (99)This is a Boussinesq problem, which can be solved using the Fourier transform method.

With equation (37) this leads to the condition

m

0

C cos(x) d = f(x). (100)

Application of the inversion theorem for the Fourier cosine integral (Sneddon, 1951) nowgives

C = 2mpi

0

f(x) cos(x) d. (101)

Using some of the Fourier integrals derived in the Appendix the value of C is found to be

C = 2R2

m + 2

{h++

R2

2mh2(h)2]

}exp(h). (102)

Although this form may be used to derive the displacements and stresses for the third partof the solution, it is assumed here that the radius of the tunnel is small, R/h 1. In thatcase equation (102) reduces to

C = 2R2

m + 2h exp(h). (103)

This is the same solution as obtained in the original problem, see equation (56), except thatthe factor m + 1 in the denominator must be replaced by m + 2.

10

It follows that the additional expressions for the displacements in this case are

ux = 4R2xh

m+ 2

{y2r42

+my(x2 3y2)2

r62

}, (104)

uy = 2R2h

m+ 2

{(m + 1)

x2 y22r42

+ 2myy2(3x2 y22)

r62

}. (105)

And the expressions for the stresses are

xx2

=4mR2hm + 2

{y2(3x2 y22)r62

+3y(x4 6x2y22 + y42)

r82

}, (106)

yy2

=4mR2hm+ 2

{y2(3x2 y22)r62

3y(x4 6x2y22 + y42)

r82

}, (107)

xy2

=48mR2hm + 2

xyy2(x2 y22)r82

. (108)

Reference

O.E. Strack, Analytic Solutions of Elastic Tunneling Problems, Ph.D. Thesis, Delft, 2002.

11

APPENDIX : SOME INTEGRAL TRANSFORMS

In this appendix a number of integral transforms are assembled, mainly from the literature.Many of these integrals are used in this paper.

A well known Laplace transform (Churchill, 1972), which can easily be derived by usingpartial differentiation, is

0

cos(at) exp(st) dt = ss2 + a2

. (109)

By a simple change of notation this integral can also be written as a Fourier cosine transform, 0

exp(h) cos(x) d = hh2 + x2

. (110)

Using the Fourier inversion theorem (Sneddon, 1951), it now follows that

2pi

0

h

h2 + x2cos(x) dx = exp(h), (111)

or 0

1h2 + x2

cos(x) dx =pi

2hexp(h). (112)

Differentiation of (112) with respect to h gives, after division by 2h, 0

1(h2 + x2)2

cos(x) dx =pi

4h3(1 + h) exp(h). (113)

Because h2 x2 = 2h2 (h2 + x2) it follows from (112) and (113) that 0

h2 x2(h2 + x2)2

cos(x) dx =pi

2exp(h). (114)

Differentiation of (113) with respect to h gives, after division by 4h, 0

1(h2 + x2)3

cos(x) dx =pi

16h5[3 + 3h+ (h)2] exp(h). (115)

Because h2 3x2 = 4h2 3(h2 + x2) it follows from (113) and (115) that 0

h2 3x2(h2 + x2)2

cos(x) dx =pi2

4hexp(h). (116)

Differentiation of (115) with respect to h gives, after division by 6h, 0

1(h2 + x2)4

cos(x) dx =pi

96h7[15 + 15h+ 6(h)2 + (h)3] exp(h). (117)

Now returning to the second integral, eq. (110), it follows, by differentiation with respect toh, that

0

cos(x) exp(h) d = h2 x2

(h2 + x2)2. (118)

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Differentiating this equation once more with respect to h gives 0

2 cos(x) exp(h) d = 2h(h2 3x2)

(h2 + x2)3. (119)

It may be noted that (118) and (119) are the inverse Fourier transforms of (114) and (116).Differentiating equation (119 with respect to h gives

0

3 cos(x) exp(h) d = 6(h4 6h2x2 + x4)(h2 + x2)4

. (120)

Another well known Laplace transform is 0

sin(at) exp(st) dt = as2 + a2

. (121)

Written in the notation of a Fourier sine transform this integral is 0

sin(x) exp(h) d = xh2 + x2

. (122)

Differentiation with respect to h gives 0

sin(x) exp(h) d = 2hx(h2 + x2)2

. (123)

Repeating the differentiation with respect to h gives 0

2 sin(x) exp(h) d = 2x(3h2 x2)

(h2 + x2)3. (124)

Another differentiation with respect to h gives 0

3 sin(x) exp(h) d = 24xh(h2 x2)

(h2 + x2)4. (125)

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Verruijt&Booker1996OriginalGeotechnique-1996Geotechnique-1996-Errata

VerruijtBooker