stress and deformation fields around a deep circular tunnel

14

Stress and deformation fields around a deepcircular tunnel

14.1 Rationale of analytical solutions

The analytic representation of stress- and deformation fields in the ground sur-rounding a tunnel succeeds only in some extremely simplified special cases,which are rather academic. Nevertheless, analytical solutions offer the follow-ing benefits:

• Being exact solutions, they provide insight into the basic mechanisms(i.e. displacements, deformation and stress fields) of the considered prob-lem.

• They provide insight of the role and the importance of the involved pa-rameters.

• They can serve as benchmarks to check numerical solutions.

In this section, some solutions are introduced which are based on Hooke’slaw, the simplest material law for solids. The underground is regarded hereas linear-elastic, isotropic semi-infinite space, which is bound by a horizontalsurface, the ground surface. The tunnel is idealised as a tubular cavity withcircular cross section. Before its construction, the so-called primary stress stateprevails. This stress state prevails also after the construction of the tunnel ina sufficiently large distance (so-called far field).

14.2 Some fundamentals

The equilibrium equation of continuum mechanics written in cylindrical coor-dinates reveals the mechanism of arching in terms of a differential equation.For axisymmetric problems, as they appear in tunnels with circular cross sec-tion, the use of cylindrical coordinates (Fig. 14.1) is advantageous. In axisym-metric deformation, the displacement vector has no component in θ-direction:uθ ≡ 0. The non-vanishing components of the strain tensor

274 14 Stress and deformation fields around a deep circular tunnel

Fig. 14.1. Cylindrical coordinates r, θ, z

εrr =∂ur

∂r, εθθ =

1r

∂uθ

∂θ+ur

r, εzz =

∂uz

∂z

εrθ = εθr =12

(1r

∂ur

∂θ− uθ

r+∂uθ

∂r

)

εrz = εzr =12

(∂ur

∂z+∂uz

∂r

)

εθz = εzθ =12

(1r

∂uz

∂θ+∂uθ

∂z

)reduce, in this case, to

εr =∂ur

∂r, εθ =

ur

r, εz =

∂uz

∂z,

where ur and uz are the displacements in radial and axial directions, respec-tively.The stress components σr, σθ, σz are principal stresses (Fig. 14.3). The equa-tion of equilibrium in r-direction reads:

∂σr

∂r+σr − σθ

r+ �g · er = 0 (14.1)

and in z-direction:

∂σz

∂z+ �g · ez = 0 . (14.2)

Herein, � is the density, �g is the unit weight, er and ez are unit vectors inr- and z-directions. The second term in equation 14.1 describes arching.This can be seen as follows: If r points to the vertical direction z (Fig. 14.2),then Equ. 14.1 reads:

14.2 Some fundamentals 275

x

g

r

z

Fig. 14.2. To the explanation of Equ. 14.3. Due to gravity g, axisymmetric condi-tions prevail only for x = 0.

dσz

dz= γ − σx − σz

r. (14.3)

Herein, the term (σx − σz)/r is responsible for that fact that σz does notincrease linearly with depth (i.e. σz = γz). In case of arching, i.e. for (σx −σz)/r > 0, σz increases underproportionally with z. Note that this term, andthus arching, exists only for σr �= σθ. This means that arching is due to theability of a material to sustain deviatoric stress, i.e. shear stress. No arching ispossible in fluids. This is why soil/rock often ’forgives’ shortages of support,whereas (ground)water is merciless.

Fig. 14.3. Components of the stress tensor in cylindrical coordinates

The equilibrium equation in θ-direction,

1r· ∂σθ

∂θ= 0, (14.4)


is satisfied identically, as all derivatives in θ-direction vanish in axisymmetricstress fields.The arching term can be easily explained as follows: Consider the volumeelement shown in Fig. 14.4.

Fig. 14.4. Equilibrium of the volume element in r-direction

The resultant A of the radial stresses reads

A = (σr + dσr) · (r + dr)dθ − σrrdθ ≈ σrdrdθ + dσrrdθ

and should counterbalance the resultant of the tangential stresses σθ. Thevectorial sum of forces shown in Fig. 14.4 yields

A

σθdr= dθ .

It then follows (for g · er = 0):

dσr

dr+σr − σθ

r= 0 .

With reference to the arching term σθ−σr

r attention should be paid to r. Atthe tunnel crown, r is often set equal to the curvature radius of the crown.However, this is not always true. If we consider the distribution of σz andσx above the crown at decreasing support pressures p, we notice that forK = σx/σz < 1 the horizontal stress trajectory has the opposite curvaturethan the tunnel crown (Fig. 14.5)Equations 14.1, 14.2 and 14.4 are the special case of the equilibrium equationsin cylindrical coordinates, which in the general case read:

14.2 Some fundamentals 277

Fig. 14.5. Stress distributions above the crown for various values of support pres-sure p, and corresponding stress trajectories. At the hydrostatic point (right), thecurvature radius of the stress trajectory vanishes, r = 0.

∂σrr

∂r+

1r· ∂σθr

∂θ+∂σzr

∂z+

1r· (σrr − σθθ) + �br = �ar

∂σrθ

∂r+

1r· ∂σθθ

∂θ+∂σzθ

∂z+

1r· (σrθ + σθr) + �bθ = �aθ

∂σrz

∂r+

1r· ∂σθz

∂θ+∂σzz

∂z+

1r· σrz + �bz = �az

b = {br, bθ, bz} and a = {ar, aθ, az} are the mass forces (i.e. force per unitmass) and acceleration, respectively.For problems with plane deformation (plane strain), the stress is usually rep-resented in Cartesian coordinates as:

⎛⎝σxx σxy 0σxy σyy 00 0 σzz

⎞⎠

or in cylindrical coordinates as

278 14 Stress and deformation fields around a deep circular tunnel⎛⎝σrr σrθ 0σrθ σθθ 00 0 σzz

⎞⎠ .

The transformation rules are:

σxx =σrr + σθθ

2+σrr − σθθ

2cos 2θ − σrθ sin 2θ

σyy =σrr + σθθ

2− σrr − σθθ

2cos 2θ + σrθ sin 2θ

σxy =σrr − σθθ

2sin 2θ + σrθ cos 2θ

and

σrr =σxx + σyy

2+σxx − σyy

2cos 2θ + σxy sin 2θ

σθθ =σxx + σyy

2− σxx − σyy

2cos 2θ − σxy sin 2θ

σrθ = −σxx − σyy

2sin 2θ + σxy cos 2θ .

14.3 Geostatic primary stress

An often encountered primary stress (for horizontal ground surface) is σzz =γz, σxx = σyy = Kσzz , whereby z is the Cartesian coordinate pointing down-ward, γ is the specific weight of the rock and K is the so-called lateral stresscoefficient. For non-cohesive materials K has a value between the active andthe passive earth pressure coefficient Ka ≤ K ≤ Kp, and we can often setK = K0 = 1 − sinϕ. The stress field around a tunnel has to fulfil the equa-tions of equilibrium

∂σzz

∂z+∂σzx

∂x= γ ,

∂σzx

∂z+∂σxx

∂x= 0 ,

as well as the boundary conditions at the ground surface (z = 0) and at thetunnel wall. We assume that the tunnel is unsupported, so the normal andthe shear stress at the tunnel wall must disappear. The analytical solutionof this problem is extremely complicated1 and consequently offers no advan-tages compared to numerical solutions (e.g. according to the method of finiteelements, Fig. 14.6).

1R.D. Mindlin: Stress distribution around a tunnel, ASCE Proceedings, April1939, 619-649

14.3 Geostatic primary stress 279

HypoplasticityMohr−Coulomb

HypoplasticityMohr−Coulomb

2)(kN/mσz2)(kN/mσh

Fig. 14.6. Distributions of vertical and horizontal stress along the vertical sym-metry axis. Circular tunnel cross section (r = 1.0m). Numerically obtained withhypoplasticity and Mohr-Coulomb elastoplasticity3

The analytical solution can be simplified, if one assumes the primary stress inthe neighbourhood of the tunnel as constant (Fig. 14.7) and not as linear in-creasing: σzz ≈ γH, σxx ≈ KγH . This approximation is meaningful for deeptunnels (H � r). The stress field around the tunnel can then be representedin polar coordinates as follows:

σrr = γH

[1 +K

2

(1 − r20

r2

)]+ γH

[1 −K

2

(1 + 3

r40r4

− 4r20r2

)cos 2ϑ

]

σϑϑ = γH

[1 +K

2

(1 +

r20r2

)]− γH

[1 −K

2

(1 + 3

r40r4

)cos 2ϑ

]

σrϑ = −γH 1 −K

2

(1 − 3

r40r4

+ 2r20r2

)sin 2ϑ

(14.5)

3Tanseng, P., Implementations of Hypoplasticity and Simulations of GeotechnicalProblems: Including Shield Tunnelling in Bangkok Clay, PhD Thesis at Universityof Innsbruck, 2004


One can easily verify that this solution fulfils the boundary conditions: Forr = r0 obviously σrr = σrϑ = 0, and for r → ∞ is

σrr = γH

(1 +K

2+

1 −K

2cos 2ϑ

)

σϑϑ = γH

(1 +K

2− 1 −K

2cos 2ϑ

)

σrϑ = −γH 1 −K

2sin 2ϑ

(14.6)

Thus, the far field is identical to the primary stress field:

σzz = γH

σxx = KγH

σxz = 0 .

Fig. 14.7. Distribution of vertical stress in the environment of a tunnel.

The problem here is known as the ’generalised Kirsch-problem’.4 From equa-tion 14.5 (2) follows that at the invert and at the crown (for r = r0 and ϑ = 0or π) the tangential stress reads σϑϑ = γH(3K − 1). Thus, for K < 1

3 oneobtains tension, i.e. σϑϑ < 0.

4For the general case that the tunnel axis coincides with none of the principaldirections of the primary stress, see the elastic solution of F.H. Cornet, Stress in Rockand Rock Masses. In: Comprehensive Rock Engineering, edited by J.A. Hudson,Pergamon Press, 1993, Volume 3, p. 309

14.4 Hydrostatic primary stress 281

14.4 Hydrostatic primary stress

The special case of equation 14.5 for K = 1 (i.e. for σxx = σyy = γH) readswith σ∞ := γH :5

σr = σ∞

(1 − r20

r2

)

σϑ = σ∞

(1 +

r20r2

)σrϑ = 0

(14.7)

This solution exhibits axial symmetry, as the radius r is the only independentvariable and ϑ does not appear.We generalise now equation 14.7 regarding the case that the tunnel wall issubjected to a constant pressure p (so-called support pressure). One obtainsthe stress field as a special case of the solution which Lame found in 1852for a thick tube made of linear-elastic material.6 ri and ra are the internal-and external radius of the tube, furthermore pi and pa are the internal- andexternal pressures, respectively. Lame’s solution reads:

σr =par

2a − pir

2i

r2a − r2i− pa − pi

r2a − r2i

r2i r2a

r2

σϑ =par

2a − pir

2i

r2a − r2i+pa − pi

r2a − r2i

r2i r2a

r2

σz = 2νpar

2a − pir

2i

r2a − r2i, σrϑ = 0 .

(14.8)

For ra → ∞, pa → σ∞ one obtains the searched stress field (based on theassumption of elastic behaviour):

σr = σ∞

(1 − r20

r2

)+ p

r20r2

= σ∞ − (σ∞ − p)r20r2

σϑ = σ∞

(1 +

r20r2

)− p

r20r2

= σ∞ + (σ∞ − p)r20r2

σrϑ = 0

(14.9)

The solution (14.8) is obtained from the equation of radial equilibrium, whichcan be written for an elastic material as:

d

dr

[1r

d

dr(ru)

]= 0 ,

5In denoting principal stresses, double indices can be replaced by single ones,i.e. σrr ≡ σr etc.

6L. Malvern, Introduction to the Mechanics of a Continuous Medium, Prentice-Hall, 1969, p. 532


where u is the radial displacement. Integration of this equation yields

u = Ar +B

r, (14.10)

where the integration constants A and B can be determined from the bound-ary conditions. From 14.10 we obtain εr = du/dr = A − B/r2 and εϑ =ur/r = A+B/r2. Introducing εr and εϑ into Hooke’s law we obtain

σr = 2Aλ+ 2GA− 2GB/r2 (14.11)σϑ = 2Aλ+ 2GA+ 2GB/r2 (14.12)

From the boundary conditions

r = a : −pi = 2A(λ+G) − 2GB/a2 (14.13)r = b : −pa = 2A(λ+G) − 2GB/b2 (14.14)

we obtain Equ. 14.8. Requiring pa → −σ∞ for b → ∞ we obtain A =

σ∞1

2(λ+G). With pi = −p and a = r0 we finally obtain B =

σ∞ − p

2Gr20 ,

hence

u =σ∞

2(λ+G)r +

σ∞ − p

2G· r

20

r. (14.15)

The first part of (14.15) is independent of p and represents the displacementwhich results from application of the hydrostatic pressure σ∞. Only the secondpart is due to the excavation of the tunnel (i.e. due to the reduction of thestress at the cavity wall from σ∞ to p).7

With the application of the pressure p the wall of the tunnel yields by theamount u|r0 . The displacement u|r0 can be computed as a function of p withthe help of the solution by Lame.8 One obtains the following linear relation-ship between u|r0 and p:

u|r0 = r0σ∞2G

(1 − p

σ∞

). (14.16)

Fig. 14.8 shows the plot of equation 14.16. If the support pressure p is smallerthan σ∞, then one obtains from Equ. (14.9) a radial stress σr which increaseswith r and a tangential stress σϑ which decreases with r, Fig. 14.9.

7For the case of the spherical symmetry the corresponding equations read:ddr

ˆ1r2

ddr

(r2u)˜

= 0, u = Ar + Br2 , σr = σ∞ − (σ∞ − p)

`r0r

´3and σφ = σθ =

σ∞ + 12(σ∞ − p)

`r0r

´3.

8thereby u is presupposed as small, so that the tunnel radius r0 may be regardedas constant

14.5 Plastification 283

Fig. 14.8. Relationship between p and u|r0 for linear elastic ground

Fig. 14.9. Stress field in linear elastic ground

14.5 Plastification

According to equation 14.9, the principal stress difference

σϑ − σr = 2(σ∞ − p)r20r2

increases when p decreases. Now we can take into account that rock is notelastic, resp. it may be regarded as elastic only as long, as the principal stressdifference σϑ − σr does not exceed a threshold which is given by the so-calledlimit condition. For a frictional material the limit condition (Fig. 14.10) reads

σϑ − σr = (σϑ + σr) sinϕ , (14.17)

where ϕ is the so-called friction angle (or the angle of internal friction). Thus,only such stress states are feasible, to which applies:

σϑ − σr ≤ (σϑ + σr) sinϕ . (14.18)

For a material with friction and cohesion c the limit condition reads

σϑ − σr = (σϑ + σr) sinϕ+ 2c cosϕ . (14.19)


The limit condition can also be represented graphically in the Mohr diagramas shown in Fig. 14.10.

Fig. 14.10. Limit condition in the Mohr diagram for non-cohesive (a) and forcohesive (b) frictional material

If, therefore, p is sufficiently small, the requirement 14.18 expressed by theequation 14.9 is violated in the range r0 < r < re (re is still to be determined).Consequently, the elastic solution 14.9 cannot apply within this range. Rather,another relation applies which is to be deduced as follows: We regard theequation of equilibrium in radial direction

dσr

dr+σr − σϑ

r= 0 (14.20)

and set into it the equation derived from relationship 14.17

σϑ = Kpσr with Kp :=1 + sinϕ1 − sinϕ

In soil mechanics Kp is called the coefficient of passive earth pressure. Oneobtains

dσr

σr= (Kp − 1)

dr

r

or

lnσr = (Kp − 1) ln r + lnC1

i.e.

σr = C1rKp−1.

The integration constant C1 follows from the requirement σr = p for r = r0,so that we finally obtain:

14.5 Plastification 285

σr = p

(r

r0

)Kp−1

σϑ = Kpp

(r

r0

)Kp−1

.

The range r0 < r < re where solution 14.21 applies, is called the plastifiedzone. Instead, the requirement 14.17 (limit condition) applies, also known asthe condition for plastic flow. The word ’plastic’ highlights that with stressstates that fulfil conditions 14.17 and 14.19, deformations occur without stresschange (so-called plastic deformations or ’flow’).If the range r0 ≤ r ≤ re is plastified, then the elastic solution 14.9 must beslightly modified: σe is the value of σr for r = re. In place of equation 14.9now applies for re ≤ r <∞:

σr = σ∞ − (σ∞ − σe)r2er2

σϑ = σ∞ + (σ∞ − σe)r2er2

σrϑ = 0

For r = re the stresses read: σr = σe, σϑ = 2σ∞ − σe. They must fulfil thelimit condition, i.e. σϑ = Kpσr or Kpσe = 2σ∞ − σe. It then follows

σe =2

Kp + 1σ∞ . (14.21)

At the boundary r = re the radial stresses of the elastic and the plastic rangesmust coincide:

p

(rer0

)Kp−1

= σe (14.22)

From (14.21) and (14.22), finally, follows the radius re of the plastic range

re = r0

(2

Kp + 1σ∞p

) 1Kp−1

. (14.23)

If we evaluate equation 14.23 for the case re = r0 we obtain the supportpressure p∗ at which plastification sets on:

r0 = r0

(2

Kp + 1σ∞p

) 1Kp−1

� p = p∗ =2

Kp + 1σ∞ = (1 − sinϕ)σ∞

The distributions of σr and σϑ in the case of plastification are represented inFig. 14.11.


Fig. 14.11. Distributions of σr and σϑ within the plastic and elastic ranges. Notethat σϑ is continuous at r = re, and therefore σr(r) is smooth at r = re (by virtueof Equ. 14.20)

14.5.1 Consideration of cohesion

If the rock exhibits friction and cohesion, one obtains from equation 14.19:

σϑ = Kp · σr + 2ccosϕ

1 − sinϕ= Kp · σr + C .

Introducing in Equ. 14.20 yields:

dσr

dr+σr(1 −Kp) − C

r= 0 .

With the substitution s := σr(1 −Kp) − C we obtain:

ds

dr+ (1 −Kp)

s

r= 0 .

Separation of variables yields:

ln s+ ln r1−Kp = const1

ors = const2 rKp−1 .

Introducing the boundary condition σr(r = r0) = p yields

s = s0

(r

r0

)Kp−1

or

σr(1 −Kp) − C = [ p(1 −Kp) − C ](r

r0

)Kp−1

14.6 Ground reaction line 287

� σr = (p+ c cotϕ)(r

r0

)Kp−1

− c cotϕ

σϑ = Kp(p+ c cotϕ)(r

r0

)Kp−1

− c cotϕ(14.24)

For r = re the elastic stresses σr = σe and σϑ = 2σ∞ − σe must also fulfil thelimit condition. Thus

σe = σ∞(1 − sinϕ) − c cosϕ . (14.25)

At the boundary r = re the radial stresses of the elastic and the plastic zonesmust coincide

(p+ c cotϕ)(rer0

)Kp−1

− c cotϕ = σ∞(1 − sinϕ) − c cosϕ .

Thus, the radius re of the plastic zone is obtained as:

re = r0

(σ∞(1 − sinϕ) − c(cosϕ− cotϕ)

p+ c cotϕ

) 1Kp−1

. (14.26)

Setting re = r0 we obtain now

p∗ = σ∞(1 − sinϕ) − c cosϕ .

Again, σϑ(r) is continuous at r = re and, consequently, σr(r) is smooth atr = re.

14.6 Ground reaction line

The linear relationship 14.16 shows how the tunnel wall moves into the cavity,if the support pressure is reduced from σ∞ to p. This relationship applies tolinear-elastic ground. Now we want to see how the relationship between p andu|r0 reads, if the ground is plastified in the range r0 ≤ r < re. Within thisrange plastic flow takes place, i.e. the deformations increase, without stresschange (Fig. 13.2 b).In order to specify plastic flow, one needs an additional constitutive relation-ship, the so-called flow rule. This is a relationship between the strains εr andεϑ (εz vanishes per definition for the plane deformation we are considering).With the volumetric strain εv := εr + εϑ the flow rule reads in a simplifiedand idealised form:

εv = bεr .


b is a material constant, which describes the dilatancy (loosening) of thematerial.9 Note, however, that several definitions of dilatancy exist. For b = 0isochoric (i.e. volume preserving, εv = 0) flow occurs. We express the strainswith the help of the radial displacement u:

εr =du

dr, εϑ =

u

r

and obtain thus

du

dr+u

r= b

du

dr,

from which follows

u =C

r1

1−b

.

The integration constant C follows from the displacement u = ue at r = reaccording to the elastic solution (equation 14.16):

ue = reσ∞2G

(1 − σe

σ∞

)=

C

r1

1−be

. (14.27)

From the two last equations we obtain

u = reσ∞2G

(1 − σe

σ∞

)(rer

) 11−b

. (14.28)

If we introduce here the relations 14.21 and 14.23 for σe and re, we obtainfinally for r = r0, c = 0 and p < p∗:

u|r0 = r0 sinϕσ∞2G

(2

Kp + 1σ∞p

) 2−b(Kp−1)(1−b)

. (14.29)

This relationship applies to p < p∗, while for p ≥ p∗ applies the elastic rela-tionship 14.16, Fig. 13.2 b. In Fig. 14.12 is shown the relationship between pand u|r0 , which is called the characteristic of the ground (also called ’Fenner-

Pacher-curve’, or ’ground reaction curve’ or ’ground line’).For the case ϕ > 0, c > 0 we can obtain the relationship between the cavitywall displacement u|r0 and p, if we use equation 14.28 with equations 14.25and 14.26:

9The angle ψ: = arctan b can be called angle of dilatancy

14.6 Ground reaction line 289

Fig. 14.12. Ground reaction line with plastification (non-cohesive ground)

Fig. 14.13. Ground reaction line with plastification (cohesive ground)

u|r0 = r0

[σ∞(1 − sinϕ) − c(cosϕ− cotϕ)

p+ c cotϕ

] 2−b(Kp−1)(1−b)

×σ∞2G

(sinϕ+

c

σ∞cosϕ

)(14.30)

Contrary to equation 14.29, equation 14.30 supplies a finite displacement u|r0

for p = 0 (Fig. 14.13). I.e., in a cohesive material the cavity can persistalso without support, whereas it closes up in non-cohesive material (if un-supported). The tunnel wall displacement u|r0 is called in tunnelling ’conver-gence’.The equations deduced in this section cannot easily be evaluated for the caseϕ = 0 and c > 0. In this case it can be obtained from (14.19) and (14.20):In the plastic range r0 < r ≤ re:

σr = 2c ln rr0

+ p

σϑ = σr + 2c

furthermore


re = r0 exp σ∞−c−p2c

σe = σ∞ − c .

With (14.28) we obtain for p < p∗ = σ∞ − c:

u|r0 = r0c

2G

[exp

(σ∞ − c− p

2c

)] 2−b1−b

. (14.31)

The ground reaction line, i.e. the dependence of p on u|r0 , clearly shows thatthe pressure exerted by the rock upon the lining is not a fixed quantity butdepends on the rock deformation and, thus, on the rigidity of the lining. Thisis a completely different perception of load and causes difficulties to manycivil engineers, who are used to consider the loads acting upon, say, a bridgeas given quantities.

14.7 Pressuremeter, theoretical background

There is an abundance of theoretical solutions of cavity expansion problems.In this section it is shown how the equations derived above can be applied tothe problem of the expansion of a cylindrical cavity within a hydrostaticallystressed elastoplastic medium.If we increase p above the value σ∞, then we obtain from the equations pre-sented in Section 14.6 a solution for the problem of the pressuremeter. Theelastic solution (14.16) still applies, whereby u|r0 becomes negative because ofp > σ∞. With the plastic solution (ϕ > 0, c = 0, Equ. 14.21) one needs onlyto replace Kp by Ka = (1− sinϕ)/(1+sinϕ) to obtain the following equationinstead of (14.29):

u|r0 = −r0 sinϕσ∞2G

(2

Ka + 1σ∞p

) 2−b(Ka−1)(1−b)

(14.32)

The rock plastifies for p > p∗ = (1 + sinϕ)σ∞. Equ. 14.32 applies for p > p∗.The relation (14.32) is shown in Fig. 14.14.The plot of u|r0 over the logarithm of 2

Ka+1σ∞p is a straight line. If one as-

sumes a value for b, then one obtains from the slope of this straight line thefriction angle ϕ.10 Thus, the evaluation of pressuremeter-tests depends on as-sumptions. Often one assumes b = 0 and regards thus the problem of theundrained cavity expansion. There is also a solution if the stress-strain curveis given by a power law before reaching ideal-plastic flow.11

10J.M.O. Hughes, C.P. Wroth, D.Windle: Pressuremeter test in sands. Geotech-nique 27, 455-477 (1977)

11R.W.Whittle, Using non-linear elasticity to obtain the engineering propertiesof clay – a new solution for the self boring pressuremeter. Ground Engineering, May1999, 30-34.

14.8 Support reaction line 291

Fig. 14.14. Pressuremeter problem (c = 0): Cavity wall pressure p as a functionof the cavity wall displacement u|r0

14.8 Support reaction line

Now we want to see, how the resistance p of the support changes with in-creasing displacement u. Considering equilibrium (Fig. 14.15) the compressivestress in the support is easily obtained as σa = pr0/d (Fig. 14.15). This stresscauses the support to compress by ε = σa/E (E is the Young’s modulus ofthe support).

Fig. 14.15. Forces acting upon and within the support

The circumference of the support shortens by the amount ε2πr0, i.e. the ra-dius shortens by the amount u = εr0. From here follows a linear relationshipbetween u and p (’characteristic of the support’, ’support reaction line’ or ’sup-port line’):

p =Ed

r20u or u =

r20Ed

p .

We assume, for simplicity, that this linear relationship applies up to the col-lapse of the support, where p = pl (Fig. 14.16).The reaction lines of the ground and the support serve to analyse the interac-tion between ground and support. To this purpose they are plotted togetherin a p-u-diagram. The characteristic of the support is represented by


Fig. 14.16. Support reaction line

u(p) = u0 +r20Ed

p .

Here, the part u0 takes into account that the installation of the support can-not take place immediately after excavation. Tunnelling proceeds with a finitespeed, and before the support can be installed, the tunnel remains temporarilyunsupported in the proximity of the excavation face, Fig. 14.17. Therefore, atthe time of the construction of the support the cavity wall has already movedby the amount u0 into the cavity. The influence of u0 is shown in Fig. 14.18: Ifu0 is small (case 1), then the support cannot take up the ground pressure andcollapses. If u0 is large (case 2) then the ground pressure decreases with defor-mation so that it can be carried by the support. According to the terminologyof the NATM (New Austrian Tunnelling Method)

. . . ground deformations must be allowed to such an extent that aroundthe tunnel deformation resistances are waked and a carrying ring inthe ground is created, which protects the cavity . . .

The influence of the stiffness (resp. Young’s modulus E) of the support canbe similarly illustrated in a diagram (Fig. 14.19). A stiff support (case 1)cannot carry the ground pressure and collapses, while a flexible support (case2) possesses sufficient carrying reserves.

14.9 Rigid block deformation mechanism for tunnels andshafts

Of particular simplicity are deformation mechanisms consisting of rigid blocksthat slip relative to each other. They are realistic, in view of the alwaysobserved localisation of deformation of soil and rock into thin shear bands.The first rigid block mechanism has been introduced into soil mechanics by

14.9 Rigid block deformation mechanism for tunnels and shafts 293

Fig. 14.17. Lack of support at the excavation face

Fig. 14.18. To the influence of u0

Fig. 14.19. Influence of the support stiffness

Coulomb in 1776, when he analysed the active earth pressure on the ba-sis of a sliding wedge. With respect to tunnels, the failure mechanism shownin Fig. 14.20 proves to be useful. It was originally introduced by Lavrikov

and Revuzhenko12 and, therefore, will be referred as L-R-mechanism. Its

geometry is determined by r0, r1 and n, where the variable n denotes thenumber of rigid blocks. The following geometric relations can be established:δ = 2π/n, κ = π · (n − 2)/(2n), a1 = 2r1 sin(δ/2), r1/r0 = sinκ/ sin(κ − α),

12S.V. Lavrikov, A.F. Revuzhenko, O deformirovanii blochnoy sredy vokrugvyrabotki, Fizikotekhnicheskie Problemy Razrabotki Poleznykh Iskopaemykh, Novosi-birsk, 1990, 7-15


α = κ−arcsin(sin δ/(2 cosκ)). The L-R-mechanism can undergo a multiplicityof various displacements, as shown in Fig. 14.21 and 14.22.

Fig. 14.20. L-R-mechanism consisting of n=6 blocks

The hodograph shows the displacements of the individual blocks. Denoting thereference ground with 0 we use the notation uij where i denotes the observerand j denotes the considered body. Thus, u43 denotes the displacement ofblock 3 as viewed by an observer situated on block 4. u03 ≡ u3 is the absolutedisplacement of block 3 with reference to the fixed ground. Obviously, uij =−uji. Relative motion implies the vectorial equation

u0i = uki + uji,

where k and j are the numbers of any two adjacent blocks.The oriented relative displacements are shown in Fig. 14.21. The cohesionforce Cij is the force exerted by the block i upon the block j. Obviously, thisforce is oriented in such a way that it counteracts the displacement uij .The L-R-mechanism does not yield closed-form solutions. However, it is use-ful for ad hoc analyses introducing appropriate geometries and shear strengthparameters. Consideration of equilibrium of the individual blocks yields esti-mation of loads at collapse.

14.10 Squeezing 295

Fig. 14.21. Possible hodographs for the L-R-mechanism

Fig. 14.22. Possible shifts of blocks

14.10 Squeezing

In so-called squeezing rock, the cavity wall displacement u|r0 resp. the supportpressure p increases with time. With constant support pressure p (e.g. p =0), u|r0 increases with time. Convergences up to amounts of meters can beobserved and make the excavation substantially more difficult. If the conver-gence is prevented, e.g. by a shield, then p can increase so that the shieldcannot be advanced any longer.

14.10.1 Squeezing as a time-dependent phenomenon

In continuum mechanics we refer to creep if the strain ε increases with con-stant stress and to relaxation if the stress decreases with constant strain. Theconsideration of time-dependent problems is difficult. So-called viscoelasticequations (Fig. 14.23) are, often, unrealistic for rock.The various suggested one-dimensional creep laws, i.e. functions ε = ε(t), aresimilarly unsatisfactory. They can hardly be represented in tensorial form,resp. they can hardly be experimentally justified for three-dimensional defor-mations. In addition, one must consider that a function ε = ε(t) depends on


Fig. 14.23. One-dimensional models (rheologic models) for viscoelastic behaviour.Upper: Kelvin-solid, lower: Maxwell-fluid. c and η are material constants.

the arbitrary definition of the time zero (t = 0) and is therefore not objective.Strictly speaking, one should differentiate between time dependence and ratedependence. The first is present when chemical changes resp. ageing occurs.The processes in squeezing rock might however belong rather to the secondcategory and be a kind of viscosity of the rock.Despite many assumptions and experimental investigations the rate depen-dence of rock is still insufficiently understood. Here we try to explain the pro-cesses in squeezing rock on the basis a simplifying model. The observationsrelated to squeezing can be described by the reduction of the stress deviator13

with time. We assume that the ground exhibits a short term cohesion. I.e., inthe course of the excavation the cohesion c0 is mobilised. Here we denote bythe term ’cohesion’ the maximum of (σ1−σ2)/2. For the reduction of cohesionwith time we meet the assumption

c = −αc , α > 0 , (14.33)

or

c = c0e−αt .

The value of α can be obtained from relaxation tests. The assumption (14.33)is meaningful, however arbitrary.14 For simplicity, we assume ϕ = 0 and β = 0.From (14.31) then follows:

u|r0 = r0c

2G

(exp

σ∞ − c− p

2c

)2

= r0c

2Gexp

σ∞ − c− p

c. (14.34)

Deriving with respect to t and using (14.33) supplies (with u|r0 = −r0) :

u|r0

(1 +

c

2Gee

σ∞−pc

)=

r0c

2Gee

σ∞−pc

(− pc

+ ασ∞ − p

c− α

)(14.35)

13i.e. the principal stress difference14With given results of measurements it can be improved, possibly as c = −α(c−

cmin)β . Appropriate data are yet missing.

14.10 Squeezing 297

For u|r0 = 0 one obtains a differential equation, which describes the increaseof p with time if convergence is inhibited:

p+ αp = α(−c0e−αt + σ∞

).

Its solution reads:

p(t) = σ∞(1 − e−αt

)− αc0te−αt (14.36)

Hence, p = σ∞ for t→ ∞.For an unsupported tunnel (p = 0) we obtain from (14.35) the law for theincrease of convergence with time:15

u|r0 =r0α

2Ge1−σ∞

c + c· (σ∞ − c) .

Taking into account that for squeezing rock σ∞ � c, we finally obtain

u|r0 ≈ r0ασ∞c

= r0ασ∞c0eαt . (14.37)

This equation agrees with the experience in as much as one observes squeezingbehaviour at high stress (σ∞ large), low strength (c0 small, e.g. in distortedzones) and creeping minerals (usually clay minerals, large α). It is also re-ported that a high pore water pressure favours squeezing.Equ. 14.37 predicts that the rate of convergence due to squeezing, i.e. u|r0 ,increases with time. This is, admittedly, unrealistic and should be attributedto the highly simplified relation (14.33). However, with equations 14.36 and14.37 we can relate the initial convergence rate w0 := u|r0(t = 0) with theinitial rate of pressure increase p0 := p(t = 0):

p0 =(

1 − c0σ∞

)c0w0

r0.

Thus, from the knowledge of w0 we can predict the initial rate of pressurethat will act upon a shield operating in the same rock.

14.10.2 Neglecting time-dependence

Many authors refer to squeezing whenever large convergences appear, no mat-ter whether they appear simultaneously with excavation or with time delay.Let us introduce the symbol ζ to denote the ratio of convergence to tunnelradius, ζ := u|r0/r0. Using equation 14.34, ζ can be expressed in dependenceof the ratio p/σ∞ (i.e. support pressure p / in situ stress σ∞) and of the ratioc

σ∞(i.e. rock mass strength qu = 2c / in situ stress σ∞) as follows:

15Note that the equations, presented here, are based on the assumption of smalldeformations. With large deformations they have to be modified.


ζ =c

2Gexp

[σ∞c

(1 − p

σ∞

)− 1

](14.38)

This equation, which is graphically shown in Fig. 14.24, presupposes ϕ = 0.The corresponding expression for ϕ > 0 can be derived from Equ. 14.30.16

0.25

0.50

0.75

0.25 0.50 0.75 1.00 1.25 1.50 1.75∞

ζ

qu σ/

Fig. 14.24. Graphical representation of Equ. 14.38 for the case p = 0.

Considering different levels of ζ, Hoek assigns to them various tunnellingproblems given in Table 14.1. In this way, squeezing is seen as large conver-gence due to low rock strength.

ζ qu/σ∞ Tunnelling problems

0 . . . 1% > 0.36 Few support problems

1 . . . 2.5% 0.22 . . . 0.36 Minor squeezing problems

2.5 . . . 5% 0.15 . . . 0.22 Severe squeezing problems

5 . . . 10% 0.10 . . . 0.15 Very severe squeezing problems

> 10% < 0.10 Extreme squeezing problems

Table 14.1. Squeezing problems related to convergence and rock strength accordingto Hoek.

In practise, squeezing is waited to fade away. The permanent lining is builtonly as soon as the rate of convergence falls below 2 mm/month in case ofconcrete lining (C20/25), or below 6-10 mm/month in case of steel reinforcedconcrete (≥50 kg/m3) or concrete C30/35.

16Similar curves can be numerically obtained if the yield criterion of Mohr-

Coulomb is replaced by that of Hoek and Brown, see E. Hoek, Big Tunnels in BadRock, 36th Terzaghi Lecture, Journal of Geotechn. and Geoenvir. Eng., Sept. 2001,726-740

14.10 Squeezing 299

14.10.3 Interaction with support

Let us now consider the interaction of a squeezing rock with the support,which is here assumed as a lining (shell) of shotcrete. We neglect hardeningand assume that shotcrete attains immediately its final stiffness and strength.Usually, the rock reaction line (i.e. the curve p vs. convergence w)17 andthe support reaction lines are plotted in the same diagram. Their intersec-tion determines the convergence and the pressure acting upon the support(Fig. 14.25).

rock

support

p

w

Fig. 14.25. The intersection of the reaction lines determines the pressure p andthe convergence w of the support (non-squeezing case)

For squeezing rock, the rock reaction line is not unique. There are infinitelymany rock reaction lines. Each of them corresponds to a particular c-value(Fig. 14.26). In this case the intersection point of rock and support linesmoves along the support reaction line and crosses the various rock reactionlines. The evolution in time of this process can be obtained if we introduceinto Equ. 14.34 the support reaction line p = ps(w), e.g. p = k · (w−w0), k =const, for w > w0:

w =c

2Gr0 exp

(σ∞ − c− ps(w)

c

)(14.39)

With c = c(t), equation 14.39 determines the relation w(t). No matter howthe relation w(t) looks like, after a sufficiently long time lapse the ultimaterock reaction line, i.e. the one corresponding to c∞, will be reached. It mayhappen, however, that the lining cannot support the corresponding load. Thiscan be the case with a brittle lining, whereas a ductile lining will yield untilan intersection with the ultimate rock reaction line is obtained (Fig. 14.27).Examples of yielding support show figures 14.28, 14.29 and 14.30 that referto the Strengen tunnel (Austria). A yielding (’elastic-ideal plastic’) supportcan be achieved by interrupting the lining with arrays of steel tubes, which

17For brevity, the symbol w is used here instead of u|r0 .


w0

c0

p

w

c(t)

c 8

Fig. 14.26. Fan of rock reaction lines for squeezing rock

w0 w1

p

w

c

w 8

8

w0 w2w3

p

w

c 8

Fig. 14.27. A brittle support collapses at w = w1(left) and the convergence contin-ues until the ultimate value w = w3. A ductile (or yielding) support remains activeuntil the ultimate convergence w = w2 is obtained.

are designed to buckle under a specific load (Fig. 14.31). The positions ofthe tube arrays within an idealised circular lining are shown in Fig. 14.32.The circumferential length of the lining is L = 2πr. The total length of theembedded steel tubes is mls (m = number of tube arrays). WithEs: Young’s modulus of shotcreted: shotcrete thicknessn: number of steel tubes per unit length of tunnelwe obtain the following relation between the thrust N and the shortening ΔLof the lining

ΔL =[(2πr −mls) · 1

d · Es+

1n

sl

Fl

]N

for N/n < Fl . With w = ΔL/(2π) and p = N/r we obtain the supportreaction line p = p(w) (Fig. 14.33) as

p = w2π

r[(2πr −mls)/(d · Es) + sl/(n · Fl)]

for p < nFl/r.

14.10 Squeezing 301

Fig. 14.28. In the upper left part is visible the array of pipes

Fig. 14.29. Anchor headplates and pipe array

The question arises whether a yielding support should be installed, which iscostly. The alternative would be to install a usual support at a later time,when the rock has converged by an amount w3 (see dotted line in Fig. 14.27).Of course, one has to determine the value of w3. This can be achieved withborehole tests, i.e. the borehole is a model of the tunnel and the convergence ismeasured within the borehole. Equation 14.39 gives a relation between w andc for an uncased borehole. Measuring w-values at various times yields (with


Fig. 14.30. Squeezed anchor headplate. The two steel tubes had initially circularcross sections. Their squeezing indicates that this anchor has been overloaded.

numerical elimination of c from this equation) the corresponding c-values.Thus, the function c(t) can be determined (for some discrete values of timet). With knowledge of c(t), the complete rock reaction line and its interactionwith the support can be determined as shown before.

Fig. 14.31. Force-displacement characteristic of embedded tube

14.11 Softening of the ground 303

N N

r

d

Fig. 14.32. Steel tubes embedded in shotcrete lining

n Fe /r

p

w

Fig. 14.33. Reaction line of yielding support

14.10.4 Squeezing in anisotropic rock

As already mentioned, squeezing rocks have low strength and are characterisedby layered silicates that predominate in schists. The orientation of schistos-ity (or foliation) imposes a mechanical anisotropy to such rocks. It appearsreasonable to assume that stress relaxation affects only shear stresses actingupon planes of schistosity. As a consequence, tunnels that perpendicular crossthe planes of schistosity (Fig. 14.34 a) are not affected by squeezing, evenat high depths. In contrast, tunnels whose axes have the same strike as theschistosity planes (Fig. 14.34 b) can be considerably affected by squeezing.Typical examples for the two cases are the Landeck and Strengen tunnels inwestern Austria. Both tunnels have been headed within the same type of phyl-litic rock. The Landeck tunnel was oriented as in Fig. 14.34 a and encountered,therefore, no squeezing problems despite a considerable overburden of up to1,300 m. In contrast, the Strengen tunnel was oriented as in Fig. 14.34 b andhad considerable problems with squeezing. Its overburden was up to 600 m.

14.11 Softening of the ground

The requirement that the lining should be flexible is limited by the fact thatdeformations that are too large imply reduction of the strength of the ground(so-called softening) so that the load upon the lining increases again. Softening


Fig. 14.34. Two different orientations of schistosity relative to tunnel axis.

is related with loosening (dilatancy, i.e. increase of volume and porosity withshear), as shown in Fig. 13.4. Softening (i.e. the reduction of stress beyond thepeak, shown in Fig. 13.4) does not comply with the concept of plastic flow,which states that the deformation increases at constant stress (cf. Fig. 13.2 b),and is responsible for the increase of the ground pressure related to the increaseof convergence u|r0 (Fig. 14.35).

Fig. 14.35. Assumed ground reaction line for softening rock (schematically). Therising branch BC is due to the reduction of the rock strength

In order to keep the load upon the lining as low as possible, the support re-action line should intersect the ground reaction line at point B (minimum ofground reaction). This is one of the main requirements of NATM. Thoughthis is in principle correct, it is hardly applicable in practise, as this mini-mum cannot be determined, neither by field measurements nor by numericalsimulation.It is worth mentioning that for a weightless rock and axisymmetric case (cir-cular cross section, hydrostatic primary stress, i.e. K = 1) a rising branch ofthe ground reaction line is not obtained, even if the strength of the rock com-

14.11 Softening of the ground 305

pletely vanishes after the peak, as shown in Fig. 14.36.18 This drastic softeningimplies σϑ−σr = 0 resp. σϑ = σr. From the equation of equilibrium (14.20) itfollows that in the plastified zone dσr/dr = 0 resp. σr = σϑ =const . The re-sulting stress distribution is shown in Fig. 14.37. Since for r = re+0 the elasticsolution (equation 14.21) must fulfil the strength condition σϑ − σr = 2c, itfollows

σe = p = σ∞ − c

Thus, the load on the lining cannot be less than σ∞ − c. This result doesnot depend on u|r0 : With increasing u|r0 the radius re of the plastified zoneincreases (according to equation 14.28), however p remains constant for r ≥ r0.Thus, we obtain the ground reaction line shown in Fig. 14.38, which does notexhibit any rising branch.

Fig. 14.36. Total loss of strength of a cohesive material

18Bliem, C. and Fellin, W. (2001): Die ansteigende Gebirgskennlinie (On the in-creasing ground reaction line) Bautechnik 78(4): 296-305.


Fig. 14.37. Stress distribution in softening rock

Fig. 14.38. Ground reaction line for a weightless rock with total loss of strength(axisymmetric case)

stress and deformation fields around a deep circular tunnel

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