CV6315 Tunnel Lecture 1

CV6315 Engineering Geology and Rock Engineering

This Lecture:

Tunnels in rockTunnels in rockby

Assoc. Prof. LOW Bak Kong

R fReferences:

• Evert Hoek, Practical Rock Engineering, Chapter 12. htt // i /h k/P ti lR kE i ihttp://www.rocscience.com/hoek/PracticalRockEngineering.asp

• Brady, B H G and Brown, E T, Rock Mechanics for Underground Mining, 3nd Ed Springer 20063nd Ed, Springer, 2006.

• Harrison, John P. and Hudson John A. (2000). Engineering rock mechanics: part 2 Illustrative worked examples Pergamonmechanics: part 2 Illustrative worked examples, Pergamon.

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Learning objectives of this lecture:

• Understand some very basic concepts of how a rock mass surrounding a tunnel deforms andsurrounding a tunnel deforms and how the support systems act to control this deformation.

• Study the variation of deformation with internal support pressure for a i l t l i l ti f tlcircular tunnel in an elastic-perfectly

plastic material under hydrostatic stress field and obeying Mohr-Coulomb failure criterion.

• Examine examples of practical support designs for different conditions.

• Use of the classical Kirsch solution for plane-strain stresses around a i l icircular opening.

• Appreciate the effects of excavation shape and in situ stresses on the boundary stresses of underground excavations

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boundary stresses of underground excavations.

Figure 1: Vertical section through a three-dimensional finite element model of the failure and deformation of the rock mass surrounding themodel of the failure and deformation of the rock mass surrounding the face of an advancing circular tunnel. The plot shows displacement vectors as well as the shape of the deformed tunnel profile.

3from Chapter 12 of Practical Rock Engineering by E. Hoek

Fi 2 P tt f d f ti i th k di

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Figure 2: Pattern of deformation in the rock mass surrounding an advancing tunnel.

from Chapter 12 of Practical Rock Engineering by E. Hoek

Tunnel deformation analysisConsider a circular tunnel subjected to a hydrostatic stress field in which the horizontal and vertical stresses are equal. Assume Mohr‐Coulomb failure criterion and that the surrounding heavily jointed rock mass behaves as an elastic‐perfectly plastic material. S t i d ll d i l t i t lSupport is modelled as an equivalent internal pressure.

Definition of failure criterion

B d M h C l b it i l ti f il h (1)Based on Mohr‐Coulomb criterion, plastic failure occurs when (1)

The uniaxial compressive strength of the rock mass σcm is: (2)p g cm ( )

and the slope k of the σ1′ versus σ3′ line is: (3)


Tunnel deformation analysis (cont.)Analysis of tunnel behaviour

(4)

(5)


Tunnel deformation analysis (cont.)


Figure 3: Plastic zone surrounding a circular tunnel

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Figure 3: Plastic zone surrounding a circular tunnel.


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Figure 4: Spreadsheet for the calculation of rock mass characteristics and the behaviour of the rock mass surrounding a circular tunnel in a hydrostatic stress field.


Dimensionless plots of tunnel deformation

Using the program @RISK in conjunction with a Microsoft g p g @ jExcel spreadsheet for estimating rock mass strength and tunnel behaviour (equations 4 to 7). Uniform distributions were sampled for the following input parameters, the two figures in brackets define the minimum and maximum values used: Intact rock strength σci (1,30 MPa), Hoek‐Brown constant mi (5,12), Geological Strength Index GSI (10,35), In situ stress (2, 20 MPa), Tunnel radius (2, 8 m).

Once the rock mass strength falls below 20% of the in situ stress level, deformations increase substantially and, unless these deformations aredeformations increase substantially and, unless these deformations are controlled, collapse of the tunnel is likely to occur.

Fi 5 T l d f ti ti f k t th t i it t b d

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Figure 5: Tunnel deformation versus ratio of rock mass strength to in situ stress based ona Monte‐Carlo analysis which included a wide range of input parameters



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Figure 6: Field observations by Chern et al (1998) from the Second Freeway, Pinglinand New Tienlun headrace tunnels in Taiwan.


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Figure 7: Ratio of tunnel deformation to tunnel radius versus the ratio of rock mass strength to in situ stress for different support pressures.


Estimates of support capacity

Figure 8: Approximate maximum capacities g pp pfor different support systems installed in circular tunnels. Note that steel sets and rockbolts are all spaced at 1 m. (Details of

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support systems for curves 1‐9 and 20‐26 are in Fig. 8 of Hoek’s Practical Rock Engineering, Chapter

12.)

Practical exampleA 4 m span drainage tunnel is to be driven in the rock mass behind the slope of an openA 4 m span drainage tunnel is to be driven in the rock mass behind the slope of an openpit mine. The tunnel is at a depth of approximately 150 m below surface and the generalrock is a granodiorite of fair quality. A zone of heavily altered porphyry associated with afault has to be crossed by the tunnel and the properties of this zone, which has beeny p p ,exposed in the open pit, are known to be very poor. Mine management has requested aninitial estimate of the behaviour of the tunnel and of the probable support requirements.The following example presents one approach to this problem, using some of thetechniques described earlier in this chapter and then expanding them to allow a morerealistic analysis of tunnel support behaviour.

Estimate of rock mass propertiesEstimate of rock mass propertiesFigures 5 and 7 show that a crude estimate of the behaviour of the tunnel can be made if the ratio of rock mass strength to in situ stress is available. For a depth of 150 m and a unit weight of 0.027 MN/m3, the vertical in situ stress is estimated to be 4 MPa. The fault g / ,material is considered incapable of sustaining high differential stress levels and it is assumed that the horizontal and vertical stresses are equal within the fault zone.

In the case of the granodiorite the laboratory uniaxial compressive strength isIn the case of the granodiorite, the laboratory uniaxial compressive strength is approximately 100 MPa. However, for the fault material, specimens can easily be broken by hand as shown in Figure 9. The laboratory uniaxial compressive strength of this material is estimated at approximately 10 MPa


estimated at approximately 10 MPa.

Based upon observations in the open pit mine slopes and utilizing the proceduresmine slopes and utilizing the procedures described in the chapter on “Rock mass properties”, the granodiorite is estimated to have a GSI value of approximately 55. The pp yfault zone, shown in Figure 9, has been assigned GSI = 15.

Th R L b (f fThe program RocLab (free from www.rocscience.com) implements the methodology described in the chapter on “Rock mass properties” and in particularRock mass properties and, in particular, the equations given in the 2002 version of the Hoek‐Brown failure criterion (Hoek et al, 2002) This program has been used to

Figure 9: Heavily altered porphyry can easily be broken by hand

2002). This program has been used to calculate the global rock mass strength σcm

for the granodiorite and the fault zone and the results are presented below: broken by hand.the results are presented below:


Support requirementsFigures 5 and 6 show that, for the granodiorite with a ratio of rock mass strength to in situ stress of 8 25 the size of the plastic zone and the induced deformations will be negligiblestress of 8.25, the size of the plastic zone and the induced deformations will be negligible.This conclusion is confirmed by the appearance of an old drainage tunnel that has stood for several decades without any form of support. Based upon this evaluation, it was decided that no permanent support was required for the tunnel in the fair quality granodiorite Spotthat no permanent support was required for the tunnel in the fair quality granodiorite. Spot bolts and shotcrete were installed for safety where the rock mass was more heavily jointed. The final appearance of the tunnel in granodiorite is shown in Figure 10.

Figure 10: Appearance of the drainage tunnel in fair quality granodiorite in which no permanent support was required. Spot bolts and shotcrete were installed for safety in jointed areas. The concrete lined drainage channel is shown in the centre of the tunnel floor.


Figure 14 Assembly of a sliding joint in a top hat section steel sethat section steel set.

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Figure 15: Installation of sliding joint top hat section steel sets immediatelybehind the face of a tunnel being advanced through very poor quality rock.



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Figure 16: Delay in the activation of passive support by the use of sliding joints.




28Figure 19: Spiling in very poor quality clay‐rich fault zone material.

Figure 20: Full face 10 m span tunnel excavation through weak rock under the protection of a f l b ll Th fi l t li i i tforepole umbrella. The final concrete lining is not included in this figure.

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Figure 21: Installation of 12 m long 75 mm diameter pipe forepoles in an 11 m span

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Figure 21: Installation of 12 m long 75 mm diameter pipe forepoles in an 11 m spantunnel top heading in a fault zone.


The next 22 slides are from:The next 22 slides are from:

• Brady B H G and Brown E T, Rock Mechanics for Underground

Mining, 3nd Ed, Springer, 2006.

- Chapter 6: Methods of Stress Analysis

- Chapter 7: Excavation design in massive elastic rock

• Harrison, John P. and Hudson John A. (2000). Engineering rock , ( ) g g

mechanics: part 2 Illustrative worked examples, Pergamon.

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(a) (b)

Figure 6.3 Problem geometry, coordinate system and nomenclature for specifying the stress and displacement distribution around a circular

ti i bi i l t fi ldexcavation in a biaxial stress field.

Brady and Brown, Chapter 6, Methods of Stress Analysis33

Kirsch equations for stress and displacement distributions around the circular opening:p g

In these expressions ur , uθ are displacements induced by excavation, hil t t l t ft ti f th iwhile σrr, σθθ, σrθ are total stresses after generation of the opening.

Brady and Brown, Chapter 6 34

By putting r = a in equation 6.18, the stresses on the excavation

bo ndar are gi en asboundary are given as

Equations 6.19 confirm that the solutions satisfy the imposed conditionEquations 6.19 confirm that the solutions satisfy the imposed condition

that the excavation boundary is traction free. Similarly, for θ = 0, and r

large, the stress components are given by

so that the far-field stresses recovered from the solutions correspond to the

imposed field stresses.


Boundary stresses. Equations 6.19 define the state of stress on the boundary of

a circular excavation in terms of the co ordinate angle θ Clearly since the surfacea circular excavation in terms of the co-ordinate angle θ. Clearly, since the surface

is traction free, the only non-zero stress component is the circumferential

component σθθ. For K < 1.0, the maximum and minimum boundary stresses occur in

the side wall (θ = 0) and crown (θ = π/2) of the excavation. Referring to Figure 6.3b,

these stresses are defined by the following:

These expressions indicate that, for the case when K = 0, i.e. a uniaxial field

di t d ll l t th i th i d i i b d tdirected parallel to the y axis, the maximum and minimum boundary stresses are

These values represent upper and lower limits for stress concentration at the

boundary. That is, for any value of K > 0, the sidewall stress is less than 3p, and the

crown stress is greater than −p The existence of tensile boundary stresses in acrown stress is greater than p. The existence of tensile boundary stresses in a

compressive stress field is also noteworthy.Brady and Brown, Chapter 6 36

In the case of a hydrostatic stress field (K = 1), equation 6.19 becomes

i.e. the boundary stress takes the value 2p, independent of the co-ordinate angle θ.

This represents the optimum distribution of local stress, since the boundary is

uniformly compressed over the complete excavation periphery.

Equations 6.18 are considerably simplified for a hydrostatic stress field, taking the

form

The independence of the stress distribution of the co-ordinate angle θ, and the

fact that σrθ is everywhere zero, indicates that the stress distribution is

axisymmetric.


Zone of influence of an excavationThe concept of a zone of influence is important in mine design, since it may provideconsiderable simplification of a design problem. The essential idea of a zone of influence is that it defines a domain of significant disturbance of the pre‐mining stress field by an excavation. It differentiates between the near field and far field of an opening. The extent of an opening’s effective near‐field domain can be explained by the following example.The stress distribution around a circular hole in a hydrostatic stress field, of magnitude

p, is given by equations 6.20. The equations indicate that the stress distribution is i i ill d i Fi 7 3 I i dil l l d h f 5 1 04axisymmetric, as illustrated in Figure 7.3a. It is readily calculated that for r = 5a, σθθ = 1.04p

and σrr = 0.96p, i.e. on the surface defined by r = 5a, the state of stress is not significantly different (within ± 5%) from the field stresses. If a second excavation (II) were generated o tside the s rface described b r 5a for the e ca ation I as sho n in Fig re 7 3b the preoutside the surface described by r = 5a for the excavation I, as shown in Figure 7.3b, the pre‐mining stress field would not be significantly different from the virgin stress field. The boundary stresses for excavation II are thus those for an isolated excavation. Similarly, if excavation I is outside the zone of influence of excavation II the boundary stresses aroundexcavation I is outside the zone of influence of excavation II, the boundary stresses around excavation I are effectively those for an isolated opening. The general rule is that openings lying outside one another’s zones of influence can be designed by ignoring the presence of all others For example for circular openings of the same radius a in a hydrostatic stress fieldothers. For example, for circular openings of the same radius, a, in a hydrostatic stress field, the mechanical interaction between the openings is insignificant if the distance DI,II between their centres is DI,II ≥ 6a

It is important to note that in general the zone of influence of an opening is related to both

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It is important to note that, in general, the zone of influence of an opening is related to both excavation shape and pre‐mining stresses.

Brady and Brown, Chapter 7 Excavation design in massive elastic rock

Figure 7 3 (a) Axisymmetric stress distribution around a circular opening in a hydrostatic stress field;

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Figure 7.3 (a) Axisymmetric stress distribution around a circular opening in a hydrostatic stress field; (b) circular openings in a hydrostatic stress field, effectively isolated by virtue of their exclusion from each other’s zone of influence.

Brady and Brown, Chapter 7

Figure 7.3b illustrates the overlap of the zones of influence of two circular openings. In

the overlap region, the state of stress is produced by the pre‐mining stresses and the

stress increments induced by each of the excavations I and II. In the other sections of

each zone of influence, the state of stress is that due to the particular excavation.each zone of influence, the state of stress is that due to the particular excavation.

Figure 7.4 illustrates a large‐diameter opening (I) with a small‐diameter opening (II) in its

f fl d h f fl f fzone of influence. Since excavation I is outside the zone of influence of excavation II, a fair

estimate of the boundary stresses around I is obtained from the stress distribution for a

single opening. For excavation II, the field stresses are those due to the presence of

excavation I. An engineering estimate of the boundary stresses around II can be obtained

by calculating the state of stress at the centre of II, prior to its excavation. This can be

introduced as the far field stresses in the Kirsch equations (Equations 6 18) to yield theintroduced as the far‐field stresses in the Kirsch equations (Equations 6.18) to yield the

required boundary stresses for the smaller excavation.

40Brady and Brown, Chapter 7

Figure 7.4 Illustration of the effect of contiguous openings of different dimensions.The zone of influence of excavation I includes excavation II but the converse does notThe zone of influence of excavation I includes excavation II, but the converse does not apply.

41Brady and Brown, Chapter 7

Harrison and Hudson (2000) Q19.7

At a depth of 450 m, a 3-m diameter circular tunnel is driven in rock having a unit

weight of 26 kN/m3 and uniaxial compressive and tensile strengths of 60.0 MPa and

3.0 MPa, respectively. Will the strength of the rock on the tunnel boundary be

reached ifreached if

(a) k= 0.3, and

(b) k = 2.5?

(Note: There are some notation differences between Brady&Brownand Harrison&Hudson.)

A second tunnel, of 6 m diameter, is subsequently driven parallel to and at the same

centre line level as the first, such that the centre line spacing of the two tunnels is 10

m. Comment on the stability of the tunnels for the field stresses given by (a) and (b) y g y ( ) ( )

above.

A19.7 This is a problem to be solved using the Kirsch

solution for circular openings in a state of plane strain.

We are asked to examine the stability of the rock on

the boundary of the tunnel. As the tunnel has neither athe boundary of the tunnel. As the tunnel has neither a

support pressure nor an internal pressure applied to it,

the rock on the boundary is subjected to a uniaxial

t t f t ith th l l 0 d l lstate of stress, with the local σ3 = σr = 0 and local

σ1 = σθ. 42

Answer to Harrison and Hudson (2000) Q19.7 (cont.)

The Kirsch solution for the circumferential stress is

and for a location on the tunnel boundary where a = r this reduces toand for a location on the tunnel boundary, where a = r, this reduces to

Assume that the vertical stress is caused by the weight of the overburden, in y g ,which case we have

The extreme values of induced stress occur at positions aligned with the principal inThe extreme values of induced stress occur at positions aligned with the principal in

situ stresses, and so in order to compute the stress induced in the crown and invert of

the tunnel we use θ = 90°, and for the sidewalls we use θ = 0°. For the case of k = 0.3,

the stresses are then found to be -1.17 MPa (i.e. tensile) and 31.59 MPa, respectively.

For the case of k = 2.5, the induced stresses are 76.05 MPa and 5.85 MPa. Thus, we

see that the compressive strength of the rock, 60 MPa, is reached at the crown and p g , ,

invert of the tunnel for the case of k = 2.5.

After the second tunnel has been driven, we can find an approximate solution to the

bl th h lti l li ti f th Ki h l ti W t t b d t i iproblem through a multiple application of the Kirsch solution. We start by determining

whether the tunnels are inside each other's zone of influence. Given the approximate43


nature of the solution, a rigorous computation of this is not justified. Instead, we take

the distance of the 5% zone of influence for the hydrostatic case and examine the

tunnel spacings on this basis.

F th fi t t l h 1 5 f hi h fi d th t (5%) √20 1 5√20For the first tunnel we have a = 1.5 m, from which we find that r(5%) = a√20 = 1.5√20

= 6.7 m. For the second tunnel, for which a = 3.0 m, we find r(5%) = 13.4 m. The

centre-to-centre spacing of the tunnels is 10 m. Thus, on the basis of the 5% zone of

influence, we see that the first tunnel is affected by the second, but the second is not

affected by the first. This means that the stresses induced on the boundary of the

second tunnel are as computed above but for the first tunnel we need to use ansecond tunnel are as computed above, but for the first tunnel we need to use an

approximation. We do this by computing the stress state induced by the second

tunnel at the position of the centre of the first tunnel, and then using this computed

stress state to determine the stresses induced on the boundary of the first tunnel.

As the centre-to-centre distance of the tunnels is 10 m, and as the two tunnels are at

the same elevation we use r = 10 m and θ = 0° in the Kirsch equations As the secondthe same elevation, we use r = 10 m and θ = 0 in the Kirsch equations. As the second

tunnel has a radius a = 3 m, for the case of k = 0.3 we find σθ = 12.48 MPa and σr =

4.20 MPa. The shear stress is zero. This means that, for the first tunnel, we have σv′ =

12.48 MPa and k′ = 4.20/12.48 = 0.336. The stress induced in the crown and invert is

then found to be44


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CHILE = acronym for Continuous, Homogeneous, Isotropic and Linearly‐Elastic46


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Excavation shape and boundary stresses

The width W of the ellipse is measured inThe width, W, of the ellipse is measured in the direction of the x axis, and the height, H, in the direction of the local z axis.

High boundary curvature (i.e. 1/ρ) leads to high boundary stresses.

For an ellipse of major and minor axes 2a and 2b, the radius of curvature at points Aradius of curvature at points A and B, ρA and ρB, is foundfrom simple analytical geometry to begeometry to be


Brady and Brown (2006), Problem 7.1

ANSWERS:

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Harrison and Hudson (2000) Q20.1

A vertical shaft is to be sunk through a rock mass in which the magnitudes of the two

horizontal principal stresses have a ratio of 3: 1. A clear space of 5 m x 5 m must be

maintained within the shaft for equipment installation. The three excavation cross‐sectionsq p

shown in the sketch are under consideration for the shaft design. Discuss the relative

design merits of each cross‐section in terms of rock mechanics principles, and hence

id d ti f th ti l h ( d i th f fil (b) d ( )provide recommendations for the optimal shape (and in the case of profile (b) and (c),

optimal orientation) for the excavation.

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Answer from Harrison and Hudson (2001) Q20.1

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Thus the design strategy is to use an elliptical shaft oriented with its major axis

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Thus, the design strategy is to use an elliptical shaft oriented with its major axis parallel to the major principal stress, and the axial ratio of the opening equal to the stress ratio, i.e. 3. Harrison and Hudson (2000)

CV6315 Tunnel Lecture 1

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