stress-controlled instability mechanisms€¦ · stress ratio: k = σ1/σ3 r r 2 2. 4 stresses...

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1 EOSC433 EOSC433 : : Geotechnical Engineering Geotechnical Engineering Practice & Design Practice & Design Lecture 7: Lecture 7: Stress Analysis around Stress Analysis around Underground Openings Underground Openings 1 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06) Stress Stress-Controlled Instability Mechanisms Controlled Instability Mechanisms 2 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06) Structurally-controlled instabilities are generally driven by a unidirectional body force, i.e. gravity. Stress-controlled instabilities, however, are not activated by a single force, but by a tensor with six independent components. Hence, the manifestations of stress-controlled instability are more variable and complex than those of structurally-controlled failures.

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Page 1: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

1

EOSC433EOSC433: :

Geotechnical Engineering Geotechnical Engineering Practice & DesignPractice & Design

Lecture 7: Lecture 7: Stress Analysis around Stress Analysis around Underground OpeningsUnderground Openings

1 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

StressStress--Controlled Instability MechanismsControlled Instability Mechanisms

2 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Structurally-controlled instabilities are generally driven by a unidirectional body force, i.e. gravity. Stress-controlled instabilities, however, are not activated by a single force, but by a tensor with six independent components. Hence, the manifestations of stress-controlled instability are more variable and complex than those of structurally-controlled failures.

Page 2: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

2

StressStress--Controlled Instability MechanismsControlled Instability Mechanisms

3 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Kaiser et al. (2000)

StressStress--Controlled Instability MechanismsControlled Instability Mechanisms

4 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Although the fundamental complexity of the nature of stress has to be fully considered in the design of an underground excavation, the problem can be initially simplified through the assumptions of continuous, homogeneous, isotropic, linear elastic behaviour (CHILE).

CHILE: Continuous, Homogeneous, Isotropic, Linear Elastic

DIANE: Discontinuous, Inhomogeneous, Anisotropic, Non-Elastic

The engineering question is whether a solution based on the CHILE assumption are of any assistance in design. In fact though, many CHILE-based solutions have been used successfully, especially in thoseexcavations at depth where high stresses have closed the fractures and the rock mass is relatively homogeneous and isotropic. However, in near-surface excavations, where the rock stresses are lower, the fractures more frequent, and the rock mass more disturbed and weathered, there is more concern about the validity of the CHILE model.

Page 3: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

3

StressStress--Controlled Instability MechanismsControlled Instability Mechanisms

5 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

A stress analysis begins with a knowledge of the magnitudes and directionsof the in situ stresses in the region of the excavation. This allows for the calculation of the excavation disturbed or induced stresses.

There exists several close form solutions for the induced stresses around circular and elliptical openings (and complex variable techniques extend these to many smooth, symmetrical geometries),and with numerical analysis techniques the values of the induced stresses can be determined accurately for any three-dimensional excavation geometry.

Brad

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Stresses & Displacements Stresses & Displacements -- Circular ExcavationsCircular Excavations

6 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

In rock mechanics, the Kirsch equations are the most widely used suite of equations from the theory of elasticity. They allow thedetermination of stresses and displacements around a circular excavation.

Pz

k·Pz

Stress ratio:k = σ1/σ3

r

r

2

2

Page 4: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

4

Stresses & Displacements Stresses & Displacements -- Circular ExcavationsCircular Excavations

7 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

From these equations we can see that the stresses on the boundary (i.e. when r = a) are given by:

Note that the radial stresses are zero because there is no internal pressure, and the shear stresses must be zero at a traction-free boundary.

σr = 0 σθ = pz[(1+k) + 2(1-k)cos2θ]

and τrθ = 0

σθmax=3σ1-σ3 = 3(k-1)σ3

σθmin=3σ3-σ1= 3(1-k)σ3

Max. Tangential Stress

Min. Tangential Stress

Stresses & Displacements Stresses & Displacements -- Circular ExcavationsCircular Excavations

8 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

In the hydrostatic case (k=1), the stress concentration around the excavation boundary is always 2·pz. The solution for stresses anywhere within the rock mass for this stress state is similarlysimplified because there are no shear stresses: the terms (1-k) are all zero. Hence the equations for radial and tangential stress reduce to:

Page 5: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

5

Example #1: Stresses around a Circular OpeningExample #1: Stresses around a Circular Opening

9 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

A.

Q. At a depth of 750 m, a 10-m diameter circular tunnel is driven in rock having a unit weight of 26 kN/m3 and uniaxial compressive and tensile strengths of 80.0 MPa and 3.0 MPa, respectively.Will the strength of the rock on the tunnel boundary be exceeded if:

(a) k=0.3, and(b) k=2.0?

Since the tunnel has neither a support pressure nor an internal pressure applied to it, the local stresses at the boundary have σ3 = σr = 0 and σ1 = σθ. The Kirsch solution for the circumferential stress is:

Example #1: Stresses around a Circular OpeningExample #1: Stresses around a Circular Opening

10 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Q. Circular tunnel: 750 m deep, 10m diameter, γrock = 26 kN/m3, σUCS = 80.0 MPa, σT = 3.0 MPa. Will the strength of the rock on the tunnel boundary be reached if: (a) k=0.3, and (b) k=2.0?

A.

For a location on the tunnel boundary, where a = r, the Kirsch equation simplifies to:

Page 6: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

6

Example #1: Stresses around a Circular OpeningExample #1: Stresses around a Circular Opening

11 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

A. We assume that the vertical stress is caused by the weight of the overburden, in which case we have:

MPazv 5.19750026.0 =×== γσ

The 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 (i.e. roof and floor) we use θ = 90°, and for the sidewalls we use θ = 0° .

Q. Circular tunnel: 750 m deep, 10m diameter, γrock = 26 kN/m3, σUCS = 80.0 MPa, σT = 3.0 MPa. Will the strength of the rock on the tunnel boundary be reached if: (a) k=0.3, and (b) k=2.0?

Example #1: Stresses around a Circular OpeningExample #1: Stresses around a Circular Opening

12 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

A. For k=0.3:

Crown and invert (θ = 90°), σθ = -1.95 MPa (i.e. tensile) Sidewalls (θ = 0°), σθ = 52.7 MPa

For k=2.0:

Crown and invert (θ = 90°), σθ = 97.5 MPaSidewalls (θ = 0°), σθ = 19.5 MPa

compressive strength is exceeded

Q. Circular tunnel: 750 m deep, 10m diameter, γrock = 26 kN/m3, σUCS = 80.0 MPa, σT = 3.0 MPa. Will the strength of the rock on the tunnel boundary be reached if: (a) k=0.3, and (b) k=2.0?

Page 7: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

7

Stress and Failure CriterionStress and Failure Criterion

13 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

We’ll see this is different when similar relationships based on Mohr-Coulomb and Hoek-Brown are incorporated into stress-strain constitutive relationships.

Brady & Brown (1993)

The key is that the failure criterion is compared to the stresses and is not part of the calculation!!

Conservation of LoadConservation of Load

14 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

… principle of conservation of load before and after excavation. The sketches show how the distribution of vertical stresses across a horizontal plane changes.

Another concept that can be elegantly demonstrated from the Kirsch equations is the principle of conversation of load.

Hudson & Harrison (1993)

Page 8: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

8

Zone of InfluenceZone of Influence

15 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

… (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.

The concept of influence is important in excavation design, since the presence of a neighbouring opening may provide a significantdisturbance to the near-field stresses to the point of causing failure.

Brady & Brown (1993)

Zone of InfluenceZone of Influence

16 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

… illustration of the effect of contiguous openings of different dimensions. The zone of influence of excavation I includes excavation II, but the converse does not apply.

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Page 9: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

9

Stresses Around Elliptical OpeningsStresses Around Elliptical Openings

17 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

The stresses around elliptical openings can be treated in an analogous way to that just presented for circular openings. There is much greater utility associated with the solution for elliptical openings than circular openings, because these can provide a first approximation to a wide range of engineering geometries, especially openings with high width/height ratios (e.g. mine stopes, power house caverns, etc.).

From a design point of view, the effects of changing either the orientation within the stress field or the aspect ratio of such elliptical openings can be studied to optimize stability.

Stresses Around Elliptical OpeningsStresses Around Elliptical Openings

18 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Assuming isotropic rock conditions, an elliptical opening is completely characterized by two parameters: aspect ratio (major to minor axis) which is the eccentricity of the ellipse; and orientation with respect to the principle stresses. The position on the boundary,with reference to the x-axis, is given by the angle χ.

Hudson & Harrison (1993)

Page 10: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

10

Stresses Around Elliptical OpeningsStresses Around Elliptical Openings

19 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

It is instructive to consider the maximum and minimum values of the stress concentrations around the ellipse for the geometry of an ellipse aligned with the principal stresses. It can be easilyestablished that the extremes of stress concentration occur at the ends of the major and minor axes.

Hudson & Harrison (1993)

Example #2: Stresses around an Elliptical OpeningExample #2: Stresses around an Elliptical Opening

20 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

A.

Q. A gold-bearing quartz vein, 2 m thick and dipping 90º, is to be exploited by a small cut-and-fill stoping operation. The mining is to take place at a depth of 800 m, and the average unit weight of the granite host rock above this level is 29 kN/m3. The strike of the vein is parallel to the intermediate stress, and the major principal stress is horizontal with a magnitude of 37.0 MPa. The uniaxialcompressive strength of the vein material is 218 MPa, and the tensile strength of the host rock is -5 MPa. What is the maximum permissible stope height before failure occurs.

We can assume that, in 2-D cross-section, the stresses induced in the sidewalls (tensile) and the crown (compressive) of the stope can be approximated using the equations for an elliptical excavation.

Page 11: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

11

Example #2: Stresses around an Elliptical OpeningExample #2: Stresses around an Elliptical Opening

21 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Q. Gold-bearing quartz vein: 2 m thick, dipping 90º. The mining stope is 800 m deep, γ = 29 kN/m3, strike parallel to σ2, σ1 = 37.0 MPa and is horizontal, σUCS

(vein) = 218 MPa, To (host rock) = -5 MPa. What is the maximum permissible stope height before failure occurs.

A. Rearranging the given equations, we can solve for the height of the excavation as the minimum of:

The maximum stress that can be sustained by the crown and the sidewall are 218 and –5 MPa, respectively. Note that the sidewall stress is negative because this represents the tensile strength.

Example #2: Stresses around an Elliptical OpeningExample #2: Stresses around an Elliptical Opening

22 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

A.

and hence the ratio of horizontal to vertical stress is:

The vertical stress is:

Q. Gold-bearing quartz vein: 2 m thick, dipping 90º. The mining stope is 800 m deep, γ = 29 kN/m3, strike parallel to σ2, σ1 = 37.0 MPa and is horizontal, σUCS

(vein) = 218 MPa, To (host rock) = -5 MPa. What is the maximum permissible stope height before failure occurs.

Page 12: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

12

Example #2: Stresses around an Elliptical OpeningExample #2: Stresses around an Elliptical Opening

23 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

A. The maximal height of a stope such that the compressive strength of the rock in the crown is not exceeded is given by:

The maximal height of a stope such that the tensile strength of the rock in the sidewall is not exceeded is given by:

Q. Gold-bearing quartz vein: 2 m thick, dipping 90º. The mining stope is 800 m deep, γ = 29 kN/m3, strike parallel to σ2, σ1 = 37.0 MPa and is horizontal, σUCS

(vein) = 218 MPa, To (host rock) = -5 MPa. What is the maximum permissible stope height before failure occurs.

Example #2: Stresses around an Elliptical OpeningExample #2: Stresses around an Elliptical Opening

24 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

A.

Thus we see that sidewall failure is the limiting condition if no stress-induced failure is acceptable in the stope design.

Q. Gold-bearing quartz vein: 2 m thick, dipping 90º. The mining stope is 800 m deep, γ = 29 kN/m3, strike parallel to σ2, σ1 = 37.0 MPa and is horizontal, σUCS

(vein) = 218 MPa, To (host rock) = -5 MPa. What is the maximum permissible stope height before failure occurs.

Page 13: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

13

Zone of InfluenceZone of Influence

25 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

… elliptical approximation to the zone of influence around an elliptical excavation.

Hudson & Harrison (1993)

Other Solutions Other Solutions -- Development of Fracture ZonesDevelopment of Fracture Zones

26 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Closed form solutions for the radial extent of the fracture zone, the stresses within it, and the stresses within the remaining elastic zone can be derived from first principles. Although these equations apply for an idealized case, they can provide guidance to intact rock failure potential and to what extent the rock might be damaged.

Hudson & Harrison (1993)

Page 14: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

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Other Solutions Other Solutions -- Slip on DiscontinuitiesSlip on Discontinuities

27 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Another possibility is that the rock has been weakened by the presence of a pre-existing discontinuity. The extent of any potential zone of instability can be established by considering whether the induced stresses locally satisfy the discontinuity shear strength criterion.

The procedure involves taking a point on the discontinuity, evaluating the stress components using the Kirsch equations, transform these components into shear and normal stresses acting on the discontinuity, and substituting them into the Mohr-Coulomb criterion.

Hudson & Harrison (1993)

Stress Analysis Stress Analysis -- Numerical Numerical ModellingModelling

28 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Many underground excavations are irregular in shape and are frequently grouped close to other excavations. These problems require the use of numerical techniques. These primarily include boundary-element, finite-element and distinct-element techniques.

Page 15: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

15

Numerical Numerical ModellingModelling

29 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

Boundary Methods – only the boundary of the excavation is divided into elements and the interior of the rock mass is represented mathematically as an infinite continuum. The solution works to find a set of approximate stresses which satisfy prescribed boundary conditions, and then uses these to calculate the stresses and displacements in the rock mass.

Eberhardt et al. (1997)

Boundary Element MethodBoundary Element Method

30 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

The figure below illustrates the trace of the surface S of an excavation to be generated in a pre-stressed medium (uniaxial in the x-direction, in this case). At any point on the surface, the pre-excavation load condition is defined by a traction, t’x(S). After excavation of the material within S, the surface opening is traction free.

In setting up the boundary element solution procedure, the requirements are to discretize and describe algebraically the surface S, and to find a method of satisfying the imposed induced traction conditions on S.

Brad

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Brow

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993)

Page 16: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

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Boundary Element MethodBoundary Element Method

31 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

In this sense, the boundary element method is an “integral method”. The characteristic of integral methods of stress analysis is that a problem is specified in terms of surface values of the field variables of traction and displacement. Since only the problem boundary is defined and discretized, the so-called boundary element methods of analysis effectively provide a unit reduction in the dimensional order of a problem. The implication is a significant advantage in computational efficiency, compared with other numerical methods (i.e. differential methods).

Brad

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Disadvantages to these methods are that they are best suited for linear material behaviour and homogeneous material properties; non-linear behaviour and material heterogeneity negate the intrinsic simplicity of a boundary element solution.

Failure CriterionFailure Criterion

32 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

2331 cc sm σσσσσ ++=Hoek-Brown failure criterion

(developed for confined conditions around tunnels)

Intact rock strength:m = lab-determineds = 1

Rock massstrength

σ1

σ3

m & s are derived from empirical charts that are related to rock mass quality RMR & Q

σc

m ~ Frictions ~ Cohesion

Page 17: Stress-Controlled Instability Mechanisms€¦ · Stress ratio: k = σ1/σ3 r r 2 2. 4 Stresses & Displacements - Circular Excavations ... The vertical stress is: Q. Gold-bearing quartz

17

Stress Analysis & Failure CriterionStress Analysis & Failure Criterion

33 of 33 Dr. Erik Eberhardt EOSC 433 (Term 2, 2005/06)

σ1

σ3

failure

in situstress

6 m

4 m

2 mm = 0s = 0.11 1.8m

σci = 100 MPa

ObservedObservedFailureFailure