International Journal of Rock Mechanics & Mining Sciences 78 (2015) 118–126

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International Journal ofRock Mechanics & Mining Sciences

http://d1365-16

n CorrE-m

journal homepage: www.elsevier.com/locate/ijrmms

Technical Note

An analytical solution for recovering the complete in-situ stress tensorfrom Flat Jack tests

Ben-Guo He, Yossef H. Hatzor n

Department of Geological and Environmental Sciences, Ben-Gurion University of the Negev, Beer-Sheva 84105, Israel

a r t i c l e i n f o

Article history:Received 5 January 2015Received in revised form15 April 2015Accepted 19 May 2015

1. Introduction

In-situ stress determination has a dominant role in the designof underground construction projects as the initial and excavationinduced stress concentrations typically affect the performance ofthe rock mass surrounding the excavation. While it is possible toestimate the initial and excavation induced stresses from theoryusing some well accepted assumptions, accurate determination ofthe stress tensor near the excavation face typically requires eitherdirect measurements or back analyses.

One of the earliest methods used in rock mechanics for in-situstress measurement is the Flat Jack method, the principles ofwhich are described in the collection of ISRM suggestedmethods[1] and reviewed among other techniques by Fairhurst.[2]

The method was first proposed by Mayer[3] and later improved byRocha[4] and Hoskins[5]. Based on the stress relief principle, thecancellation pressure is used to obtain the in-situ stress compo-nent acting normal to the Flat Jack plate[6,7].

The Flat Jack method is a “cancellation” in-situ stress testmethod, in which a thin slot is cut into the rock surrounding anunderground opening and a pressure gage is inserted into the slot(Fig. 1a). The pressure in the flat jack is increased until the re-corded strain relaxation of the sidewall is cancelled (Fig. 1b). Thejack pressure at which the strain relaxation is cancelled, is as-sumed to be the normal stress acting on the face of the flat jack.Therefore, with known solutions for stress concentrations aroundthe opening, the initial in situ stress field can be recovered bymeans of inversion, if a sufficient number of flat jack tests areperformed around the opening, at different orientations. The fol-lowing assumptions are made in the solution process: the width ofthe slot is very small with respect to the span of the opening, the

x.doi.org/10.1016/j.ijrmms.2015.05.00709/& 2015 Elsevier Ltd. All rights reserved.

esponding author.ail address: hatzor@bgu.ac.il (Y.H. Hatzor).

elastic modulus of the rock is linear and reversible in the range ofstresses of interest, the rock mass is isotropic, there is perfectcoupling between the flat jack and the rock which is typicallyachieved using cement filling between the jack and the rock.

The method has been used in many case studies around theworld. A set-up of a total of twelve slots cut along the wall of anunderground opening was recommended by Pinto and Cunha[8] toobtain the stress tensor. Franco el al.[9] used the method to backanalyze the 3D stress tensor via the least square method in theSerra Sa Mesa Hydroelectric Power Plant in Brazil. Results oftwelve Flat Jack tests performed in an underground hydroelectricpower project in northern Portugal have been integrated into astress model that took into account both topography and tectoniceffects[10]. Similarly, the in-situ stress field of the Gardanne basin inFrance was assessed using the Flat Jack method where elevenstress measurements were carried out at different depths[11]. InItaly, 60 tests were carried out during six geomechanical in-vestigation campaigns performed for construction of five largeunderground power plants by the Italian Electricity Board[12].

Most of the current procedures for determination of the in-situstress tensor from the Flat Jack method are based on a two-di-mensional analytical solution, in which the orientation of oneprincipal stress is usually assumed to be parallel to the axis of theinvestigated gallery. If the far field principal stresses are assumedto be horizontal and vertical, then only two flat jack tests arenecessary to obtain the magnitude of the principal stresses bydirect application of Kirsch solution [13], as shown by Goodman[7];see Fig. 2. Amadei and Stephansson[6] proposed a three-dimen-sional solution for a transversely isotropic medium where theyassume that the stress at every point has one elastic axis of sym-metry which is aligned with the tunnel axis direction (Fig. 3). Thesolution is provided in complex variable functions. Ultimately, thesolution for the isotropic case is provided by Amadei andStephansson[6] for three out of the six components of the far field

p

Kp

u

v

θ

a

τrθ

Fig. 2. Stresses around a circular hole in polar coordinates under horizontal andvertical far field Principles stresses (modified after Ref. [7]).

τxy

τxz

τyx τyz

τzy

τzx

z x

y

σz

σx

σy

σy

σx

Fig. 3. Three–dimensional far field stress tensor and a circular tunnel aligned withthe z axis.

σθi

σy

σxσx

τxy

τxy

θi

τyx

σy

τxy

Fig. 4. Sign convention used in Ref. [6] for Flat Jack test i (i¼1, 2, 3) performedaround a circular opening subjected to far field stress tensor.

Jack

σ0

d

Pump Oil

Pin

sepa

ratio

n

d0

Time Jack pressure pc

σ0 ≈ pc

Fig. 1. Principles of the Flat Jack test: (a) Plan view and cross section of flat jack.(b) Cross section through the side wall of a tunnel showing the flat jack test as-sembly. (c) Pin separation curve (modified after Ref. [7]).

B.-G. He, Y.H. Hatzor / International Journal of Rock Mechanics & Mining Sciences 78 (2015) 118–126 119

stress tensor, where θi (i¼1, 2, 3) is the angle defining the positionof the Flat Jack test from the x-axis (Fig. 4):

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1 cos 2 1 cos 2 4 sin 2

1 cos 2 1 cos 2 4 sin 2

1 cos 2 1 cos 2 4 sin 2 1

x

y

xy

1

2

3

1 1 1

2 2 2

3 3 3

σσσ

θ θ θθ θ θθ θ θ

σστ

=− + −− + −− + −

⋅

( )

θ

θ

θ

A complete three-dimensional solution, the derivation of whichis not provided, for the far field stress tensor was suggested byPinto and Cunha[8] where at every point θ around the tunnel axisfour tests must be performed at four inclination angles α withrespect to the direction vector r, separated at 45° intervals (Fig. 5).

In this approach, 12 tests are needed to resolve the complete farfield stress tensor, to be performed at three different θ positionsaround the tunnel axis. This procedure requires at every θ positionfour slots one close to another and this may adversely affect theaccuracy of the solution due to possible interactions between theslots.

In this note we present an analytical solution for recovering thesix components of the far field stress tensor in an isotropic rockmass using only six Flat Jack tests to be performed at a minimumof three different θ positions around the tunnel axis and a mini-mum of three α inclinations around the tunnel radius vector. Noassumptions need to be made with respect to the orientation ofthe initial far field stresses. The complete derivation of the ana-lytical solution we propose is provided below.

σy

σxσz

τxy τzy

τxz

τyz τyx

A

B

C

1, α=0

56

78

9

1011

1245

2, α=45

3, α=904, α=135

Fig. 5. Procedure proposed in Ref. [8] for determination of the far field stress tensorin which 12 tests are required.

θ

x

y

z

r

α

Fig. 6. Definition of the rotation angle α around tunnel radius vector r. Note that αis measured clockwise from tunnel axis (dashed line) when viewing the tunnelfrom the outside. From viewed from the inside of the tunnel (the field case) α ismeasured counterclockwise with respect to tunnel axis.

r(r'') α

z

θ

α

Fig. 7. Coordinate transformation rotation around r axis.

B.-G. He, Y.H. Hatzor / International Journal of Rock Mechanics & Mining Sciences 78 (2015) 118–126120

2. Stress distribution around a circular hole in the x–y plane

It can be shown (for complete derivation see Appendix A) thatthe stress distribution around a circular hole in cylindrical co-ordinates may be expressed as

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎫

⎬

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

ar

ar

ar

ar

ar

ar

a

r

a

r

ar

ar

ar

ar

21

21 1

3

cos 2 1 13

sin 2

21

21

3cos 2

13

sin 2

21 1

3sin 2

1 13

cos 22

rx y x y

x y x y

x y

xy

2

2

2

2

2

2

xy

2

2

2

2

2

2

4

4

xy

4

4

r r

2

2

2

2

2

2

2

2

σσ σ σ σ

θ τ θ

σσ σ σ σ

θ

τ θ

τ τσ σ

θ

τ θ

=+

− +−

− −

+ − −

=+

+ −−

+

− +

=

= −−

− +

+ − +( )

θ

θ θ

where θ is measured as in Fig. 4, a is the radius of circular opening.Letting r equal a we obtain the same solution for σθ as proposed byAmadei and Stephansson[6] for the immediate wall of the tunnel:

⎡

⎣⎢⎢

⎤

⎦⎥⎥

3

2 cos 2 4 sin 2

1 2 cos 2 1 2 cos 2 4 sin 2

x y x y xy

x

y

xy

( )[ ]

( )

σ σ σ σ σ θ τ θ

θ θ θσστ

= + − − −

= − + −

⋅

θ

The complete analytical solution which is derived in the ap-pendix and shown in Eq. (2) is identical to Eq. (1) suggested byAmadei and Stephansson[6], provided that the slots of the jack areparallel to tunnel axis, thus yielding xσ , yσ and xyτ in the x–y plane(the tunnel cross section; see Fig. 4) based on three different tests.To obtain the remaining three components of the stress tensor, atleast three more tests are necessary. In this note we propose aprocedure in which the jack slots in the additional three tests arenot necessarily parallel to the tunnel axis but rather form an angleα with respect to tunnel axis, as originally proposed by Pinto andCunha[8]. The angle α can be viewed as a rotation angle around thetunnel radius vector. Note that the angle α around the tunnel ra-dius vector is measured clockwise from the tunnel axis to the slotwhen the system is viewed from outside the tunnel (Fig. 6). When

viewed from the inside of the tunnel, as would be the case inpractice, α is measured counterclockwise with respect to thetunnel axis. Also note that here the slots are always perpendicularto the tunnel wall, as in previous procedures.

In our proposed procedure a total of only six tests is required, incontrast to a minimum of 12 tests that are required in the proce-dure proposed by Pinto and Cunha[8]. We believe this represents asignificant improvement, as possible interferences between thefour slots in each measurement point in Pinto and Cunha proce-dure are avoided.

3. Finding the stress tensor components in the in z-direction

To obtain the stress components in the z-direction ( zσ , zxτ andzyτ ) we rotate the coordinate system around the r axis (Fig. 7) toestablish the relationship between σθ and the far field stress in thez-direction, thus the six components of the in-situ stress tensor canbe determined.

3.1. Transformation of field variables from Cartesian to Cylindricalcoordinates

We begin by rotating the x–y coordinate system around the zaxis (Fig. 8). The rotation matrix for this case is

θ

y

x

z(z')

θ

Fig. 8. Second order tensor coordination transformation.

``

z

θ

α1=45 α2 = 0

α3=75

α4 = 0

α5=135 α6 = 0

θ1 = θ2 =55

θ3 =90

θ4 =120

θ5 = θ6 =335

θ

α5=135

α6 = 0

α1=45

α2 = 0α3=75α4 = 0

y

x

θ1 = θ2

θ3θ4

θ5 = θ6

r

Fig. 9. Definition of angles to locate the slots of the Flat Jack tests: α inclinationaround the tunnel radius vector r; θ dip around the tunnel.

B.-G. He, Y.H. Hatzor / International Journal of Rock Mechanics & Mining Sciences 78 (2015) 118–126 121

⎡

⎣⎢⎢

⎤

⎦⎥⎥Q

cos sin 0sin cos 00 0 1 4

ij

θ θθ θ= −

( )

Employing the second-order matrix transformation of isotropictensor between Cartesian coordinate systems [e.g. Ref. [14]], weobtain:

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥⎡

⎣⎢⎢

⎤

⎦⎥⎥

5

z Q y z Q

symm

symm

r, , x, ,

cos sin 0sin cos 0

0 0 1

cos sin 0sin cos 0

0 0 1

cos sin cos sin cos sin

sin cos sin cos sin cos

cos sin 0sin cos 00 0 1

T

r r rz

z

z

x xy xz

y yz

z

x yx xy y xz yz

x yx xy y xz yz

zx zy z

ij ij

T

( )

σ θ σ

σ τ τσ τ

σ

θ θθ θ

σ τ τσ τ

σ

θ θθ θ

θσ θτ θτ θσ θτ θτ

θσ θτ θτ θσ θτ θττ τ σ

θ θθ θ

( ) = × ( ) ×

× = −

× × −

=+ + +

− + − + − +

×−

θ

θ θ

Using Eq. (5), we can now express zτθ as

sin cos 6z zx zyτ θτ θτ= − + ( )θ

3.2. Local change in sz due to tunneling

Assuming there is no excavation induced deformation alongthe tunnel axis in the z-direction, we may write

E/ 0z z x yε σ ν σ σΔ = [Δ − Δ( + )] = , where ν is Poisson's ratio. We cannow find the excavation induced change in zσ near the tunnel:

⎡⎣ ⎤⎦ 7z x y x y x y( )σ ν σ σ ν σ σ σ σΔ = Δ( + ) = + − ( ‵ + ‵) ( )

where xσ ′and yσ ′ are the local stresses near the tunnel after ex-cavation. Employing the first stress invariant, we may write:

8y x rσ σ σ σ′ + ′ = + ( )θ

Substituting Eq. (8) into Eq. (7), we can find the change in zσdue to tunneling:

⎡⎣ ⎤⎦⎡⎣ ⎤⎦

⎡⎣⎢

⎤⎦⎥

ar

ar

2 cos 2 4 sin 29

z x y

x y x y

r x y

x y xy

2

2

2

2

( )( )

( )

σ ν σ σ

ν σ σ σ σ

ν σ σ σ σ

ν σ σ θ τ θ

Δ = Δ( + )

= ( ‵ + ‵) − +

= ( + ) − +

= − − +( )

θ

Thus, the normal stress in the z direction near the tunnel afterthe excavation can be presented by the original stress components

xσ , yσ , zσ and xyτ as follows:

⎡⎣⎢

⎤⎦⎥

ar

ar

ar

ar

ar

2 cos 2 4 sin 2

2 cos 2 2 cos 2 4 sin 210

z z z z x y xy

x y xy z

2

2

2

2

2

2

2

2

2

2

( )σ σ σ σ ν σ σ θ τ θ

ν θσ ν θσ ν θτ σ

‵ = + Δ = − − +

= − + − +( )

Note that the value of Poisson's ratio is needed for character-izing the change in stress in the z-direction induced by tunneling.

3.3. Rotation of the cylindrical coordinate system about the r axis

From Eqs. (2) and (10), the stress components at the wall of acircular tunnel of radius a (r¼a) are

Table 1Concentrated results of six theoretical flat jack tests performed around a circularopening.

Test no. P1 P2 P3 P4 P5 P6

Rotation angle θ/° 55 55 90 120 335 335Rotation angle α/° 45 0 75 0 135 0Measured cancellation pressure inslot/MPa

�4.5 �6.0 �4.5 �4.0 �5.0 �3.5

B.-G. He, Y.H. Hatzor / International Journal of Rock Mechanics & Mining Sciences 78 (2015) 118–126122

⎡⎣ ⎤⎦

⎫

⎬⎪⎪

⎭⎪⎪

0

2 cos 2 4 sin 2

0

2 cos 2 4 sin 2 11

r

x y x y xy

r r

z z x y xy

σσ σ σ σ σ θ τ θ

τ τ

σ σ ν σ σ θ τ θ

== + − ( − ) −

= =

′ = − ( − ) + ( )

θ

θ θ

The stress matrix in the tunnel cross section where r¼a is

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

r zsymm symm

, ,0 0

12

r r rz

z

z

rz

z

z

σ θσ τ τ

σ τσ

τσ τ

σ( ) =

‵=

‵ ( )

θ

θ θ θ θ

We now rotate the cylindrical coordinate system about thetunnel radius vector r by an angle α (Fig. 7) in order to be able toretrieve the additional three components of the stress tensor inthe z-direction ( zσ , zxτ , zyτ ). The rotation matrix for this case is

⎡

⎣⎢⎢

⎤

⎦⎥⎥Q

1 0 00 cos sin0 sin cos 13

ij α αα α

= −( )

According to the second-order tensor (matrix) transformationrule of isotropic Cartesian coordinates, we obtain:

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

14

z Q z Q

symm

symm

r , , r, ,

1 0 00 cos sin0 sin cos

0 0 1 0 00 cos sin0 sin cos

0 0 0sin cos sin cos sin

cos sin cos sin cos

1 0 00 cos sin0 sin cos

T

r r rz

z

z

rz

z

z

zr z z z

rz z z z z

ij ij

T

( )

σ θ σ θ

σ τ τσ τ

σα αα α

τσ τ

σα αα α

ατ ασ ατ ατ ασατ ασ ατ ατ ασ σ

α αα α

( ′′ ′′ ′′) = × ( ) ×

′′ ′′ ′′′′ ′′

′′= −

×′

× −

= − − − ′+ + ′

×−

θ

θ θ

θ θ

θ θ θ

θ θ θ

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

17

1 2 cos 2 cos 2

cos 2 sin

1 cos 2 cos 2

cos 2 sin

sin 4 sin

2 cos sin

sin sin 2 cos sin 2

1 2 cos 2 cos 2

cos 2 sin

1 cos 2 cos 2

cos 2 sin

sin 4 sin

2 cos sin

sin sin 2 cos sin 2

1 2 cos 2 cos 2

cos 2 sin

1 cos 2 cos 2

cos 2 sin

sin 4 sin

2 cos sin

sin sin 2 cos sin 2

1 2 cos 2 cos 2

cos 2 sin

1 cos 2 cos 2

cos 2 sin

sin 4 sin

2 cos sin

sin sin 2 cos sin 2

1 2 cos 2 cos 2

cos 2 sin

1 cos 2 cos 2

cos 2 sin

sin 4 sin

2 cos sin

sin sin 2 cos sin 2

1 2 cos 2 cos 2

cos 2 sin

1 cos 2 cos 2

cos 2 sin

sin 4 sin

2 cos sin

sin sin 2 cos sin 2

x

y

z

xy

zx

zy

1

2

3

4

5

6

1 2 1

12

1

1 2 1

12

1

21

1 2 12

1

1 1 1 1

2 2 2

22

2

2 2 2

22

2

22

2 2 22

2

2 2 2 2

3 2 3

32

3

3 2 3

32

3

23

3 2 32

3

3 3 3 3

4 2 4

42

4

4 2 4

42

4

24

4 2 42

4

4 4 4 4

5 2 5

52

5

5 2 5

52

5

25

5 2 52

5

5 5 5 5

6 2 6

62

6

6 2 6

62

6

26

6 2 62

6

6 6 6 6

( )

( )

( )

( )

( )

( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )( )

σσσσσσ

θ α ν

θ α

θ α ν

θ α

α

θ α ν α

θ α θ α

θ α ν

θ α

θ α ν

θ α

α

θ α ν α

θ α θ α

θ α ν

θ α

θ α ν

θ α

α

θ α ν α

θ α θ α

θ α ν

θ α

θ α ν

θ α

α

θ α ν α

θ α θ α

θ α ν

θ α

θ α ν

θ α

α

θ α ν α

θ α θ α

θ α ν

θ α

θ α ν

θ α

α

θ α ν α

θ α θ α

σσστττ

=

− − + + −

+

−

− − + + −

+

−

− − + + −

+

−

− − + + −

+

−

− − + + −

+

−

− − + + −

+

−

θ

θ

θ

θ

θ

θ

Solving Eq. (14), the desired σ ′′θ can therefore be expressed as

⎡⎣ ⎤⎦⎡

⎣⎢⎢

⎤

⎦⎥⎥

15

sin cos sin cos sin0cos

sin

cos sin 2 sin

zr z z z

z z2 2 ( )

σ ατ ασ ατ ατ ασ αα

σ ασ ατ ασ

′′ = − − − ′ ⋅−

′′ = − + ′

θ θ θ θ

θ θ θ

Substituting Eq. (6) and the expressions for σθ and zσ ′ shown inEq. (11) into Eq. (15), we obtain the desired solution for σ ′′θ which

includes all six components of the stress tensor:

⎡⎣ ⎤⎦⎡⎣ ⎤⎦

16

1 2 cos 2 cos 2 cos 2 sin

1 cos 2 cos 2 cos 2 sin sin 4 sin

2 cos sin sin sin 2 cos sin 2

z

xy zx zy

2 2x

2 2y

2

2 2( ) ( )

σ θ α ν θ α σ

θ α ν θ α σ ασ

θ α ν α τ θ ατ θ ατ

′′ = ( − ) −

+ ( + ) + + −

× + + −

θ

To conclude this section, we show that by performing six flatjack tests that provide σ ′′θ at different positions and orientationsaround the tunnel we can recover the six components of the in-situ stress tensor ( xσ , yσ , zσ , xyτ , zxτ , zyτ ) by means of the standardFlat Jack method. Although from a theoretical standpoint only sixtests are required, it would be prudent to perform more than sixtests in field applications, in case one or two tests were not per-formed strictly according to the suggested methods, or have failed.

4. Worked example

Consider the flat jack testing array shown in Fig. 9. Assumingthe material is isotropic we need to solve a six-variable linearequation as shown in Eq. (17) below utilizing a software packagesuch as MATLAB, for example.

In Table 1, the results of six theoretical flat jack tests are pre-sented for four different θ angles and four different α angles. Wewill assume here that Poisson's ratio is ν¼0.3. Note that the po-sitions and rotation angles (θ, α) are displayed in Fig. 9.

Using the data obtained in the six measurement points aroundthe tunnel the complete far field stress tensor may be recovered bysolving a system of linear equations as shown in Eq. (17). For thetest results shown in Table 1 the obtained in-situ stress tensor is

-6

-5

-4

-3

-2

-1

00.1 0.2 0.3 0.4

Prin

cipa

l Str

esse

s σ(M

pa)

Poisson's Ratio ν

σ1 σ2 σ3

Fig. 10. Sensitivity of analytical solution to the value of Poisson's ratio.

B.-G. He, Y.H. Hatzor / International Journal of Rock Mechanics & Mining Sciences 78 (2015) 118–126 123

⎡

⎣⎢⎢⎢

⎤

⎦⎥⎥⎥symm

2.4993 0.2889 0.02082.2318 1.4312

4.2953 18

σ =−

− −− ( )

The magnitude and orientation of the principal stresses arereadily determined by finding the eigenvalues and eigenvectors ofEq. (18):

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥ 19

n

n n n

5.0369 0 00 2.5487 00 0 1.4409

0.0595 0.9716 0.22920.4585 0.1773 0.87080.8867 0.1569 0.4350

1,2,3

1

2

3

1,2,3

1 2 3

( )

σσ

σσ

=− ( )

− ( )− ( )

= −−

−

A sensitivty analysis of the analytical solution with respect tothe value of Poisson's ratio reveals that it is not very sensitive tothe value of ν (Fig. 10).

5. Summary and conclusions

A complete analytical solution has been derived for obtainingthe six components of the far field in-situ stress tensor based onsix flat jack tests performed around a circular tunnel. In the ana-lysis, we assume a Continuous, Homogenous, Isotropic, LinearElastic (CHILE) rock mass. No assumptions need to be made withrespect to the orientation of the initial far field stresses and theslots do not have to perpendicular to the tunnel axis. We haveshown that a minimum of three different θ positions around thetunnel axis and a minimum of three α inclinations around thetunnel radius vector are required for obtaining the complete stresstensor. The proposed analytical solution is not very sensitive to the

y

σx

σy

x

τxy

θ

y

σx

xθ

Fig. A1. Superposition of stresses. (a) Natural stress, (b) h

assumed value of Poisson’s ratio for the rock mass near the tunnel.

Acknowledgments

We thank the Israel Commission for Higher Education for a post-doctoral fellowship awarded for excellent doctoral students fromChina (No. 850203241) through Ben-Gurion University of the Negev.Two anonymous reviewers are thanked for critical reading of an earlyversion of this technical note and for their insightful comments.

Appendix. A. Analytical solution for the stress components incross section of a circular tunnel using superposition of farfield stresses

To obtain the stress distribution around a hole in an infiniteelastic plate consisting of isotropic material we will investigate thecontribution of each of the stresses separately using the principleof superposition (Fig. A1).

A.1.. Horizontal tension

Fig A1(b) represents a plate with a hole submitted to a hor-izontal uniform tension xσ in the x-direction. The original andprimed coordinate systems shown in Fig. A2 establish the anglesbetween the various axes. The rotation matrix is given by [e.g. Ref.[14]]:

⎡⎣⎢

⎤⎦⎥Q cos sin

sin cos A. 1ij

θ θθ θ

=− ( )

Transformation of field variables between rectangular and cy-lindrical coordinates [14] gives in our case:

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢⎢

⎤⎦⎥⎥

⎡⎣⎢

⎤⎦⎥ A. 2

Q Q

cos sinsin cos

00 0

cos sinsin cos

cos 0

sin 0cos sinsin cos

r r

rx y ij

T

x

x

x

ij ,

T

( )

σ ττ σ σ

θ θθ θ

σ θ θθ θ

θσθσ

θ θθ θ

= × ×

=− −

=−

−

θ

θ θ ( )

Solving Eq. (A.2) for the case of horizontal uniform tension, weobtain:

cos 1 cos 2 cos 2

sin cos sin 2 A. 3

r x x x x

r x x

2 12

12

12

12

σ θσ θ σ σ θσ

τ θ θσ θσ

= = ( + ) = +

= − = − ( )θ

The result obtained in Eq. (A.3) can be arrived at by super-position of two cases, the first due to the normal constant com-ponent 1/2 xσ( ) (Fig. A3b) and the other due to the normal stress

yσy

xθ

y

x

τxy

θ

orizontal stress, (c) vertical stress, and (d) Pure shear.

y

θ x

θ

B.-G. He, Y.H. Hatzor / International Journal of Rock Mechanics & Mining Sciences 78 (2015) 118–126124

1/2 cos 2 xθσ( ) and the shear stress 1/2 sin 2 xθσ( − ) components(Fig. A3c).

The equation of compatibility [15] is as follows:

⎛⎝⎜

⎞⎠⎟

ddr r

ddr r

dd

1 10

A. 44

2

2 2

2

2

2

ϕθ

ϕ∇ = + + =( )

where ϕ is a general stress function. In the case of horizontaluniform tension (Fig. A3(b)) where the stress function depends onr only, the general solution of Eq. (A.4) can be resolved by means ofEuler differential equation and MATLAB to obtain the followinggeneral stress function:

A r Br r Cr Dlog log A. 52 2ϕ = + + + ( )

Using the stress function, we can obtain the stress and thedisplacement components:

⎫

⎬⎪⎪⎪

⎭⎪⎪⎪

r r rAr

B r C

rAr

B r C

1 11 2 log 2

3 2 log 2

0 A. 6

r

r

2

2

2 2

2

2 2

σ φ φθ

σ φ

τ

= ∂∂

+ ∂∂

= + ( + ) +

= ∂∂

= − + ( + ) +

= ( )

θ

θ

uBrE

Hr I K4

sin cos A.7θ θ θ= + − + ( )θ

Note that in the displacement uθ , the term Br E4 /θ is not single-valued after making a complete circle around the ring. Therefore,we let B═0 and rσ in Eq. (A.6) reduces to A r C/ 2r

2σ = + . From theboundary conditions in Fig. A3(b) ( 0r r aσ =( = ) ; 1/2r r b xσ σ= ( )( = ) ) we

obtain the constants A a /2 x2 σ= − ( ) and C 1/4 xσ= ( ) for this case.

For the far field stresses we get:

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

ar

ar2

12

1 0A. 8

rx

rx 2

2

2

2σ

σσ

στ= − = + =

( )θ θ

Considering now the contribution of the normal force1/2 cos 2 xθσ( ) and the shearing force 1/2 sin 2 xθσ−( ) in Fig. A3(c),we may assume a stress function such as:

f r cos 2 A. 9φ θ= ( ) ( )

Substituting Eq. (A.9) into the general form of the compatibility

Fig. A2. Coordination transformation.

Fig. A3. Stress state of circular hole under horizontal stress in plate. (a) Str

Eq. (A.4), we can express the general solution of Eq. (A.9) as fol-lows:

⎛⎝⎜

⎞⎠⎟Ar Br C

rD

1cos 2

A. 102 4

2ϕ θ= + + +

( )

Combined with a stress function [15], the corresponding stresscomponents are:

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎫

⎬

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

r r rA

C

r

Dr

rA Br

C

r

r rA Br

C

r

Dr

1 12

6 4cos 2

2 126

cos 2

12 6

6 2sin 2

A. 11

r

r

2

2

2 4 2

2

22

4

24 2

σ φ φθ

θ

σ φ θ

τ φθ

θ

= ∂∂

+ ∂∂

= − + +

= ∂∂

= + +

= ∂∂

∂∂

= + − −( )

θ

θ

The constants of integration can be determined by using theboundary conditions as shown in Fig. A3(c), 0r r aσ =( = ) ;

1/2 cos 2r r b xσ θσ= ( )( = ) ; 0r r aτ =θ ( = ) ; 1/2 sin 2r r b xτ θσ= − ( )θ ( = ) ):

A B Ca

Da1

40

4 2 A. 12x x x

2 2σ σ σ= − = = − = ( )

Substituting these constants into Eq. (A.11), and adding thestresses produced by the uniform tension 1/2 xσ( ) on the outerboundary calculated from Eq. (A.8), we obtain the stress compo-nents of Fig. A3(b) due to horizontal tension:

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎫

⎬

⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪

ar

ar

ar

ar

ar

ar

ar

21

21 1

3cos 2

21

21

3cos 2

21 1

3sin 2

A. 13

rHorizontalstress x

2

2

x2

2

2

2

Horizontalstress x2

2x

2

2

rHorizontalstress

ra x

2

2

2

2

σσ

σθ

σσ σ

θ

τ τσ

θ

= −

+ − −

= + − +

= = − − +( )

θ

θ θ

A.2. Vertical tension

Consider now a plate with a hole submitted to a vertical uni-form tension yσ (Fig. A1c). Transformation of field variables be-tween rectangular and cylindrical coordinates (Fig. A2), gives inthis case [14]:

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢⎢

⎤⎦⎥⎥

⎡⎣⎢

⎤⎦⎥ A. 14

Q Q

cos sinsin cos

0 00

cos sinsin cos

0 sin

0 coscos sinsin cos

r r

rij x y ij

T

y

T

y

y

,

( )

σ ττ σ σ

θ θθ θ σ

θ θθ θ

θσ

θσθ θθ θ

= × ×

=− −

= −

θ

θ θ ( )

ess at radius b. (b) Normal forces. (c) Normal force and shearing force.

Fig. A4. Stress state of circular hole under vertical stress in plate. (a) Stress at radius b. (b) Normal forces. (c) Normal force and shearing force.

Fig. A5. Corresponding stress as uniform far-field tension and shearing loading.

B.-G. He, Y.H. Hatzor / International Journal of Rock Mechanics & Mining Sciences 78 (2015) 118–126 125

Solving Eq. (A.14) in terms of cylindrical coordinates, we get:

sin12

1 cos 212

12

cos 2

sin cos12

sin 2 A. 15

r y y y y

r y y

2σ σ θ σ θ σ σ θ

τ σ θ θ σ θ

= = ( − ) = −

= = ( )θ

The result obtained in Eq. (A.15) can be arrived at by super-position of two contributions as shown in Fig. A4(b) and(c) namely, the first due to the normal component 1/2 yσ( ) and theremaining part due to the normal stress 1/2 cos 2yσ θ−( ) and theshear stress 1/2 sin 2yσ θ( ) components.

The constant component 1/2 yσ( ) of the normal forces in Fig. A4(b) is similar to Eq. (A.8) in Fig. A3(b). Therefore, we obtain:

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

ar

ar2

12

1 0A. 16

ry

ry 2

2

2

2σ

σσ

στ= − = + =

( )θ θ

Considering now the contribution of the normal force1/2 cos 2yσ θ−( ) and the shearing force 1/2 sin 2yσ θ( ) in Fig. A4(c),

we assume stresses that may be derived from a stress functionwhich is of the same form as of Eq. (A.9). Therefore, the corre-sponding stress components around the hole are identical to Eq.(A.11).

The constants of integration for this case can be determined bysolving the boundary conditions as shown in Fig. A4(c), 0r r aσ =( = ) ;

1/2 cos 2r r b yσ θσ= − ( )( = ) ; 0r r aτ =θ ( = ) ; 1/2 sin 2r r b yτ θσ= ( )θ ( = ) :

A B Ca

Da1

40

4 2 A. 17y y y

4 2σ σ σ= = = = − ( )

Substituting the constants of integration into Eq. (A.11) andadding the stresses produced by the uniform tension 1/2 yσ( ) onthe outer boundary calculated from Eq. (A.16), we obtain the stresscomponents due to vertical tension in Fig. A1(c):

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎫

⎬

⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪

ar

ar

ar

ar

ar

ar

ar

21

21 1

3cos 2

21

21

3cos 2

21 1

3sin 2

A. 18

rVerticalstress y

2

2

y2

2

2

2

Verticalstress y2

2y

2

2

rVerticalstress y

2

2

2

2

σσ

σθ

σσ σ

θ

τσ

θ

= −

− − −

= + + +

= − +( )

θ

θ

A.3. Pure shear

Fig. A1(d) represents a plate with a hole submitted to pureshear xyτ . Transformation of field variables between rectangularand cylindrical coordinates (Fig. A2) gives [14]:

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢⎢

⎤⎦⎥⎥⎡⎣⎢

⎤⎦⎥

⎡⎣⎢⎢

⎤⎦⎥⎥

⎡⎣⎢

⎤⎦⎥

A. 19

cos sinsin cos

0

0cos sin

sin cos

sin cos

cos sincos sinsin cos

sin 2

cos 2

r r

r

xy

yx

T

yx xy

yx xy

r xy

r xy } ( )

σ ττ σ

θ θθ θ

τ

τθ θ

θ θ

θτ θτ

θτ θτθ θθ θ

σ τ θ

τ τ θ

=−

×−

=−

−

=

=

θ

θ θ

θ

Fig. A5 shows a large plate in a state of pure shear xyτ , perturbedby a hole of radius a. Due to the normal stress sin 2r xyσ τ θ= andshearing stress cos 2r xyτ τ θ=θ in Fig. A5, we assume stresses thatmay be derived from a stress function of the form:

f r sin 2 A. 20ϕ θ= ( ) ( )

Substituting the stress function from Eq. (A.20) into the com-patibility Eq. (A.4), we can express the general solution of Eq.(A.20) as:

⎛⎝⎜

⎞⎠⎟Ar Br C

rD

1sin 2

A. 212 4

2ϕ θ= + + +

( )

Combined with a stress function [15], the corresponding stresscomponents are:

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎫

⎬

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

r r rA

C

r

Dr

rA Br

C

r

r rA Br

C

r

Dr

1 12

6 4sin 2

2 126

sin 2

12 6

6 2cos 2

A. 22

r

r

2

2

2 4 2

2

22

4

24 2

σ φ φθ

θ

σ φ θ

τ φθ

θ

= ∂∂

+ ∂∂

= − + +

= ∂∂

= + +

= − ∂∂

∂∂

= + − −( )

θ

θ

The constants of integration can be determined by the

B.-G. He, Y.H. Hatzor / International Journal of Rock Mechanics & Mining Sciences 78 (2015) 118–126126

boundary conditions shown in Fig. A5, 0r r aσ =( = ) ;sin 2r r b xyσ τ θ=( = ) ; 0r r aτ =θ ( = ) ; cos 2r r b xyτ τ θ=θ ( = ) :

A B Ca

D a12

02 A. 23xy xy xy

42τ τ τ= − = = − = ( )

Substituting the constants of integration into Eq. (A.22), weobtain the stress components due to pure shear stress:

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

ar

ar

a

r

ar

ar

1 13

sin 2

13

sin 2

1 13

cos 2A.24

rPureshear

xy

Pureshearxy

rPureshear

xy

2

2

2

2

4

4

2

2

2

2

σ τ θ

σ τ θ

τ τ θ

= − −

= − +

= − +( )

θ

θ

The result obtained here for pure shear is identical to the ex-pression obtained in Ref. [16] using the Fourier method.

Finally, subsequent to the analysis of horizontal, vertical andshear stress respectively, the state of stress around the hole can beobtained by superposition as follows:

Eqs. A. 13 Eqs. A.18 Eqs. A. 24 A. 25

Horizontalstress Verticalstress Purestressσ σ σ σ= + += ( ) + ( ) + ( ) ( )

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎫

⎬

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪A. 26

a

r

a

r

a

r

a

r

a

r

a

r

a

r

a

r

a

r

a

r

a

r

a

r

21

21 1

3cos 2

1 13

sin 2

21

21

3cos 2

13

sin 2

21 1

3sin 2 1

13

cos 2

r

xy

xy

xy

x y 2

2

x y 2

2

2

2

2

2

2

2

x y 2

2

x y 2

2

4

4

r r

x y 2

2

2

2

2

2

2

2 ( )

σσ σ σ σ

θ

τ θ

σσ σ σ σ

θ

τ θ

τ τ

σ σθ τ

θ

=+

− +−

− −

+ − −

=+

+ −−

+

− +

=

= −−

− + + −

+

θ

θ θ

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