stress n rock masses

Stresses in Rock Stresses in Rock MassesMasses

Geomechanics (EBS 4173)

Hareyani Zabidi

Types of stresses in rockTypes of stresses in rock

Stress determination Stress determination methodsmethods

Results from stress Results from stress determinationdetermination

To learnTo learn

Why stress?Why stress?

Pre-existing stress in the groundPre-existing stress in the ground, , and need to understand it; and need to understand it; duringduring engineeringengineering projectproject ( (tunnelingtunneling), ), pre-pre-existingexisting stressesstresses isis disturbeddisturbed

Engineering applied – stress state Engineering applied – stress state can be changed dramaticallycan be changed dramatically, the , the rock previously contained stresses , has rock previously contained stresses , has been removed and taken to somewhere been removed and taken to somewhere elseelse

Stress is a tensor quantityStress is a tensor quantity, not , not familiar subject to encounteredfamiliar subject to encountered

In-situIn-situ Stress Stress The importance for rock The importance for rock

engineeringengineeringWhy to determine?Why to determine?

Basic knowledge of the stress Basic knowledge of the stress state for engineering (direction, state for engineering (direction, magnitude)magnitude)

Effect of stress against Effect of stress against engineering structuresengineering structures

Direction of breakingDirection of breaking Groundwater flowGroundwater flow

In-situIn-situ Stress Stress

Determination of Determination of in-situin-situ stress stress Expectation of Expectation of in-situin-situ stress stress

STRESSES IN ROCK STRESSES IN ROCK MASSESMASSES

•Rock at depth is subjected to Rock at depth is subjected to stresses resulting from the weight of stresses resulting from the weight of the overlying strata and from locked the overlying strata and from locked in stresses of tectonic origin.in stresses of tectonic origin.

•When an opening is excavated in When an opening is excavated in this rock, the stress field is locally this rock, the stress field is locally disrupted and a new set of stresses disrupted and a new set of stresses are induced in the rock surrounding are induced in the rock surrounding the opening.the opening.

Figure of stresses induced in the rock Figure of stresses induced in the rock surrounding a horizontal circular tunnel. surrounding a horizontal circular tunnel. (ref : (ref : in situ in situ and induced stresses)and induced stresses)

Excavation of Excavation of tunneltunnel – no new – no new loads are applied in loads are applied in unsupported unsupported excavationsexcavations

Pre-existing stresses Pre-existing stresses are distributed by are distributed by engineering activityengineering activity

Result- stresses Result- stresses increase in some increase in some areas, and contrast in areas, and contrast in other areasother areas

Knowledge of the magnitude and Knowledge of the magnitude and direction of these direction of these in situin situ and and induced stresses is an essential induced stresses is an essential component of underground component of underground excavation design.excavation design.

In many cases, the strength of the In many cases, the strength of the rock is exceeded and the resulting rock is exceeded and the resulting instability can have serious instability can have serious consequences on the behaviour of consequences on the behaviour of the excavations. the excavations.

TYPES OF STRESSES IN TYPES OF STRESSES IN ROCKROCK

Stresses in rock can be Stresses in rock can be grouped according to grouped according to origin into:origin into:

i.i. Natural stressesNatural stresses

ii.ii. Induced stressesInduced stresses

Natural stressesNatural stresses are those are those stresses found in rock before stresses found in rock before excavation :excavation :

1-1- gravitational stressesgravitational stresses

2-2- tectonic stressestectonic stresses

3-3- residual stressesresidual stresses

4-4- thermal stressesthermal stresses

Induced stressesInduced stresses are those are those that occured as the result of that occured as the result of stress changes due to stress changes due to manmade excavations.manmade excavations.

1.1. Gravitional StressesGravitional Stresses

Gravity stresses (Gravity stresses (σvv)/vertical )/vertical stress resulted from the weight of stress resulted from the weight of the overburden/overlying stratathe overburden/overlying strata

σvv = = ρρggzz

where where σv v = = vertical stressvertical stress

ρρ = density = mass/volume = density = mass/volume z = depthz = depth g = accelerationg = acceleration

Vertical stress component Vertical stress component increases in magnitude as the increases in magnitude as the depth below the ground surface depth below the ground surface increases, weight of the overburden increases, weight of the overburden At shallow depths, the actual value At shallow depths, the actual value is much lessis much less

•1 MPa 1 MPa is induced by is induced by 40m40m of of overlying rockoverlying rock• 1 psi 1 psi is induced by is induced by 1 ft 1 ft of of overlying rock overlying rock

The density of common rocks such The density of common rocks such as quartz-sandstone, limestone, as quartz-sandstone, limestone, quartz-rich magnetic rocks = quartz-rich magnetic rocks = 2670 2670 kg/mkg/m33

The vertical component/stress at a The vertical component/stress at a depth of 1000 m, as follows;depth of 1000 m, as follows;

σvv = = ρρggz z

= 2670 x 9.8 x 1000 = 2670 x 9.8 x 1000

= = 26 MPa26 MPa

The gradient of stress over 1000 The gradient of stress over 1000 m due to gravity = m due to gravity = 0.026 MPa/m.0.026 MPa/m.

For rocks such as For rocks such as basic basic magmatic rocksmagmatic rocks, metamorphic , metamorphic rocks having a density of 3000 rocks having a density of 3000 kg/mkg/m33, the gradient will increase to , the gradient will increase to = = 0.029 MPa/m.0.029 MPa/m.

Horizontal stresses

2. 2. Horizontal stressesHorizontal stresses

The horizontal stresses acting on an The horizontal stresses acting on an element of rock at a depth of, element of rock at a depth of, zz below below the surface are much more difficult to the surface are much more difficult to estimate then the vertical stresses. estimate then the vertical stresses.

If the material in the earth crust is If the material in the earth crust is considered strictly elastic and no considered strictly elastic and no lateral strain was permitted during lateral strain was permitted during formation of the overburden, the formation of the overburden, the horizontal stresses =horizontal stresses =

σHH = v = v σvv

1-v1-v v = Poisson’s ratiov = Poisson’s ratio

For example For example σHH = 0.25 = 0.25 σHH = 1 = 1 σvv

1- 0.25 3 1- 0.25 3 For most rock, poisson’s ratio varies For most rock, poisson’s ratio varies between 0.15 – 0.35. between 0.15 – 0.35.

The common value is 0.25 which The common value is 0.25 which defines the defines the σHH as 1/3 of the as 1/3 of the σvv

The rock material cannot sustain The rock material cannot sustain shear stresses on a long term basis.shear stresses on a long term basis.

The horizontal stress will reach the The horizontal stress will reach the magnitude of the vertical stress after magnitude of the vertical stress after a period of time. This is known as a period of time. This is known as ‘lithostatic’.‘lithostatic’.

σσHH = = σσvv v = 0.5, v = 0.5, σHH == 0.5 ; 0.5 ; σH = H =

σvv

1- 0.51- 0.5

Lithostatic stress can be found Lithostatic stress can be found

• In areas where sedimentation is In areas where sedimentation is ongoing.ongoing.

• The sediments with high water The sediments with high water contentcontent

• At great depth in the earth crustAt great depth in the earth crust

The force exerted on a rock buried deep within the Earth by overlying rocks.

Because lithostatic pressure is exerted equally from all sides of a rock, it compresses the rock into a smaller, denser form without altering the rock's shape.

Thus the horizontal stress derived from Thus the horizontal stress derived from gravity can be expressed by the factor gravity can be expressed by the factor ‘k’ ‘k’ (lithology factor).(lithology factor).

σHH = = kk..σvv

with 0 < with 0 < kk < 1 < 1

kk varies from 0 with no lateral restraint to varies from 0 with no lateral restraint to kk = 1 for a lithostatic stress field = 1 for a lithostatic stress field

kk = 1 for fluids = 1 for fluidsk k < 1 = rigid material such as rock; shale , < 1 = rigid material such as rock; shale , k k = 0.37= 0.37

High horizontal stressHigh horizontal stress

Caused by many factors: Caused by many factors: erosion, erosion, tectonics, rock anisotropy, local tectonics, rock anisotropy, local effect of discontinuities effect of discontinuities

The earth is not static. It always The earth is not static. It always experiences movement in the earth’s experiences movement in the earth’s crust continuously. crust continuously.

E.g; (Mid-Atlantic Range) E.g; (Mid-Atlantic Range) continents of both sides of the atlantic continents of both sides of the atlantic are moving apart away from the Atlantic are moving apart away from the Atlantic rift systemrift system

Tectonic stresses result from tectonic Tectonic stresses result from tectonic activitiesactivities

High horizontal stress is due to High horizontal stress is due to tectonic forcestectonic forces

Earthquake along the San Andreas (1906 – 1989), recorded high shear stress and result from tectonic activity

Subduction zone off the coast of Chile, near the Andes, expected to have high horizontal normal stresses. Proven by rock slopes instability & rockburst – surface & underground mines

Subduction Subduction zone off zone off the west the west coast of coast of South South America – America – high high horizontal horizontal in-situ in-situ stressesstresses

Normal faultNormal faultVertical stress (Vertical stress (σv) is the ) is the max principal stress (max principal stress (σ1))σv > σH > σh

Thrust faultThrust faultHorizontal stress (Horizontal stress (σH) is ) is the max principal stress the max principal stress ((σ1))σH > σh > σv Strike-slip faultStrike-slip faultHorizontal stress (Horizontal stress (σH) is ) is the max principal stress the max principal stress ((σ1))σH > σv > σh

σ1=

v σ3=

h

σ2=

H

σv=

2 σ1=

H σ3=

h

σ3=

v

σ2=

h

σ1=

H

3.3. Residual StressesResidual Stresses

Residual stresses are stresses Residual stresses are stresses that remain ‘locked in’ after that remain ‘locked in’ after rock is removed from the rock is removed from the groundground

ErosionErosion

An increase in the An increase in the KK-value or -value or horizontal stress caused by erosionhorizontal stress caused by erosion

Removal of the overburden and the Removal of the overburden and the consequential effect on both the consequential effect on both the vertical and horizontal stresses – vertical and horizontal stresses – result high result high kk-value-value

Locked horizontal stresses/lower Locked horizontal stresses/lower vertical stresses (near to the vertical stresses (near to the surface) = higher surface) = higher kk-value-value

4.4. Thermal StressesThermal Stresses

Stresses induced by natural or Stresses induced by natural or man-made phenomena that man-made phenomena that cause thermal expansion or cause thermal expansion or contraction of the rockcontraction of the rock

5.5. Induced Stresses Induced Stresses

Induced stresses are the result Induced stresses are the result of excavation activity and of excavation activity and therefore are of great concern therefore are of great concern in underground designin underground design

σ1

STRESS STRESS

Stress Stress is defined asis defined as force over force over areaarea and according to Newton, a and according to Newton, a force force (F)(F) is defined as the product of is defined as the product of mass mass (m)(m) times times accelaration accelaration (a)(a)..

F = m.aF = m.a

In the International System In the International System of Units (SI units), of Units (SI units), forceforce is is defined in defined in Newton (N).Newton (N).

F = 1 N = kg . F = 1 N = kg . mm ss2 2

On earth, the acceleration On earth, the acceleration due to gravity of a = g = due to gravity of a = g = 9.8 m/s9.8 m/s22, and a kg of mass , and a kg of mass create a force (weight) of:create a force (weight) of:

F(earth) = 1 kg x 9.8 F(earth) = 1 kg x 9.8 m/sm/s22 = = 9.8 N9.8 N

If a force of 1 N is acting over an If a force of 1 N is acting over an area of 1 marea of 1 m22, this stress is called 1 , this stress is called 1 pascal (Pa). pascal (Pa).

From engineering point of view From engineering point of view this is relatively small stress. this is relatively small stress. Generally, it is preferable to deal Generally, it is preferable to deal with 10with 1066 Pa = 1 Mpa. Pa = 1 Mpa.

Stress componentsStress components

FFFFnn

FFss

Existance of normal forces, Existance of normal forces, Fn and and shear forces, shear forces, Fs on a plane on a plane

Shear force + normal force = stress Shear force + normal force = stress tensortensor

The normal and shear stress – the The normal and shear stress – the normal and shear forces per unit areanormal and shear forces per unit area

Force component, Force component, FFnn in direction in direction өө from from FF

FFnn = F cos = F cos өө && FFss = F sin = F sin өө; only force ; only force resolvedresolved

Force vs. Force vs. stressstress

Normal stress component, Normal stress component, σnn in same in same directiondirection

σnn = = σ cos cos22 өө; force and area resolved; force and area resolved

Shear force + normal force = stress Shear force + normal force = stress tensortensor

Force vs. Force vs. stressstress

Stress components on a Stress components on a small cubesmall cube

Normal stress – directly evident; shear stress - Normal stress – directly evident; shear stress - indirectindirect

Shear force resolved into 2 perpendicular Shear force resolved into 2 perpendicular components; aligned with 2 axes parallel to components; aligned with 2 axes parallel to the edgesthe edges

Resolved = 3 normal components ; 6 shear Resolved = 3 normal components ; 6 shear componentscomponents

Consider normal & shear components in rectangular x-y-z

3 visible faces of a cubeEquilibrium state

Stress components on a Stress components on a small cubesmall cube

Development of two shear stressesDevelopment of two shear stresses

Shear force resolved into 2 perpendicular Shear force resolved into 2 perpendicular components; aligned with 2 axes parallel to components; aligned with 2 axes parallel to the edgesthe edges

Resolved = 3 normal components ; 6 shear Resolved = 3 normal components ; 6 shear componentscomponents

Stress matrixStress matrix

Stress components:Stress components: Row = components on any plane; axis perpendicular Row = components on any plane; axis perpendicular

to it, e.g., to it, e.g., ττzyzy acts on a plane perpendicular to the z-axis

Column = components acting in any given directionColumn = components acting in any given direction

σxxxx ττxyxy

ττyy

zzττzxzx

ττxzxz

ττyzyz

ττyy

zz

σzzzz

σyyyy

ττxy xy x =x = plane on which the comp actsplane on which the comp actsy =y = direction in which the stress compdirection in which the stress comp actsacts

Stress equilibriumStress equilibrium

Nine separate stress components at a pointNine separate stress components at a point Body is in equilibrium; forces and moments at all Body is in equilibrium; forces and moments at all

equilibriumequilibrium Should inspect the equilibrium of forces at apoint in Should inspect the equilibrium of forces at apoint in

terms of these stress componentsterms of these stress components 4 stress components acting on the edges of a small 4 stress components acting on the edges of a small

squaresquare

Principal stressesPrincipal stresses

Principal stressesPrincipal stresses Stress components = 3 normal Stress components = 3 normal

stresses + 3 shear stressesstresses + 3 shear stresses

Max and min values of normal Max and min values of normal stresses = shear components equal stresses = shear components equal to zeroto zero

Principal stresses = Principal stresses = normal normal components of stress act at planes, components of stress act at planes, that the shear stress components that the shear stress components are zeroare zero

σσ1, 1, σσ2, 2, σσ3 = principal stresses3 = principal stressesUnsupported excavation surfaces Unsupported excavation surfaces == pricipalpricipalstress planesstress planes

Unsupported excavated Unsupported excavated surfacesurface

Force component, Fn in direction Force component, Fn in direction from F = cos from F = cos өө

Shear force + normal force = stress Shear force + normal force = stress tensortensor

In-situ stress In-situ stress determinationdetermination

ISRM method of stress ISRM method of stress determinationdetermination1. 1. Flatjack test Flatjack test

2. 2. Hydraulic fracturing Hydraulic fracturing

3. 3. United States Bureau of Mines United States Bureau of Mines (USBM) (USBM) overcoring torpedoovercoring torpedo

4. 4. Commonwealth Scientific and Commonwealth Scientific and Industrial Industrial Research Organization Research Organization (CSIRO) overcoring (CSIRO) overcoring gauge gauge

The four ISRM suggested methods for rock stress

determination and their ability to determine the components of

the tensor with one application of the particular method

σxxxxττxyxy ττxzxz

ττyzyz

σzzzz

σyy

yy

σxxxxττxyxy ττxzxz

ττyzyz

σzzzz

σyy

yy

σxxxxττxyxy ττxzxz

ττyzyz

σzzzz

σyy

yy

σxxxxττxyxy ττxzxz

ττyzyz

σzzzz

σyy

yy

One normal One normal stress stress component component determined, determined, say parallel to say parallel to xx-axis-axis

Principal Principal stresses stresses assumed assumed parallel to parallel to axes, i.e. axes, i.e. plane of the plane of the fracture, say fracture, say σ11& & σ3,3, one one estimatedestimated, , σ22All 6 All 6 components components determined determined from 6 (or from 6 (or more) more) measurements measurements of strain at of strain at one timeone time

3 components 3 components in 2-D in 2-D determined determined from 3 from 3 measurements measurements of borehole of borehole diameter diameter changechange

1. Flatjack1. Flatjack

3. USBM overcoring 3. USBM overcoring torpedotorpedo

2. Hydraulic 2. Hydraulic fracturingfracturing

4. CSIRO overcoring 4. CSIRO overcoring gaugegauge

STRESS DETERMINATION STRESS DETERMINATION METHODSMETHODS

1.1. Stress Compensation Methods / Stress Compensation Methods / Flatjack testFlatjack test

2 pins are installed at a suitable point in 2 pins are installed at a suitable point in the wall of an underground excavation the wall of an underground excavation and the distance, and the distance, dd, between those pins , between those pins is measuredis measured

Then, a slot is made with overlapping Then, a slot is made with overlapping holes or a rock saw and as the slot holes or a rock saw and as the slot closes, closes,

If normal stress is compressive, these If normal stress is compressive, these pins converge and their final position is pins converge and their final position is measured again.measured again.

A flatjack or hydraulic cushion of mild A flatjack or hydraulic cushion of mild steel, comprised of 2 metal sheets steel, comprised of 2 metal sheets placed together.placed together.

The tool is installed in the slot and The tool is installed in the slot and tightly packed with cement. It carries tightly packed with cement. It carries two connections for hydraulic lines so two connections for hydraulic lines so that a Bourdon gauge can be fittedthat a Bourdon gauge can be fitted

On pressurizing the flatjack, the pins On pressurizing the flatjack, the pins will move apartwill move apart

Pin separation distance reaches the Pin separation distance reaches the value it had before the slot was cut – value it had before the slot was cut – pre-existing normal stresspre-existing normal stress

The flatjact is set under enough The flatjact is set under enough pressure to move the previously pressure to move the previously installed pins back to their original installed pins back to their original position.position.

Figure of flatjack methodFigure of flatjack method

flatjack test configurationflatjack test configurationThe flatjack test in The flatjack test in progressprogress

2.2.Hydraulic FracturingHydraulic Fracturing

Provide 2 items of information – the breakdown Provide 2 items of information – the breakdown pressure (major principal stress: pressure (major principal stress: σσ11)and the )and the shut-in pressure (minor principal stress: shut-in pressure (minor principal stress: σσ33))

Advantage Advantage - allows the determination of - allows the determination of ground stress levels in deep drill holes. ground stress levels in deep drill holes. Originated from the oil industry.Originated from the oil industry.

DisadvantegeDisadvantege – assumptions have to be – assumptions have to be made in order to complete stress tensor: made in order to complete stress tensor:

1.1. Principal stress Principal stress – parallel and perpendicular – parallel and perpendicular to the boreholes axis to the boreholes axis

2.2. Vertical principal stress Vertical principal stress – can be – can be estimated from the depth of overburdenestimated from the depth of overburden

This method allows the determination of ground This method allows the determination of ground stress levels in deep drill holes. Originated stress levels in deep drill holes. Originated from the oil industry.from the oil industry.

The equipment comprises bore hole tools to set The equipment comprises bore hole tools to set packers, a flow pump, injection fluids and packers, a flow pump, injection fluids and borehole inspection units to determine borehole inspection units to determine orientation of cracks in the pressurized orientation of cracks in the pressurized section. section.

Hydraulic fracturingHydraulic fracturing

A borehole subjected to fluid pressure will A borehole subjected to fluid pressure will develop an extension fracture if the tensile develop an extension fracture if the tensile stresses developed by the fluid pressure (stresses developed by the fluid pressure (pp) ) exceed the external stresses on the borehole exceed the external stresses on the borehole wall and the tensile strength (wall and the tensile strength (TToo) of the ) of the material.material.

If If TToo is zero, as in the case of an old fracture is zero, as in the case of an old fracture intersecting the drill hole wall which is held intersecting the drill hole wall which is held closed by the normal stress acting on it, the closed by the normal stress acting on it, the fracture will open.fracture will open.

3. Overcoring 3. Overcoring

When a piece of rock is drilled out of the When a piece of rock is drilled out of the surrounding material (overcored), it expands surrounding material (overcored), it expands due to its elasticity. due to its elasticity.

If this elastic strain recovery is measured, If this elastic strain recovery is measured, and the elastic properties of the overcored and the elastic properties of the overcored material are determined, then the stresses material are determined, then the stresses which were acting on the overcored material which were acting on the overcored material can be calculated according to Hooke’s Lawcan be calculated according to Hooke’s Law

Overcoring in-situ stresses Overcoring in-situ stresses testtest