AGS A1 Rock Mechanics Basics

Lenhardt Rock Mechanics/ Basics 1

Rock MechanicsRock Mechanics--

BasicsBasicsby by

Wolfgang A. Wolfgang A. LenhardtLenhardt

[email protected]@zamg.ac.at

Department of GeophysicsDepartment of GeophysicsCentral Institute for Meteorology and GeodynamicsCentral Institute for Meteorology and Geodynamics

HoheHohe WarteWarte 3838AA--1190 Vienna1190 Vienna

AustriaAustria

Phone: +43 1 36 026 ext. 2501Phone: +43 1 36 026 ext. 2501TelefaxTelefax: +43 1 368 6621: +43 1 368 6621

www.zamg.ac.atwww.zamg.ac.at


Content

TerminologyStress & Strain

Mohr CircleState of Stress

Stress ConcentrationsStrain Energy Density

ClosureDiscontinuities

Excess Shear Stress


Youngs ModulusE - modulus of elasticity[GPa]

normalstress

normal strain

Terminology

Uniaxial Compressive StrengthUCS stress at failure[MPa]

Shear ModulusG - modulus of rigidity[GPa]

shearstress

shear strain

slope = G

slope = E


Terminology - contd

Poissons Ratio - Ratio of radial/axial deformation[1]

CohesionSo - Intrinsic shear strength[MPa]

=-x/z

AdhesionTo Tensile strength[MPa]

Tonormal stress

shearstress

SoCoefficient of Friction Tangent of angle between shear and normal stress[1]

slope =


Stress and Strain

d1

l1

lo

do

Area

Force

Stress = Force / Area

Radial strain = r = (do- d1)/doAxial strain = a = (lo- l1)/lo

< 0in compression !

> 0

Poissons ratio = - r / aYoungs modulus E = a / a

72 GPa0.22

Typical values of Witwatersrand quartzite

E

Note: Poissons ratio and the Youngs modulus are material-dependent and may vary in different directions.Special cases of : 0.25 ... purely elastic, 0.5 ... liquid (incompressible)


Mohrs Circle

The Mohr-Circle diagram connects the state of deformation and the currently applied stress regime in a graphical way.

Let us consider a cube, which is exposed to external forces per area or stresses. Within this cube we would like to determine these stresses. For reasons of simplicity, lets assume, that the largest stress is acting vertical, and the smallest stress is acting horizontal:

Definition:These major two stresses are called the

1. Major principal stress, and2. Minor principal stress

The stress, which acts orthogonal to these stresses, is named intermediate stress.

In these three distinct directions, only normal stresses are acting. No shear stresses are present. Why?

Each stress-system can be represented by three principal (normal) stresses and their directions only! We will see now, how this works.


Mohrs Circle Derivation contdLets consider a pressure p acting at an arbitrary angle on an inclined plane in respect to a Cartesian co-ordinate system.

The force-equilibrium

requires, the following:

Consider a stress px in x-direction due to a stress acting at an angle to the inclined surface A:

px = x . . . . cos + yx . sin


Mohrs Circle Derivation contd

Now, these components can be separated into a normal and a shear stress acting on the plane:

is the angle between the normal stress on a plane and the major principal stress. Rotation of the arbitrary chosen coordinate system by would eliminate the shear stresses.The angle , at which shear stresses vanish, is given by:

xy= yx

Since the forces exerted on a square must be in equilibrium, we set:


Mohrs Circle StressesThe derived relation between shear- and normal stresses can be expressed graphically by a circle the Mohr Circle, which can be completely described by the principal stresses.


Mohrs Circle Strains

Note: Only half of the shear strains are plotted in the diagram, for the other half is consumed by rotation.


Mohrs Circle Summary

Stress acting normalto the plane

Shear stress acting indirection of the plane


Mohrs Circle Summary contd

2

2

Stress

Strain

General case:3 normal stresses (x ,,,,y ,,,,z) or (xx ,,,,yy ,,,,zz) 3 shear stresses (xy , , , ,yz ,,,,zx) or (xy , , , ,yz ,,,,zx)

Transformed:3 principal stresses (1... ... ... ... 3)3 principal directions (1... ... ... ... 3)

or strains ...


Mohrs Circle Examples

1 = 200 MPa2 = 0 MPa3 = 0 MPa

1 = 400 MPa2 = 50 MPa3 = 0 MPa

1 = 350 MPa2 = 10 MPa3 = 10 MPa


Mohrs Circle Examples

1 = 50 MPa2 = 5 MPa3 = -10 MPa

1 = 0 MPa2 = 0 MPa3 = 0 MPa

1 = 50 MPa2 = 50 MPa3 = 50 MPa


Mohrs Circle Exercise 1

1 = 210 MPa

2 =10 MPa = 60

Task:

Determine the stresses normal and shear stresses acting on the inclined plane!

Hint:1. Determine 2. Plot principal stresses on x-axis3. Draw Mohrs Circle4. Introduce in Mohrs Circle5. Calculate the average principal stress6. Calculate the maximum shear stress7. Determine normal and shear stress

Solution:Normal stress =average stress +

cos(2)*maximum shear stressShear stress = sin(2)*maximum shear stress


Mohrs Circle Exercise 1 Solution

Result:

= average + cos(2) * max = 60 MPa

= sin(2) * max = 86.6 MPa

Solution:

= = 60 (just in this case, for the angle was given from 2, which is the same as the angle between the plane normal and the maximum principal stress!)

average = (1 + 2) / 2 = 110 MPa

max = (1 - 2) / 2 = 100 MPa

normal stress1 = 210 MPa

2 = 10 MPa

= ? = ? = ? = ?

=? =? =? =?

2 = 120

shear stress

max

shear stress


Mohrs Circle Exercise 2

y = 100 MPa

x = 50 MPa

Task:

Determine the magnitude and orientation of the principal stresses!

Hint:1. Plot normal stresses on x-axis2. Determine average normal stress3. Plot shear stress on y-axis4. Draw Mohrs Circle5. Determine 6. Calculate principal stresses

Solution:Orientation = = arctan (yx/(y- average))/2Determine average normal stressDetermine maximum shear stress1 = average stress + max2 = average stress - max

yx = 25 MPa

Note sign convention!


Mohrs Circle Exercise 2 Solution

normal stress

1 ????2 ???? 2

x

y

xy

yx

shear stress

shear stress

Result:

= (arctan(yx/(y- average)))/2 = 22.5

1111 = average + max = 110.4 MPa

2222 = average - max = 39.6 MPa

2 =39.6 MPa

= 22.51 =110.4 MPa

Solution:xy = - yx = - 25 MPa

average = (x + y) / 2 = 75 MPa

max = = 35.35 MPa


3-D Mohr Space

From: Jaeger & Cook (1978). In Pollard, D.D. & Fletcher, R.C. 2005. Fundamentals of Structural Geology. Cambridge.


Failure Criteria

normal stress

max= So

shear stress

No failure

Failure

MAXIMUM SHEAR STRESS

normal stress

shear stress

No failure

Failure

COULOMB-NAVIER

normal stress

shear stress

No failure

Failure

HOEK-BROWN

= So + n . tan

1111 = = = = 3((((m. UCS. 3 + UCS s)m, s... describe the integrity of the rock mass (Hoek & Brown parameters), s = integrity (0-1 = degree of fracturing), m = type of rock


Failure Criteria contd

normal stress

shear stress

No failure

Failure

COULOMB-NAVIER

minor stress

major stress

No failure

Failure

Compressive StrengthUCS

Tensile StrengthTo

Intrinsic StrengthSo


Failure Criteria contd

The following relations between the different parameters are useful:


Pre- and Post Failure Behaviour


Pre- and Post Failure Behaviour contdWhether the rock behaves brittle or ductile depends mainly on the confining stress.

After: Wawersik, W.R. & Fairhurst C.A, 1970: A study of brittle rock fracture in laboratory compression experiments. Int. J. Rock Mech. Min. Sci. 7, 561 675.

Confin

ing

Stres

s

Yielding(resisting constant stress)

Not yielding(loosing load

carrying capacity)


Pre- and Post Failure Behaviour contdIf the confining stress is low such as in slim pillars shear failure is likely to take place, especially when transgressive fractures (due to blasting or geology) are present.


Strain Rate

Materials do respond different to high and low strain rates:

UCS... Uniaxial compressive strengthE... Youngs modulusTd... Duration of test

From: Chong & Boresi, 1990: Strain rate properties of New Albany Reference shale. Int. J. Rock Mech. Min. Sci. & Geomech. Abstr., Vol. 27, No.3, 199 - 205.


State of Stress

k= horizontal st ressvert ical st ressdepth

surface

k < 1 (deep)

k > 1 (shallow)


State of Stress contd

From: Brady, B.H.G. & Brown, E.T., 1985. Rock Mechanics for Underground Mining, George Allen & Unwin Publishers.

Terminology:

VIRGIN STRESS Original stress prior to miningINDUCED STRESS Additional change in stress due to miningFIELD STRESS VIRGIN + INDUCED STRESS = actual stress acting around the excavation

k = 0.5 (pxx + pyy) / pzz


State of Stress 2D Plane Stress

An example of a plane stress regime is the surface of an underground excavation. Perpendicular to the free surface at the stope face, no stress is acting only strain.

The basic equations for calculating the principal strains and stresses during the linear-elastic state of deformation are given here:

An example of a plane strain condition is a section through a tunnel. In direction of the axis of a tunnel only the stress remains, - the strain equals zero. The equivalent formulas can be derived from the 3D state of stress.


State of Stress 3D

The general case in three dimensions for a linear-elastic state of deformation is given by:


Stress ConcentrationsStress concentrations do occur around all excavations. The actual field stress depends on:

1. Depth2. Shape of excavation3. Mining layout4. Support5. Rock characteristics

The rock mass has a very low tensile strength. Whether an excavation is subjected to tensile stresses depends on its shape (e.g. height to width ratio) and the ratio of the horizontal to the vertical stress (k - ratio).


Stress Concentrations in a Tunnel Exercise


Stress Concentrations in a Tunnel contd

Tensile stresses may easily develop in the crown of a haulage (rockfall !) if the

k-ratio is low and/or theheight/width ratio is small

After: Brady, B.H.G. & Brown, E.T., 1985. Rock Mechanics for Underground Mining, George Allen & Unwin Publishers.


Stress Concentrations contd

High stresses such as inside or below a pillar may lead to disking of borehole-samples. The example shown, was taken from a pillar 3000 m belowsurface, intersecting the reef (pebbles). The disk at the right end originated from the edge of the pillar.


Stress Shadow

The presence of nearby mining openings affect the field stresses and thus the intensity and orientation of induced fractures in e.g. tunnels.

stope

From: An Industry Guide to Methods of Ameliorating the Hazards of ROCKFALL and ROCKBURSTS, Chamber of Mines of South Africa, 1988.


Mining Conditions

Conditions in deep mines differ from shallow mines in many ways.



Mining Conditions contd



Strain Energy DensityStrain energy density represents the energy stored in the rock per volume:

W=i= 1

3 i i

2

i... principal stressi... principal strain

The total stored strain energy U in the rock mass with volume V is therefore

U = W * V

The major principal stress dominates the stored strain energy density once the k-ratio is less than 0.5 (error < 10 %, exact if k = 0):

Example:Uniaxial Compressive Test (k-ratio = 0)

220 MPa at failure

E = 72 GPa(Youngs modulus)

/2 = W

Strain Energy Density:


Strain Energy - ExerciseCalculate the strain energy density as function of depth, while considering Poissons ratio, the k-ratio and the material properties.


Strain Energy Exercise contd


Closure


Closure is the amount by which the original stoping width is reduced.

Closure = Elastic deformation + Inelastic deformation

convergence

Principle of beamBed separation

Opening of jointsSwelling

Slip along plane

Inelastic deformation exceeds by far elastic deformation!

Examples:


Closure contd

Deformation of the rock mass around deep stopes.

From: Jager, A.J. & Ryder, J.A. 1999. A Handbook for Rock Engineering Practice for tabular hard rock mines. The Safety in Mines Research Advisory Committee (SIMRAC), Johannesburg, South Africa.


Closure contd

Inelastic deformation manifests itself in face-parallel fracturing of the footwall.

Face advance

Stope face

Direction of advance

Old mine workings


Closure contd

Timber support in an old stope, at which total closure took place.


Rock MassThe description of discontinuities is a matter of scale!


Types of Fractures

Triaxial compression Shear(offset between surfaces)

Special caseCross Shear

Uniaxialcompression

Uniaxial tension

Special casesIntrusion, Uplift

Longitudinal splitting during perfect compression (no end-effects)(irregular surface)

Extension(no shear-offset between

surfaces)


Rock Quality

Barton, N. 2007. Rock quality, seismic velocity, attenuation and anisotropy. Taylor & Francis/Balkema, The Netherlands, 729 pp.


DiscontinuitiesRock characteristics depend on the number of discontinuities existing inside of the rock mass. A very small sample (laboratory test) usually contains no discontinuities. A largerock sample may contain numerous discontinuities, however.

FRACTURE plane which separates the rock materialJOINT break of geological origin, no shear displacement visibleFAULT fracture with identifiable shear displacementDYKE long narrow vertical intrusionSILL near horizontal intrusionBEDDING PLANE separates sedimentary rocks into beds or strataSHEAR ZONE bonds of material in the order of metres in which local shear

displacement took place


Discontinuities contd

Shear fracture Joints & plumoses



Dyke-contact

Dyke



Fault Movement along a dyke-contact, as seen through the

shot (gunned) concrete

Dyke Contact contd

Dyke Quartzite

After: Cook, N.G.W. et al. 1966.52


Types of Faults

Reverse or thrust faultNormal fault

Strike-slip fault

There are

Normal,Reverse/thrust andStrike-slip faults

Sometimes, their effects daylight on surface.


Types of FaultsThree types of faults can be distinguished:

Reverse or thrust faultNormal fault Strike-slip fault


Characteristics of FaultsCoefficients of sliding-friction differ significantly (based on faults at shallow depth, < 1000 m below surface):

From: Jaeger, J.,C. & Cook, N.G.W. 1979. Fundamentals of Rock Mechanics, 3rd edition, Chapman and Hall, London, pages 377-379 and 425 427.

Under mining conditions faults may slip reverse to their geological sense of displacement!

> 10Extremely highAAA1 10Very highAA0.1 1HighA

0.01 0.1ModerateB

0.001 0.01LowC< 0.001Extremely lowD

Slip rate (cm/year)Fault activityclass

From: Bonilla M. G. 1982. Evaluation of potential surface faulting and other tectonic deformation. 8Th World Conference on Earthquake Engineering, Vol.1, 65. Note: No distinction is made here for aseismic slip (creep).

0.202 (-8.0)0.235 (+6.9)0.3830.220Strike-slip faulting0.381 (-10.9)0.476 (+11.4)0.7900.427Thrust faulting0.521 (-16.6)0.746 (+19.6)0.8480.625Normal faulting0.471 (-9.6)0.577 (+10.5)0.7860.522Combined data

Lower(per cent)

Upper(per cent)

Correlation coefficient

Coefficient of friction from

regression line

98 per cent confidence limits


Tectonic UpliftPost-glacial rebound leads to a dome-shaped uplift of the crust, along with an alteration of the prevailing stress-regime in the dome and the surrounding forebuldge.

From: Muir Wood, R. 1995. Deglaciation seismotectonics: A principal influence on intraplate seismogenesis at high latitudes? In Proc. of 2nd France-United States Workshop Earthquake Hazard Assessment in Intraplate Regions (G. Mohammadioun, ed.), Quest Editions, Press

Acamedique, 85 103.


Excess Shear Stress - ESS

For details see: Ryder, J.A. 1988. Excess shear stress in the assessment of geologically hazardous situations. J.S.Afr.Inst.Min.Metall., Vol.88, pages 27-39.

All parameters in this Mohr-diagram refer to conditions along a geological feature and not to solid (undisturbed) rock.

ESS =

normal stress

shearstress

So slope =

ESS

Stress prior to slip

Stress after slip


Excess Shear Stress contd

ESS-lobes above and below a stope



Blasting contd

Blasting induced fractures are orientated perpendicular

to the minor principal stress and the borehole-axis, or follow the stratification of the rock

mass.

drill hole

1 - major principal stress

3 -minor principal stress


Tectonic versus Induced/Trigggered Events

Tectonic Induced /TriggeredVirgin stresses High Low

Short term stress changes Low HighDepth < 3 km > 3 kmMechanism Shear Slip Fracture and Shear SlipMax. magnitude 9.4 6.5 (?)Rupture length 1.400 km Few kms


SummaryTerminology

Stress & StrainMohr Circle

State of StressStress ConcentrationsStrain Energy Density

ClosureDiscontinuities

Excess Shear Stress

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AGS A1 Rock Mechanics Basics

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