Rock Mechanics Basics

7/27/2019 Rock Mechanics Basics

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Lenhardt Rock Mechanics/ Basics 1

Rock MechanicsRock Mechanics

--

BasicsBasics

byby

Wolfgang A.Wolfgang A. LenhardtLenhardt

[email protected]@zamg.ac.at

Department of GeophysicsDepartment of Geophysics

Central Institute for Meteorology and GeodynamicsCentral Institute for Meteorology and Geodynamics

HoheHohe WarteWarte 3838

AA--1190 Vienna1190 Vienna

AustriaAustria

Phone: +43 1 36 026 ext. 2501Phone: +43 1 36 026 ext. 2501

TelefaxTelefax: +43 1 368 6621: +43 1 368 6621

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


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Content

Terminology

Stress & Strain

Mohr Circle

State of StressStress Concentrations

Strain Energy Density

Closure

DiscontinuitiesExcess Shear Stress


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Youngs ModulusE - modulus of elasticity

[GPa]

normalstress

normal strain

Terminology

Uniaxial Compressive Strength

UCS stress at failure

[MPa]

Shear ModulusG - modulus of rigidity

[GPa]

shearstress

shear strain

slope = G

slope = E


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Terminology - contd

Poissons Ratio

- Ratio of radial/axial deformation

[1]

CohesionSo - Intrinsic shear strength

[MPa]

=-x/z

Adhesion

To Tensile strength

[MPa]

Tonormal stress

shearstress

SoCoefficient of Friction Tangent of angle betweenshear and normal stress

[1]

slope =


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Stress and Strain

d1

l1

lo

do

Area

Force

Stress = Force / Area

Radial strain = r = (do- d1)/do

Axial strain = a = (lo- l1)/lo

< 0in compression !> 0

Poissons ratio = - r / a

Youngs 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 indifferent directions.

Special cases of : 0.25 ... purely elastic, 0.5 ... liquid (incompressible)


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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, and

2. 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.


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Mohrs Circle Derivation contd

Lets 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


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Mohrs Circle Derivation contd

Now, these components can be

separated into a normal and a shearstress acting on the plane:

is the angle between the normal stress on a plane and the majorprincipal 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, weset:


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Mohrs Circle Stresses

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


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Mohrs Circle Strains

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


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Mohrs Circle Summary

Stress acting normalto the plane

Shear stress acting in

direction of the plane


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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 ...


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Mohrs Circle Examples

1 = 200 MPa2 = 0 MPa3 = 0 MPa

1 = 400 MPa2 = 50 MPa

3 = 0 MPa

1 = 350 MPa2 = 10 MPa3 = 10 MPa


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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 Circle

4. Introduce in Mohrs Circle5. Calculate the average principal stress6. Calculate the maximum shear stress

7. Determine normal and shear stress

Solution:

Normal stress =average stress +cos(2)*maximum shear stressShear stress = sin(2)*maximum shear stress


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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 wasgiven from 2, which is the same as the anglebetween the plane normal and the maximumprincipal stress!)

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

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

normal stress

1 =210 MPa2 = 10 MPa

= ? = ? = ? = ?

=? =? =? =?

2 = 120

shear stress

max

shear stress


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Mohrs Circle Exercise 2

y = 100 MPa

x =50 MPa

Task:

Determine the magnitude and orientation ofthe principal stresses!

Hint:

1. Plot normal stresses on x-axis2. Determine average normal stress3. Plot shear stress on y-axis

4. Draw Mohrs Circle5. Determine 6. Calculate principal stresses

Solution:

Orientation = = arctan (yx/(y- average))/2Determine average normal stressDetermine maximum shear stress

1 = average stress + max2 = average stress - max

yx = 25 MPa

Note sign convention!


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Mohrs Circle Exercise 2 Solution

normal stress

1

????

2 ???? 2x

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


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3-D Mohr Space

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


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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


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Failure Criteria contd

normal stress

shear stress

No failure

Failure

COULOMB-NAVIER

minor stress

major stress

No failure

Failure

Compressive StrengthUCS

Tensile Strength

To

Intrinsic Strength

So


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Pre- and Post Failure Behaviour


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Pre- and Post Failure Behaviour contd

Whether 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

ingStre

ss

Yielding

(resistingconstantstress)

Not yielding(loosing load

carryingcapacity)


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Pre- and Post Failure Behaviour contd

If the confining stress is low such as in slim pillars shear failure is likely to take place, especiallywhen transgressive fractures (due to blasting or geology) are present.


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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.


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State of Stress

k=horizontal stress

vertical stressdepth

surface

k < 1 (deep)

k > 1 (shallow)


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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


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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 ofdeformation 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 derivedfrom the 3D state of stress.


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State of Stress 3D

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


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Stress Concentrations

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

1. Depth2. Shape of excavation3. Mining layout

4. Support

5. 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).


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Stress Concentrations in a Tunnel Exercise


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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 the

height/width ratio is small

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


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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.


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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 ofAmeliorating the Hazards of ROCKFALL andROCKBURSTS, Chamber of Mines of SouthAfrica, 1988.


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Mining Conditions

Conditions in deep mines differfrom shallow mines in manyways.

From: An Industry Guide to Methods ofAmeliorating the Hazards of ROCKFALL andROCKBURSTS, Chamber of Mines of SouthAfrica, 1988.


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Mining Conditions contd

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


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

W=

i=1

3

ii

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 lessthan 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:


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Strain Energy - Exercise

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


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Strain Energy Exercise contd

Cl


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Closure


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

Closure = Elastic deformation + Inelastic deformation

convergence

Principle of beam

Bed separationOpening of joints

Swelling

Slip along plane

Inelastic deformation exceeds by far elastic deformation!

Examples:

Cl td


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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 ResearchAdvisory Committee (SIMRAC), Johannesburg, South Africa.

Cl td


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Closure contd

Inelastic deformation manifests itself in face-

parallel fracturing of the footwall.

Faceadvanc

e

Stope face

Direction ofadvance

Old mineworkings

Closure contd


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Closure contd

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

Rock Mass


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Rock Mass

The description of discontinuities is a matter of scale!

Types of Fractures


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Types of Fractures

Triaxial compression Shear(offset between surfaces)

Special case

Cross Shear

Uniaxialcompression

Uniaxial tension

Special cases

Intrusion, Uplift

Longitudinal splittingduring perfectcompression (no end-effects)(irregular surface)

Extension

(no shear-offset betweensurfaces)

Rock Quality


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Rock Quality

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

Discontinuities


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Discontinuities

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

FRACTURE plane which separates the rock material

JOINT break of geological origin, no shear displacement visible

FAULT fracture with identifiable shear displacement

DYKE long narrow vertical intrusion

SILL near horizontal intrusionBEDDING PLANE separates sedimentary rocks into beds or strata

SHEAR ZONE bonds of material in the order of metres in which local shear

displacement took place

Discontinuities contd


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Discontinuities cont d

Shear fracture Joints & plumoses



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Dyke-contact

Dyke



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FaultMovement along a dyke-

contact, as seen through theshot (gunned) concrete

Dyke Contact contd


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y

Dyke Quartzite

After: Cook, N.G.W. et al. 1966.

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Types of Faults


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yp

Reverse or thrust faultNormal fault

Strike-slip fault

There are

Normal,Reverse/thrust andStrike-slip faults

Sometimes, their effects daylight on surface.

Types of Faults


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yp

Three types of faults can be distinguished:

Reverse or thrust faultNormal fault Strike-slip fault

Characteristics of Faults


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Coefficients of sliding-friction differ significantly (based on faults at shallow depth, < 1000 m belowsurface):

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 highAAA

1 10Very highAA

0.1 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 faulting

0.381 (-10.9)0.476 (+11.4)0.7900.427Thrust faulting

0.521 (-16.6)0.746 (+19.6)0.8480.625Normal faulting

0.471 (-9.6)0.577 (+10.5)0.7860.522Combined data

Lower

(per cent)

Upper

(per cent)

Correlationcoefficient

Coefficient offriction from

regression line

98 per cent confidence limits

Tectonic Uplift


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Post-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 2ndFrance-United States Workshop Earthquake Hazard Assessment in Intraplate Regions (G. Mohammadioun, ed.), Quest Editions, Press

Acamedique, 85 103.

Excess Shear Stress - ESS


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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, pages27-39.

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


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ESS-lobes above and below a stope


Blasting contd


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Blasting induced fractures

are orientated perpendicularto the

minor principal stress and

the borehole-axis, or followthe stratification of the rock

mass.

drill hole

1 - major principalstress

3 -

minorprincipalstress

Tectonic versus Induced/Trigggered Events


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Tectonic Induced /Triggered

Virgin stresses High Low

Short term stress changes Low High

Depth < 3 km > 3 km

Mechanism Shear Slip Fracture and Shear Slip

Max. magnitude 9.4 6.5 (?)

Rupture length 1.400 km Few kms

Summary


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Terminology

Stress & Strain

Mohr Circle

State of Stress

Stress Concentrations


Closure

Discontinuities

Excess Shear Stress

Rock Mechanics Basics

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