Rock Mechanics Basics

Rock Mechanics Basics by

Wolfgang A. Lenhardt [email protected] Department of Geophysics Central Institute for Meteorology and Geodynamics Hohe Warte 38 A-1190 Vienna Austria Phone: +43 1 36 026 ext. 2501 Telefax:: +43 1 368 6621 Telefax www.zamg.ac.at

Content Terminology Stress & Strain Mohr Circle State of Stress Stress Concentrations Strain Energy Density Closure Discontinuities Excess Shear Stress

Content Terminology Stress & Strain Mohr Circle State of Stress Stress Concentrations Strain Energy Density Closure Discontinuities Excess Shear Stress

Terminology

Uniaxial Compre Uniaxial Compressive ssive Strength ‚UCS ‚UCS‘‘ – stre stress ss at at failu failure re [MPa]

Young‘s Modulus ‚E‘ ‚E‘ - modu modulu lus s of elast elastici icity ty

normal stress

slope = E

[GPa] normal strain

Shear Modulus ‚G‘ ‚G‘ - modu modulu lus s of rig rigid idit ity y

shear stress

slope = G

[GPa] shear strain

Terminology - cont‘d

Poisson‘s Ratio

=-∆x/∆z

ν‘ - Ratio of radial/axial deformation ‚ ν [1]

Cohesion ‚So‘ - Intrinsic shear strength‘ [MPa] Adhesion ‚To‘ – Tensile strength [MPa]

α

shear stress

slope = µ So

Coefficient of Friction ‚µ‘ – Tangent of angle ‚ α‘ between shear and normal stress [1]

To

normal stress

Stress and Strain Force Stress = Force / Area Area lo Radial strain = εr = (do- d1)/do do

Axial strain = εa = (lo- l1)/lo

<0 in compression ! >0

Poisson‘s ratio ‚ ν ν‘ = - εr / εa Young‘s modulus ‚E‘ = σa / εa

l1

d1

Typical values of Witwatersrand quartzite

νν

E

0.22

72 GPa

Note: Poisson‘s ratio and the Young‘s modulus are material-dependent and may vary in different directions.

Mohr‘s 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, let‘s 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.

Mohr‘s Circle – Derivation cont‘d Let‘s consider a pressure ‚p‘ acting at an arbitrary angle on an inclined plane in respect to a Cartesian co-ordinate system.

The force-equilibrium

Consider a stress p x in x-direction due to a stress acting at an angle θ to the inclined surface ‚A‘:

requires, the following:

px = σx . cos θ + τyx . sin θ

Mohr‘s Circle – Derivation cont‘d

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

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

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

= −

xy

τ

yx

Mohr‘s Circle – Stresses The 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.

Mohr‘s Circle – Strains

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

Mohr‘s Circle – Summary

θ Stress acting normal to the plane

Shear stress acting in direction of the plane

Mohr‘s Circle – Summary – cont‘d

Stress

2∗θ General case: 3 normal stresses (σx ,σ y ,σz) (σxx ,σ yy ,σzz) 3 shear stresses (τxy ,τyz ,τzx) (σxy ,σ yz ,σzx) Transformed: 3 principal stresses (σ1... σ3) 3 principal directions (θ1... θ3)

2∗θ

Strain

or strains ...

or or

Mohr‘s Circle – Examples σ1 = 200 MPa σ2 = 0 MPa σ3 = 0 MPa

σ1 = 400 MPa σ2 = 50 MPa σ3 = 0 MPa

σ1 = 350 MPa σ2 = 10 MPa σ3 = 10 MPa

Mohr‘s Circle – Examples σ1 = 50 MPa σ2 = 5 MPa σ3 = -10 MPa

σ1 = 0 MPa σ2 = 0 MPa σ3 = 0 MPa

σ1 = 50 MPa σ2 = 50 MPa σ3 = 50 MPa

Mohr‘s Circle – Exercise 1

σ1 = 210 MPa

Task: Determine the stresses – normal and shear stresses – acting on the inclined plane!

δ = 60°

σ2 =

10 MPa

Hint: 1. Determine ‚θ‘ 2. Plot principal stresses on x-axis 3. Draw Mohr‘s Circle 4. Introduce ‚θ‘ in Mohr‘s Circle 5. Calculate the average principal stress 6. Calculate the maximum shear stress 7. Determine normal and shear stress Solution: Normal stress =average stress + cos(2θ)*maximum shear stress Shear stress = sin(2 θ)*maximum shear stress

Mohr‘s Circle – Exercise 1 Solution Solution:

τshear stress

δ = θ = 60°(just in this case, for the angle was given from σ2, which is the same as the angle between the plane normal and the maximum principal stress!)

σ =?

σaverage = (σ1 + σ2) / 2 = 110 MPa τmax

τ=? 2θ = 120° σ2 = 10 MPa

σ1 =

τmax = (σ1 - σ2) / 2 = 100 MPa σnormal stress

210 MPa Result:

σ = σaverage + cos(2θ) * τmax = 60 MPa τ = sin(2θ) * τmax = 86.6 MPa −τshear stress

Mohr‘s Circle – Exercise 2

Task:

σy = 100 MPa

Determine the magnitude and orientation of the principal stresses!

τyx = 25 MPa σx =

Hint: 1. Plot normal stresses on x-axis 2. Determine average normal stress 3. Plot shear stress on y-axis 4. Draw Mohr‘s Circle 5. Determine ‚θ‘ 6. Calculate principal stresses

50 MPa Solution: Orientation = θ = arctan (τyx/(σy- σaverage))/2 Determine average normal stress Determine maximum shear stress σ1 = average stress + τmax σ2 = average stress - τmax

Mohr‘s Circle – Exercise 2 Solution τshear stress

Solution: τxy = - τyx = - 25 MPa

σaverage = (σx + σy) / 2 = 75 MPa τmax =

= 35.35 MPa

Result:

τyx σ2 ?

σx

σnormal stress

2θ σy

−τxy

θ = (arctan(τyx/(σy- σaverage)))/2 = 22.5°

σ1 ?

σ1 = σaverage + τmax = 110.4 MPa σ2 = σaverage - τmax = 39.6 MPa

θ = 22.5°

σ1 =

110.4 MPa

−τshear stress

σ2 =

39.6 MPa

3-D Mohr Space

Failure Criteria MAXIMUM SHEAR STRESS

τshear stress

τshear stress

Failure

No failure

COULOMB-NAVIER

τ = So + σn . tan α

Failure

τmax

= So

No failure

σnormal stress

τshear stress

σnormal stress HOEK-BROWN

Failure

σ1 = σ3√(m. UCS. σ3 + UCS² s)

No failure

σnormal stress

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 – cont‘d σmajor stress

COULOMB-NAVIER

τshear stress

Failure

ψ ψ

α

Failure No failure

No failure

Intrinsic ‚Strength‘

σnormal stress

Compressive ‚Strength‘ UCS

So

σminor stress

Failure Criteria – cont‘d The following relations between the different parameters are useful:

Pre- and Post Failure Behaviour

Pre- and Post Failure Behaviour – cont‘d Whether the rock behaves brittle or ductile depends mainly on the confining stress.

Yielding (resisting constant stress) s s t r e S g i n i n f n C o

Not yielding (loosing load carrying capacity)

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.

Pre- and Post Failure Behaviour – cont‘d If 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... E... Td...

Uniaxial compressive strength Young‘s modulus 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

surface

k > 1 (shallow)

k=

depth k < 1 (deep)

horizontal stress vertical stress

State of Stress – cont‘d k = 0.5 (p xx + p yy) / p zz

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

Terminology: VIRGIN STRESS INDUCED STRESS FIELD STRESS

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

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

1. 2. 3. 4. 5.

Depth Shape of excavation Mining layout Support 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 – cont‘d

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

Stress Concentrations – cont‘d

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 below surface, intersecting the reef (‚pebbles‘). The disk at the right end originated from the edge of the pillar.

Stress Shadow

stope

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

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 – cont‘d


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

σ∗ε/2 = W

3

σi... εi...

σ

principal stress principal strain

∑ W=

σ ε i

i= 1

i

2

ε 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 Strain Energy Density: E = 72 GPa (Young‘s modulus)

Strain Strain Energ Energy y - Exerci Exercise se Calculate the strain energy density as function of depth, while considering Poisson‘s ratio, the kratio and the material properties.

Strain Strain Ener Energy gy – Exercis Exercise e cont‘d cont‘d

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

Closure =

Elastic deformation

+

Inelastic deformation

‚convergence‘

Principle of beam Examples:

Bed separation Opening of joints Swelling Slip along plane

Inelastic deformation exceeds by far elastic deformation!


Closure – cont‘d

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


Direction advance Old mine workings

of

Stope face

e d v a n c F a c e a

Inelastic deformation manifests itself in faceparallel fracturing of the footwall.


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

Rock Mass The description of discontinuities is a matter of scale!

Types of Fractures Triaxial compression

Shear (offset between surfaces)

Special case

‚Cross Shear‘

Uniaxial compression

Longitudinal splitting during perfect compression (no ‚endeffects‘) (irregular surface)

Uniaxial tension

Extension (no shear-offset between surfaces)

Special cases

‚Intrusion‘, ‚Uplift‘

Rock Quality

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

Discontinuities Rock characteristics depend on the number of discontinuities existing inside of the rock mass. A very small sample (laboratory test) usually contains no discontinuities. A large rock 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 intrusion

BEDDING 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 – cont‘d

Shear fracture

Joints & plumoses


Dyke

Dyke-contact


Fault

Movement along a dykecontact, as seen through the

Dyke Contact – cont‘d

Dyke

Quartzite

Types of Faults Normal fault

Reverse or thrust fault

There are Normal, Reverse/thrust and Strike-slip faults Sometimes, their effects daylight on surface.

Strike-slip fault

Types of Faults Three types of faults can be distinguished:

Normal fault

Reverse or thrust fault

Strike-slip fault

Characteristics of Faults Coefficients of sliding-friction differ significantly (based on faults at shallow depth, < 1000 m below surface): 98 per cent confidence limits Coefficient of friction from regression line

Correlation coefficient

Upper

Lower

(per cent)

(per cent)

Combined data

0.522

0.786

0.577 (+10.5)

0.471 (-9.6)

Normal faulting

0.625

0.848

0.746 (+19.6)

0.521 (-16.6)

Thrust faulting

0.427

0.790

0.476 (+11.4)

0.381 (-10.9)

Strike-slip faulting

0.220

0.383

0.235 (+6.9)

0.202 (-8.0)

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! class

Fault activity

Slip rate (cm/year)

AAA

Extremely high

> 10

AA

Very high

1 – 10

A

High

0.1 – 1

B

Moderate

0.01 – 0.1

C

Low

0.001 – 0.01

D

Extremely low

< 0.001 Th

Tectonic Uplift 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 2nd France-United States Workshop ‚Earthquake Hazard Assessment in Intraplate Regions‘ (G. Mohammadioun, ed.), Quest Editions, Press

Excess Shear Stress - ESS ESS = τ − µ σ

shear stress

τ

Stress prior to slip

ESS

So

slope = µ Stress after slip normal stress σ

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

Excess Shear Stress – cont‘d

ESS-lobes above and below a stope From: An Industry Guide to Methods of Ameliorating the Hazards of ROCKFALL and ROCKBURSTS, Chamber of Mines of South Africa, 1988.

Blasting – cont‘d σ1 - major principal stress

drill hole

σ3 -

minor principal stress

Blasting induced fractures are orientated perpendicular to the minor principal stress and the borehole-axis, or follow the stratification of the rock mass.

Tectonic versus Induced/Trigggered Events

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 km‘s

Rock Mechanics Basics

Recommend Documents