Fracture mechanics, Mohr circles, and the Coulomb criterion (Stress and failure)

Fracture mechanics, Mohr circles, and the Coulomb criterion

(Stress and failure)

Introduction to Fracture Mechanics

In this lecture, we will be focusing on faults:How they form

Definitions

Stress states

Fault strain

How we can use faults to tell us things about the geologic history

Representative structures on planetary bodies

Definitions

Fracture: a pair of distinct surfaces that are separated in a material; ‘‘a structure defined by two surfaces or a zone across which displacement occurs

A joint is a fracture that exhibits only opening displacement (Mode I)

A fault is a planar zone along which shear displacement occurs (Mode II)

Fracture modesMode I: opening displacement (joints)

Mode II: in-plane shear (faults)

Mode III: out-of-plane shear (scissors, tearing)

Stress

Stress is force/area, units of pressure (most often bars or pascals)

σ = F/A

Normal stress (σ) acts perpendicular to a surface

Shear stress (τ) acts parallel to a surface

StressThe “on-in” convention says that the stress component σij acts normal to the ‘i’ direction and parallel to the ‘j’ direction:

Normal stresses i = j

Shear stresses i ≠ j

StressPrinciple stresses have magnitudes and directions

σ1: maximum compressive stress

σ2: intermediate compressive stress

σ3: minimum compressive stress

Principle stresses act on planes that do not ‘feel’ shear stress, i.e., they are normal stresses

StressCalculate principle stresses from an arbitrary remote stress state with normal stresses σxand σy, parallel to the x and y axis, respectively

And the orientation of the 2 mutually perpendicular principle planes

Stress exampleGiven values of σx = 1.2 MPa (compression positive), σy= –0.85 MPa, and τxy = 0.45 MPa, find (a) the principal stresses, and (b) predict the orientations of the principal planes on which the principal stresses act.

SOLUTION

Substitute the values of normal and shear stress into equation 1 to obtainσmax,min=1.294MPa and –0.944MPa

Now substitute the same initial values of stress into equation 2 to obtain θp=11.6° and 101.6°

Add 90º to get second plane orientation

Resolving stresses onto planes

To resolve the stresses on a plane is to calculate the magnitude and direction of the normal and shear stresses onto a plane of particular orientation (here, a fault plane) due to the principle stresses

Example

σ1 = 5 MPa, σ2 = 1 MPa, α = 30¡ (to the plane), and θ = 60º (to the plane’s normal).

What is the magnitude of the normal and shear stresses?

σn = 2 MPa (compressive) and τ = 1.7 MPa

Which way is the fault going to slip?Left-lateral

Mohr CirclesMean stress: describes the confining stress experienced by rock at some depth

(σ1 + σ3)/2

Differential stress: describes the greatest amount of stress change that a rock can withstand without breaking

(σ1 – σ3)

Mohr circles is a geometric representation of these equations used to determine when rock will fracture or when faults will slip

Mohr circles

Mohr circles

X and Y axes are normal and shear stress, respectively.

This method only works for compressional normal stress (i.e., compression vs. tension; faults)

Plot σ1 and σ3 on the X axis as points. The difference between these values is the differential stress.

We’ll revisit this when we talk about the Coulomb Criterion.

Mohr circles and effective stress

Effective stress is the normal stress reduced by the pore fluid pressure

σn*=σn – pf

Pore pressure counteracts the effects of normal stress, reducing the magnitude of the principle stresses, but not of the differential stress.

Pore fluid must not support shear stress (i.e., is a fluid like water or air)

Coulomb criterionDefines the amount of shear stress needed to overcome the frictional resistance of a fault, leading to slip

Where C0 is the rock cohesion, μ is the coefficient of friction (typically between 0.6 and 0.85), ρ is the material density, g is the gravitational acceleration, and z is the depth

Cohesion describes the minimum amount of shear stress needed to start slip when the normal stress is tiny but compressive.

Coulomb criterion

Equation of a line!C0 is the y-intercept

μ is the slope of the line

If shear stress is greater than frictional resistance, the fault will slip. If not, no slip.

Coulomb criterion and Mohr circles

Combining the Coulomb criterion and Mohr circles allows you to predict if failure will occur with given stress magnitudes and if so, what the orientation of the plane will be when failure occurs.

Coulomb criterion practice

Given the values of remote principal normal stresses σ1 = 4.8 MPa (compression positive), σ2 = 1.15 MPa, and an angle to the normal to the plane from the σ1 direction of 58°, with values for cohesion of 0.001 MPa and friction coefficient of 0.58, determine whether the fracture implied will slip (using the Coulomb criterion), and if so, what its sense of shear will be.

SOLUTION

We find that σn = 2.175 MPa (compressive) and τ = 1.640 MPa (left-lateral).

Plugging these numbers into the Coulomb criterion, we find that 1.640 MPa > 0.001 + (0.58) (2.175) Mpa

1.640 MPa > 1.363 MPa. So the left-hand side (the driving forces) is greater than the right-hand side (the resisting forces) and the surface will fail by frictional sliding, and in a left-lateral sense because the calculated shear stress was positive.

Griffith CriteriaTensile failure (i.e., Mode I) occurs when the local stress at the most optimally oriented flaw attains a value characteristic of the material

Uniaxial tensile failure strength:E: Young’s modulus

ν: Poisson’s ratio

γ: energy required to create new crack walls

a: flaw half length

Gc: critical strain energy release rate

Griffith CriteriaThe Griffith criterion and Mohr circles showing (a) Uniaxial tensile failure, σ1 = 3T0 (at σ2 = T0); (b) uniaxial compressive failure, σ1 = 8T0 (at σ2 = 0); and (c) transition stress from crack growth to frictional sliding, σ1 ≅ 4.5T0.

Pure Mode I failure is predicted where the Mohr circle touches the Griffith criterion at exactly one point

Andersonian Fault Mechanics

E. M. Anderson’s first work on the subject of faulting was in 1905, but he is probably best known for his 1951 book (at right).

First proposed that faults are brittle fractures that occur according to the Coulomb criterion.

Divided faults into 3 classes that form depending on the ratio and orientation of principle stresses:

Normal

Thrust or reverse

Strike-slip


For all faults:σ1 > σ2 > σ3

Fault forms at an angle θ from σ1

Thrust fault:σ1 = σH

σ2 parallel to fault strike

σ3 = σv

Strike-slip faultσ1 = σH

σ2= σv

σ3 = σh

Normal faultσ1 = σv

σ2 parallel to fault strike

σ3 = σh


σv = weight of overburden or ρgz

Cube of rock subjected to vertical stress from overburden will extend in x and y directions (horizontal): Poisson ratio

ν = horizontal expansion/vertical shortening

ν< 0.5

The coefficient of friction and fault dip are related in order to minimize horizontal stress resolved on the fault plane.

To determine the fault dip from vertical for each dip-slip fault type, we use this equation:

Tan2θ = 1/μ

θ = 0.5(tan-1(1/μ))

This plot shows the minimum differential stress required to initiate sliding on a normal, strike-slip, and thrust fault.

Which line is which? Why?

Thrust faults require 16x more stress to rupture than normal faults and 2.6 x more stress than strike-slip faults.

How does this relate to earthquakes?

THRUST

STRIKE-SLIP

NORMAL

StrainStress σ causes strain ε

Strain is non-dimensional

Strain is any change in shape, volume, or orientation of a rock volume

Contractional normal strain perpendicular to σ1 and extensional normal strain perpendicular to σ3

Normal faulting results in extensional strain (horizontal)

Thrust faulting results in contractional strain (horizontal)

Extensional strain (vertical)

StrainStrain can be calculated by doing a 1D traverse across a set of faults

Calculate strain based on fault geometry

StrainExtension

Need depth of graben (d)

Fault dip angle (θ)

h = d/tanθ

Add these up for every fault in the traverse

Strain

This method of calculating strain can miss some faults.

In addition, the traverse may miss locations of maximum fault displacement (usually near the center of the fault, but not always).

Thus, better, more accurate methods exist to calculate strain from fault populations.

Seismology to the rescue!

Strain, the right way

Where D is the average displacement, L is the fault length, and H is the down-dip fault height (faulting depth/sinθ), V is the volume of the deformed region (depth of faulting*area of faults), and δ is the fault dip angle

This method includes all mapped faults and is not dependent on the location of 1D traverses.

Tour of Tectonics

http://mesic.astronomie.cz/Prohlidka/mesicni-brazda.php

NORMAL FAULTING

Moon

Planetary Tectonics:Normal Faults

Primary extensional morphologies on the planets:• graben• single normal faults

Graben are the down-dropped blocks between two antithetic NF.

Single normal faults are rare.

Where are NF found?• Moon• Mars• Venus• Earth• Mercury• Asteroids• Icy satellites


How do these NF form?• rifting• dike intrusion• basin loading• regional-scale uplift• tidal stresses• impacts


• Rare individual NF

• Basin loading or from Imbrium?

• Grew from small segments from S to N• opposite direction if formed by Imbrium

impact

• Likely not from basin loading, since it is on the edge of Nubium, too young, and straight

Moon


• Valles Marineris: the largest canyon system in the solar system

• rift valley• 10 km wide• 5 km deep• 3000 km long

• Crustal extension along large-scale normal faults related to Tharsis volcanism and heat production

Mars


• Pantheon Fossae • set of ~radial graben near

center of Caloris basin, Mercury

• Hypotheses:• Formed as a result of the

impact• Dike intrusion• Basin interior uplift

• PF formed in response to a dome with a R = 300 km, T = 150 km, and a maximum uplift of 10 km [Klimczak et al. (2011)]

Mercury

40 km


http://apod.nasa.gov/apod/image/0911/PSP_007769_9010_IRB_Stickney.jpg

• Grooves• long• parallel• narrow

• tidal or thermal stresses?• impact-related?

Phobos (Mars)


http://nssdc.gsfc.nasa.gov/imgcat/hires/mgn_c130n279_1.gif

• Venusian chasma (rift)• very long• high relief• relatively young features?

• Regional-scale crustal extension along antithetic normal faults

• Associated with volcanoes, likely due to uplift from underlying plume

Venus

~100 km


Basin and Range province formed by crustal extension along antithetic normal faults forming graben (basin) and horsts (range) from subduction of the East Pacific Rise ~15 Ma

http://rst.gsfc.nasa.gov/Sect6/nv.jpg http://rst.gsfc.nasa.gov/Sect6/38Basin_Range_aerialsm.jpg

Earth


THRUST FAULTING

Mercury

~40 km

Primary contractional morphologies on the planets:• wrinkle ridges• lobate scarps

Wrinkle ridges are blind thrust faults, often in areas of interbedded lava and regolith and/or pyroclastic material.

Lobate scarps are surface-cutting thrust faults that occur in mechanically homogenous terranes.

http://www.nasa.gov/images/content/475507main_lobate_scarp_thrust_fault_graphic.jpg

Planetary Tectonics:Thrust Faults

Where are TF found?• Moon• Mars• Venus• Earth• Mercury• Asteroids


How are these TF formed?• basin loading• lava cooling and contraction• impacts• (global contraction)


http://www.lpi.usra.edu/publications/slidesets/redplanet2/slide_9.html

• Hesperia Planum• type locality for Hesperian

epoch• Lava plains likely

interbedded with pyroclastic deposits (from Tyrrhena Patera)

• Multiple generations of WR indicate several temporally and spatially distinct episodes of compression

Mars


~25 km

• Wrinkle ridges in Mare Crisium

• Show a buried crater• Can give an idea of depth

of mare fill in Crisium Basin

Moon

~15 km


http://spacefellowship.com/news/art19041/crisium-s-region-of-interest.html

http://www.sciencemag.org/content/286/5437/87/F1.large.jpg

• Venusian wrinkle ridges appear to preferentially form in topographic lows

• Compressional stress likely from cooling and contraction in a topographic low

• Some WR terranes show more than one orientation of WR• more complicated and varied

stress field

Venus


~40 km

Watters et al., GRL 2011

• Lobate scarp (Hinks Dorsum) on the asteroid Eros

• Modeling gives fault parameters• Depth = 250 m• 90 m of offset

• Near-surface shear strength ~1 – 6 MPa

• Formed by impact-induced compression

Eros


~6 km

http://www.visit-himalaya.com/gifs/nepal-everest2.jpg

• Himalayan fold-and-thrust belt

• Large-scale continental collision

• Began ~50 Ma• Peak of Everest is

LIMESTONE• ~2400 km of India already ‘lost’• 1500 km of India subducted

over the next 10 Myr

Earth


~300 km

STRIKE-SLIP FAULTING

Earth

~ 110 km

http://photojournal.jpl.nasa.gov/catalog/PIA06661

Strike-slip fault morphology• linear• long

Strike-slip faults are rare on planets other than Earth.

Planetary Tectonics:Strike-slip Faults

Where are SSF found?• Earth• Mars• Venus• Icy satellites


How are these SSF formed?• plate motion (Earth only)• lateral movement of the lithosphere

• tidal stresses• accommodation structures w/WR• impacts?


• San Andreas: best-studied fault zone on Earth

• Farallon plate and spreading center subducted under N. American plate

• development right-lateral transform (SSF) that propagated along the continental margin

• Formed • ~28 Ma• 470 km of offset inferred

Earth


Andrews-Hanna et al., JGR 2008

• Martian strike-slip faults• Noachian in age and

continued for ~ 1 Gyr• used to estimate crustal

deviatoric stress magnitude and orientation during early Mars’ history

• Strike-slip faulting caused by loading and a background compressional stress

Mars


• Ovda Regio, in Aphrodite Terra

• Formed as a transform fault from collisional zone in the north (‘escape tectonics,’ a la Tibet)

• ~75 km of left-lateral strike-slip motion

http://www.lpi.usra.edu/publications/slidesets/venus/images/sven_s13.gif

Venus


• Androgeos Linea• 350 m high

• Strike-slip movement from tidal deformation and internal convection

http://www.lpi.usra.edu/galileoAnniv/img/hiRes/eplains-mos-KH.jpg

Europa


• Tiger stripes • 130 km long• 35 km apart• 2 km wide• 500 m deep

• Strike-slip faults with displacements of 0.5 m/event (Smith-Konter and Pappalardo, 2008)

• Associated with high heat flow and water-vapor plumes

• Tidal stresses from Saturn at periapsis

Enceladus

http://photojournal.jpl.nasa.gov/catalog/PIA06247


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Fracture mechanics, Mohr circles, and the Coulomb criterion (Stress and failure)

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