Presentation on theme: "Introduction to Fracture Mechanics"— Presentation transcript:
1Fracture mechanics, Mohr circles, and the Coulomb criterion (Stress and failure)
2Introduction to Fracture Mechanics In this lecture, we will be focusing on faults:How they formDefinitionsStress statesFault strainHow we can use faults to tell us things about the geologic historyRepresentative structures on planetary bodies
3DefinitionsFracture: a pair of distinct surfaces that are separated in a material; ‘‘a structure defined by two surfaces or a zone across which displacement occursA joint is a fracture that exhibits only opening displacement (Mode I)A fault is a planar zone along which shear displacement occurs (Mode II)
5StressStress is force/area, units of pressure (most often bars or pascals)σ = F/ANormal stress (σ) acts perpendicular to a surfaceShear stress (τ) acts parallel to a surface
6StressThe “on-in” convention says that the stress component σij acts normal to the ‘i’ direction and parallel to the ‘j’ direction:Normal stresses i = jShear stresses i ≠ j
7Stress Principle stresses have magnitudes and directions σ1: maximum compressive stressσ2: intermediate compressive stressσ3: minimum compressive stressPrinciple stresses act on planes that do not ‘feel’ shear stress, i.e., they are normal stresses
8StressCalculate principle stresses from an arbitrary remote stress state with normal stresses σxand σy, parallel to the x and y axis, respectivelyAnd the orientation of the 2 mutually perpendicular principle planes
9Stress 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.SOLUTIONSubstitute the values of normal and shear stress into equation 1 to obtainσmax,min=1.294MPa and –0.944MPaNow 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
10Resolving 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
11Exampleσ1 = 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 MPaWhich way is the fault going to slip?Left-lateral
12Mohr CirclesMean stress: describes the confining stress experienced by rock at some depth(σ1 + σ3)/2Differential 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
14Mohr 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.
15Mohr circles and effective stress Effective stress is the normal stress reduced by the pore fluid pressureσn*=σn – pfPore 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)
16Coulomb criterionDefines the amount of shear stress needed to overcome the frictional resistance of a fault, leading to slipWhere 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 depthCohesion describes the minimum amount of shear stress needed to start slip when the normal stress is tiny but compressive.
17Coulomb criterion Equation of a line! C0 is the y-interceptμ is the slope of the lineIf shear stress is greater than frictional resistance, the fault will slip. If not, no slip.
18Coulomb 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.
19Coulomb 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 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.SOLUTIONWe find that σn = MPa (compressive) and τ = MPa (left-lateral).Plugging these numbers into the Coulomb criterion, we find that1.640 MPa > (0.58) (2.175) Mpa1.640 MPa > 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.
20Griffith CriteriaTensile failure (i.e., Mode I) occurs when the local stress at the most optimally oriented flaw attains a value characteristic of the materialUniaxial tensile failure strength:E: Young’s modulusν: Poisson’s ratioγ: energy required to create new crack wallsa: flaw half lengthGc: critical strain energy release rateEnded here
21Griffith 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
22Andersonian 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:NormalThrust or reverseStrike-slip
23Andersonian Fault Mechanics For all faults:σ1 > σ2 > σ3Fault forms at an angle θ from σ1Thrust fault:σ1 = σHσ2 parallel to fault strikeσ3 = σvStrike-slip faultσ2= σvσ3 = σhNormal faultσ1 = σv
24Andersonian Fault Mechanics σv = weight of overburden or ρgzCube of rock subjected to vertical stress from overburden will extend in x and y directions (horizontal): Poisson ratioν = horizontal expansion/vertical shorteningν< 0.5The 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/μθ = 0.5(tan-1(1/μ))
25This 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?THRUSTSTRIKE-SLIPNORMAL
26Strain Stress σ causes strain ε Strain is non-dimensional Strain is any change in shape, volume, or orientation of a rock volumeContractional normal strain perpendicular to σ1 and extensional normal strain perpendicular to σ3Normal faulting results in extensional strain (horizontal)Thrust faulting results in contractional strain (horizontal)Extensional strain (vertical)
27StrainStrain can be calculated by doing a 1D traverse across a set of faultsCalculate strain based on fault geometry
28Strain Extension Add these up for every fault in the traverse Need depth of graben (d)Fault dip angle (θ)h = d/tanθAdd these up for every fault in the traverse
29Strain 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!
30Strain, the right wayWhere 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 angleThis method includes all mapped faults and is not dependent on the location of 1D traverses.
33Planetary Tectonics: Normal Faults Primary extensional morphologies on the planets:grabensingle normal faultsGraben are the down-dropped blocks between two antithetic NF.Single normal faults are rare.33
34Planetary Tectonics: Normal Faults Where are NF found?MoonMarsVenusEarthMercuryAsteroidsIcy satellites34
35Planetary Tectonics: Normal Faults How do these NF form?riftingdike intrusionbasin loadingregional-scale uplifttidal stressesimpacts35
36Planetary Tectonics: Normal Faults Rare individual NFBasin loading or from Imbrium?Grew from small segments from S to Nopposite direction if formed by Imbrium impactLikely not from basin loading, since it is on the edge of Nubium, too young, and straightMoon36
37Planetary Tectonics: Normal Faults Valles Marineris: the largest canyon system in the solar systemrift valley10 km wide5 km deep3000 km longCrustal extension along large-scale normal faults related to Tharsis volcanism and heat productionMars37
38Planetary Tectonics: Normal Faults Pantheon Fossaeset of ~radial graben near center of Caloris basin, MercuryHypotheses:Formed as a result of the impactDike intrusionBasin interior upliftPF 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)]40 kmMercury38
39Planetary Tectonics: Normal Faults Grooveslongparallelnarrowtidal or thermal stresses?impact-related?Phobos (Mars)39
40Planetary Tectonics: Normal Faults Venusian chasma (rift)very longhigh reliefrelatively young features?Regional-scale crustal extension along antithetic normal faultsAssociated with volcanoes, likely due to uplift from underlying plume~100 kmVenus40
41Planetary Tectonics: Normal Faults EarthBasin 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 Ma41
43Planetary Tectonics: Thrust Faults Primary contractional morphologies on the planets:wrinkle ridgeslobate scarpsWrinkle 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.43
44Planetary Tectonics: Thrust Faults Where are TF found?MoonMarsVenusEarthMercuryAsteroids44
45Planetary Tectonics: Thrust Faults How are these TF formed?basin loadinglava cooling and contractionimpacts(global contraction)45
46Planetary Tectonics: Thrust Faults Hesperia Planumtype locality for Hesperian epochLava plains likely interbedded with pyroclastic deposits (from Tyrrhena Patera)Multiple generations of WR indicate several temporally and spatially distinct episodes of compression~25 kmMars46
47Planetary Tectonics: Thrust Faults Wrinkle ridges in Mare CrisiumShow a buried craterCan give an idea of depth of mare fill in Crisium Basin~15 kmMoon47
48Planetary Tectonics: Thrust Faults Venusian wrinkle ridges appear to preferentially form in topographic lowsCompressional stress likely from cooling and contraction in a topographic lowSome WR terranes show more than one orientation of WRmore complicated and varied stress field~40 kmVenus48
49Planetary Tectonics: Thrust Faults Lobate scarp (Hinks Dorsum) on the asteroid ErosModeling gives fault parametersDepth = 250 m90 m of offsetNear-surface shear strength ~1 – 6 MPaFormed by impact-induced compression~6 kmWatters et al., GRL 2011Eros49
50Planetary Tectonics: Thrust Faults Himalayan fold-and-thrust beltLarge-scale continental collisionBegan ~50 MaPeak of Everest is LIMESTONE~2400 km of India already ‘lost’1500 km of India subducted over the next 10 Myr~300 kmEarth50
52Planetary Tectonics: Strike-slip Faults Strike-slip fault morphologylinearlongStrike-slip faults are rare on planets other than Earth.52
53Planetary Tectonics: Strike-slip Faults Where are SSF found?EarthMarsVenusIcy satellites53
54Planetary Tectonics: Strike-slip Faults How are these SSF formed?plate motion (Earth only)lateral movement of the lithospheretidal stressesaccommodation structures w/WRimpacts?54
55Planetary Tectonics: Strike-slip Faults San Andreas: best-studied fault zone on EarthFarallon plate and spreading center subducted under N. American platedevelopment right-lateral transform (SSF) that propagated along the continental marginFormed~28 Ma470 km of offset inferredEarth55
56Planetary Tectonics: Strike-slip Faults Martian strike-slip faultsNoachian in age and continued for ~ 1 Gyrused to estimate crustal deviatoric stress magnitude and orientation during early Mars’ historyStrike-slip faulting caused by loading and a background compressional stressAndrews-Hanna et al., JGR 2008Mars56
57Planetary Tectonics: Strike-slip Faults Ovda Regio, in Aphrodite TerraFormed as a transform fault from collisional zone in the north (‘escape tectonics,’ a la Tibet)~75 km of left-lateral strike-slip motionVenus57
58Planetary Tectonics: Strike-slip Faults EuropaAndrogeos Linea350 m highStrike-slip movement from tidal deformation and internal convection58
59Planetary Tectonics: Strike-slip Faults Tiger stripes130 km long35 km apart2 km wide500 m deepStrike-slip faults with displacements of 0.5 m/event (Smith-Konter and Pappalardo, 2008)Associated with high heat flow and water-vapor plumesTidal stresses from Saturn at periapsisEnceladus59