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

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Presentation on theme: "Fracture mechanics, Mohr circles, and the Coulomb criterion (Stress and failure)"— Presentation transcript:

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

2 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

3 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)

4 Fracture modes Mode I: opening displacement (joints) Mode II: in-plane shear (faults) Mode III: out-of-plane shear (scissors, tearing)

5 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

6 Stress The “on-in” convention says that the stress component σ ij acts normal to the ‘i’ direction and parallel to the ‘j’ direction: Normal stresses i = j Shear stresses i ≠ j

7 Stress Principle stresses have magnitudes and directions σ 1 : maximum compressive stress σ 2 : intermediate compressive stress σ 3 : minimum compressive stress Principle stresses act on planes that do not ‘feel’ shear stress, i.e., they are normal stresses

8 Stress Calculate principle stresses from an arbitrary remote stress state with normal stresses σ x and σ y, parallel to the x and y axis, respectively And the orientation of the 2 mutually perpendicular principle planes

9 Stress example Given values of σ x = 1.2 MPa (compression positive), σ y = – 0.85 MPa, and τ xy = 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 obtain σ max,min =1.294MPa and –0.944MPa Now substitute the same initial values of stress into equation 2 to obtain θ p =11.6° and 101.6° Add 90º to get second plane orientation

10 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

11 Example σ 1 = 5 MPa, σ 2 = 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? σ n = 2 MPa (compressive) and τ = 1.7 MPa Which way is the fault going to slip? Left-lateral

12 Mohr Circles Mean stress: describes the confining stress experienced by rock at some depth ( σ 1 + σ 3 )/2 ( σ 1 + σ 3 )/2 Differential stress: describes the greatest amount of stress change that a rock can withstand without breaking ( σ 1 – σ 3 ) Mohr circles is a geometric representation of these equations used to determine when rock will fracture or when faults will slip

13 Mohr circles

14 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 σ 1 and σ 3 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.

15 Mohr circles and effective stress Effective stress is the normal stress reduced by the pore fluid pressure σ n * = σ n – p f 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)

16 Coulomb criterion Defines the amount of shear stress needed to overcome the frictional resistance of a fault, leading to slip Where C 0 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.

17 Coulomb criterion Equation of a line! C 0 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.

18 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.

19 Coulomb criterion practice Given the values of remote principal normal stresses σ 1 = 4.8 MPa (compression positive), σ 2 = 1.15 MPa, and an angle to the normal to the plane from the σ 1 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. SOLUTION We find that σ n = MPa (compressive) and τ = MPa (left-lateral). Plugging these numbers into the Coulomb criterion, we find that MPa > (0.58) (2.175) Mpa 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.

20 Griffith Criteria Tensile 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

21 Griffith Criteria The Griffith criterion and Mohr circles showing (a) Uniaxial tensile failure, σ 1 = 3 T 0 (at σ 2 = T 0 ); (b) uniaxial compressive failure, σ 1 = 8 T 0 (at σ 2 = 0); and (c) transition stress from crack growth to frictional sliding, σ 1 ≅ 4.5 T 0. Pure Mode I failure is predicted where the Mohr circle touches the Griffith criterion at exactly one point

22 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

23 Andersonian Fault Mechanics For all faults: σ 1 > σ 2 > σ 3 Fault forms at an angle θ from σ 1 Thrust fault: σ 1 = σ H σ 2 parallel to fault strike σ 3 = σ v Strike-slip fault σ 1 = σ H σ 2 = σ v σ 3 = σ h Normal fault σ 1 = σ v σ 2 parallel to fault strike σ 3 = σ h

24 Andersonian Fault Mechanics σ v = weight of overburden or ρ gz 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: Tan2 θ = 1/ μ θ = 0.5(tan -1 (1/ μ ))

25 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

26 Strain Stress 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 σ 1 and extensional normal strain perpendicular to σ 3 Normal faulting results in extensional strain (horizontal) Thrust faulting results in contractional strain (horizontal) Extensional strain (vertical)

27 Strain Strain can be calculated by doing a 1D traverse across a set of faults Calculate strain based on fault geometry

28 Strain Extension Need depth of graben (d) Fault dip angle ( θ ) h = d/tan θ Add these up for every fault in the traverse

29 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!

30 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/sin θ ), 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.

31 Tour of Tectonics


33 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.

34 Where are NF found? Moon Mars Venus Earth Mercury Asteroids Icy satellites Planetary Tectonics: Normal Faults

35 How do these NF form? rifting dike intrusion basin loading regional-scale uplift tidal stresses impacts Planetary Tectonics: Normal Faults

36 Rare individual NFRare individual NF Basin loading or from Imbrium?Basin loading or from Imbrium? Grew from small segments from S to NGrew from small segments from S to N opposite direction if formed by Imbrium impactopposite direction if formed by Imbrium impact Likely not from basin loading, since it is on the edge of Nubium, too young, and straightLikely not from basin loading, since it is on the edge of Nubium, too young, and straight Moon Planetary Tectonics: Normal Faults

37 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 Planetary Tectonics: Normal Faults

38 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 Planetary Tectonics: Normal Faults

39 Grooves long parallel narrow tidal or thermal stresses? impact-related? Phobos (Mars) Planetary Tectonics: Normal Faults

40 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 Planetary Tectonics: Normal Faults

41 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 Earth Planetary Tectonics: Normal Faults

42 THRUST FAULTING Mercury ~40 km

43 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. Planetary Tectonics: Thrust Faults

44 Where are TF found? Moon Mars Venus Earth Mercury Asteroids Planetary Tectonics: Thrust Faults

45 How are these TF formed? basin loading lava cooling and contraction impacts (global contraction) Planetary Tectonics: Thrust Faults

46 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 Planetary Tectonics: Thrust Faults ~25 km

47 Wrinkle ridges in Mare Crisium Show a buried crater Can give an idea of depth of mare fill in Crisium Basin Moon ~15 km Planetary Tectonics: Thrust Faults

48 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 Planetary Tectonics: Thrust Faults ~40 km

49 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 Planetary Tectonics: Thrust Faults ~6 km

50 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 Planetary Tectonics: Thrust Faults ~300 km

51 STRIKE-SLIP FAULTING Earth ~ 110 km

52 Strike-slip fault morphology linear long Strike-slip faults are rare on planets other than Earth. Planetary Tectonics: Strike-slip Faults

53 Where are SSF found? Earth Mars Venus Icy satellites Planetary Tectonics: Strike-slip Faults

54 How are these SSF formed? plate motion (Earth only) lateral movement of the lithosphere tidal stresses accommodation structures w/WR impacts? Planetary Tectonics: Strike-slip Faults

55 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 Planetary Tectonics: Strike-slip Faults

56 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 Planetary Tectonics: Strike-slip Faults

57 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 Venus Planetary Tectonics: Strike-slip Faults

58 Androgeos Linea 350 m high Strike-slip movement from tidal deformation and internal convection Europa Planetary Tectonics: Strike-slip Faults

59 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 Planetary Tectonics: Strike-slip Faults

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