Mechanics of Earthquakes and Faulting Thursday 27 Aug. 2015 - PowerPoint PPT Presentation

Mechanics of Earthquakes and Faulting Thursday 27 Aug. 2015 www.geosc.psu.edu/Courses/Geosc508 •Surface and body forces •Tensors, Mohr circles. •Theoretical strength of materials •Defects •Stress concentrations •Griffith failure criteria

Dislocation model for fracture and earthquake rupture Dislocation model, circular crack For an increment of stress ( Δσ ), how much slip occurs between the crack faces ( Δ u), and how does that slip vary with position (x, y) and crack radius (c) Δ u ( x , y ) = 24 7 π Δ σ 2 − x 2 + y ( 2 ) c µ c or r Relation between stress drop and slip for a circular dislocation (crack) with radius r For ν =0.25, Chinnery (1969) •Importance of slip: e.g., M o = µ A u

Body forces act on every mass element of a body. Surface forces, or tractions, act only along boundaries of a body. Stress and Transformation of Stress (From One Coordinate System to Another) In general we have 9 components of stress in 3d; and six of these are independent. Why only 6 independent? z τ zz Need 9 components to fully specify the τ yz stress state. τ xz τ zy These components make up a tensor. Stress is a 2 nd rank tensor. τ yy τ zx Vector is a 1 st rank tensor τ xy y Scalar is a 0 th rank tensor τ yx τ xx x Right-handed cartesian system and a cube of dimensions dx, dy, dz

Nine components of the stress tensor z τ zz τ xx , τ xy , τ xz τ yx , τ yy , τ yz τ zy τ yz τ zx τ zx , τ zy , τ zz τ xz τ yy y τ xy τ yx Convention: first index refers to plane (face perpendicular to that axis), second index τ xx x refers to resolved direction of force, τ yx , τ 12 We can apply an independent force to each of the surfaces. F y is the force on the surface perpendicular to the y face. Force is a vector, so it can be decomposed into it ’ s components in the x, y, and z directions.

Transformation of Stress From One Coordinate System to Another •Resolving the applied stress onto a plane, or set of planes, in a different orientation y Segment dy of area: A cos α Plane P of area: A τ xy τ xx α x Segment dx of area: A sin α τ yx τ yy The forces on plane A must balance those on segments dx and dy

Stress Transformation y Segment dy of area: A cos α The forces on plane A τ xy must balance those on Plane P of area: A segments dx and dy τ xx α x Segment dx of area: A sin α τ yx τ yy The force in a direction normal to P ( σ A) has contributions from each of the four stress components: • Force = Stress x Area • 1) the shear force along dx is and it ’ s component normal to P is τ yx A sin α cos α • 2) the normal force along dx is and it ’ s component normal to P is τ yy A sin α sin α • 3) the shear force along dy is and it ’ s component normal to P is τ xy A cos α sin α • 4) the normal force along dy is and it ’ s component normal to P is τ xx A cos α cos α Force Normal to P: A σ = + τ yy + τ xy + τ xx τ yx

Stress Transformation Force Normal to P: A σ = τ yx A sin α cos α + τ yy A sin α sin α + τ xy A cos α sin α + τ xx A cos α cos α This can be simplified by eliminating A, using τ xy = τ yx and using the identity 2 sin α cos α = sin 2 α Normal Stress on Plane P: σ = τ xx cos 2 α + τ xy sin 2 α + τ yy sin 2 α Shear Force on P: A τ = τ yx - τ yy - τ xy + τ xx This can be simplified to: Shear Stress on Plane P: τ = ( τ xx - τ yy ) cos α sin α + τ xy (sin 2 α - cos 2 α ) •Stress components are a function of coordinate frame and orientation in 2D •Principal Stresses τ xx , 0 •Shear stresses vanish, only normal stresses 0 , τ yy, •By convention, maximum principal stress is σ 1 and σ 1 > σ 2 > σ 3 , compression is positive

Stress Transformation Shear and Normal Stress on a Plane of Arbitrary Orientation --written in terms of P rincipal Stresses : σ = τ xx cos 2 α + τ xy sin 2 α + τ yy sin 2 α 2D τ = ( τ xx - τ yy ) cos α sin α + τ xy (sin 2 α - cos 2 α ) τ xx , 0 0 , τ yy, σ = τ xx σ = σ 1 cos 2 α + σ 2 sin 2 α , Normal Stress + τ yy τ = ( τ xx - τ yy ) τ = ( σ 1 - σ 2 ) cos α sin α , Shear Stress Use trig. identities such as cos 2 α = 1 – 2 sin 2 α and sin 2 α = 2 sin α cos α ( ) ( ) σ 1 + σ 2 σ 1 − σ 2 Note that these relations make use of the σ = cos2 α + 2 2 mean stress and the differential stress

Shear and Normal Stress on a Plane of Arbitrary Orientation --written in terms of P rincipal Stresses : Mohr Circle . ( ) ( ) σ 1 + σ 2 σ 1 − σ 2 σ = cos2 α + σ 1 2 2 σ 2 ( ) σ 1 − σ 2 plane P 2 τ P ( ) Length = σ 1 − σ 2 sin 2 α 2 2 α σ 2 σ 1 σ Length = ( ) σ 1 + σ 2 2

Shear and Normal Stress on a Plane of Arbitrary Orientation --written in terms of P rincipal Stresses : Mohr Circle . ( ) ( ) σ 1 + σ 2 σ 1 − σ 2 σ 1 σ = cos2 α + 2 2 σ 2 ( ) σ 1 − σ 2 plane P 2 τ P Length = 2 α 2 α σ 2 σ 1 σ Length = ( ) σ 1 + σ 2 2

Coulomb-Mohr Failure Criterion τ = τ o + µ ’ σ n where τ is shear stress τ o is ‘ cohesion, ’ σ 1 µ ’ is the coefficient of internal friction and σ n is normal stress σ 2 plane P τ τ = τ o + µ ’ σ n The parameter µ ’ describes the τ o 2 α effect of normal stress on shear 2 α strength. σ 2 σ 1 σ Pressure-dependent brittle failure Failure stress is higher for things under higher normal stress.

Theoretical strength of materials •Defects •Stress concentrations •Griffith failure criteria •Energy balance for crack propagation •Stress intensity factor Start by thinking about the theoretical strength of materials –and take crystals as a start. The strength of rocks and other polycrystalline materials will also depend on cementation strength and grain geometry so these will be more complex.

Theoretical strength, σ t , of simple crystals: Bonds must break along a lattice plane a λ /2 Consider a tensional stress field, and take a as the equilibrium lattice spacing. Approximate the region around the peak strength as a sinusoid, wavelength λ Then, for small changes in lattice spacing: the rate of stress change is related to E.

σ t = E λ Tensile Strength of single x ’ l, by our approximation: 2 π a The strain energy and stress is zero at thermodynamic equilibrium, which occurs at r= 3a/2 and since a ≈ λ , the theoretical strength is about E/2 π . (See Scholz, Ch. 1.1 for additional details). σ t ≈ E 2 π •This type of calculation was carried out in the early 1900 ’ s and people immediately realized that there was a problem. •Experiments showed that E was on the order of 10 ’ s of GPa, whereas the tensile strength of most materials is closer to 10 ’ s of MPa. •Griffith proposed a solution in two classic papers in the early 1920 ’ s –but the proof of his ideas had to wait until the invention of the electron microscope. Bottom line: Defects. Defects severely reduce the strength of brittle materials relative to the theoretical estimate. Flaws exist at all scales from atomic to the specimen size (laboratory sample size or continent scale, in the case of plate tectonics)

Stress concentrations around defects cause the local stress to reach the theoretical strength. Two types of defects cause two types of deformation: • cracks and crack propagation lead to brittle deformation; • dislocations and other types of atomic misregistration lead to plastic flow and ‘ ductile ’ deformation. Brittle deformation generally leads to catastrophic failure and separation of lattice elements. Plastic flow produces permanent deformation without loss of lattice integrity. Scholz generalizes these modes of deformation to make a connection with lithospheric deformation. The upper lithosphere deforms primarily by brittle mechanisms and can be referred to as the schizosphere (lit. the broken part), whereas the lower lithosphere deforms by ductile mechanisms and can be classified as the plastosphere .

Rheology and Deformation. Definitions. The terms brittle and ductile can be defined in a number of ways. One def. is given above. Another important operational definition involves the stress-strain characteristics and the dependence of strength on mean (or normal stress). Ductile, pressure insensitive Yield strength, σ y Brittle, pressure sensitive Mean (normal) stress Brittle and Ductile (or plastic) deformation can be distinguished on the basis of whether the yield strength depends on pressure (mean stress or normal stress).

Rheology and Deformation. Definitions. The term ‘ brittle ’ is also used to describe materials that break after very little strain . Shear or differential stress, σ <1% Strain, ε , Fracture toughness describes a material ’ s ability to deform without breaking. •Brittle materials (like glass or ceramics) have low toughness. •Plastics have high toughness