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CU-Boulder GEOL 5700 - Rock Rheology

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Rock RheologyGEOL 5700 Physics and Chemistry of the Solid EarthReferences:• Turcotte and Schubert, Geodynamics, Sections 2.1,-2.4, 2.7, 3.1-3.8, 6.1, 6.2, 6.8, 7.1-7.4.• Jaeger and Cook, Fundamentals of Rock Mechanics, Chapman and Hall, gives a rathermore complete overview of elasticity (most of this in Ch. 2, 4).• Also Sleep and Fujita’s Principals of Geophysics provides this in tensor notation (Ch. 9)• Section 1.3 of Ka rato is relevant to the discussion of viscosity.This will be a quick overview of the pieces we need to sensibly discuss the deformation of theEarth. We start with the easiest piece, elasticity, and move on to more fluid-like rheologies.ElasticityThe simplest form of elastic behavior is a spring. As you should recall from physics, the forceexerted by a spring is proportional to the displacement of the end of the spring:F = k x (1)Of course, within the Earth we need to deal with stresses, which is simply the force per unit area,and strain, which is the fractional change in length relative to an unstressed starting length.Stresses are usually denoted by σ (sometimes by τ), strains by ε, and subscripts denote thecomponent. Thus the normal stress on a face normal to the x axis (which is therefore directedalong the x axis as well) is denoted σxx. Stresses are a force per unit area and so are (in SA units)N/m2, which is called a Passcal (Pa). Strains are a ratio and therefore are dimensionless. Strainsare often most easily thought of in terms of the displacements. If the displacement field w(x,y,z)has components wx, wy, and wz, then€ εxx=∂wx∂x,εyy=∂wy∂y,εzz=∂wz∂zεxy=εyx=12∂wx∂y+∂wy∂x      εxz=εzx=12∂wx∂z+∂wz∂x      εyz=εzy=12∂wy∂z+∂wz∂y      (2)Note that these are ignoring higher order terms, so these are correct for infinitesimal objects.Also note that the shear strains εxy, εxz and εyz are angular changes as a rock is deformed; exy canbe thought of as motion of the x face in the y direction. Note that rotational terms are ignored formost applications. The rotational terms (ωxy, ωxz, ωyz) have a minus sign in (2) where theequivalent shear strains have a plus.Under these terms, the equivalent expression to (1) for a rock under uniaxial stress isσxx = E εxx(3)where E is Young’s modulus (which is defined as the ratio of stress to strain for a rock underuniaxial compression). Of course, a rock is a three dimensional object, and when you squish it inone direction, you expect that the other directions might change. How much the rock squishesout is determined by Poisson’s ratio, which is the ratio of lateral expansion to longitudinalcontraction, -εyy/εxx if we have an isotropic medium under uniaxial compression along the x axis.If a material is incompressible (its volume doesn’t change) the Poisson’s ratio is 0.5.A more general version of basic isotropic elasticity has to include shear stresses as well, whichare the forces parallel to the face. So σxy is the shear stress on the face normal to x directed alongthe y axis. Because we are assuming equilibrium, there can be no net torque on the rocks, and soσxy = σyx.€ Eεxx=σxx−ν σyy+σzz( )Eεyy=σyy−ν σxx+σzz( )Eεzz=σzz−ν σxx+σyy( )Eεxy= 1+ν( )σxyEεxz= 1+ν( )σxzEεyz= 1+ν( )σyz(4)Note that it is possible for all the coefficients to vary with orientation (including some that arezero in this isotropic formulation). Thus it is possible in a highly anisotropic medium to need 21coefficients instead of the 2 terms E and Poisson’s ratio (ν) to describe the relationship of stressto strain.Our isotropic case can also be expressed as stresses defined by the strains:€ σxx= 2G +λ( )εxx+λ εyy+εzz( )σyy= 2G +λ( )εyy+λ εxx+εzz( )σzz= 2G +λ( )εzz+λ εxx+εyy( )σxy= 2Gεxyσxz= 2Gεxzσyz= 2Gεyz(5)where λ and G are Lame’s parameters, but G is specially known as the shear modulus (ormodulus of rigidity). With some juggling it is straightforward to find that€ λ=Eν1+ν( )1− 2ν( )G =E2 1 +ν( )(6)A final important parameter is the bulk modulus (or incompressibility). It controls how thevolume of the rock changes with pressure and is defined as€ K =σxx+σyy+σzzεxx+εyy+εzz=λ+23G=E3 1 − 2ν( )(7)Aside: One has to be a bit careful in conventions, as earth scientists routinely use positive forcompressional stresses, and occasionally the polarity for the shear stresses reverses (engineersroutinely use positive for tensile stresses). A related problem is that the shear strain, εxy, is halfthe common engineering shear strain, γxy.Stresses and strains can be rotated into a frame where the shear stresses and shear strains go tozero. The axes in this case become the principal stress (or strain) axes; the magnitudes of thesestresses (or strains) are denoted by subscripts 1 to 3 (e.g., σ1, σ2, σ3). Usually the mostcompressive stress is σ1, the least σ3, but again this convention is occasionally reversed.As stresses increase, at some point the rock fails. Failure generally depends on the increase inthe deviatoric stress, |σ1- σ3|, and failure occurs at a stress termed the yield stress. (You shouldknow that the deviatoric stress is twice the maximum shear stress). A simple case is Coulombfailure, which is a modified version of simple frictional laws you may have seen in physics(we’ll come back to that later in the course). Beyond failure, many things can happen. First andmost generally, deformation becomes permanent: remove the stress and the rock fails to return toits original shape. Elastic deformation is always reversible: remove the stress and strains go tozero. But other deformations are not so easily reversed. Material beyond the yield stress canlose a fair amount of strength, but a more common case is one where deformation continues aslong as stress is applied. This case is called plastic deformation; a perfectly plastic deformationwill have the strain continue to as long as the yield stress is applied. Material that behaveselastically until a yield stress is applied and beyond that behaves plastically is termed an elastic-plastic material.Seismology. Before we leave elasticity, one quick note. You can take Hooke’s Law (eqn. 4) andcombine it with the equations for equilibrium (F=ma, in essence, but allow a to not be 0 as weusually do in tectonophysical calculations) and you will derive the wave equation, which


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