Rock mechanics. Master 1, Course PAX7STAD. University of Grenoble I Mai-Linh Doan

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1 Rock mechanics Master 1, Course PAX7STAD University of Grenoble I Mai-Linh Doan Chapter I : Stress & Strain (M.-L. Doan) Tutorial I : Stress & Strain (M.-L. Doan) Chapter II : Elasticity (M.-L. Doan) Tutorial II : Elasticity (M.-L. Doan) Partial exam + Lecture III : Fracture & Friction (M.-L. Doan) Chapter III : Fracture & Friction (M.-L. Doan) Lab visit + Tutorial III : Fracture & Friction (M.-L. Doan) Chapter IV : Other Rheologies (M.-L. Doan) ?? Final Exam

2 Chapitre 1 Force, Stress, Strain Objectives of this chapter : (1) Understand the concepts of strain and stress, (2) Be able to switch from forces to stress, (3) Learn how to use the Mohr diagram to express stresses on a plane (or a fault), whatever its direction. 1.1 Mechanics of continuous media In high school, you have learned how to predict and describe the movement of a particle. But can you apply the formalism of point mechanics to objects like the folds near L Alpes d Huez (Figure 1.1? (answer : no!, why?) To describe the deformation of homogeneous media, we shall divide the medium into small sub- Figure 1.1 Deformation patterns (folds and faults) in the Bourg d Oisans massif, France. particles (mesoscopic particles, as in fluid mechanics) and consider how these particles move and deform. To do so, two new concepts stress instead of force, and strain instead of displacement need to be introduced. 1.2 Force and stress The concept of force The concept of force is usually introduced in the framework of point mechanics when studying the 2 nd Newton s law that relates the acceleration to the force applied to a particle. For a particle of mass m and acceleration a, this law states that m a = Σ F (1.1) 1

3 where F denotes one of the forces applied on the particle. Therefore, comes the definition of a force : A force is a mechanical action that modifies the motion of a particle. One may distinguish between : Body forces : they act throughout the volume of a solid (proportional to its volume or mass). One famous example is weight. W = m g, where g is the gravity acceleration. Another example is the electromagnetic force that applies within a whole body containing electrical charges. Surface forces : they act on the surface area bounding an element of volume. They arise from interatomic forces exerted by material on one side of the surface onto material on the opposite face The concept of stress Let consider a very small surface S on which a surface force applies. If the surface is small, the load is homogeneous. The magnitude of the surface force is directly proportional to the area of the surface on which it acts. The proportionality factor is called stress. Stress is a microscopic description of surface forces. We will present later a more rigorous mathematical presentation of stress (equation 1.2). But before we have to be able to describe a surface in a 3D space. To do so, we need to introduce some sign conventions Conventions Consider a solid with an external surface which is submitted to a distribution of forces. Convention for surface orientation When zooming on a surface, a tangent plane at every point can be defined. This plane is defined by a normal vector labeled n. There are two possibilities to orient this surface ; we follow the usual convention where the normal of the surface n exits the surface. As the surface force is dependent of the surface area S, it is convenient to introduce the surface vector S, which is the product of the surface area S > 0 and the surface vector : S = S n. Convention for stress orientation The definition of stress is : F σ = lim S 0 S = d F ds (1.2) The minus sign comes from the convention in geophysics, for which positive normal stress are compressive, whereas the surface normal vector "exits" from the surface. You may have noticed the double arrow above the stress tensor in equation 1.2. One arrow is inherited from the force, the other from the surface normal vector. It means that stress is a matrix (= tensor of rank 2). Hence, stress depends on the direction of the force, and on surface orientation. Stress is independent of the shape of the surface of the mesoscopic particle (it is a derivative). Therefore, we have efficiently replaced the concept of force by the concept of stress. That is the very neat feature of the concept of stress. 2

4 1.2.4 Normal stress and tangential stress When discussing stress applying on a specific surface, one may distinguish between normal and tangential stresses. The magnitude of the normal stress σ n is computed as σ n = lim F n S 0 S = df n ds Note that σ n is an algebraic quantity. It means it can be positive (i.e. compression), or negative (i.e. extension). In geophysics, positive normal stress means compression. That is why we have all the "minus" signs in the equations above. In a general 3D problem, the tangential stress (also called shear stress), noted τ, is a vector quantity. However, in 2D, it reduces to a scalar like σ n. As for σ n, there is a sign convention relative to τ (see Figure 1.2). Remark : Physics and mechanics scientists tend to use an opposite convention than geophysicists. Our geophysical convention is made to get positive stresses when they are compressive (and they are % of the cases you will encounter). (1.3) Figure 1.2 Surface forces acting on a surface (mechanics convention and geophysics convention). For a positive geophysical normal stress, force is compressive, acting in a reverse direction compared to n Stress tensor in 2D General expression of the stress tensor in 2D Stress components can be defined at any point into a material. To illustrate this point, let consider a small rectangular element with dimensions δx, δy, δz (Figure 1.3). The normal stresses are σ xx and σ yy, and the shear stresses are σ xy and σ yx. By convention, the second subscript gives the direction of the force, and the first subscript the direction of the normal to the surface on which the force acts. The force F exerted on a surface ds of normal tensor n is expressed as : that is ( Fx F y ) F = σ n ds (1.4) ( σxx σ = xy σ yx σ yy ) ( nx n y or 1 F x = (σ xx n x + σ xy n y ) ds ) ds (1.5) F y = (σ yx n x + σ yy n y ) ds (1.6) 1. You may notice that the " " operator enables to reduce the ranks of the tensors. A dot product reduces two vectors (2 arrows) into one scalar (0 arrow), and a tensor and a vector (3 arrows overall) into one vector (1 arrow). 3

5 In 2D and for geophysical conventions, the pressure is defined as : Figure 1.3 2D stress applied on the surfaces of a cube. From Turcotte and Schubert, P = σ xx + σ yy 2 Exercise 1.1 : Calculate the force you exert on the Earth s surface when you stand? the normal stress? the pressure? Exercise 1.2 : Calculate the pressure at 10 m depth in a lake containing fresh water? Symmetry of the stress tensor This section uses notions on torque, that are recalled in the appendix. The tangential stresses, also called shear stresses, σ xy and σ yx have associated surfaces forces that tend to rotate the element around the z-axis. The moment (couple in French, torque, angular moment in English, see appendix) exerted by the surface force σ yx δzδx is the product of the force and the moment arm (lever) δy, that is : -σ yx δzδxδy. Note that a torque is positive if the object rotates according to the trigonometric convention. At equilibrium, the element does not rotate, and this first moment is counteracted by a surface force σ xy δzδy with a moment arm δx. This implies that the sum of the torques must be equal to zero : σ yx δzδxδy + σ yx δzδyδx = 0, or that : (1.7) σ xy = σ yx (1.8) Thus, the stress tensor is symmetric. Only three independent components of stress, σ xx, σ yy, and σ xy must be specified in order to prescribe the two-dimensional state of stress. Principal stresses A set of two orthogonal directions (x, y) are principal stress direction if σ xy = 0. We assume that there are only 2 principal stress directions, that are orthogonal to each other in 2D. The existence of the principal stress directions are usually derived from linear algebra (for mathoriented people, principal stresses are the eigendirections of a symmetric real matrix). In TD1, we see a non mathematical, more physical method to prove it. 4

6 $,%(D(,%(,>&)"ED",(,>-&"B&->5(--.&! ==.&! &EA->&?(&-D()$B$(%&$,&"5%(5&>"&D5(-)5$?(&>*(& >K"6%$E(,-$",+#&->+>(&"B&->5(--3& & 4(E+5OH& >(,%& >"& A-(& +,">*(5& )",G(,>$",& >*+,& NA5& By convention, the principal stresses are labeled by decreasing magnitude. In three dimension, they are noted σ 1, σ 2, σ 3, with σ 1 > σ 2 > σ 3. The principal "B&>*(&)+-(&@"A&K$##&(,)"A,>(5<3& stress directions are relative easy to get experimentally (e.g., borehole breakouts, hydraulic fracturing, statistics on focal mechanism, see TD on world stress map). The amplitude of the principal stresses aremechanics much harder to derive (usually by direct Geophysics hydraulic testing, or with a lot of assumptions on borehole == & Rotation of axes! The surfaces on which it may be =@& &! interesting to apply stress are =@ not necessary well oriented relative to arbitrary (x,y) directions. The state=& of stress is dependent on the orientation =& of the coordinate system. We can calculate the three components of stress in a coordinate system x, y rotated by an & angle θ with respect to the (x, y) coordinate system (see TD1) : b. Stress in 2D: General formulation L*(&->+>(&"B&->5(--&$-&%(D(,%(,>&",&>*(&"5$(,>+>$",&"B&>*(&)""5%$,+>(&-@->(E3&R(&)+,&)+#)A#+>(& σ x x = σ xx cos 2 θ + σ yy sin 2 θ + σ xy sin 2θ σ x y = σ yy σ xx >*(&>*5((&)"ED",(,>-&"B&->5(--&$,&+&)""5%$,+>(&-@->(E&=S.&@S&5">+>(%&?@&+,&+,;#(&#&K$>*&5(-D()>& sin 2θ + σ xy cos 2θ (1.9) 2 >"&>*(&=.&@&)""5%$,+>(&-@->(E3& σ y y = σ xx sin 2 θ + σ yy cos 2 θ σ xy sin 2θ With& this expression, & & it is& easy to find the angle θ for which (x,y ) are principal stress directions. == & /! x ' x' %! xx cos # $! yy sin # $! xy sin /# & Figure 1.4 Change of axes during a rotation. / & We just write that the shear stress σ xy is null and solve for θ : 6&9&6& σ x y = 0 = 1 2 (σ yy σ xx ) sin 2θ + σ xy cos 2θ tan 2θ = 2σ xy σ xx σ yy (1.10) The direction θ is known as a principal stress axis. If θ is a principal stress direction, ( θ + π 2 ) is also a principle direction of stress, as tan ( 2 ( θ + π 2 )) = tan (2θ + π) = tan (2θ). Mohr circle Let now assume that (x,y) are principal stress directions and that σ xx = σ 3 and σ yy = σ 1. By definition of a principal stress direction, σ xy = 0. In that case, equations 1.9 simplifies to : σ x x = σ xx cos 2 θ + σ yy sin 2 θ = σ 1 + σ 3 2 σ x y = σ 1 σ 3 2 σ yy σ xx 2 sin 2θ = σ 1 σ 3 sin 2θ 2 σ y y = σ xx sin 2 θ + σ yy cos 2 θ = σ 1 + σ σ 1 σ 3 2 cos 2θ cos 2θ 5

7 This is the equation of a circle in the plane (σ y y,σ x y ). The circle has a radius σ 1 σ 3 2 and a center at the point ( σ 1 +σ 3 2, 0 ). This circle is called the Mohr circle. The Mohr circle is a graphical representation of any 2-D stress state. It was proposed by Christian Otto Mohr in It can be applied to many engineering quantities such as stresses, strains, and moments. The Mohr circle is a very powerful tool. Suppose that you have a state of stress (σ xx,σ yy,σ xy ) at a point P in a body. If you rotate the axis by an angle θ (or consider a plane with a normal vector n tilted by an angle θ), you can predict the new stresses as they lie on the Mohr circle. The use of the equations 1.9 is therefore advantageously replaced by the geometrical construction of the Mohr circle (cf. TD2). A possible way to construct the Mohr circle knowing a state of stress (σ xx,σ yy,σ xy ) is to compute the principal stresses. However, the Mohr circle may be constructed in a more simple way with the following geometrical construction : 1. Draw two perpendicular axes, with the horizontal axis representing normal stress, while the vertical axis represents the shear stress. 2. Plot the state of stress on the x-plane as the point A, whose abscissa (x value) is the magnitude of the normal stress, σ x (tension is negative), and whose ordinate (y value) is the shear stress (anticlockwise shear is positive). 3. Mark the magnitude of the normal stress σ yy on the horizontal axis (tension being negative). 4. Mark the midpoint of the two normal stresses, O (step 1). 5. Draw the circle with radius OA, centered at O (step 2). 6. A point on the Mohr circle represents the state of stress on a particular plane at the point P. Of special interest are the points where the circle crosses the horizontal axis, for they represent the magnitudes of the principal stresses (step 3). The Mohr circle can be used to find the planes of maximum normal and shear stress, as well as the state of stress on known weak planes. Mohr circle on a plane (fault) The Figure 1.6 shows how a Mohr circle helps to assess the stress applied on the surface. The angle θ denotes here the angle relative to the largest principal stress direction σ 1. If you do not know initially the principal stress directions, but have the complete stress tensor, you can use the previous section to build the Mohr circle. σ n = σ 1 + σ σ 1 σ 3 2 cos 2θ τ = σ 1 σ 3 2 sin 2θ Stress tensor in 3D General expression of stress tensor in 3D In 3D, there are six independent quantities σ ij which are the components of a third-order symmetric tensor called the stress tensor. Additional stress components must be defined (those involving the z direction) : σ zz, σ xz, and σ yz. The force F exerted on a surface ds with a normal vector n is expressed as : F = σ n ds (1.11) that is F x F y F z = σ xx σ xy σ xz σ yx σ yy σ yz σ zx σ zy σ zz n x n y n z ds (1.12) 6

8 Figure 1.5 Construction of a Mohr circle in three steps. n n 1 Figure 1.6 Mohr diagram used on a predefined surface. Note that the two parts of the figure are to be kept together, as the right part sets the convention used on the left part. or F x = (σ xx n x + σ xy n y + σ xz n z ) ds F y = (σ yx n x + σ yy n y + σ yz n z ) ds (1.13) F z = (σ zx n x + σ zy n y + σ zz n z ) ds 7

9 Figure 1.7 3D stress tensor applied on a cube. From Turcotte and Schubert, 2002 Symmetry Applying equilibrium on the torque relative to the axis z, one gets as in the 2D case : σ yx = σ xy. Applying equilibrium on the torque relative to the axis y, one gets as in the 2D case : σ xz = σ zx. Applying equilibrium on the torque relative to the axis x, one gets as in the 2D case : σ yz = σ zy. Therefore, σ zx = σ xz, and σ zy = σ yz, so that the matrix is effectually symmetric. The Cauchy stress tensor σ has therefore only 6 independent components, and is reduced to Principal stress directions σ = σ xx σ xy σ xz σ xy σ yy σ yz σ xz σ yz σ zz As in 2D, the transformation of coordinates to principal axes can also be carried out in three dimensions. Three orthogonal axes can always be chosen such that all shear stress components are equal to zero. The normal stresses on planes perpendicular to these axes are the principal stresses, usually noted and sorted as σ 1 > σ 2 > σ 3. When the three principal stresses are equal, the state of stress is isotropic and the principal stress can be identified as the pressure. When the three principal stresses are not equal, the pressure is defined as their mean value p = σxx+σyy+σzz 3. The pressure is invariant to the choice of coordinates. To calculate the principal stresses σ 1, σ 2 and σ 3, one often uses the three invariants of the 3D stress tensor : I 1 = I 2 = I 3 = σ xx + σ yy + σ zz σ xx σ yy + σ yy σ zz + σ zz σ xx σ 2 xy σ 2 yz σ 2 zx σ xx σ yy σ zz + 2σ xy σ yz σ zx σ xx σ 2 yz σ yy σ 2 zx σ zz σ 2 xy Then the characteristic equation of 3-D principle stresses is expressed as : σ 3 I 1 σ 2 + I 2 σ I 3 = 0 (1.14) 8

10 The three roots of this equation are the principle stresses σ 1, σ 2 and σ 3. When they are found it can be shown that the three invariants can be expressed in terms of principle stresses : Mohr circle I 1 = σ 1 + σ 2 + σ 3 I 2 = σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1 I 3 = σ 1 σ 2 σ 3 The Mohr circle may also be applied to three-dimensional stress. In this case, the diagram has three circles, with two inner circles touching the third one, and delimiting a "forbidden zone". The 3D Mohr diagram is much more difficult to use, and is, in fact, scarcely used. It is often reduced to the 2D Mohr diagram, by restricting it to the (σ 1,σ 3 ) plane. That is, we use only the outer circle. Figure 1.8 Mohr circle in 3D. 1.3 Strain Stresses cause solids to deform : stresses produce changes in the distances separating neighboring small elements of the solid. This is the major difference with the particle mechanics you studied in high schools : the system can deform internally. We will first describe two kinds of deformation (longitudinal and tangential). We will then see that in the case of small deformation, we can unify both strain types within a single strain tensor, analog to the stress tensor Longitudinal strain This is the simple strain, which you can visualize as stretching (of a bungee, for instance). Suppose you extend a core (or a bungee) of initial length l 0 by a length l. The strain is then ɛ = dl l 0. (The minus sign is due to the geophysics convention, for which compression is positive, and extension negative). What is the unit of strain? Intuitively, both stretching and compressing induce some volume change. We will show that small volume change can be described as the combined effect of longitudinal strains. Let consider a pure shear strain, as in the Figure 1.9. A specimen of dimensions x 0 and y 0 is stretched by dx in the x direction, and squeezed by dy in the y direction. Both dx and dy are 9

11 positive. For the x direction, l 0 = x 0 and dl = dx. We then get a negative strain, which makes sense, Figure 1.9 Pure shear. since the specimen experiences expansion in the x direction : ɛ xx = dx x 0. (1.15) For the y direction, l 0 = y 0 and dl = dy (the specimen retracts in this direction). We then get a positive strain, which makes sense, since the specimen experiences compression in the y direction ɛ yy = dy y 0 = dy y 0. (1.16) The former surface area was V 0 = x 0 y 0. The new surface is V 1 = (x 0 + dx)(y 0 dy) = x 0 y 0 + y 0 dx x 0 dy dx dy. The last term is supposed negligible (small strain condition) and we obtain the relative volume variation dv V 0 = V 1 V 0 V 0 More generally, for small strains, the change in volume is given by : Shear strain dv V 0 = dx x 0 dy y 0 = ɛ xx ɛ yy. (1.17) = i ɛ ii. (1.18) Shear strain is a strain that acts parallel to the surface of a material that it is acting on. Normal strain, in contrast, acts perpendicular to the surface. There are two definitions of shear strain : the average shear strain (ɛ), and the engineering shear strain (γ). The engineering strain is the most intuitive one of the two. It corresponds to a simple shear strain. For instance, let move tangentially a stack of cards by applying a tangential force at its top. The shear stress will displace the top of the stack of cards of height( y 0 by a very small distance dx. The total shear strain (or engineering strain) is defined as γ = tan dx y 0 ) dx y 0. However, this is not the engineering shear strain we will use to compare with the longitudinal strain. We will introduce a strain by choosing a set of axes so that σ xx = σ yy. With this choice, the symmetry is kept (and this choice highlights the principal stress directions). One sees in the Figure 1.11 that simple shear is indeed a pure shear strain combined with a rotation. In a free body, the application of a shear stress against just one set of parallel surfaces results in a net torque on the body and, for 10

12 "#!!!!!! $%! Figure 1.10 Simple shear. Due to the effect of the tangential force (blue arrow), the rectangle tilts. Note that the stress follows geophysical conventions, and so does the strain. Figure 1.11 Graphical definition of shear strain. equilibrium, i.e. no rotation, an equal and opposite shear stress must exist. The two shear stresses result in equal shear in two directions. In order for the body to exhibit the same overall engineering shear strain, the actual shear strain on each of the two sides must be γ/ Strain tensor We will generalize the two kinds of strain defined above. A deformed solid is one which has experienced a change of shape. When stresses are applied, different points of a solid move with respect to each other and their separation distance changes. Let consider a 2D example. The strain ɛ xx is the change in length of the side parallel to the x-axis divided by the original length of the side. The strain components of a small element of solid can be related to the displacements of the element. Angles between straight lines, such as AB and AC in the undeformed solid, are also modified. These changes in lengths and angles are the basic parameters which allow calculating how ds 2 = AB 2 = dx 2 +dy 2 +dz 2 transforms into (A B ) 2 after deformation. Point A is transformed into A (i. e. AA = u ) and point B, which is infinitesimally close to A, is transformed into B (i. e. BB = u + d u). The displacement vector u varies with position (x, y, z) : u = u(x, y, z). We assume however that u is always small compared to the size of the object considered. We want to calculate : A B 2 = (dx + du x ) 2 + (dy + du y ) 2 + (dz + du z ) 2. Noting that 11

13 Figure 1.12 Undeformed and deformed solid. Deformation is expressed through a change in length (AB A B ) and a change in angle ( ABC A B C ). From Gué guen and Palciauskas. du x = ux x ux ux dx + y dy + z dz and using similar expressions for du y and du z, one derives : A B 2 = dx 2 + dy 2 + dz u x x dx2 + 2 u y y dy2 + 2 u z z dz2 ( ux + y + u ) ( y uy dxdy + x z + u ) ( z uz dydz + y x + u ) x dzdx z + negligible terms ( ) The neglected quantities are of the form ux u y y x dxdy and contain the product of two derivatives u x y and uy x. Because we have assumed that u is a small vector, we can consider that ux y 1, so that partial derivatives are first-order quantities and the neglected terms are of second order. With these assumptions, one finds the following expression for the change in separation between two adjacent points in the medium : A B 2 AB 2 = i j ( ui j + u ) j dx i dx j = i i ( 2ɛ ij ) dx i dx j The minus sign comes again from the geophysical convention : strain is negative in case of an extension, and so the strain for geophysical convention is ɛ ij = 1 2 j ( ui j + u ) j. (1.19) i 1.4 Constitutive laws (introduction to the next lessons) Solids, liquids and gases have stress fields. Static fluids support normal stress (hydrostatic pressure) but will flow under shear stress. Moving viscous fluids can support shear stress (also called dynamic pressure). When deforming a rock sample, strain and stress evolve jointly. The typical curve in a stress-strain diagram is similar to figure We can see different behaviors that will be discussed in the three next chapters : Elasticity is discussed in chapter 2. The elastic regime is characterized by a linear relationship between stress and strain, denoted linear elasticity. This idea was first stated by Robert Hooke in This deformation is reversible and continuous. The classic model of linear elasticity is the perfect linear spring. Although the general proportionality constant between stress and strain in three dimensions is a 4 th order tensor ; when considering simple situations of higher symmetry such as a rod in one dimensional loading, the relationship may often be reduced to applications of Hooke s law with two independent elastic constants. Geological examples are, for example, the bending of the lithosphere (subduction, oceanic crust) or the elastic tectonic energy released in earthquakes. 12

14 Figure 1.13 Typical strain-stress curve for a rock sample. From Wikimedia Commons. Fracture and friction are discussed in chapter 3. If stress is too high, the sample fails and deformation is no more continuous as fractures or shear bands appear. Movement can happen along the newly created fractures (i.e. earthquake), and friction controls the deformation of the sample. The rupture is quite sudden. Otherwise, the deformation can become localized along well-defined bands of deformation (example : compaction bands, shear bands). Plasticity is discussed in chapter 4. Plastic deformation is continuous, but irreversible (example : mantle flow). Ductile materials can sustain large plastic deformations without fracturing. However, even ductile metals will fracture when the strain becomes large enough - this is a result of work-hardening of the material, which causes it to become brittle. Heat treatment such as annealing can restore the ductility of a worked piece, so that shaping can continue. Usually plastic deformation is ratedependent, and instead of defining a relationship between stress and strain, the constitutive relationship relates stress and strain rate. 13

15 Chapitre 2 Elasticity Objectives of this chapter : (1) Assimilate the definition of elasticity, (2) Understand the various elastic coefficients. 2.1 Notions Context When a strain is gradually imposed on a rock sample, the strain-stress curve of Figure 1.13 is typically observed. The linear domain corresponds to the elastic behavior. Experimentally, the elastic domain corresponds to small deformations. When the stress-strain starts to deviate from linearity, a domain of irreversible deformation is reached. The peak of the curve, before failure, is called the strength of the material Definition Rheology expresses how stress and strain are linked. Elasticity is a rheology law that depends on strain only, for which strain is reversible. Usually, elasticity refers to linear elasticity, for which : 1. stress and strain are proportional, 2. strain is reversible Reminder of high school lessons l 0 x l F Figure 2.1 An extended spring retracts with a force proportional to the extension. You have already learned an example of elastic object : the spring. When you extend a spring, a force tends to pull the spring back to its initial length l 0. The distance of pulling is proportional 14

16 to the force F with a proportionality constant k called the stiffness. We consider here the retraction force that opposes the pulling force : F = k(l l 0 ) = k x (2.1) What is the unit of k? For this simple elastic object, the elastic energy can be calculated as : E = W ork = Integral of (F orce Displacement) = 1 2 k (l l 0) 2 = 1 2 k x2 (2.2) Stress and strain in a spring During the previous lesson, we learned about stress and strain tensor : F σ = F Cross area of the spring (2.3) x ɛ = l l 0 l 0 (2.4) However, the stress and strain tensors are not unidimensional, they are described by 2x2 tensors in 2D or 3x3 tensors in 3D. Can we then only use a single coefficient k? What will be the unit of the equivalent to k coefficient? Microscopic interpretation of elasticity Figure 2.2 Potential energy for inter-atomic chemical bonds (Lenhard-Jones model). Let us take the Lenhard-Jones model of chemical bond energy that describes the equilibrium distance between two atoms (Figure 2.2), (e.g., for Si-O : 0.16 nm). Around the equilibrium distance between an internuclear repulsion force and an attractive force, the potential can be represented by a cup-like form that we approximate as a paraboloid 1. By analogy with equation (2.2), we can describe the force between two atoms as an elastic spring with stiffness k. By extending this computation to the whole crystal, we get an estimate of the elastic 1. This is a Taylor expansion of order 2 (Développement limité d ordre deux). 15

17 constants. Note that, if the displacement is large enough, we cannot describe anymore the Lenhard- Jones potential with a parabola curve because the linearity condition is relaxed. Moreover, if the displacement is too large, bonds can break, and will not be recovered : the reversibility condition is lost. That is why it is often said that elasticity describes small displacements, as seen in Figure Elastic constants How many constants? σ xx σ yy σ zz σ xy σ yz σ zx = C xx,xx C xx,yy... C yy,xx C yy,yy ɛ xx ɛ yy ɛ zz ɛ xy ɛ yz ɛ zx (2.5) This is a simplified way to write : σ xx = C xx,xx ɛ xx + C xx,yy ɛ yy + C xx,zz ɛ zz + C xx,xy ɛ xy + C xx,yz ɛ yz + C xx,zx ɛ zx (2.6) This is the famous Hooke s law of linear elasticity 2. In the most general case, there are 6*6=36 elastic constants for a homogeneous medium. To simplify the problem, we impose some constraints. Isotropy The problem is independent of the direction considered. The proportionality matrix can be described with only two parameters. Take any plane, you need to consider the effect when : 1. you push (see section 2.2.2) 2. you shear (see section 2.2.4) In the following section, we will restrict ourselves to isotropic materials. Transverse anisotropy The problem is the same whatever the direction considered along a given plane orientation. Typically, this kind of problem applies to layered media, such a sediments. Then x = y z. The proportionality matrix can be described with only five parameters. Take a horizontal plane, you need to consider the effect when : 1. you push perpendicular to the plane 2. you shear parallel to the plane Take a vertical plane, you need to consider the effect when : 3. you push perpendicular to the plane 4. you shear along a horizontal direction 5. you shear along a vertical direction 2. Robert Hooke ( ) : British bio-astro-physicist who experimentally discovered the linear relationship between stress and strain. His famous Latin citation was "ceiiinosssttuv", an anagram for "ut tensio, sic vis" (such elongation, such force). Despite his huge scientific work, he is best remembered for his rivalty with Newton over the gravitation law. 16

18 2.2.2 Young s Modulus Module de Young 3. F= σ x S δ Suppose you apply a stress σ normal to the face of a solid. If you pull, the solid expands ; if you push, it contracts and the strain is defined as : ɛ = δ l 0 (2.7) l 0 For a compression along the z direction, the Young s modulus is defined as : E = σ zz ɛ zz (2.8) What is the unit of E? Experimental determination of the Young s modulus of a bungee. If the section is 1 cm 2, l 0 = 1 m, l = 1.10 m, M = 10 kg, g = 10, the deformation is equal to 0.1, the stress is 10 6 Pa, and E = 10 7 Pa. Order of magnitude : E rock = GPa (2.9) E soil = GPa (2.10) E granite = 60 GPa (2.11) E limestone = 30 GPa (2.12) The Young s modulus is usually measured with a triaxial rig. In the field, the Young s modulus can be estimated with a Schmidt Hammer Poisson s ratio Coefficient de Poisson 4 In the previous section, we focused on ɛ zz. We now concentrate on the other strain components. First, from the isotropy assumption : ɛ xx = ɛ yy (2.13) When a sample is compressed along one direction, it expands in the other directions. We therefore define the Poisson s ratio with a minus sign. For a uniaxial compression σ zz along the z direction, the Poisson s ratio is defined as : ν = ɛ xx ɛ zz (2.14) For most rocks, 0.1 ν 0.4. Softer rocks have larger Poisson s ratio. An incompressible solid (or a fluid) has a Poisson s ratio equal to ν = 0.5. In most cases, ν 0.3 (in seismology a value ν = 0.25 is often used). We have now the two coefficients necessary to express the relationship between stress and strain. To 3. Thomas Young ( ) is a British physician, physicist and specialist of ancient Egypt (he challenged Champollion in deciphering the hieroglyphs). He introduced the Young s modulus to study the deformation of the eye s cornea. 4. Siméon Denis Poisson ( ) was a French mathematician and physicist. His important works on electricity, magnetism, astronomy, mechanics, probability,... should have made his name already familiar to you. 17

19 simplify, let suppose that the axes are oriented along the principal directions (1,2,3) : or simply : ɛ 1 = σ 1 E ν σ 2 E ν σ 3 E ɛ 2 = ν σ 1 E + σ 2 E ν σ 3 E ɛ 3 = ν σ 1 E ν σ 2 E + σ 3 E (2.15) (2.16) (2.17) ɛ 1 = 1 E (σ 1 ν(σ 2 + σ 3 )) (2.18) ɛ 2 = 1 E (σ 2 ν(σ 3 + σ 1 )) (2.19) ɛ 3 = 1 E (σ 3 ν(σ 1 + σ 2 )) (2.20) Shear modulus Module de cisaillement This modulus links σ xy and ɛ xy. G = σ xy 2ɛ xy (2.21) Why a factor of 2? The shear modulus G was first defined when deforming a sample under a simple shear strain. The amplitude of shear deformation is given by tan(γ). When the deformation is small, tan(γ) = γ. The shear strain is expressed as : F= σ xy x S ɛ xy = 1 ( ux 2 y + u ) y (2.22) x γ and we get simply = 1 (γ + 0) (2.23) 2 G = σ xy γ (2.24) Relationship between G and (E, ν) We have seen that for an isotropic solid, only two parameters describe the elastic behavior of the material. Thus, G can be expressed with E and µ. To do so, we use the fact that simple shear can be described as the combination of pure shear and rotation. Let take a square of size 1 1, whose sides are parallel to the x and y directions. The diagonal has a length l = 2. For pure shear, the principal directions are such that σ l xy is along the direction l and we have σ xy along a direction perpendicular to l (if one squeezes a square along two diagonal corners γ with σ xy and stretch it along the two other corners with σ xy, the deformation is similar to coupling simple shear and rotation). Then, the elongation dl/l is easy to compute : l+dl dl l = σ xy E ν σ xy E = 1 + ν E σ xy = ɛ (2.25) 18

20 We have also : l = = 2 (2.26) l + dl = (1 + γ) γ (1 + γ/2) (2.27) dl l = γ 2 (2.28) Using the definition of equations (2.24) and (2.26) : G = E 2 (1 + ν) (2.29) Bulk modulus Module d incompressibilité volumique This convenient quantity links pressure and volume. It is often introduced in fluid mechanics where the pressure is an usual parameter. Pressure can be linked to the stress tensor, by redefining the pressure as the average confining stress : This will induce a change in volume : p = σ xx + σ yy + σ zz 3 (2.30) dv V = ɛ xx + ɛ yy + ɛ zz = ɛ ii = (2.31) Note that we used here the repeated index notation : when an index is repeated, it means that you have to consider the sum of the three components, with the repeated index replaced by the component. Note again that, in geophysical conventions, compression is considered positive. The bulk modulus is defined as : p K = (2.32) dv/v Relationship between K, E and ν Adding equations (2.18) to (2.20) gives : dv V dv V Combining equations (2.32) and (2.34), we get the relationship : = 1 (3 p ν(6 p)) (2.33) E = 3 p (1 2ν) (2.34) E K = E 3 (1 2ν) (2.35) Important note For incompressible fluids : dv = 0, so that K = and 1 2ν = 0. ν = 1 2 (2.36) 19

21 2.2.6 Lamé coefficients The two Lamé coefficients λ and µ are often used by theoreticians in fundamental mechanics 5. You have to link the stress and strain with a formulation that holds whatever the axes chosen to express them. There are only two ways to express a linear relationship independently of the axes chosen : by using a linear invariant of the strain tensor (i.e. the trace of the tensor) multiplied by the identity matrix ; by simply using the strain tensor itself. The Lamé coefficients are the coefficients in front of these two quantities. On each axis direction : σ = λ (ɛ xx + ɛ yy + ɛ zz ) µ ɛ (2.37) σ xx = λ (ɛ xx + ɛ yy + ɛ zz ) + 2µ ɛ xx = (λ + 2µ)ɛ xx + λɛ yy + λɛ zz (2.38) σ yy = λ (ɛ xx + ɛ yy + ɛ zz ) + 2µ ɛ yy = λɛ xx + (λ + 2µ)ɛ yy + λɛ zz (2.39) σ zz = λ (ɛ xx + ɛ yy + ɛ zz ) + 2µ ɛ zz = λɛ xx + λɛ yy + (λ + 2µ)ɛ zz (2.40) σ xy = 2µ ɛ xy (2.41) σ yz = 2µ ɛ yz (2.42) σ zx = 2µ ɛ zx (2.43) We can write equations (2.37) to (2.43) into a single equation by using the Kronecker index : Equation (2.37) can conveniently be expressed as : δ ij = 1, if i = j (2.44) = 0, if i j (2.45) σ ij = λɛ kk δ ij + 2µɛ ij (2.46) Relationship between (λ, µ) and (E, ν) By combining equations (2.21) and (2.41), we get : µ = G (2.47) The shear modulus is often denoted with µ. As in section 2.2.5, if we add the strain-stress relationship along each principal direction, we get : σ xx + σ yy + σ zz = 3p = (λ + 2µ) + λ + λ = (3λ + 2µ) (2.48) From the definition of the bulk modulus (2.21), we get : and inversely : K = λ = K 2/3µ = 3λ + 2µ 3 (2.49) ν (1 + ν)(1 2ν) E (2.50) We deduce from these equations that the Lamé coefficients have the same order of magnitude as K and E : typically a few tens of GPa. 5. Gabriel Lamé ( ) was a French mathematician who was the first to define the stress ellipsoid. 20

22 Air Water Clay Sandstone Granite Lower mantle V P (m/s) V S (m/s) Table 2.1 Seismic velocities of common fluids and rocks. For fluids, temperature and pressure affects the seismic velocities, whereas for rock weathering, fracturing and confining pressure are the main parameters affecting the seismic velocities Seismic velocities You will see in the seismology course the wave equations for the propagation of elastic waves into a solid or a fluid. Two kinds of waves can propagate, each with a specific velocity : 1. The P -waves are longitudinal waves, with volumetric strain change. ɛ xx 0; ɛ yy = ɛ zz =0, so that σ xx = (λ + 2µ)ɛ xx. With the conservation law, we can derive a wave equation 6, and we get the P wave velocity : λ + 2µ V P = (2.51) ρ 2. The S-waves are shear waves. As for pure shear deformation : σ xy = 2µɛ xy = µγ, we get : V S = µ ρ (2.52) Measuring seismic velocities is relatively easy and is often used to estimate elastic properties of remote materials (see table 2.1) Summary of elastic constants We have seen many elastic coefficients, but we have also demonstrated that only two are necessary to fully express the elastic properties of an isotropic material. Thus, any elastic constant can be written thanks to two other independent elastic constants. Table 2.2 summarizes these relationships for the most used pairs of elastic constants (you do not need to learn by heart all the formulas, but you must know how to use them). Constants E = ν = K = G=µ=ρ V 2 s = λ = ρ V 2 P = (λ, µ) (λ, G) (E, ν) µ 3λ+2µ λ+µ λ 2 (λ+µ) λ + 2µ 3 λ + 2µ E 3 (1 2ν) E 2 (1+ν) E ν (1+ν) (1 2ν) (K, ν) 3 K (1 2ν) 3 K 1 2ν 2 (1+ν) 3 K ν E (E, µ) 2µ 1 µ E 3µ E (K, µ) 9 K µ 3K+µ (1+ν) 3 (3µ E) µ E 2µ E (1 ν) (1+ν) (1 2ν) 3 K 1 ν (1+ν) µ 4µ E 3µ E 3K 2µ 2 (3K+µ) K 2µ 3 K + 4µ 3 (µ, ν) 2 µ (1 + ν) µ 2(1+ν) 3 (1 2ν) µ 2ν 1 2ν Table 2.2 Correspondence between elastic constants. µ 2 2ν 1 2ν 6. This is done in the next chapter. 21

23 2.3 How to solve an elastic problem? An elastic problem consists of solving a set of partial differential equations. This requires also to constrain the boundary conditions and the initial condition Partial differential equations We know how to link displacement u and strain ɛ (chapter 1), strain ɛ and σ (this chapter). We now close the problem by examining how we can link stress and displacement. The governing equation is obtained by considering the conservation of momentum (also known as Newton s third law) on a cube of size (dx,dy,dz) located between x and x + dx. List all the forces applied on this cube. Do not hesitate to draw them. We neglect the volumetric inner forces, and we project the sum of all forces along the x direction : Σf x = σ xx (x) dy dz σ xx (x + dx) dy dz + σ xz (z) dx dy σ xz (x + dz) dx dy + σ xy (y) dx dz σ xy (y + dy) dx dz (2.53) = σ xx x dx dy dz σ xy y dx dy dz σ xz dx dy dz (2.54) z ( σxx = x + σ xy y + σ ) xz dx dy dz (2.55) z To simplify the equation, we use the repeated index convention : σ xi x i = σ xx x + σ xy y + σ xz z (2.56) Static equilibrium We can now write the equation for static equilibrium : j σ ij x j + f i = 0 (2.57) Dynamic motion And when the solid is moving : j σ ij x j + f i = ρ 2 u i dt 2 (2.58) Boundary conditions Free surface condition At a free surface, the normal stress and all the shear stresses parallel to the surface are null 7 Thus, the normal to a free surface is a principal stress direction, with a principal stress equal to zero and a free surface condition give constraints on the stress tensor. 7. The atmospheric pressure (10 5 Pa) is neglected relative to the lithospheric stress (the lithostatic stress is 26 MPa= Pa at 1 kilometer depth, for a rock density of 2.6). So in geophysics problems, the Earth s surface is considered a free surface. 22

24 Plane stress Let consider the case of a horizontal thin plate, then σ zz = σ xz = σ yz = 0. z is a principal stress direction. σ xx σ xy 0 σ = σ xy σ yy 0 (2.59) The number of unknown parameters of the problem is reduced from 6 to 3. Clamped surface If no motion is possible along a surface (for instance, when a pole is attached to the ground), then u i = 0 along a surface, and consequently ɛ ij = 0 at the boundary. Plain strain We can get a similar boundary condition if we suppose the material is infinite along a direction (for instance, the x direction). We suppose that the system is invariant along the x direction and all x derivative are null, especially ɛ xx = 0. The x direction is a principal stress direction ɛ = 0 ɛ yy ɛ yz (2.60) 0 ɛ yz ɛ zz The plane strain assumption applies for elongated geometries, such as subduction trenches or along valleys. It is a frequent situation in geophysics. The number of unknown parameters of the problem is reduced from 6 to 3. The stress component σ xx is then related to σ yy and σ zz with equation (2.15). σ xx = ν (σ yy + σ zz ) (2.61) 23

25 Chapitre 3 Friction, fracturation Objectives of this chapter : (1) Understand the different criteria for fracture and friction (2) Use the Mohr Diagram to infer the fault planes favorable to fractures. Preliminary remark : Whereas the first two chapters dealt with concepts established and formalized from the beginning of the 20 th century, the concepts discussed here are still areas of active research. 3.1 Introduction This chapter will deal with friction and fracturation. These are two different concepts : 1. Friction controls sliding on a preexisting surface, 2. Fracturation is the creation of a new discontinuity within a material, but it happens that they use the same onset criterion : the "Coulomb law". 3.2 Friction Friction laws differ depending one is dealing with onset of sliding (static friction) or with an object already sliding (dynamic friction). Note that we focus here only on solid-solid friction. There is a viscous friction for fluids, but we do not deal with aerodynamics in this lesson. Figure 3.1 A simple experiment of static friction (From Turcotte and Schubert, 2002). 24

26 3.2.1 Onset of friction Experiment Extensive laboratory studies have been carried out to determine when slip does initiate on a contact surface. A simple example is a block of mass m sitting on an inclined surface (Figure 3.1). The angle θ is increased until the block begins to slip. The normal stress exerted on the surface is σ n = m g A cos(θ) and the shear stress is τ f = m g A sin(θ). Slip will occur if τ f τ fs, the frictional static stress. One notices : 1. The force of friction is directly proportional to the applied load (Amonton s 1 st 1 Law), 2. The force of friction is independent of the apparent area of contact (Amonton s 2 nd Law). Amonton s Law Under a wide variety of conditions, a general law is found experimentally : τ fs = µ σ n + τ 0 (3.1) where µ is called the friction coefficient and τ 0 is the cohesion. This relation is known as Amonton s law. The greater the normal stress, the harder it is to initiate sliding. Exercise 3.1 : How is this condition expressed on a Mohr diagram? Most experimental data performed on rocks of very different types collapse on the same experimental law. This is the Byerlee 2 law. { τ = 0.85 σn for σ n < 200 MPa τ = 0.60 σ n [10 3 (3.2) bar] for σ n 200 MPa Figure 3.2 Byerlee s results on friction. From Scholz, The mechanics of earthquakes, Guillaume d Amontons is a French engineer who wrote a famous scientific article on friction in Byerlee was an American geophysicist, famous for performing extensive experiments of rock deformation. 25

27 Friction coefficient In rocks, the friction coefficient varies between 0.5 and 0.85 (usually this value of 0.85 is taken in the upper crust). Note however, that values as low as 0.1 can be observed (for instance in talc). Exercise 3.2 : Assuming Amonton s law to be applied with µ = 0.8, determine the angle at which the block illustrated above will begin to slip. Result : tan(θ) = 0.8, implying that θ = Cohesion The cohesion represents the strength necessary to break the interface before sliding is possible. At shallow depth, the cohesion is hardly measurable. At greater depth, the cohesion is about MPa in sedimentary rocks and 50 MPa in crystalline rocks. It is still small compared to the ambient stress, and it is often neglected. Effective stress The presence of water that wets the Earth s crust affects the frictional behavior of a fault. The pressure of water in a porous medium (or a fault zone) is referred to as the pore pressure p w. The effective normal stress acting on a wet fault is the actual normal stress minus the pore pressure. Therefore, on a wet fault, Amonton s law can be written τ = µ (σ p w ). If the water is freely connected to the surface, and there is no flow loss, the water pressure is hydrostatic (p w = ρ w g z) and is generally 35% to 50% of the overburden or lithostatic pressure. In some cases, however, water is trapped, and the pore pressure can nearly equal or even exceed the overburden pressure. In these cases the shear stress resisting motion on a fault is low Dynamic friction When a block is moving on a surface, friction forces still apply. The Amonton s laws is still used, except that the friction coefficient µ d is different and that cohesion is ignored. In the first experiment, when the block starts sliding, it is not stopped anymore by friction. The static friction coefficient µ s is greater than the dynamic friction coefficient µ d. µ s µ d (3.3) Coulomb found that the dynamic friction coefficient did not depend on speed. However, more precise experiments 3 showed a slight velocity weakening. Following Coulomb s observation, the dynamic friction coefficient depends poorly on the sliding velocity. This is Coulomb s law. Yet, precise experiment performed since the 1960 s show a slight dependence on friction. It is small but it is important to explain the various modes of sliding observed Dynamic modes of sliding Some faults move smoothly and continuously (Example : Creeping segment of the San Andreas Fault). Others make earthquakes. These are the two major modes of sliding 4. Experiment : Acoustic differentiation of the two modes for a chalk rod on a blackboard. Stable slip We assume that a fault is moving at a constant velocity. A slight change in velocity is accommodated by the fault, and it keeps moving smoothly. This mode is also called a stable mode (and the fault is called a creeping fault). If a fault that moves at a velocity v now switches to a velocity v + dv, then if µ(v + dv) < µ(v), the 3. Dieterich-Ruina law was settled in early 1980 s only. It is a favorite friction law among modelers. 4. Observations on subduction (Dragert, Nature, 2001) zones showed a third sliding mode : a slow slip. This is a "hot topic" these days. 26

28 friction diminishes and the fault is able to slip faster, leading to a runaway acceleration. This is the stick-slip regime. On the other hand, if µ(v + dv) > µ(v), the friction increases and the velocity goes back to v, the velocity is stable, and it goes to a stable slip regime. Stick-slip Figure 3.3 The seismic cycle. From Turcotte and Schubert, 2002 The fault locks and a displacement occurs when the stress across the fault builds up to a sufficient level to cause rupture of the fault. This is known as the stick-slip behavior. A simple model for the stick-slip behavior of a fault is illustrated on Figure 3.3. We assume that the behavior of the fault is uniform with depth. The stress across the fault is τ fb, a uniform velocity u 0 is applied at a distance b from the fault, and the shear strain increases with time according to ɛ(t) = u 0 t 4 b. The shear stress on the fault as a function of time t since the last displacement on the fault is therefore : τ = τ fb + G u 0 t 2 b where G is the shear modulus. The locked fault can transmit any shear stress less than the static frictional stress τ fs. When this stress is reached, slip occurs. Therefore, the time t when the next displacement occurs on the fault is t = 2 b G u 0 (τ fs τ fb ). The quantity (τ fs τ fb ) is called the stress drop on the fault during an earthquake. After the earthquake, the fault locks and the cycle repeats. The displacement on a fault during an earthquake can be measured from the surface rupture or from inversion of seismological data. A typical value for large earthquakes is 5 m, up to 30 m for the largest earthquakes. Estimating the stress drop during large earthquakes range from (τ fs τ fb )= 1 to 100 MPa. Taking G = 30 GPa, we find that b lies in the range 750 m to 75 km. Quick calculation : (τ fs τ fb )= 1 MPa, µ = 0.85, G = 30 GPa. The recurrence time is 67 years (be carefull with this recurrence time). 27

29 Mode I: Opening Mode II: In-plane shear Mode III: Out-of-plane shear Figure 3.4 The three different modes of fracture propagation. 3.3 Fracture Fracture mode Figure 3.5 Typical failure modes of triaxial experiments. There are three different way to propagate a fracture (Figure 3.4) : 1. Mode I (extension fracture) : for instance under extension (very rare in geophysics), hydraulic fracturing (dike opening) or near cavities ; 2. Mode II (in-plane shear fracture) : for instance for strike-slip fault motion and earthquakes ; 3. Mode III (out-of-plane shear fracture) : for instance for earthquakes in subduction zones. Experiment : fracture opening in mode I, II et III with a paper sheet Results of laboratory triaxial experiments Rock experiments under shear stress show that new fractures initiate in a plane perpendicular to the maximum principal stress σ 1 and minimal stress direction σ 3 (Figure 3.5) 5. The intermediate principal stress σ 2 is often considered to play a minor role. Hence, only mode I and mode II fractures are initiated 6. Mode I happens for low confining pressure, or at greater depth if there is a cavity. Hence, for standard geophysical applications, we will consider the criterion for the creation of mode II fractures. 5. You may wonder how straight mode II fracture can be created. Acoustic emissions performed before failure shows extensive microfracturing before localization on a single fracture. Hence, the mode II fracture is due to coalescence of microfractures. 6. They can afterwards propagate in mode III, as it seems to happen for large subduction ruptures. 28

30 Figure 3.6 Experimental results for rock fracturing in triaxial tests. (a) Ambient pressure. (b) σ c = 3.5 MPa, (c) σ c = 35 MPa, (d) σ c = 100 MPa. From Paterson, Coulomb criterion of faulting In Figure 3.6, the maximum differential stress (σ 1 σ 3 ) depends markedly on the confining pressure σ c. For a moderate range of confining pressure σ c = σ 3, there is a linear relationship : σ 1 σ 3 = σ 0 + µ m σ c (3.4) Exercise 3.3 : Draw three Mohr circles from the linear part of the experimental data of Figure 3.7. Show that the Mohr circles have a common tangent line, as in Figure 3.8. It can be drawn in terms of a Mohr diagram by a straight-line envelope (Figure 3.8). τ = τ 0 + µ m σ n (3.5) The sample breaks as soon the Mohr circle touches the Coulomb limit envelope. The angle ±2θ of the contact points give the orientation ±θ of the two conjugate fracture planes that will nucleate relatively to the principal stress direction σ 1. One can show that : τ 0 = σ 0 2 µ m (3.6) µ m µ = µ m (3.7) This is the Coulomb criterion for faulting. Note the strong analogy with the Amonton s law seen in section 3.2. µ m is called the "internal coefficient of friction" 7. One also defines the internal angle of friction. Exercise 3.4 : From the experimental data of figure 3.7, find the internal friction angle. Find the orientation of the new fractures relative to σ By pure analogy with the friction. But it is not friction! 29

31 Figure 3.7 Fracturation data for a granite. From Scholz, Figure 3.8 Coulomb criterion for fracturation in a Mohr diagram. 30

32 Figure 3.9 Geometry of the problem of section From Turcotte and Schubert, Anderson s law of faulting We now use the results of the previous sections and determine the angle of dip β of normal and thrust faults in terms of the coefficient of "internal friction" µ. We assume that the vertical stress σ yy is the lithostatic pressure and that the horizontal stress σ xx is the sum of the lithostatic pressure and a tectonic deviatoric stress σ xx : σ yy = ρ g y (3.8) σ xx = ρ g y + σ xx (3.9) For thrust faulting σ xx is positive, and for normal fault σ xx is negative. To apply Coulomb s law, it is necessary to relate σ xx and σ yy to σ n and τ. The result is : σ n = 1 2 (σ xx + σ yy ) (σ xx σ yy ) cos(2θ) (3.10) τ = 1 2 (σ xx σ yy ) sin(2θ) (3.11) where θ is the angle 8 of the fault with respect to the vertical direction θ = π 2 we find that the normal and tangential stresses on the fault are : β. By substitution, σ n = ρ g y + σ xx 2 τ = σ xx 2 (1 + cos(2θ)) (3.12) sin(2θ) (3.13) If we include the effect of pore pressure on the fault, these stresses are related by the Coulomb s law in the form : ± σ ( xx sin(2θ) = µ ρ g y p w + σ ) xx (1 + cos(2θ)) (3.14) 2 2 where the positive sign applies for thrust faults, and the negative sign for normal faults. Rearrangement gives an expression for the tectonic stress σ xx in terms of the angle of the fault with the vertical : 2 µ (ρ g y p w ) σ xx = (3.15) ± sin(2θ) µ (1 + cos 2θ) We hypothesize that under a tectonic stress, these pre-existing zones of weakness will be reactivated from a dip-slip fault at an angle requiring the minimum value of the tectonic stress. In other words, 8. Beware the sign issue. Look at the orientation of θ. 31

33 thrust and normal faulting will occur at angles that minimize σ xx. The angle that minimizes σ xx is determined by σxx θ = 0 with the result tan(2θ) = ± 1 µ. This can be rewritten as tan(2β) = ± 1 µ, implying that 2β = atan ± 1 µ for compression and π 2β = atan ± 1 µ for extension, giving two possible values for β, depending of the way the system is stressed. Figure 3.10 shows Figure 3.10 Dependence of dip angle of a new fault on the coefficient of friction. From Turcotte and Schubert, how the dip angle of normal and thrust faults depends on the friction coefficient : thrust faults dip less steeply than normal faults. Applying these results with the following values : p w = ρ w g y, ρ = 2700 km/m 3, ρ w = 1000 km/m 3, g = 10 m/s 2 and y = 5 km. Thrust faulting requires somewhat larger stresses, in absolute magnitude, than does normal faulting. Based on laboratory measurements, a typical value for µ is The corresponding angle of dip for a thrust fault is β = 24.8 and at a depth of y = 5 km, σ xx = 305 MPa. The angle of dip of a normal fault is β = 65.2 and the stress σ xx = 65 MPa. This derivation was developed by E. Anderson and is referred to the Anderson theory of faulting. Exercise 3.5 : Wind river thrust fault (Figure 3.11). This mountain chains is 220 km long and 70 km large. Its elevation is up to 4267 m. Clearly these mountains are the result of crustal compression. The angle of the fault up to 24 km depth was imaged using seismic profiles. The angle is nearly constant, equal to Calculate the friction coefficient? 2. Could you give an interpretation of why the coefficient of friction has deviated from the laboratory value of 0.85? 3. What would be the angle of a normal fault? Answer : 1. The coefficient of friction corresponds to the angle of dip : 1/f s = +/- tan(2*β), f s = This low value is probably associated with high pore fluid pressure. 3. With the dip of a normal fault, the angle would be 55 (typical measurements for normal faults indicate dip angles of 55 to 65 ). 32

34 3.4 Friction or fracturing? Figure 3.11 Sketch of Wind River fault. The experimental data of Figure 3.7 show than it is easier to slip on a preexisting surface rather than breaking a rock to create a new surface. In the two previous section, we distinguished the two processes of friction and fracturation. On a smaller scale, the distinction is not so clear. Especially, during friction between two rough surfaces, the asperities break and wear products ("gouge") are produced. On the other hand, during the propagation of a mode II fracture,the two fracture surfaces are sliding, and friction also occurs. Remember that this lesson deals with open questions! Fortunately, the two processes use the same kind of "failure" criterion. And the two notions are assimilated when discussing the brittle regime of rocks. 3.5 Strength envelope The Anderson s theory of faulting can also be used to find a strength envelope for the Earth s crust. For example, let us consider the oceanic lithosphere and assume that the failure stress is given by the Anderson s theory of faulting : p w = ρ w g y, ρ = 3300 km/m 3, ρ w = 1000 km/m 3, g = 10 m/s 2 and µ = 0.6. In Figure 3.12 we plot the stress σ xx as a function of depth for compressional and tensional failure. We further assume that the failure stress in the lower lithosphere is given by the solid state creep law for the upper mantle and define a strength envelope. 33

35 Figure 3.12 Strength envelope for the oceanic lithosphere.! 34

36 Chapitre 4 Other rheologies Objectives of this chapter : (1) Understand the concept of plasticity and viscosity, (2) Predict the behavior of simple viscoelastic models, (3) Understand the strength envelopes of the lithosphere. 4.1 Introduction So far, we discussed mainly elasticity as a rheological law. Why is Coulomb law not a rheological law? Elasticity was defined as a rheological law that is : reversible, independent of time, an elastic response being instantaneous. We focused on homogeneous isotropic linear elasticity in chapter 2. It was a simple lesson! Here we discuss two cases that are not elastic : when there is some permanent deformation plasticity, when there is a time dependence viscosity. We will discuss in the next section these rheological laws, again focusing on simple cases. 4.2 Plasticity Loading curve Stress Plastic region Fracture Elastic region εp εel Strain Figure 4.1 Typical strain-stress curve for a rock sample. Once inside the plastic domain, there is a permanent strain ɛ p, even after relaxation. 35

37 Permanent strain Let load uniaxially a solid. At small strain, the solid deforms elastically. ɛ = σ/e (4.1) When going beyond the elastic limit, the solid keeps a permanent deformation ɛ p after loading. The solid behaves plastically. ɛ = ɛ el + ɛ p (4.2) Experimentally, the unloading line is parallel to the loading line in the elastic domain and the elastic strain is ɛ el = σ/e. The plastic strain can be quite large. Application : Load a marshmallow. Apply a small loading an check that there is reversibility. Load more, and check there is a permanent strain after relaxation. Strain hardening - strain softening Note that the slope of the ɛ σ curve changes when one reaches the plastic limit. One can defines a local Young modulus as E = σ ɛ. Depending on the sign of E, one distinguishes : Strain hardening (also called work hardening) if E > 0, Strain softening if E < 0. The reason of this denomination can be understood if one considers the reloading of the sample after the relaxation. After relaxation, the sample will reach the elastic part of the diagram until a new threshold is reached, that is higher than the initial one if E > 0. This indicates that memory effects do occur when considering plasticity Plasticity threshold Such as for fracturation or sliding, there is a plasticity criterion. If a certain level of stress is exceeded, permanent deformation accumulates. However, there are different yield criteria. Tresca criterion Tresca criterion is similar to Coulomb criterion. It can be described as : max i j ( σ i σ j ) > σ 0 (4.3) In rock mechanics, a traditional convention is to denote the principal stress in decreasing order σ 1 > σ 2 > σ 3 and : σ 3 σ 1 > σ 0 (4.4) Figure 4.2 Two plasticity threshold criteria. 36

38 Von Mises criterion The Von Mises criterion is an energy criterion. It uses the second invariant of the stress tensor. (σ1 σ 2 ) 2 + (σ 2 σ 3 ) 2 + (σ 3 σ 1 ) 2 > σ 0 (4.5) Note that with this criterion, σ 2 plays a role Mechanisms Micro mechanisms for large strains in rocks will be presented in the Tectonophysics course. The following is only for reference. Dislocations (high temperature, high confining stress Predominant mechanism for a single mineral. Microfracturing (low temperature) Predominant mechanism for rock at low temperature (except limestones). 4.3 Viscosity Viscosity is a time dependent behavior. It is frequent in fluids, but it happens also for solids Linear viscosity A viscous law is dependent on the strain rate ɛ = ɛ t. In fluid mechanics, linear viscosity is very common (Newtonian fluids). σ = 2η ɛ (4.6) with So that for simple shear : ɛ = ɛ t τ = η v x y (4.7) (4.8) 37

39 Activation exponential log 10 (T/T threshold ) Figure 4.3 Activation of viscous strain as a function of temperature (effect of the activation energy). Note that velocity gradient appears here, and not displacement gradient. η is the dynamic viscosity 1. What is its unit? A specific unit is also specified in Poise (after the French physicist J.-M. Poiseuille). 1 Poise = 0.1 Pa s. Here are some orders of magnitude of viscosity for some common materials. Material Dynamic Viscosity (Pa s) Air 10 5 Water 10 3 Honey 1 10 Basaltic magma Silicic magma Earth mantle Note also that viscosity depends strongly on temperature General case In general, rocks behave not so nicely. They depend on strain according to a non linear law, typically : ( ɛ = C σn d b exp E ) threshold (4.9) R T where d is the grain size, b is in the range 2 4, indicating that the viscous dynamics depends much on the grain size. Figure 4.3 shows that the mechanism is suddenly activated once T > E threshold R Mechanisms Mechanisms for dynamics of large strains in rocks will be presented in the Tectonophysics course. The following is only for reference. The viscosity is due to the diffusion of the defects inside the structure. The effects can be : 0D : like vacancies [lacunes] and intersticial atoms, 1D : like dislocation lines, 1. Kinematic viscosity is ν = µ/ρ. It is expressed in m 2 /s, like a diffusivity (viscosity in fluids is a diffusion of momentum). This quantity appears also in dimensionless numbers like the Reynold number Re = vl/ν 38

40 Cobble creep This is the diffusion of matter along the boundaries of the crystal. There is a strong grain size dependence, but otherwise the relationship between stress and strain rate is linear (like for Newtonian fluids). ɛ = C σ ( d 3 exp E ) (4.10) R T Nabarro-Herring creep This is the diffusion of vacancies or interstitial atoms within the volume of the crystal. There is a strong grain size dependence, but otherwise the relationship between stress and strain rate is linear. ɛ = C σ ( d 2 exp E ) (4.11) R T Dislocation creep Within this process, dislocation diffuses within the volume. There is not a so strong grain size dependence, but the stress-strain rate relationship is no more linear. ( ɛ = C σ n exp E ) (4.12) R T Deformation mechanism regimes: Temperature and stress where n typically is between 2 and 4. Phase diagram for the various creep mechanisms Temperatures at which creep begins at geologic differential stresses: Calcite: o C Quartz: a-slip 275 o C c-slip wet o C c-slip dry 700 o C K-feldspar 450 o C Pyroxene: 500 o C Plagioclase 550 o C Creep One important application for geotechnics and slope stability is creep. Even at constant stress, the material still deforms and goes to failure suddenly. 4.4 Some classical rheological laws We combine the following rheological laws : elasticity, viscosity, plasticity. 39

41 Figure 4.4 Creep curve. Even at constant time, the strain evolves with time, until final failure. There is a graphical representation of the above laws : Elasticity spring, Viscosity hydraulic piston or dash-pot (it is supposed Newtonian), Plasticity clutch ("embrayage"). These elements can be combined within "mechanical circuits" (just like for electric components 2 ). Some classical laws are presented below Maxwell material Figure 4.5 Maxwell model of viscoelastic material. An elastic component and a viscous component are put in series (figure 4.5). ɛ = ɛ el + ɛ visc (4.13) with ɛ el = σ E ɛ visc = σ 2η (4.14) (4.15) This material relaxes stress with time as the viscous ( fluid) part accommodates it. This is a common situation in geophysics (for example, the mantle that behaves elastically for seismic waves and flows at longer time scales). For more, see TD. 2. Elasticity = spring = resistance and Viscosity = hydraulic piston = capacitance 40

42 Figure 4.6 Kelvin-Voigt model of viscoelastic material and a classical example : the shock absorber (here from Honda Civic car) Kevin-Voigt material with It can be modeled by an elastic and a viscous component put in parallel (Figure 4.6) : σ = σ el + σ visc (4.16) σ el = E ɛ (4.17) σ visc = 2η dɛ (4.18) dt The material progressively accommodates strain. For example, car struts (Figure 4.6, on the right) help attenuate the shaking when driving rough roads. This model also fits the phase I and II of creeping Elastoplastic material 4.5 Dependence of rheological laws on thermodynamic conditions We have seen several rheological laws. Remember that the choice of a rheological law does not depend only on the material, but also on temperature and pressure. Example : Fragile-ductile transition of chocolate. This is important to determine the strength envelope of the lithosphere. 41

43 Figure 4.7 Strength envelope for the oceanic lithosphere.! 42

44 Annexe A Presenting a numerical data Warning : inadequate units and excessive digits may alter the notation of your numerical results you produce in exams (and may reduce your grade). A.1 Units Force : 1 kg m s 2 = 1 N Stress : 1 N m 2 = 1Pa One may write 1 pascal, 1 newton (invariable), but 1 Pa, 1 N (capitalized letters). Angles : radian is the SI unit. Distance : m Mass : kilogra Time : second is the SI unit. In geology the year (annum) can be used. Temperature : 1 C = 1 K A.2 Significant digits When multiplying or dividing two quantities in scientific notations, the number of significant figures of the result is the same as that of the quantity with the least significant figures. When adding or subtracting a set of numbers, they are arranged by place (hundreds, tens, ones) and the result is rounded off to the least place that contains significant figures in all the number of the set. 43

45 Table A.1 From Fletcher and Pollard, Fundamentals of Structural Geology,

46 Annexe B Tangential force, moment, and torque B.1 Torque In French : moment d une force Take a ruler in equilibrium against a table (Figure B.1. To disequilibrate it, we can put a load on the hanging part. The ability of the load to topple the ruler depends on how the ruler is laying on the table. Also, if you put a larger load, it becomes more easy to destabilize the ruler. The ability to topple depends not only on the force intensity, but also on the relative position of the application point and of the rotation point. Theses parameters are included in the concept of the torque (also called moment), which is the ability of a force to make a mechanical system turn around a given point, called the pivot point. B.2 Moment relative to a point The moment relative to the point O of a force f applied at a point M is defined as M( f)/o = OM f (B.1) The intensity of M = OM f sin α depends on : OM : the distance between the application point and the pivot point. It explains why handles are located at the far end of doors. In French, OM is called bras de levier, in English moment arm. f : the larger the force, the greater the torque. α : if f were parallel to OM, you could not rotate the system. Figure B.1 Will it topple or not? 45

47 B.3 Angular momentum exerted by a couple of forces In French : Couple de force If a force f is applied at a point A and a force f is applied at a point B, then the sum of the forces is f f = 0, but the sum of the moments is not null : M( f/a)/o + M( f/b)/o = OA f + OB f = BA f The sum of the forces is independent of the chosen pivot point O. This quantity is called angular momentum (and very often also a torque). An angular momentum is null if the too forces are on the same line of action ((AB) parallel to f). It increases with the distance of the application point, and with the intensity of the force f. 46

48 Annexe C Conventions We use geological conventions throughout the text. A reminder of the main conventions is given in this page. This is meant as a reference within the text. The 3 conventions are gathered in the figures below. C.1 Convention 1 : Forces applied on a plane C.2 Convention 2 : Physical meaning of a 2D stress tensor y σ yy σ xy z σ xy σ xy σ xx σ xx σ xy x σ yy C.3 Convention 3 : Mohr circle n n 1 47