Continuum mechanism: Stress and strain

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1 Continuum mechanics deals with the relation between forces (stress, σ) and deformation (strain, ε), or deformation rate (strain rate, ε). Solid materials, rigid, usually deform elastically, that is the relation between force (stress) and strain is linear, ε = µσ. Fluids, and viscous materials, that have a linear relation between stress and strain rate, ε = ησ, are called Newtonian fluids and linearly viscous materials, respectively. Most materials show more complicated behavior, non-linear viscosity, combination of elastic and viscous behavior, and so on. A Stress Stress, σ = F/A, is one of the fundamental variables of continuum mechanics. The units of stress are force per area, or Pascal, [N m 2 = Pa]. Body and surface forces The forces acting on a parcel of material are of two types: Body forces. Affect the whole volume, and the size of the force is proportional to the volume. An example is the force of gravity. Surface forces. Act on the surface of the material. Friction is an example. In continuum mechanics surface forces, stress, which is force per area, play an important role. Definition of stress Consider a force t(n), acting on a surface with a unit normal n (Figure 1a). Considering a small parcel of the material the stress vectors have equal and opposite signs on the opposite sides (Figure 1b). The stress vector can be resolved into a component normal to the surface, extension, and parallel to the surface, shear (Figure 1c). We therefore have, F t(n) =lim δ 0 A (total force on area δ with normal n) =. (1) (area δ) The stress vector depend on the particular surface orientation expressed, therefore the state of stress is described by a second order tensor. Throstur Thorsteinsson 1

2 Figure 1: (a) Relation between unit surface normal n, and the stress vector t(n). (b) The stress vectors on opposite sides of the parcel are equal in magnitude. (c) Stress vectors (dashed arrows) are resolved into components of shear (parallel to the surface) and extension (normal to the surface). Stress tensor Stress is a tensor quantity, σ xx σ xy σ xz σ ij = σ yx σ yy σ yz. (2) σ zx σ zy σ zz The first indices refers to the direction the force is acting, and the second to the surface the force is action on, i.e. σ xz refers to the stress acting in direction ˆx on a surface with normal ẑ. Of the 9 components, only 6 are independent (when there is no acceleration), since σ ij = σ ji. If that wasn t true, the little parcel in Figure 1 would rotate. Principal stress Principal directions and values. λ (a real number) is a principal value of A if and only if there is a vector n ( n =1) such that A ij n j = λn i. n is called a principal direction of A belonging to λ. A real number λ is a principal value of A if and only if it is a real root of the cubic equation, det(a ij λδ ij )= λ 3 + φ 1 λ 2 φ 2 λ + φ 3 =0, (3) where φ 1 = A ii, φ 2 = 1 2 (A iia jj A ij A ji ), φ 3 = det(a ij ). Throstur Thorsteinsson 2

3 The φ s are the scalar invariants of the tensor. Principal stress and the associated directions play an important role in seismology, for instance. The principal stress components are the eignvalues of the stress tensor. They are usually written as σ 1 >σ 2 >σ 3. In the coordinate system defined by the eigenvectors there are no shear stresses (Turcotte and Schubert, 2002, section 2-3). In two dimensions this can be visualized as rotating the coordinate system until the shear stress is zero. If we call the rotated coordinate system x y and the angle between the axes θ, then we get: τ x y = 1 2 (σ yy σ xx )sin2θ + τ xy cos 2θ. (4) When τ x y = 0 we get, tan 2θ = 2τ xy. (5) σ xx σ yy The principal stress is then, Specific stress states σ 1,2 = σ xx + σ yy 2 (σxx σ yy ) ± 2 + τxy 4 2. (6) Hydrostatic pressure: All principal stresses are equal, p = σ 1 = σ 2 = σ 3. An example is the state of stress in stationary fluids. Lithostatic pressure: All principal stress components equal to the load of the overlying rock. Mean pressure: p =(σ 1 + σ 2 + σ 3 )/3. Deviatoric stress: σ ij = σ ij pδ ij, for example σ xx = σ xx p, and σ xz = σ xz. B Strain Strain refers to the change in shape of a material. Consider a deformable string, such as shown in Figure 2. The string is attached to the origin, and two arbitrary points, one at x and the other at x+ x, a distance x away from each other, are marked on the string in the un-stretched state (Fig. 2a). After deformation the points are at x + u and x + u + x + u, where the length of the element is now x + u (Fig. 2b). The strain is then, change in length original length The strain at point P is defined as x + u x = = u x x. (7) u ε = lim x 0 x = du dx. (8) Throstur Thorsteinsson 3

4 a b 0 0 x x+u x P x+ u P Figure 2: Deformable string, (a) before deformation, and (b) after stretching. READING ASSIGNMENT Introduction Read chapter 1 in T+S. Review basic linear algebra, rotation of coordinate systems, and tensor (subscript, indices) notation. Chapter 2, and also 1, in M+W is a good reference. This handout also has some of the basic relations. Stress and strain Read about the definition of stress and strain. Familiarize yourself with the notation used, and what principal stress is. You can go quickly through the Mohr circle construction. Good sections to read are: Turcotte and Schubert (2002):chapter 2 M+W: Chaper , and Chapter 7 Nye (1985): p Lay and Wallace (1995): p PROBLEMS 1 What is A ij in the equation for transformation of axes? Show how that relation is used when the x, y-axes are rotated by an angle θ (rotation of z). 2 Write out (the first part at least) of the products listed (similar to the example shown for the summation convention). 3 Consider the oceanic and continental structures shown in Figure 3. The continental crust has a thickness h c and the density ρ c ; its upper surface is at sea level. The oceanic crust is covered with water of depth h w and density ρ w. The oceanic crust Throstur Thorsteinsson 4

5 has a thickness h o and density ρ o. The mantle density is ρ m. Apply the principle of isostacy or hydrostatic equilibrium to show that the depth of the ocean basin relative to the continent is given by, h w = ρ m ρ c ρ m ρ w h c ρ m ρ o ρ m ρ w h o. (9) Calculate h w for h c = 35 km, h o = 6 km, ρ m = 3300 kg m 3, ρ w = 1000 kg m 3, ρ c = 2800 kg m 3, and ρ o = 2900 kg m 3. Ocean Oceanic Crust Continental Crust h c h w h o Mantle h m Figure 3: Continental-ocean crust structure for problem. 4 A mountain range has an elevation of 5 km. Assuming that ρ m = 3300 kg m 3, ρ c = 2800 kg m 3, and that the reference or normal continental crust has a thickness 35 km, determine the thickness of the continental crust beneath the mountain range. Assume that hydrostatic equilibrium is applicable. 5 Show that the maximum shear strain is given by 1 2 (ε 1 ε 2 ). What is the direction in which the shear strain is maximum? 6 Show that the permutation symbol gives the determinates, u 1 u 2 u 3 ɛ ijk u i v j w k = v 1 v 2 v 3 w 1 w 2 w 3. 7 Write the scalar invariants (Eq. 3) in terms of principal values. Lay, T. and T. C. Wallace Modern Global Seismology. Academic Press Inc., U.S. Nye, J. F Physical properties of crystals. Oxford, 1st edition. Turcotte, D. L. and G. Schubert Geodynamics: Application of continuum physics to geological problems. Cambridge University Press, 2nd edition. Throstur Thorsteinsson 5

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