Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Defining Stress in a Solid Imagine applying external forces to a solid. Make an imaginary cut inside, and apply a set of forces carefully chosen so that the rest of the solid does not feel the cut. These forces will depend on the position and orientation of the surface. These forces that we must apply to the cuts are exactly the forces that act within the original solid. In general these forces depend on: • the position within the solid • the orientation of the surface 30 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Why are surfaces important? • When we push on something, we apply forces to a surface. • When we break something, we create surfaces. Things break along surfaces. • When we repair something, we reattach or glue surfaces. 31 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Look more closely at the forces acting on a surface tangential component normal component ν Force F unit normal area dA F ν The orientation of the surface is specified by the outward normal vector ν The normalized force per unit area is called the Traction Vector or Stress Vector. 32 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Tractions and Stresses We define the traction vector as the force per unit area acting on a surface. We define the stress as the set of components of tractions acting on the various surfaces. In cartesian coordinates: σ zz σ zy σ zx σ yz σ xz σ xx z σ yy y σ yx σ xy x 33 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Units of Stress 1 bar = 106 dyne/cm2 = 14.50 psi 10 bar = 1 MPa = 106 N/m2 Mudweight to Pressure Gradient 1 psi/ft = 144 lb/ft3 = 19.24 lb/gal = 22.5 kPa/m 1 lb/gal = 0.052 psi/ft 34 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Traction on a Surface of Arbitrary Orientation z T ν = (βx, βy, βz) y x Cauchy’s Formula: Tx σ xx σ xy σ xz Ty = σ yx σ yy σ yz σ zx σ zy σ zz Tz ≈ T = σν 35 νx νy νz Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis The Stress Tensor is Symmetric We must also balance the torques applied to the cube and we find that: σxy = σyx σxz = σzx σyz = σzy σ xx σ xy σ xz σ yx σ yy σ yz σzx σzy σzz Stress tensor is symmetric. There are only 6 independent terms. 36 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Einstein Summation Convention It is customary in stress and strain analysis to use a shorthand notation, sometimes called the Einstein summation convention. Any repeated index within a term implies summation over that index with a range of 1 to 3. For example: c ijkl σ kl 3 3 c ijkl σkl Σ Σ k=1l=1 and are equivalent. Here, k is repeated so it implies a summation; l is also repeated, so it also implies a summation. i and j are not repeated so there are no summations. There is no summation in this expression: σ ij + σ jk because there is no repeated summation in any of the terms. The two occurrances of the index j are in two different terms. 37 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Transformation of Coordinates We found it convenient to define the elements of the stress tensor in a given coordinate system. We did so by labeling the various components of forces that act on the faces of orthogonal planes in that coordinate system. We now want to be able to write the stress tensor in terms of another coordinate system. That is, for the same physical configuration of forces, how do we simply label or describe these in terms of another set of axes? Or, how do we find the components of force acting on the orthogonal planes in the new coordinate system? We do it in two steps: 1. First find the coordinate transformation that relates the two sets of axes. 2. Then use it to re-express the stress tensor in the new coordinate system. z r z' x' y x y' 38 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Coordinate Transformations A real physical vector can be expressed in any two coordinate systems: r = x 1e 1 + x 2e 2 + x 3e 3 r = x ′1e ′1 + x ′2e ′2 + x ′3e ′3 It is important to remember that we are simply giving the same physical vector different names in different coordinate systems. x ′1 β 11 β 12 β 13 x ′2 = β 21 β 22 β 23 β 31 β 32 β 33 x ′3 Or in index notation: 3 x ′i = j=1 Σ x1 x2 x3 β ijx j = β ijx j where β ij is the direction cosine of the new ith axis relative to the old jth axis. So for any vector we can write: ′ ≈ r = βr 39 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis So now go back to Cauchy’s formula Told = σoldν old Now transform T and ν to the new coordinates β ij TTnew = σold β ij Tν new Rewrite as: Tnew = β ij σ old β ij T ν new Rename term in brackets: Tnew = σ new ν new β ij σold β ij T = σnew Formula for coordinate transformation σij = β ikβ jlσ kl 40 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis The hydrostatic stress or mean stress is the average of the three normal stresses. σxx + σ yy + σ zz σ αα = σ0 = 3 3 This is an invariant. The mean stress is the same in any coordinate system. 41 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Principal Stresses Stress is a real physical state or quantity: we have forces, surfaces, and stress whether or not we choose a coordinate system. Now, if we do define a coordinate system, then we assign numerical values to the components σ ij in the matrix: σ ij In general, different choices of coordinate systems cause the numerical values in the matrix (stress tensor) to vary. Now recall Cauchy’s formula: T = σ ij ν Can we find directions such that the traction vector is normal to the surface? T = σ ij ν = σν where T is the magnitude of the traction vector, ν is the direction, and σ is a scalar. This is a standard problem of finding the eigenvalues and eigenvectors of a matrix. We call the three eigenvalues the principal stresses, and the three eigen vectors the principal directions. 42 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Principal Stresses Let’s emphasize the physical interpretation: The traction T is parallel to ν , which is always perpendicular to the surface. If T is normal to the surface then there is no component of T parallel to the surface. Therefore, there is no shear stress. The simplest and most important thing to remember about the principal axes is that there are no shear stresses on surfaces perpendicular to the three principal axes. If we find these directions and make them axes then we know how to transform the stress tensor to these axes and we will find: σ1 0 0 0 σ2 0 0 0 σ3 43 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Infinitesimal strain P' P Q' Q We can describe the deformation of a continuum with a vector displacement field U . This displacement gives the vector motion of each point in the body. It is important to note that the displacement can vary with position. 44 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Rigid Body Translation The vector displacement field is a constant. 45 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Rigid Body Rotation P' P Q' Q u=ω×r ω = rotation vector • direction is along axis of rotation • ω is the angle of rotation 46 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Strain If a body undergoes a displacement that cannot be simulated point-by-point as a rigid body translation or rotation, then we say it is strained. With strain we have deformation or distortion. The size or shape is changed. U(x,y,z) U(x + dx, y + dy, z + dz) (x,y,z) (x + dx, y + dy, z + dz) 47 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Ux Uy Uz ∂U x ∂x ∂U y ∂x ∂Uz ∂x Ux = Uy Uz r + dr ∂Ux ∂y ∂Uy ∂y ∂Uz ∂y ∂U x ∂z ∂U y ∂z ∂U z ∂z + r ∂U x ∂x ∂Uy ∂x ∂Uz ∂x ∂U x ∂y ∂U y ∂y ∂U z ∂y ∂Ux ∂z ∂Uy ∂z ∂Uz ∂z dx dy dz 0 –ω 3 ω2 ε xx ε xy ε xz = ε yx ε yy ε yz + ω3 0 –ω 1 ε zx ε zy ε zz –ω 2 ω1 0 strain= symmetric part rotation= antisymmetric part We usually work in the range of infinitesimal strain • mathematically • geophysically ∂Ux << 1 ∂x ∂Ux << 10 –4 ∂x 48 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Exploring the Physical Interpretation of Strain 1. Extensional Strain U(x) U(x+dx) L Lo The relative change in length can be described as ∆L = L – L 0 = U x + dx – U x ≈ ∂U L 0 ∂x ∆L ≈ ∂U ≈ ε xx L 0 ∂x The strain terms along the diagonal describe relative changes in length. 49 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Exploring the Physical Interpretation of Strain 2. Volumetric Strain The original volume is V0 = L x0L y 0L z0 The volume after straining is V0 + ∆V = L x 0 + ∆L x L y 0 + ∆L y L z0 + ∆L z V0 + ∆V ≈ L x 0L y0L z0 ∆L x ∆L y ∆L z 1+ + + L x0 L y0 L z0 ∆V ≈ ε + ε + ε xx yy zz V0 The volumetric strain is the trace of the strain tensor or is the sum of the three extensional strains. This is sometimes called the dilatation. 50 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Exploring the Physical Interpretation of Strain 3. Pure Shear Strain x α α y When we deform the square like this, notice that the corner changes from a 90˚ angle to a lesser angle 90˚ - 2α. Note that ∂Ux ∂U y tan α = = ∂y ∂x For small angles: ∂Ux ∂Uy 1 tan α ≈ α ≈ + ∂x 2 ∂y α = ε xy The shear strain describes a change in angle. 51 Stanford Rock Physics Laboratory - Gary Mavko Stress and Strain Analysis Exploring the Physical Interpretation of Strain 4. Simple Shear Strain x 2α y In this case the angle change is again exactly 2α. But now: ∂U ∂U x ∂y Therefore: y =0 , ∂x = 2α ∂Ux ∂U y 1 ε xy = + ≈α ∂x 2 ∂y ∂Uy ∂Ux 1 ωxy = – ≈α ∂y 2 ∂x Hence the strain in this example is identical to the previous case (pure shear). However, the rotation α is superimposed. α 2α = + α 52 α