GG611 Structural Geology Sec1on Steve Martel POST 805 smartel@hawaii.edu Lecture 5 Mechanics Stress, Strain, and Rheology 11/6/16 GG611 1 Stresses Control How Rock Fractures hkp://hvo.wr.usgs.gov/kilauea/update/images.html 11/6/16 GG611 2 1
Predicted Coulomb Stress Changes Caused by 1992 Landers Earthquake, California From King et al., 1994 (Fig. 11) Coulomb stress change caused by the Landers rupture. The ley-lateral ML=6.5 Big Bear rupture occurred along doked line 3 hr 26 min ayer the Landers main shock. The Coulomb stress increase at the future Big Bear epicenter is 2.2-2.9 bars. hkp://earthquake.usgs.gov/research/modeling/papers/landers.php 11/6/16 GG611 3 I. Stress at a point II. Strain at a point III. Rheology Outline IV. Applica1ons: Displacement and Stress Fields V. Appendix: Eigenvectors, eigenvalues, and principal stresses 11/6/16 GG611 4 2
Stress at a Point Stresses refer to balanced internal "forces (per unit area)". They differ from force vectors, which, if unbalanced, cause accelera1ons "On -in conven1on": The stress component σ ij acts on the plane normal to the i- direc1on and acts in the j- direc1on 1 Normal stresses: i=j 2 Shear stresses: i j 11/6/16 GG611 5 Dimensions of stress: force/unit area Conven1on for stresses Tension is posi1ve Compression is nega1ve Stress at a Point Follows from on-in conven1on Consistent with most mechanics books Counter to most geology books 11/6/16 GG611 6 3
σ ij = σ xx σ yx σ xy σ yy Stress at a Point 2-D 4 components σ xx σ xy σ xz σ ij = σ yx σ yy σ yz σ zx σ zy σ zz 3-D 9 components For rota1onal equilibrium, σ xy = σ yx, σ xz = σ zx, σ yz = σ zy ; stress matrix is symmetric In nature, the state of stress can (and usually does) vary from point to point 11/6/16 GG611 7 Stress at a Point Analogy with vectors The components of a vector vary with the reference frame, even though the vector does not For certain reference frame orienta1ons, some vector/ tensor components are zero The non-zero components are meaningful and illumina1ng in a reference frame where some components are zero 11/6/16 GG611 8 4
Principal Stresses Have magnitudes and orienta1ons Principal stresses act on planes which feel no shear stress Principal stresses are normal stresses Principal stresses act on perpendicular planes owing to symmetry of stress tensor The maximum, intermediate, and minimum principal stresses are usually designated σ 1, σ 2, and σ 3, respec1vely Designated by a single subscript 11/6/16 GG611 9 Principal Stresses F Principal stresses represent the stress state most simply G σ ij = σ 1 0 0 σ 2 2-D 2 components H σ 1 0 0 σ ij = 0 σ 2 0 0 0 σ 3 3-D 3 components * If σ 1 = σ 2 = σ 3, the state of stress is called isotropic. This occurs beneath a s1ll body of water. 11/6/16 GG611 10 5
Applica1on: Ver1cal Strike-slip Faults 11/6/16 GG611 11 Applica1on: Dip-slip Faults 11/6/16 GG611 12 6
Strain at a point: Basic Concepts Normal strain (ε): change in rela1ve line length Shear strain (γ): change in angle between originally perpendicular lines Volumetric strain (Δ): change in rela1ve volume Based on rates of change of displacement Strains are dimensionless 11/6/16 GG611 13 Finite strain at a point Chain rule relates difference in ini1al posi1ons (dx and dy) of neighboring points to difference in displacements (dx and dy ) At a point, displacement deriva1ves are constants Matrix rela1ng difference in displacement [du] to difference in ini1al posi1on [dx] contains constants Unit circles (spheres) deform to ellipses (ellipsoids) du x = u x x dx + u x y dy du y = u y x dx + u y y dy u x u y = u x x u y x u x y u y y dx dy [J u ] du [ ] = [ J u ][ dx] 11/6/16 GG611 14 7
Infinitesimal strain at a point In infinitesimal strain, displacement deriva1ves are small rela1ve to 1 The infinitesimal strain matrix contains normal strains (on main diagonal) and shear strains (offdiagonal terms) The infinitesimal strain matrix is symmetric Infinitesimal principal strains are perpendicular ε ij = ε xx ε yx ε xy ε yy u x x = 1 u x 2 y + u y x ε 1 0 ε ij = 0 ε 2 1 u x 2 y + u y x u y y 11/6/16 GG611 15 Rheology The rela1onship between the flow or deforma1on of a material and the loads causing the flow or deforma1on Typically relates stress to strain, or stress to strain rate In reality, rheology is a complicated func1ons of pressure, temperature, fluid content, etc. We generally use simple rheologic models 11/6/16 GG611 16 8
Rheology: Viscous (fluid) behavior Shear strain rate is propor1onal to shear stress hkp://manoa.hawaii.edu/graduate/content/slide-lava 11/6/16 GG303 17 Rheology: Duc1le (plas1c) behavior No strain un1l stress reaches a cri1cal level hkp://www.hilo.hawaii.edu/~csav/gallery/scien1sts/lavahammerl.jpg hkp://hvo.wr.usgs.gov/kilauea/update/images.html 11/6/16 GG303 18 hkp://upload.wikimedia.org/wikipedia/commons/8/89/ropy_pahoehoe.jpg 9
11/6/16 Rheology: BriKle behavior (fracture) Deforma1on is discon1nuous 11/6/16 GG303 19 hkp://upload.wikimedia.org/wikipedia/commons/8/89/ropy_pahoehoe.jpg Rheology: Linear elas1c behavior Deforma1on reverses when stress is relieved Stress and infinitesimal strain are linearly related Principal strains and principal stresses are parallel in isotropic materials [σ] = [C][ε] hkps://thegeosphere.pbworks.com/w/page/24663884/sumatra hkp://www.earth.ox.ac.uk/ data/assets/image/0006/3021/seismic_hammer.jpg 11/6/16 GG303 20 10
Applica1ons: Displacement and Stress Fields Displacement and stresses over a region can be obtained by solving boundary value problems Requirements Body geometry Boundary condi1ons (e.g., stress components or displacements ac1ng on a boundary surface) Rheology Governing equa1on(s) for equilibrium and compa1bility General solu1on Specific solu1on that honors boundary condi1ons 11/6/16 GG611 21 Geologic Problem: Radia1ng Dikes Shiprock, New Mexico Aerial view showing radial dikes Both images from hkp://en.wikipedia.org/wiki/shiprock#images 11/6/16 GG611 22 11
Displacement Field Around a Pressurized Hole in an Elas1c Plate Geometry and boundary condi>ons Displacement field For u 0 > 0 ( u r u 0 ) = ( a r) 11/6/16 GG611 23 Displacement and Stress Fields Around a Pressurized Hole Displacement field Stress field σ 2 σ 1 Foru 0 > 0 :( u r u 0 ) = ( a r) For P < 0 : ( σ 1 P) = a r ( ) 2 ( σ 2 P) = ( a r) 2 11/6/16 GG611 24 12
Dikes Open in Direc1on of Most Tensile Stress, Propagate in Plane Normal to Most Tensile Stress Most tensile stress hkp://hvo.wr.usgs.gov/kilauea/update/images.html 11/6/16 GG611 25 Modeled vs. Measured Displacement Fields Around a Fault Model for Fric1onless, 2D Fault in an Elas1c Plate Model Displacement field GPS Displacements, Landers 1992 11/6/16 GG611 26 13
Modeled vs. Measured Co-seismic Displacement, Landers Earthquake Co-seismic displacement field due to the 1992, Landers EQ" Co-seismic deforma1on can be modeled using elas1c disloca1on theory (based on Massonnet et al., 1993) 11/6/16 GG611 27 Stress Fields Around a Fric1onless, 2D Model Fault in an Elas1c Plate vs. Observa1ons Model stress field: Most tensile stress & stress trajectories Tail cracks at end of lek-lateral strike-slip fault Note loca1on and orienta1on of tail cracks 11/6/16 GG611 28 14
Appendix: Eigenvectors, eigenvalues and principal stresses 11/6/16 GG611 29 I Stress! vector! (trac1on) τ = lim F / A A 0 Trac1on Vector on a Plane Trac1on vectors can be added as vectors A trac1on vector can be resolved into normal (τ n ) and shear (τ s ) components A normal trac1on (τ n ) acts perpendicular to a plane A shear trac1on (τ s ) acts parallel to a plane Local reference frame n-axis is normal to plane s-axis is parallel to plane 11/6/16 GG611 30 15
Cauchy s Formula Transforms stress state at a point to the trac1on ac1ng on a plane with normal n Transforms normal vector n to the trac1on vector τ τ j = n i σ ij Trac1on component that acts in the j-direc1on Dimensionless weigh1ng factor (cosine between the n- and i- direc1ons;) Expansion (2D) τ x = n x σ xx + n y σ yx τ y = n y σ xy + n y σ yy Stress component that acts on a plane with its normal in the j-direc1on, and that acts in the j-direc1on 11/6/16 GG611 31 Cauchy s Formula E Deriva1on Contribu1ons to τ x 1 τ x = w ( 1) σ xx + w ( 2) σ yx Based on a force balance Note that all contribu1ons must act in x-direc1on 2 F x = A n A x A n F x + A x A y A n F x A y 3 τ x = n x σ xx + n y σ yx Similarly 4 τ y = n x σ xy + n y σ yy n x = cosθ nx = cosθ x n y = cosθ ny = cosθ y 11/6/16 GG303 32 16
Principal Stresses III Eigenvectors and eigenvalues A τ has a magnitude and direc1on B τ from Cauchy s formula C τ x τ y τ x τ y σ xx σ xy = τ n x n y = σ xx σ yx σ xy σ yy σ yx σ yy n x n y Let λ = τ = λ n x n y n x n y n is unit normal to plane A n The form of (C ) is [A][X]=λ[X], and [σ] is symmetric 11/6/16 GG303 33 Principal Stresses 1 [σ][x]=τ[x] 2 This is an eigenvalue problem (e.g., [A][X]=λ[X]) A [σ] is a stress tensor (represented as a square matrix) B τ is a scalar C [X] is a vector D [X], [σ][x], and τ[x] all point in the same direc1on 3 Solving for τ yields the principal stress magnitudes (Most tensile σ 1, Intermediate σ 2, least tensile σ 3 ) 4 Solving for [X] yields the principal stress direc1ons Principal stresses are normal stresses and mutually perpendicular ([σ] is symmetric) 11/6/16 GG611 34 17