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I. Introduction and Basic Concepts A. Stress: force applied to rock unit, that results in deformation (strain) B. Definitions
- Force: vector with magnitude and direction a. compressional vs. tensional forces in geology (1) squeezing vs. pulling apart b.c. magnitude = how much force?direction = direction of force?
d. Force = Mass x Acceleration
- Traction = force distributed per unit area a. given a constant magnitude... (1) > area, < traction (lesser concentration of force) (2) < area, > traction (greater concentration of force b. Stress = "traction" = Force / Area (1) e.g. force applied to a fracture plane or bedding plane
- Force Components a. 2-D Analysis (1) Force may be broken into 2 vector components oriented atright angles
b. 3-D Analysis (1) Force may be broken into 3 vector components oriented atright angles
c. Force distributed over an area(1) Force component normal to surface ("normal stress) (2) Force component parallel to surface ("shear stress")
- Surface Stress Equilibrium a. Traction force applied to surface (1) Equilibrium condition: a pair of equal and opposite tractions acting across a surface of given orientation
- Vector Review
a. Vector: quantity with magnitude and direction (1) e.g. Velocity, Force (a) e.g. car travels 40 mi/hr in east direction (2) Graphical depiction (a) arrow shows direction (b) length of arrow scaled to magnitude b. "Scalar Quantity": magnitude only (a) e.g. area, temperature, density c. Vectors in 2-D (1) Parallelogram method of vector resolution (a) Vector addition: V + W = R d. Vectors in 3-D (1) Orthogonal Cartesian Coordinate System (a) x-y-z axes mutally perpendicular (also known as x 1 , x 2 and x 3 respectively) (2) Resolution of force F in 3-D (a) F = F1+F2+F3 where F1, F2 and F3 = force components parallel to x,y,z reference axesrespectively
- Remember Your Trigonometry!!!a. Triangles
(1) All interior angles of any triangle must = 180 degrees (2) Right Triangle: one of the angles of triangle = 90 degrees b. Right Triangles and Trig. Functions (1) theta = θ = given interior angle of right triangle, not the 90 degree angle (2)(3) "hypotenuse" = line opposite right angle of right triangle"adjacent" = line forming ray of angle θ (4) (5) "opposite" = line opposite angle 2 θ = 2 times the angle of θ θ
c. Basic Trig. Functions (1) Sin θ = length opposite / length hypotenuse
(1) F (top of surface)^ = -F (bottom of surface) c. Principle carries on to resulting stresses over unit areas (1) σntop^ = -σnbottom (2) σstop^ = -σsbottom
- Normal Stress Relations a. Compressive Stress:pointing toward one another σntop^ and σnbottom^ are equal and opposite,
(1) squeezing or compressive action (2) **** by definition compressive stress = positive magnitude** b. Tensile Stress: σntop^ and σnbottom^ are equal and opposite, pointing away from one another (1) pull apart action (2) by definition tensile stress = negative magnitude
- Shear Stress Relations a. Shear couple: σstop^ and σsbottom^ are equal and opposite. (1) clockwise shear couple ("right lateral") = negative magnitude by definition (2) counterclockwise shear couple ("left lateral") = positivemagnitude by definition
C. Two Dimensional Stress At a Point (Stress Ellipse)
- Identifying stress σ at a point a. pass imaginary plane (surface) through point b. identify normal and shear stress, σn and σs
- 2-D Plane of Reference a. "Stress Elipse": complete graphical representation of total stress σ at a point in space (1) Principal Stresses (a) Maximum and Minimum stresses acting on planes through a given point at right angles to one another
in the stress elipse (b) Maximum Stress (highest magnitude) = σ 1 (c) Minimum Stress (lowest magnitude) = σ 3 (d) By definition, σ 1 >/= σ 3 (2) Requirements of Principal Stress Field (a) Magnitude and directions ofdefine stress field in 2D at a given point. σ 1 and σ 3 uniquely (b) Since σ 1 and σ 3 act perpendicular to principal surfaces, the component of shear stress in each is = 0 (i.e.components) σ 1 and σ 3 are comprised totally of normal (c) principal planes are those which the principle stresses are acting upon i) principal stress acts perpendicular to principal planes (d) System must be at mechanical equilibrium, with principal stresses acting in equal and oppositemanner
D. Three Dimensional Stress At a Point (Stress Ellipsoid)
- Expand stress ellipse into third dimension = "principal stress ellipsoid" a. 3-mutually perpendicular axes to ellipsoid b. 3-mutually perpendicular principal stresses acting normal to principal surfaces of cube in 3-D (shear stress component of eachprincipal stress = 0 by definition) (1) σ 1 = maximum principal stress (long axis of magnitude on ellipsoid) (2) σmagnitude on ellipsoid) 2 = intermediate principal stress (intermediate axis of (3) σellipsoid) 3 = minimum principal stress (short axis of magnitude on
c. By definition: σ 1 >/= σ 2 >/= σ 3 d. Vectoral Components ofsystem σ 1 , σ 2 and σ 3 relative to x,y,z coordinate
(1)(2) σσ 1 : divided into x,y,z subvectors (3) σ^23 : divided into x,y,z subvectors: divided into x,y,z subvectors
b. Vertical axis ("y axis") = shear stress component σs (1) counterclockwise shear stress (left lateral) = positive (2) clockwise shear stress (right lateral) = negative
- Mohr Diagrams and Experimental Rock Mechanics a. Mohr diagram derived from lab testing of rock strength b. Method (1) place rock core of given diameter and length in triaxial press (2)(3) rock core under confining pressure from sides withStress applied vertically in press σ 2 = σ 3 (a) compression: σ 1 > σ 2 = σ 3 > 0 (b) tension: σ 1 = σ 2 > σ3 ; σ1, σ 2 > 0; σ 3 < 0 (4) Stress applied until rock core undergoes failure (a) confining stress recorded (b) active stress at point of failure recorded (5) Complete multiple runs on rock core, varying confining pressure, actively stressing until rock fails. (a) record data c. Plot of "Mohr Circle" (1) Mohr Circle: a circular plot of stress applied to a given point, that defines stress components acting on all possible planes passing through that point (a) Center of Mohr circle lies on x-axis, or normalstress axis of Mohr Diagram
(2) Principal stress components (a)(b) stress difference =σ σ 1 - σ 3 Mohr diagram (i.e. plotted on normal stress axis)^1 and^ σ^3 plotted as points on the x-axis of the (c) σenvelope of the Mohr Circle, and on the 1 and σ 3 define unique points that lie on the σ where σs = 0. n^ axis, i) hence σ 1 and σ 3 are entirely normal in composition at these unique points (3) Surface stress and orientation of planes
(a) pass a plane in space through stress point i) apply σ 1 to the plane ii) resolve normal and shear stress components operating on this plane iii) normal and shear stress components a function of angle between plane and σ 1. a) at 90 degrees, σ 1 =σn and σs = 0. b) at 0 degrees, σ 1 = σs , and σn = 0 c) between 0-90 degrees, varying components ofaccordingly σn and σs ,
(b) Back to the Mohr circle... i) angle thetaoperating on plane θ = angle between σ 1 and σn
ii) On mohr circle,represented by 2 θθ is doubled and a) θ has values of 0-180 in physical space b) 2 θ has values of 0-360 in terms of generating the Mohr circle (4) Resolving normal and shear stress components acting on a plane (a) Mohr circle: i) circle drawn with center lying on x-axis or σn axis of Mohr Diagram ii) Diameter and position of circle defined by σ 1 and^ σ 3 a) (^) axis of Mohr diagram whereσ 1 and σ 3 plotted as points on σ σn s = 0. b) Mohr circle drawn through σ 1 and σ 3 , with diameter of circle = σ 1 - σ 3 (b) center of mohr circle on σn axis
a. σ 1 > σ 2 = σ 3 = 0
- Uniaxial Tension a. 0 = σ 1 = σ 2 > σ 3 (i.e. σ 3 is negative) C. Confined Compression
- σ 1 > σ 2 = σ 3 > 0
- uniaxial compression + hydrostatic stress D. Extensional Stress
- σ 1 = σ 2 > σ 3 > 0
- uniaxial tension + hydrostatic stress
