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Mohr-Coulomb failure Goal: To understand relationship between stress, brittle failure, and frictional faulting and to use this relationship to predict rock behavior Stress review • Stress = Force/Area • 3 principal vectors: σ1, σ2, and σ3 at right angles to each other • σ 1 ≥ σ2 ≥ σ3 • σ1 is the maximum principal stress direction, σ2 is the intermediate principal stress direction, and σ3 is the minimum principal stress direction We also define: • Static stress as σ1 = σ2 = σ3 • Lithostatic stress as static stress generated by mass of overlying rocks • Differential stress (σd) as (σ1 - σ3) • Confining pressure as σ2 = σ3 for the conditions σ1 > σ2 = σ3 Shear stress and normal stress For any plane in a stress field defined by σ1, σ2, and σ3 with strike parallel with σ2: σ1 θ σ3 σ3 σ1 The stress is resolved into 2 components: 1. Shear stress (σs), acting parallel with the plane 2. Normal stress (σn), acting perpendicular to the plane σ1 σ n θ σ3 σ3 σs σn σ1 σs Stress components are related by: 1. σs = ½(σ1 - σ3)sin(2θ) 2. σn = ½(σ1 + σ3) - ½(σ1 - σ3)cos(2θ) where θ = angle between plane and σ1 σ1 σ n θ σ3 σ3 σs σn σ1 σs Mohr diagram for stress Relationship between σ1, σ3, σs, and σn is plotted graphically in Cartesian coordinates σs σn Mohr circle for stress: circle with diameter = σd plotted on mohr diagram Center on the σn-axis at point = ½(σ1 + σ3) σs σ3 σ1 σn ½(σ1 + σ3) Finding σs, and σn Can use a Mohr circle to find σs, and σn for any plane σs σ3 σ1 σn Plot a line from center to edge of circle at angle 2θ-clockwise from σn-axis σs 2θ σ3 σ1 σn X- and y-coordinates of intersection of line and circle define σs and σn for the plane (σs, σn) of plane σs σ3 σ1 σn Coulomb’s failure criterion • Every homogeneous material has a characteristic failure envelope for brittle shear fracturing • Combinations of σs and σn outside of the envelope result in fracture Determining failure envelope Experimental rock deformation The Coulomb envelope Tensile Fracture σs Shear Fracture Stable 2θ σ3 σ1 Stable Shear Fracture σn Coulomb law of failure σs σc = σ0 + tan(φ)σn φ σ 0 σn σc = σ0 + tan(φ)σn Formula defines shear stress under which rocks will fracture σc = critical shear stress — σs at failure σ0 = cohesive strength — σs when σn = 0 φ = angle of internal friction — φ ≈ 90 - 2θ • For most rocks, angle of internal friction ≈ 30° • Therefore, θ at failure is also ≈ 30° • σs is greatest when θ = 45° Failure envelopes for different rocks Slip on pre-existing fractures Pre-existing fractures have no cohesive strength, σ0 = 0 Failure envelopes for pre-existing fractures derived experimentally Envelope of sliding friction σs φf = angle of sliding friction σn Byerlee’s law Describes frictional sliding envelope σc = tan(φf)σn φf ≈ 40° for low confining pressures and ≈ 35° for high confining pressures Byerlee’s law for different rock types Effect of pore-fluid pressure Pore fluid pressure (Pf) effectively lowers the stress in all directions The effective stresses (σ1eff, σ2eff, and σ3eff) = principal stresses - Pf σ1eff = σ1 - Pf σ2eff= σ2 - Pf σ3eff = σ3 - Pf σs Stable stress conditions σ1 σ3 σn σs Increase in pore fluid pressure can drive faulting!! σ1eff σ3eff σ1 σ3 σn
