1. Mohr-Coulomb failure Goal: To understand relationship between stress, brittle failure, and frictional faulting and to use this relationship to predict rock behavior

2. Stress review Stress = Force/Area 3 principal vectors: σ 1, σ 2, and σ 3 at right angles to each other σ 1 ≥ σ 2 ≥ σ 3 σ 1 is the maximum principal stress direction, σ 2 is the intermediate principal stress direction, and σ 3 is the minimum principal stress direction

3. Static stress as σ 1 = σ 2 = σ 3 Lithostatic stress as static stress generated by mass of overlying rocks Differential stress (σ d ) as (σ 1 - σ 3 ) Confining pressure as σ 2 = σ 3 for the conditions σ 1 > σ 2 = σ 3 We also define:

4. Shear stress and normal stress For any plane in a stress field defined by σ 1, σ 2, and σ 3 with strike parallel with σ 2 : θ σ1σ1 σ1σ1 σ3σ3 σ3σ3

5. The stress is resolved into 2 components: 1.Shear stress (σ s ), acting parallel with the plane 2.Normal stress (σ n ), acting perpendicular to the plane θ σ1σ1 σ1σ1 σ3σ3 σ3σ3 σnσn σnσn σsσs σsσs

6. Stress components are related by: 1.σ s = ½(σ 1 - σ 3 )sin(2θ) 2.σ n = ½(σ 1 + σ 3 ) - ½(σ 1 - σ 3 )cos(2θ) where θ = angle between plane and σ 1 θ σ1σ1 σ1σ1 σ3σ3 σ3σ3 σnσn σnσn σsσs σsσs

7. Mohr diagram for stress Relationship between σ 1, σ 3, σ s, and σ n is plotted graphically in Cartesian coordinates σsσs σnσn

8. Mohr circle for stress: circle with diameter = σ d plotted on mohr diagram Center on the σ n -axis at point = ½(σ 1 + σ 3 ) σsσs σnσn σ1σ1 σ3σ3 ½(σ 1 + σ 3 )

9. Finding σ s, and σ n Can use a Mohr circle to find σ s, and σ n for any plane σsσs σnσn σ1σ1 σ3σ3

10. Plot a line from center to edge of circle at angle 2θ-clockwise from σ n -axis σsσs σnσn σ1σ1 σ3σ3 2θ2θ

11. X- and y-coordinates of intersection of line and circle define σ s and σ n for the plane σsσs σnσn σ1σ1 σ3σ3 (σ s, σ n ) of plane

12. Coulomb’s failure criterion Every homogeneous material has a characteristic failure envelope for brittle shear fracturing Combinations of σ s and σ n outside of the envelope result in fracture

13. Determining failure envelope Experimental rock deformation

14. The Coulomb envelope σsσs σnσn σ1σ1 σ3σ3 2θ2θ Stable Tensile Fracture Shear Fracture Shear Fracture

15. Coulomb law of failure σsσs σnσn σ0σ0 φ σ c = σ 0 + tan(φ)σ n

16. Formula defines shear stress under which rocks will fracture σ c = critical shear stress — σ s at failure σ 0 = cohesive strength — σ s when σ n = 0 φ = angle of internal friction — φ ≈ 90 - 2θ σ c = σ 0 + tan(φ)σ n

17. For most rocks, angle of internal friction ≈ 30° Therefore, θ at failure is also ≈ 30° σ s is greatest when θ = 45°

18. Failure envelopes for different rocks

19. Slip on pre-existing fractures Pre-existing fractures have no cohesive strength, σ 0 = 0 Failure envelopes for pre-existing fractures derived experimentally

20. Envelope of sliding friction σsσs σnσn φ f = angle of sliding friction

21. Byerlee’s law Describes frictional sliding envelope σ c = tan(φ f )σ n φ f ≈ 40° for low confining pressures and ≈ 35° for high confining pressures

22. Byerlee’s law for different rock types

23. Effect of pore-fluid pressure Pore fluid pressure (P f ) effectively lowers the stress in all directions The effective stresses (σ 1 eff, σ 2 eff, and σ 3 eff ) = principal stresses - P f σ 1 eff = σ 1 - P f σ 2 eff = σ 2 - P f σ 3 eff = σ 3 - P f

24. σsσs σnσn σ1σ1 σ3σ3 Stable stress conditions

25. σsσs σnσn σ1σ1 σ3σ3 σ 1 eff σ 3 eff Increase in pore fluid pressure can drive faulting!!