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Stress and Deformation: Part I (D&R, 122-126; 226-252)

Stress and Deformation: Part I (D&R, 122-126; 226-252). The goal for today is to explore the stress conditions under which rocks fail (e.g., fracture), and the orientation of failure with respect to the principal stress directions. 1. Coulomb law of failure 2. Byerlee's law.

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Stress and Deformation: Part I (D&R, 122-126; 226-252)

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  1. Stress and Deformation: Part I(D&R, 122-126; 226-252) The goal for today is to explore the stress conditions under which rocks fail (e.g., fracture), and the orientation of failure with respect to the principal stress directions.1. Coulomb law of failure2. Byerlee's law

  2. Experimental studies are fundamental in the study of rock failure

  3. Common types of deformation experiments

  4. Compressive strength tests: The Goal

  5. Compressive strength tests: The Approach #3 #2 #1

  6. Compressive strength tests: The resultsLinear envelope of failure. The fractures form at angles of 25 to 35 degrees from s1- very consistent!

  7. Coulomb's Law of Failure sc = s0 + tanf(sn) sc = critical shear stress required for failures0 = cohesive strengthtanf = coefficient of internal friction (f = 90 - 2q) sN = normal stress

  8. Tensile strength tests with no confining pressureApproach: Similar to compressive strength testsResults: (1) Rocks are much weaker in tension than in compression (2) Fracture oriented parallel to s1 (q = 0)

  9. Tensile + Compressive strength testsResult: Failure envelope is parabolic0 < q < 30

  10. Failure envelopes for different rocks: note that slope of envelope is similar for most rocks sc = s0 + tanf(sn) sc = critical shear stress required for failures0 = cohesive strengthtanf = coefficient of internal frictionsN = normal stress

  11. Example: calculating compressive failure for a limestone

  12. The effect of mean stress:

  13. The effect of differential stress

  14. Byerlee's Law Question: How much shear stress is needed to cause movement along a preexisting fracture surface, subjected to a certain normal stress?Answer: Similar to Coulomb law without cohesionFrictional sliding envelope: sc = tanf(sN), where tanf is the coefficient of sliding friction

  15. Preexisting fractures of suitable orientation may fail before a new fracture is formed

  16. What about pore fluid pressure? Increasing pore fluid pressure favors failure!-Also may lead to tensile failure deep in crustEffective stress = sn – fluid pressure

  17. What is it? Tensile fracture filled with vein during dilation What is it?s1 is parallel to the structure. What does this suggest about the magnitude of effective stress?What mechanism may help produce this structure within the deeper crust? very low high fluid pressure to counteract lithostatic stress

  18. What happens at higher confining pressures? Von Mises failure envelope- Failure occurs at 45 degrees from s1

  19. Next Lecture Stress and Deformation II...A closer look at fault mechanics and rock behavior during deformation( D&R: pp. 304-319; 126-149)

  20. Important terminology/concepts Uniaxial vs. axial states of stress Coulomb law of failure: known how it is determined and equation q values for compression q values for tension Cohesive strength Coefficient of internal friction Byerlee's Law / frictional sliding envelope- know equation Important role of pore fluid pressure

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