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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),

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Presentation on theme: "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),"— Presentation transcript:

1 Stress and Deformation: Part I (D&R, ; ) 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

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   

6

7 Compressive strength tests: The results Linear envelope of failure. The fractures form at angles of 25 to 35 degrees from  1- very consistent!

8  c = critical shear stress required for failure  0 = cohesive strength tan  = coefficient of internal friction (  )  N = normal stress Coulomb's Law of Failure  c =  0 + tan  (  n )

9 Tensile strength tests with no confining pressure Approach: Similar to compressive strength tests Results: (1) Rocks are much weaker in tension than in compression (2) Fracture oriented parallel to  1 (  = 0)

10 Tensile + Compressive strength tests Result: Failure envelope is parabolic 0 <  < 30

11 Failure envelopes for different rocks: note that slope of envelope is similar for most rocks  c =  0 + tan  (  n )  c = critical shear stress required for failure  0 = cohesive strength tan  = coefficient of internal friction  N = normal stress

12 Example: calculating compressive failure for a limestone

13 The effect of mean stress:

14 The effect of differential stress

15 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 cohesion Frictional sliding envelope:  c = tan  (  N ), where tan  is the coefficient of sliding friction

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

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18 Increasing pore fluid pressure favors failure! -Also may lead to tensile failure deep in crust Effective stress =  n – fluid pressure What about pore fluid pressure?

19 What is it? What is it?  1 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? Tensile fracture filled with vein during dilation very low high fluid pressure to counteract lithostatic stress

20 What happens at higher confining pressures? Von Mises failure envelope - Failure occurs at 45 degrees from  1

21 Next Lecture Stress and Deformation II...A closer look at fault mechanics and rock behavior during deformation ( D&R: pp ; )

22 Important terminology/concepts Uniaxial vs. axial states of stress Coulomb law of failure: known how it is determined and equation  values for compression  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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