Announcements This week's lab: 1-3 PM with Andrew McCarthy. Please come prepared with specific questions. There will be no lecture this Wednesday! Please use the time to: 1) Study important terms/concepts listed at end of Powerpoint files 2) Study the practice problems 3) Fault project Reminder: Midterm Oct. 14
21
Embed
Announcements This week's lab: 1-3 PM with Andrew McCarthy. Please come prepared with specific questions. There will be no lecture this Wednesday! Please.
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Announcements
This week's lab: 1-3 PM with Andrew McCarthy. Please come prepared with specific questions.
There will be no lecture this Wednesday!
Please use the time to:1) Study important terms/concepts listed at end of
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
Experimental studies are fundamental in the study of rock failure
Common types of deformation experiments
Compressive strength tests: The Goal
Compressive strength tests: The Approach
Compressive strength tests: The resultsLinear envelope of failure. The fractures form at angles of 25 to 35 degrees from 1- very consistent!
c = critical shear stress required for failure0 = cohesive strengthtan = coefficient of internal friction () N = normal stress
Coulomb's Law of Failure
c = 0 + tan(n)
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 1 (= 0)
Tensile + Compressive strength tests
Result: Failure envelope is parabolic0 < < 30
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
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: c = tan(N), where tan is the coefficient of sliding friction
Preexisting fractures of suitable orientation may fail before a new fracture is formed
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?
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
What happens at higher confining pressures?
Von Mises failure envelope- Failure occurs at 45 degrees from 1
Next Lecture
Stress and Deformation II
...A closer look at fault mechanics and rock behavior during deformation
( D&R: pp. 304-319; 126-149)
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