ResGeo 202 Reservoir Geomechanics Spring 2015, OpenEdX Homework 8 – Identifying Critically-Stressed Faults Due June 2, 2015 (7:30 UTC) Please direct any questions to the forum on the OpenEdX Courseware page
Part 1: Stress State Information Part 2: Shear Slip on Faults
The red line is the failure line, where ! = µ " n , and each fracture fracture is a point plotted by its shear stress and effective normal stress, colored it’s dip angle. On this graph, that means the horizontal distance between that point, and the red line. As you can see, the three fractures dipping at about 60 degrees are critically stressed.
Histogram of amount of pore pressure that it would take to bring each fracture to failure in PSI. This is calculated as the horizontal distance from the fracture to the red diagonal line in the above plot. The analytical solution is: Pp_to_failure = " n - (! / µ ). On a third figure, plot depth (in ft) on the vertical axis, and pore pressure to failure (in PSI) on the horizontal axis. If possible, color these points by fracture dip angle.
How many fractures are critically stressed, (within 100 PSI of the failure envelope)? 3 fractures are critically stressed, they each have a red X on them in the above plot. They are critically stressed because they are near the ideal dip angle for slipping in a normal faulting regime.
How many fractures are within 1000 PSI of failure? 23, See histogram above.
What are the dip angles of the critically stressed fractures? 2 are 57 degrees, 1 is 59 degrees. In what formation are the critically stressed fractures (see homework 3 for formation depth intervals). They are in the Barnett, which is at 5514 to 6195 ft. What coefficient of friction ( µ= ! /" n) would be required for the fracture least likely to slip in the current stress state, to be critically stressed? The fracture least likely to slip is at a depth of 6164 ft, and ! /" n = 0.065774. Thus, if the coefficient of friction were (hypothetically) that low, then that fracture would be able to slip without a perturbation in pore pressure. For µ= 0.65, it would require 3144 PSI of pore pressure to reach failure Part 4: Answer the questions on the page below Use the plots and the calculations from Parts 1-3 to answer the questions on the page below. The answers will be posted after the due date. Numerical entry type responses have only a range of acceptable values and are graded electronically, so please adhere to the value of constants given here to prevent misgrading of your submissions.
Refer to your plot of the Mohr-Coulomb failure envelope in 2D to answer the questions below. >>How many fractures are critically stressed (within 100 psi of failure)?<< = 3.0 +- 0.3 [explanation] 3 fractures have resolved shear stresses within 100 PSI of the frictional failure line and are therefore within measurement error of being critically-stressed. [explanation] >>At what dip angles are the critically-stressed fractures?<< ( ) 0-10° ( ) 10-20° ( ) 20-30° ( ) 30-40°
( ) 40-50° (x) 50-60° ( ) 60-70° ( ) 70-80° ( ) 80-90° [explanation] The 3 active fractures are oriented between 50-60°, 2 are 57 degrees, 1 is 59 degrees. 60° is the ideal dip angle for slip in a normal faulting regime. [explanation]
>>How many fractures are within 1,000 PSI (pore pressure) of failure?<< = 23.0 +- 0.1 >>What coefficient of friction ( µ) would be required for the fracture least likely to slip in the current stress state and pore pressure, to be critically stressed?<< = 0.065774 +- 0.01
>>In what formation are the critically stressed fractures? (Refer to homework 3 data for formation depth intervals)<< ( ) Marble Falls ( ) Duffer (x) Barnett
