The strength of fractured rock

1. The strength of fractured rock Erling Fjær SINTEF Petroleum Research

2. Strength Porosity, Density, Sonic, . . . . Challenge: Estimation of rock strength from log data Traditional approach: correlations Wanted Available

3. Strength Porosity, Density, Sonic, . . . . Challenge: Estimation of rock strength from log data Brandås et al. (2012) Wanted Available

4. Alternative approach: • Establish a constitutive model for static and dynamic moduli of rocks • Use the measured dynamic moduli (i.e. velocities) to calibrate the model • Use the calibrated model to simulate a test where strength can be measured

5. Dry, weak sandstone static moduli vs dynamic moduli Rock mechanical testincluding acoustic measurementson a dry sandstone static moduli  dynamic moduli The differences changes with stress and strain

6. Dry, weak sandstone static moduli vs dynamic moduli Rock mechanical testincluding acoustic measurementson a dry sandstone We are seeking mathematical relations between the static and the dynamic moduli static moduli  dynamic moduli The differences changes with stress and strain

7. Building relations Hydrostatic test We introduce a parameter P, defined as: Pis a measure ofthe inelastic part of the deformation caused by a compressive hydrostatic stress increment. v- total volumetric strain - elastic strain

8. Building relations Hydrostatic test We introduce a parameter P, defined as: Pis a measure ofthe inelastic part of the deformation caused by a compressive hydrostatic stress increment. v- total volumetric strain - elastic strain  K = Static bulk modulus Ke = Dynamic bulk modulus

9. Observations Hydrostatic test

10. Observations Hydrostatic test

11. Building relations Uniaxial loading test We introduce a parameter F, defined as: Fis a measure ofthe inelastic part of the deformation caused by a shear stress increment. z- total axial strain - elastic strain

12. Building relations Uniaxial loading test We introduce a parameter F, defined as: Fis a measure ofthe inelastic part of the deformation caused by a shear stress increment. z- total axial strain - elastic strain  E = Static Young’s modulus Ee = Dynamic Young’s modulus

13. Observations Uniaxial loading test

14. Observations Uniaxial loading test

15. Discussion: the F - parameter Note: Since E  (1 - F) whenF =1 thenE = 0  peak stress F = 1 rock strength

16. Discussion: the F - parameter Griffith’s failure criterion: Our model: If we can assume that: (1 - 3) (1 - 3) then we could state that F = 1  Fulfilment of the Griffith criterion

17. Discussion: the F - parameter (1 - 3) (1 - 3) ? OK for a purely elastic material Also OK at the intact parts of the material even after local failure has occurred elsewhere Local (1 - 3) Global (1 - 3) !

18. Discussion: the F - parameter The development of F can be seen as a gradual fulfillment of the Griffith criterion May be associated with local failure at various places in the rock, triggered at different stress levels due to variable local strength

19. We have a set of equations…… These represent a constitutive model for the rock We may use it to predict rock behavior, and thereby derive mechanical properties for the rock

20. Constitutive model Strength Porosity, Density, Sonic, . . . . Simulates rock mechanical test on fictitious core Application for logging purposes

21. … an example: Courtesy of Statoil Prediction from logs Core measurements

22. Challenge: What is the impact of the intermediate principal stress on rock strength? In the field In the lab s1s2s3in general s2 = s3

23. Most convenient description: -plane cross sections (planes normal to the hydrostatic axis) Hydrostatic axis -plane Projections of the principal axes Cross section of the failure surface

24. Failure criteria (-plane): No impact of the intermediate stress Empirical Assumption: Rotational symmetry in -plane (No physical argument)

25. s3 s2 s1 Basic theory on shear failure: Shear failure occurs when the shear stress over some plane within the rock exceeds the shear strength of the rock  The intermediate principal stress (2) has no impact Stress symmetry is not important

26. Experimental observations: No impact of intermediate stress

27. Experimental observations: Takahashi & Koide (1989)

28. Numerical simulations: Fjær & Ruistuen (2002)

29. Experimental observations: Mohr-Coulomb Drucker-Prager -plane

30. Question: What is similar when 2 = 3 and 2 = 1 but different when 1 > 2 > 3 ? Tetragonal Tetragonal Orthorhombic It’s the stress symmetry!

31. How can stress symmetry affect the strength? - It’s because it affects the probability for failure!  s3 s2 s1

32.  1  0 m Classical picture Probability for failure  s3 s2 s1 

33.  1  0 m Classical picture Probability for failure  s3 s2 s1 

34.  1  0 m Classical picture Probability for failure  s3 s2 s1 

35.  1 0 m Classical picture Probability for failure   s3 s2 s1 

36.   1 0 m Classical picture Probability for failure  s3 s2 s1 

37.  1 0 m Classical picture Probability for failure   s3 s2 s1 

38.   1 0 m Classical picture Probability for failure  s3 s2 s1 

39. 1 0 Classical picture Classical picture: Failure occurs if the shear stress across any planein the rock sample exceedsSo +  – otherwise not. Probability for failure Introducing fluctuations:The shear strength varies from plane to plane. The rock fails when  exceeds the shear strength for one of them.  So +  The probability for failureincreases when   So + 

40. s3 = s2 Classical picture All planes oriented at an angle  relative to the 1 axis   Many potential failure planes in a critical state  High probability for failure 2 s1

41. Classical picture Only planes oriented at an angle  relative to the 1 axis, and parallel to the 2 axis   Few potential failure planes in a critical state  Low probability for failure 2 s3 s2 s1

42. s1 = s2 Classical picture All planes oriented at an angle /2 - relative to the 3 axis   Many potential failure planes in a critical state  High probability for failure 2 s3

43. Mathematical model Probability for failure of a plane with orientation specified by (,): (n    classical Mohr-Coulomb) Overall probability for failure: Expected strength of the material:

44. Mathematical model Probability for failure of a plane with orientation specified by (,): (n    classical Mohr-Coulomb) Overall probability for failure: Expected strength of the material:

45. Mathematical model

46. Mathematical model The impact of the intermediate principal stress is directly linked to the non-sharpness of the failure criterion(represented by 1/n) i.e. to the rock heterogeneity

47. Comparing model and observations Takahashi and Koide, 1989 n = 30

48. Comparing model and observations Numerical model n = 25

49. Fractures are planes with largely reduced or no strength Outcrop from a Marcellus shale formation Han, 2011

50. Borehole breakouts in a non-fractured rock