advertisement

The strength of fractured rock Erling Fjær SINTEF Petroleum Research SINTEF Petroleum Research 1 Challenge: Estimation of rock strength from log data Porosity, Density, Sonic, . ... Strength Traditional approach: correlations UCS a tPb Wanted Available SINTEF Petroleum Research 2 Challenge: Estimation of rock strength from log data Porosity, Density, Sonic, . ... Available Strength Brandås et al. (2012) SINTEF Petroleum Research Wanted 3 Alternative approach: 1. Establish a constitutive model for static and dynamic moduli of rocks 2. Use the measured dynamic moduli (i.e. velocities) to calibrate the model 3. Use the calibrated model to simulate a test where strength can be measured SINTEF Petroleum Research 4 Dry, weak sandstone static moduli vs dynamic moduli 160 140 Axial Stress (MPa) Rock mechanical test including acoustic measurements on a dry sandstone 120 Peak stress 100 80 60 40 Radial 20 0 static moduli dynamic moduli The differences changes with stress and strain Young's modulus (GPa) 20 0.005 Dynamic 0.010 0.015 0.020 0.025 0.015 0.020 0.025 15 10 5 Static 0 -5 0.005 0.010 Axial strain SINTEF Petroleum Research 5 Dry, weak sandstone static moduli vs dynamic moduli 160 140 120 Rock mechanical test including acoustic measurements on a dry sandstone Stress (MPa) Axial Peak stress 100 80 60 40 Radial 20 static moduli dynamic moduli The differences changes with stress and strain 0 20 0.005 Young's modulus (GPa) We are seeking mathematical relations between the static and the dynamic moduli Dynamic 0.010 0.015 0.020 0.025 0.015 0.020 0.025 15 10 5 Static 0 -5 0.005 0.010 Axial strain SINTEF Petroleum Research 6 Hydrostatic test Building relations We introduce a parameter P, defined as: P v - total volumetric strain v v ,e 3 v ,e P is a measure of the inelastic part of the deformation caused by a compressive hydrostatic stress increment. - elastic K e strain SINTEF Petroleum Research 7 Hydrostatic test Building relations We introduce a parameter P, defined as: P v - total volumetric strain v v ,e 3 v ,e P is a measure of the inelastic part of the deformation caused by a compressive hydrostatic stress increment. - elastic K e strain Ke K 1 3 PKe K = Static bulk modulus Ke = Dynamic bulk modulus SINTEF Petroleum Research 8 Hydrostatic test Observations 25 1/P [GPa-1] 20 15 10 5 0 0 5 10 15 20 25 30 Hydrastatic stress [MPa] SINTEF Petroleum Research 9 Hydrostatic test Observations 25 1/P [GPa-1] 20 15 P 10 g T 5 0 0 5 10 15 20 25 30 Hydrastatic stress [MPa] SINTEF Petroleum Research 10 Uniaxial loading test Building relations We introduce a parameter F, defined as: F z - total axial strain z z ,e z , p z F is a measure of the inelastic part of the deformation caused by a shear stress increment. z ,e - elastic Ee strain z , p Pz z SINTEF Petroleum Research 11 Uniaxial loading test Building relations We introduce a parameter F, defined as: F z - total axial strain z z ,e z , p z F is a measure of the inelastic part of the deformation caused by a shear stress increment. z ,e - elastic Ee strain z , p Pz z Ee E 1 F 1 Pz Ee E = Static Young’s modulus Ee = Dynamic Young’s modulus SINTEF Petroleum Research 12 Uniaxial loading test Observations F* F z r S 0.04 F* 0.03 0.02 0.01 0.00 0.000 0.005 0.010 0.015 Shear strain SINTEF Petroleum Research 13 Uniaxial loading test Observations F* F z r S 0.04 F* 0.03 z r o FA z r S 0.02 0.01 0.00 0.000 0.005 0.010 0.015 Shear strain SINTEF Petroleum Research 14 Discussion: the F - parameter Ee E 1 F 1 Pz Ee 160 140 120 Note: Since E (1 - F) when F =1 then E = 0 Stress (MPa) Axial Peak stress 100 80 60 40 Radial 20 peak stress 0 0.005 0.010 0.015 0.020 0.025 F = 1 rock strength SINTEF Petroleum Research 15 Discussion: the F - parameter 1 3 1 8To 1 3 2 Griffith’s failure criterion: 1 3 o 1 3 S 2 Our model: A 2 If we can assume that: F2 ( 1 - 3) ( 1 - 3 ) then we could state that F = 1 Fulfilment of the Griffith criterion SINTEF Petroleum Research 16 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) ! SINTEF Petroleum Research 17 Discussion: the F - parameter The development of F can be seen as a gradual fulfillment of the Griffith criterion 0.40 F* 0.30 0.20 May be associated with local failure at various places in the rock, 0.10 triggered at different stress levels due to variable local strength Peak stress 0.00 0.000 0.005 0.010 0.015 0.020 0.025 Shear strain SINTEF Petroleum Research 18 We have a set of equations…… Ke K 1 3 PKe These represent a constitutive model for the rock E We may use it to predict rock behavior, and thereby derive mechanical properties for the rock P Ee 1 F 1 Pz Ee g T z r o FA z r S SINTEF Petroleum Research 19 Application for logging purposes Porosity, Density, Sonic, . ... Strength Constitutive model 12 10 Stress (MPa) 8 6 4 2 0 -5 0 5 10 15 Strain (mStrain) Simulates rock mechanical test on fictitious core SINTEF Petroleum Research 20 … an example: Strength (MPa) @ 2MPa 0 25 50 75 100 125 Depth (m - from a reference point) 0 Prediction from logs Courtesy of Statoil 50 100 150 200 Core measurements SINTEF Petroleum Research 21 Challenge: What is the impact of the intermediate principal stress on rock strength? In the field 1 2 in general In the lab 3 = 2 SINTEF Petroleum Research 3 22 Most convenient description: -plane Hydrostatic axis -plane cross sections (planes normal to the hydrostatic axis) Projections of the principal axes Cross section of the failure surface SINTEF Petroleum Research Failure criteria (-plane): No impact of the intermediate stress Empirical Assumption: Rotational symmetry in -plane (No physical argument) SINTEF Petroleum Research 24 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 3 2 Stress symmetry is not important 1 SINTEF Petroleum Research 25 Experimental observations: 2 = 3 Largest principal stress 1 [MPa] 350 2 = 1 300 250 200 150 No impact of intermediate stress 100 50 0 0 50 100 150 200 250 300 350 Intermediate principal stress 2 [MPa] SINTEF Petroleum Research 26 Experimental observations: 2 = 3 Largest principal stress 1 [MPa] 350 2 = 1 300 250 200 150 100 50 0 0 50 100 150 200 250 300 350 Intermediate principal stress 2 [MPa] Takahashi & Koide (1989) SINTEF Petroleum Research 27 Numerical simulations: 2 = 3 Largest principal stress 1 [MPa] 3 2 = 1 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 2 2.5 3 Intermediate principal stress 2 [MPa] Fjær & Ruistuen (2002) SINTEF Petroleum Research 28 Experimental observations: 2 = 3 Largest principal stress 1 [MPa] 350 2 = 1 300 MohrCoulomb 250 DruckerPrager 200 150 100 50 0 0 50 100 150 200 250 300 350 Intermediate principal stress 2 [MPa] SINTEF Petroleum Research -plane 29 Question: 2 = 3 Largest principal stress 1 [MPa] 3 2 = 1 What is similar when 2.5 2 = 3 Tetragonal 2 and 1.5 2 = 1 Tetragonal but different when 1 1 > 2 > 3 Orthorhombic 0.5 ? 0 0 0.5 1 1.5 2 2.5 3 Intermediate principal stress 2 [MPa] It’s the stress symmetry! SINTEF Petroleum Research 30 How can stress symmetry affect the strength? - It’s because it affects the probability for failure! 3 2 1 SINTEF Petroleum Research 31 Classical picture Probability for failure 1 3 2 0 1 m SINTEF Petroleum Research 32 Classical picture Probability for failure 3 2 1 0 1 m SINTEF Petroleum Research 33 Classical picture Probability for failure 3 2 1 0 1 m SINTEF Petroleum Research 34 Classical picture Probability for failure 3 2 1 0 1 m SINTEF Petroleum Research 35 Classical picture Probability for failure 1 3 2 0 1 m SINTEF Petroleum Research 36 Classical picture Probability for failure 1 3 2 0 1 m SINTEF Petroleum Research 37 Classical picture Probability for failure 1 3 2 0 1 m SINTEF Petroleum Research 38 Classical picture Classical picture: Failure occurs if the shear stress across any plane in the rock sample Probability for failure exceeds So + – otherwise not. Introducing fluctuations: The shear strength varies from plane to plane. The rock fails when exceeds the shear strength for one of them. 1 0 So + The probability for failure increases when So + SINTEF Petroleum Research 39 Classical picture All planes oriented at an angle relative to the 1 axis Many potential failure planes in a critical state 2 3 2 High probability for failure 1 SINTEF Petroleum Research 40 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 2 3 2 Low probability for failure 1 SINTEF Petroleum Research 41 Classical picture All planes oriented at an angle /2 - relative to the 3 axis Many potential failure planes in a critical state 2 3 High probability for failure 1 2 SINTEF Petroleum Research 42 Mathematical model Probability for failure of a plane with orientation specified by (,): p f , n n So n (n classical Mohr-Coulomb) Overall probability for failure: Expected strength of the material: Pfailure 1 1 p , all , f Pfailure 1 exp 1 SINTEF Petroleum Research N T d 1 43 Mathematical model Probability for failure of a plane with orientation specified by (,): (n classical Mohr-Coulomb) p f , f ( 1 , 1+ 1 ) Probability distribution 0.015 n So n 2 = 0.01 2 = 1 Overall probability for failure: 2 = 0.5( 1 + 3) 0.005 0 0.75 n 1 1.25 1/ M-C 1.5 Pfailure 1 1 p , all , N T f Expected strength of the material: Pfailure 1 exp 1 SINTEF Petroleum Research d 1 44 Mathematical model Pfailure f 1 , 1 1 1 1 f ( 1 , 1+ 1 ) Probability distribution 0.015 2 = 0.01 2 = 1 2 = 0.5( 1 + 3) 0.005 0 0.75 1 1.25 1.5 1/ M-C SINTEF Petroleum Research 45 Mathematical model 8 2 = 3 2 = 1 6 exp 4 n = 30 n = 175 n = 1000 Mohr-Coulomb 2 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 0 0 2 4 6 8 2 SINTEF Petroleum Research 46 Comparing model and observations 2 = 1 Largest principal stress 1 [MPa] 350 300 Takahashi and Koide, 1989 250 200 150 n = 30 100 50 0 0 50 100 150 200 250 300 350 Intermediate principal stress 2 [MPa] SINTEF Petroleum Research 47 Comparing model and observations 2 = 1 Largest principal stress 1 [MPa] 3 2.5 Numerical model 2 1.5 1 n = 25 0.5 0 0 0.5 1 1.5 2 2.5 3 Intermediate principal stress 2 [MPa] SINTEF Petroleum Research 48 Fractures are planes with largely reduced or no strength Outcrop from a Marcellus shale formation Han, 2011 SINTEF Petroleum Research Borehole breakouts in a non-fractured rock SINTEF Petroleum Research Borehole breakouts in a non-fractured rock Shear failure planes SINTEF Petroleum Research Borehole breakouts in a non-fractured rock Shear failure planes Failure would be more easily triggered by a set of suitably oriented SINTEF Petroleum Research Simple example Mudweight [sg] 1 1.2 1.4 1.6 1.8 1500 No fractures 1505 1510 1515 1520 1525 1530 SINTEF Petroleum Research 53 Simple example Mudweight [sg] 1 1.2 1.4 1.6 1.8 1500 No fractures 1505 1510 Sealed fractures 1515 || borehole 1520 1525 1530 SINTEF Petroleum Research 54 Simple example Mudweight [sg] 1 1.2 1.4 1.6 1.8 1500 No fractures 1505 1510 Sealed fractures 1515 || borehole 1520 Open fractures || borehole 1525 1530 SINTEF Petroleum Research 55 Several fracture sets complicates the situation. Blocks may become detached at washed away by the circulating mud. More fractures will be exposed to the drilling fluid. SINTEF Petroleum Research Other possible failure modes – bedding plane splitting SINTEF Petroleum Research Other possible failure modes – bedding plane splitting SINTEF Petroleum Research Other possible failure modes – bedding plane splitting SINTEF Petroleum Research Other possible failure modes – bedding plane splitting SINTEF Petroleum Research Other possible failure modes – bedding plane splitting SINTEF Petroleum Research Other possible failure modes – bedding plane splitting SINTEF Petroleum Research Other possible failure modes – bedding plane splitting SINTEF Petroleum Research Økland and Cook 1998 SINTEF Petroleum Research 20 To avoid the problem: Økland and Cook 1998 The “angle of attack” between the well and the bedding plane should be at least SINTEF Petroleum Research 20. Challenge: What is the strength of a fractured rock (if we consider it as homogeneous)? Available alternative: Hoek-Brown Purely empirical criterion Hoek & Brown (1980) SINTEF Petroleum Research 66 Geologocal Strength Index GSI SINTEF Petroleum Research 67 Rocks are heterogeneous – treating them as homogeneous comes at a price….. SINTEF Petroleum Research 68 The strength of a homogeneous material is size invariant. Rocks, on the other hand, - Hoek & Brown (1980) SINTEF Petroleum Research 69 Current work: Relate the failure probability model to Hoek-Brown SINTEF Petroleum Research 70 Challenge: Match with observations 60 50 1 40 30 20 10 0 0 5 10 15 20 3 Data from Hoek; Kaiser (2008) Failure probability model SINTEF Petroleum Research 71 Consideravble scatter in measured strength Kaiser (2008) SINTEF Petroleum Research 72 60 50 1 40 Probability distribution 0.16 30 0.14 20 0.12 0.1 10 0.08 0.06 0 0.04 0 5 10 15 20 3 0.02 0 0 10 20 30 1 40 50 Failure probability model SINTEF Petroleum Research 73 Conclusions: Physics helps us to make better tools for rock mechanics applications There is still room for more physics in rock mechanics SINTEF Petroleum Research 74