Formulas - Rock Mechanics

BS”D Reservoir Rock Mechanics – Formulas 1. Index Properties of Reservoir Rocks Volume [𝒎𝟑 ] 𝑉𝑔 𝑉𝑝 𝑉𝑓 𝑉𝑡 𝑉𝑠 Mass [𝒌𝒈] 𝑀𝑓 𝑀𝑠 𝑀𝑡 Parameter Formula 𝑉𝑝 𝑉𝑡 𝑉𝑝 𝑒= 𝑉𝑠 𝑉𝑓 𝑆= 𝑉𝑝 𝑀𝑓 𝑊= 𝑀𝑠 𝑀𝑠 𝜌𝑑𝑟𝑦 = 𝑉𝑡 (𝑀𝑠 + 𝜌𝑓 𝑉𝑝 ) 𝜌𝑠𝑎𝑡 = 𝑉𝑡 𝑀 𝑘𝑔 𝜌= [ 3] 𝑉 𝑚 𝑀𝑔 𝑘𝑁 𝛾 = 𝜌𝑔 = [ ] 𝑉 𝑚3 Porosity 𝑛= Void ratio Saturation Water content Dry density (S=0%) Saturated density (S=100%) Density Unit weight Unit weight if the rock is below the ground water table 𝛾𝑏 = (𝜌𝑡 − 𝜌𝑤 )𝑔 𝐺= Specific gravity of solid 𝜌𝑠 𝜌𝑓 Useful values: 𝑘𝑁 𝛾𝑤 = 9.81 [ 3 ] 𝑚 𝑘𝑔 𝜌𝑤 = 1000 [ 3 ] 𝑚 If 𝑉𝑡 is not given, assume 𝑉𝑡 = 1𝑚3 𝑉 If the rock is composed of several minerals, 𝐺 = ∑ni=1 𝑉𝑖 ⋅ 𝐺𝑖 𝑠 𝑘 𝑑𝑝 Permeability (Darcy’s law): 𝑄 = 𝜇 𝑑𝑙 𝐴 → 𝑘 = 𝑄𝜇 𝐴 𝑑𝑝 −1 ( 𝑑𝑙 ) [𝐿2 ] 5783 Matrix of direction cosines for a given dip and strike x – east, y – north, z – upwards. x’ – normal to the dip-strike plane. y’ – upwards on the dip-strike plane. z’ – coalesces with the strike. Directions are determined by the right-hand rule. 𝑙 𝑥 ′ 𝑚𝑥 ′ 𝑛 𝑥 ′ 𝑙 = cos 𝛿 cos 𝛽 𝐿 = (𝑙𝑦′ 𝑚𝑦′ 𝑛𝑦′ ) , 𝑚 = cos 𝛿 sin 𝛽 𝑙 𝑧 ′ 𝑚𝑧 ′ 𝑛 𝑧 ′ 𝑛 = sin 𝛿 1. Find the 𝛽’s of the axes on the x-y plane. For instance, if 𝛽𝑥′ = −110°, then 𝛽𝑦′ = 70° and 𝛽𝑧′ = 160°: Sound velocity measurement: 𝑉𝑡𝑝/𝑠 = 𝑡 𝐿 𝑝/𝑠 , 𝑡𝑝/𝑠 = 𝑡𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 − 𝑡𝑒𝑛𝑑𝑐𝑎𝑝𝑠 2. Stress Transformations – 2D 𝜎𝑥𝑥 𝜎 = (𝜏 𝑦𝑥 𝜏𝑥𝑦 𝜎𝑦𝑦 ) , 𝐿=( cos 𝜃 − sin 𝜃 2. Find the 𝛿’s of the axes. For instance, if the dip magnitude is 70°, then 𝛿(𝑥) = 20° and 𝛿𝑦′ = 70°. z’ is parallel to the x-y plane (as it’s completely horizontal), thus 𝛿𝑧′ = 0°. 𝑙 𝑚 𝑛 Line Bearing 𝛽 𝛿 = cos 𝛿 cos 𝛽 = cos 𝛿 sin 𝛽 = sin 𝛿 𝑥′ East −110° 20° −0.321 −0.883 0.342 𝑦′ West 70° 70° 0.117 0.321 0.940 South 𝑧′ 160° 0° −0.939 0.342 0 Maximum shear stress on the x’ plane Magnitude: 𝜏𝑥′ ,𝑚𝑎𝑥 = √𝜏𝑥2′ 𝑦′ + 𝜏𝑥2′ 𝑧 ′ Direction (clockwise for the negative direction 𝜏 ′ ′ 𝜎 ′ = 𝐿𝜎𝐿𝑇 𝑥′ 𝑦′ 0 0 0 0 0 0 0 0 0 0 0 0 0 2(1 + 𝜈) 𝐸 0 0 0 0 0 2(1 + 𝜈) 𝐸 0 0 𝜆 𝜆 + 2𝐺 𝜆 0 0 0 0 𝜆 𝜆 𝜆 + 2𝐺 0 0 0 2(1 + 𝜈) ) 𝐸 0 0 0 0 𝐺 0 0 0 𝜎𝑥 𝜎𝑦 𝜎𝑧 𝜏𝑥𝑦 𝜏𝑦𝑧 (𝜏𝑧𝑥 ) 0 0 0 0 𝐺 0 𝜀𝑥 0 𝜀𝑦 0 𝜀𝑧 0 𝛾𝑥𝑦 0 𝛾 𝑦𝑧 0 𝐺 ) (𝛾𝑧𝑥 ) 𝜎1 = (𝜆 + 2𝐺)𝜀1 + 𝜆𝜀2 + 𝜆𝜀3 𝜎2 = 𝜆𝜀1 + (𝜆 + 2𝐺)𝜀2 + 𝜆𝜀3 𝜎3 = 𝜆𝜀1 + 𝜆𝜀2 + (𝜆 + 2𝐺)𝜀3 Δ𝑉 Volumetric Strain: Δ = 𝜀1 + 𝜀2 + 𝜀3 = 𝑉 0 𝜎1 = 𝜆Δ + 2𝐺𝜀1 𝜎2 = 𝜆Δ + 2𝐺𝜀2 𝜎3 = 𝜆Δ + 2𝐺𝜀3 3𝜎𝑚 = 𝜎1 + 𝜎2 + 𝜎3 = (3𝜆 + 2𝐺)Δ 2 1 𝜎𝑚 = 𝐾Δ, 𝐾 ⏟ = 𝜆 + 𝐺, ⏟ 3 𝐾 𝑏𝑢𝑙𝑘 𝑏𝑢𝑙𝑘 𝑐𝑜𝑚𝑝𝑟𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 1 𝜆+𝐺 𝜆 𝜆 (𝜎 − 𝜈(𝜎2 + 𝜎3 )) = 𝜎 − 𝜎 − 𝜎 𝐸 1 𝐺(3𝜆 + 2𝐺) 1 2𝐺(3𝜆 + 2𝐺) 2 2𝐺(3𝜆 + 2𝐺) 3 1 𝜆 𝜆+𝐺 𝜆 𝜀2 = (𝜎2 − 𝜈(𝜎1 + 𝜎3 )) = − 𝜎1 + 𝜎2 − 𝜎 𝐸 2𝐺(3𝜆 + 2𝐺) 𝐺(3𝜆 + 2𝐺) 2𝐺(3𝜆 + 2𝐺) 3 1 𝜆 𝜆 𝜆+𝐺 )) 𝜀3 = (𝜎3 − 𝜈(𝜎1 + 𝜎2 = − 𝜎 − 𝜎 + 𝜎 𝐸 2𝐺(3𝜆 + 2𝐺) 1 2𝐺(3𝜆 + 2𝐺) 2 𝐺(3𝜆 + 2𝐺) 3 𝜀1 = 4. Strain Analysis 2D: 𝜀 = ( 𝜀𝑥 Γ𝑥𝑦 Γ𝑥𝑦 cos 𝜃 sin 𝜃 ), 𝐿 = ( ) 𝜀𝑦 − sin 𝜃 cos 𝜃 1 𝜕𝑢𝑥 𝜕𝑢𝑦 1 1 1 = 2𝛾𝑥𝑦 = 2𝛾𝑦𝑥 = ( + ) = (tan 𝜙1 + tan 𝜙2 ) 2 𝜕𝑦 𝜕𝑥 2 𝜀′ = 𝐿𝜀𝐿𝑇 Δ𝐿 𝑙 −𝑙 𝜀 = 𝑙 = 0𝑙 (shortening is positive!) 0 0 𝐼1 = 𝜎𝑥 + 𝜎𝑦 = 𝜎𝑥′ + 𝜎𝑦′ The strains can be found with the following equations as well: 𝜀𝑥′ = 𝜀𝑥 cos 2 𝜃 + 𝜀𝑦 sin2 𝜃 + Γ𝑥𝑦 sin 2𝜃 2 2 {𝜀𝑦′ = 𝜀𝑥 sin 𝜃 + 𝜀𝑦 cos 𝜃 − Γ𝑥𝑦 sin 2𝜃 Γ𝑥′𝑦′ = 2(𝜀𝑦 − 𝜀𝑥 ) sin 2𝜃 + Γ𝑥𝑦 cos 2𝜃 2Γ𝑥𝑦 Principal strains: tan 2𝜃 = 𝜀 Γ𝑚𝑎𝑥 1 (𝜀 2 𝑥 2 + + 𝜀𝑦 ) ± √𝛤𝑥𝑦 𝜀1 − 𝜀3 = 2 𝜀1,3 = 𝐼1 = 𝜎𝑥 + 𝜎𝑦 = 𝜎𝑥′ + 𝜎𝑦′ The stresses can be found with the following equations as well: 𝜎𝑥′ = 𝜎𝑥 cos 2 𝛼 + 𝜎𝑦 sin2 𝛼 + 𝜏𝑥𝑦 sin 2𝛼 2 2 {𝜎𝑦′ = 𝜎𝑥 sin 𝛼 + 𝜎𝑦 cos 𝛼 − 𝜏𝑥𝑦 sin 2𝛼 𝜈 𝐸 𝜈 − 𝐸 1 𝐸 − 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 of y’): 𝜃 = tan−1 (𝜏 𝑥 𝑧 ) 1 sin 𝜃 ) ↺ cos 𝜃 𝜈 𝐸 1 𝐸 𝜈 − 𝐸 − ( 0 Γ𝑥𝑦 = Γ𝑦𝑥 Load at failure: 𝑃 = 𝐴𝑝𝑖𝑠𝑡𝑜𝑛 𝜎𝑝𝑖𝑠𝑡𝑜𝑛 UCS: 𝜎𝑐 = 𝐹 ⋅ 𝐼𝑠 (F depends on lithology, assume 𝐹 = 24 if there’s no info) 2𝑃 Brazilian tensile strength: 𝜎𝑡 = 𝜋𝐷𝑡 𝜀𝑥 𝜀𝑦 𝜀𝑧 𝛾𝑥𝑦 = 𝛾𝑦𝑧 (𝛾𝑧𝑥 ) 1 𝐸 𝜈 − 𝐸 𝜈 − 𝐸 𝜎𝑥 𝜆 + 2𝐺 𝜎𝑦 𝜆 𝜎𝑧 𝜆 𝜏𝑥𝑦 = 0 𝜏𝑦𝑧 0 ( 𝜏𝑧𝑥 ) ( 0 𝑃 Point load index: 𝐼𝑠 = 𝐷2 The generalized Hooke’s law for an isotropic material (Same 𝐸 and 𝜈 for all directions): −𝜈𝜀1 + 𝜈𝜀2 = 0,Δ = 𝜀1 + 𝜀2 + 𝜀3 = 0 Plane Stress: 𝜎1 ≠ 0, 𝜎2 ≠ 0, 𝜎3 = 0 → 𝜀1 ≠ 0, 𝜀2 ≠ 0, 𝜀3 ≠ 0 Plane Strain: 𝜀1 ≠ 0, 𝜀2 ≠ 0, 𝜀3 = 0 → 𝜎1 ≠ 0, 𝜎2 ≠ 0, 𝜎3 ≠ 0 6. Friction on Rock Surfaces 𝑥−𝜀𝑦 1 (𝜀 4 𝑥 Special Cases Uniaxial Stress: 𝜎1 ≠ 0, 𝜎2 = 𝜎3 = 0, 𝜀1 ≠ 0, 𝜀2 = 𝜀3 Hydrostatic Stress: 𝜎1 = 𝜎2 = 𝜎3 = 𝑃, 𝜀1 = 𝜀2 = 𝜀3 ≠ 0 Uniaxial Strain: 𝜀1 ≠ 0, 𝜀2 = 𝜀3 = 0, 𝜎1 ≠ 0, 𝜎2 = 𝜎3 ≠ 0 𝜎 −𝜎 Pure Shear: 𝜎1 = 𝑆 = −𝜎3 , 𝜎2 = 0, 𝜏𝑚 = 1 2 3 = 𝑆 → 𝜀1 = −𝜀3 , 𝜀2 = − 𝜀𝑦 ) Calculation of maximal and residual shear strength from a direct shear test: 2 Mohr Circle: 2 (𝜀𝜃 − 𝜀𝑚 )2 + Γ𝜃2 = Γ𝑚𝑎𝑥 𝜀𝑥 + 𝜀𝑦 𝜀𝑚 = 2 Coordinates: (𝜀𝑥𝑥 , Γ𝑥𝑦 ), (𝜀𝑦𝑦 , −Γ𝑥𝑦 ) Stress cube ↺ = 𝛼 → Line in MC ↻ = 2𝛼 𝜏 In 𝜏 − 𝜎𝑛 space it is possible to find 𝜇 = tan 𝜙 = 𝜎 assuming there’s no cohesion. Finding the failure envelope of the joint according to its failure 𝝈𝟏 , 𝝈𝟑 values: build the Mohr circles → draw lines from 𝜎3 with an angle of 𝜓 → draw a tangent line between the intersection points of those lines with the corresponding Mohr circles. 1 𝜏𝑥′𝑦′ = 2(𝜎𝑦 − 𝜎𝑥 ) sin 2𝛼 + 𝜏𝑥𝑦 cos 2𝛼 2𝜏𝑥𝑦 Principal stresses: tan 2𝜃 = 𝜎 𝑥 −𝜎𝑦 1 1 2 2 + (𝜎 − 𝜎 ) 𝜎1,2 = 2(𝜎𝑥 + 𝜎𝑦 ) ± √𝜏𝑥𝑦 𝑦 4 𝑥 𝜎1 − 𝜎3 𝜏𝑚𝑎𝑥 = 2 Mohr circle: 2 (𝜎𝜃 − 𝜎𝑚 )2 + 𝜏𝜃2 = 𝜏𝑚𝑎𝑥 𝜎𝑥 + 𝜎𝑦 𝜎𝑚 = 2 Coordinates: (𝜎𝑥𝑥 , 𝜏𝑥𝑦 ), (𝜎𝑦𝑦 , −𝜏𝑥𝑦 ) Stress cube ↺ = 𝛼 → Line in MC ↻ = 2𝛼 Finding the angle between 𝝈𝒏 and the sample axis (𝜷) that will cause the joint failure, according to a certain failure envelope and a Mohr circle: 5. Linear Elasticity 𝜈=− 𝜀𝑥 𝜎𝑥 ,𝜀 = , 𝜀𝑦 𝑥 𝐸𝑥 𝐺= 𝜏𝑥𝑦 𝐸 = , 𝛾𝑥𝑦 2(1 + 𝜈) 𝜆= 𝐸𝜈 (1 + 𝜈)(1 − 2𝜈) 3. Stress Transformations – 3D 𝜎𝑥𝑥 𝜎 = (𝜏𝑦𝑥 𝜏𝑧𝑥 𝜏𝑥𝑦 𝜎𝑦𝑦 𝜏𝑧𝑦 𝑙𝑥′ = cos(𝑥 ′ 𝑥) 𝑙𝑦′ = cos(𝑦 ′ 𝑥) 𝑙𝑧 ′ = cos(𝑧 ′ 𝑥) 𝜏𝑥𝑧 𝜏𝑦𝑧 ) , 𝜎𝑧𝑧 𝑙𝑥 ′ 𝐿 = ( 𝑙𝑦 ′ 𝑙𝑧 ′ 𝑚𝑥′ = cos(𝑥 ′ 𝑦) 𝑚𝑦′ = cos(𝑦 ′ 𝑦) 𝑚𝑧 ′ = cos(𝑧 ′ 𝑦) 𝑚𝑥 ′ 𝑚𝑦 ′ 𝑚𝑧 ′ 𝑛𝑥 ′ 𝑛𝑦 ′ ) 𝑛𝑧 ′ 𝑛𝑥′ = cos(𝑥 ′ 𝑧) 𝑛𝑦′ = cos(𝑦 ′ 𝑧) 𝑛𝑧 ′ = cos(𝑧 ′ 𝑧) 𝜎 ′ = 𝐿𝜎𝐿𝑇 ⟨𝜎𝑥′ 𝜏𝑥′ 𝑦′ 𝜏𝑥′ 𝑧 ′ ⟩ = 𝐿𝑥′ 𝜎𝐿𝑇 Principal stresses (eigenvalues) and their directions (eigenvectors) 𝜎𝑛 = 𝜎 ∗ 𝑛 → (𝜎 − 𝜎 ∗ 𝐼)𝑛 = 0 𝜎𝑥𝑥 − 𝜆 𝜏𝑥𝑦 𝜏𝑥𝑧 𝜎𝑦𝑦 − 𝜆 𝜏𝑦𝑧 | = 0 | 𝜏𝑦𝑥 𝜏𝑧𝑥 𝜏𝑧𝑦 𝜎𝑧𝑧 − 𝜆 𝜆3 − 𝐼1 𝜆2 − 𝐼2 𝜆 − 𝐼3 = 0 𝐼1 = 𝜎𝑥𝑥 + 𝜎𝑦𝑦 + 𝜎𝑧𝑧 2 2 2 𝐼2 = 𝜏𝑥𝑦 + 𝜏𝑥𝑧 + 𝜏𝑦𝑧 − 𝜎𝑥𝑥 𝜎𝑦𝑦 − 𝜎𝑥𝑥 𝜎𝑧𝑧 − 𝜎𝑦𝑦 𝜎𝑧𝑧 2 2 2 𝐼3 = 𝜎𝑥𝑥 𝜎𝑦𝑦 𝜎𝑧𝑧 + 2𝜏𝑥𝑦 𝜏𝑥𝑧 𝜏𝑦𝑧 − 𝜎𝑥𝑥 𝜏𝑦𝑧 − 𝜎𝑦𝑦 𝜏𝑥𝑧 − 𝜎𝑧𝑧 𝜏𝑥𝑦 𝜎1 , 𝜎2 , 𝜎3 = 𝜆1 , 𝜆2 , 𝜆3 from largest to smallest Example of finding a direction for any principal stress (after gaussian elimination): 1 −1 0 1 𝑛𝑥 = 𝑛𝑦 1 1 (0 0 1) → ( 1) → (1) 𝑛𝑥 , 𝐿 = √12 + 12 = √2 → 𝑛𝑧 = 0 √2 0 0 0 0 0 7. Induced Seismicity Coulomb-Mohr criterion – a formula to move from 𝜏 − 𝜎 to 𝜎1 − 𝜎3 : 𝜙 𝜙 𝜎1 = 2𝑆0 tan (45° + ) + 𝜎3 tan2 (45° + ) ⏟ 2 2 𝜎𝑐 For the 𝜎1 − 𝜎3 coordinate system: tan 𝜓 = 1 + sin 𝜙 2𝑐 ⋅ cos 𝜙 ,𝜎 = 1 − sin 𝜙 𝑐 1 − sin 𝜙 Tensile strength of a rock according to Grifitth’s theory 𝑈𝐶𝑆 8 𝑇0 = Proof example: 1 1 𝐶𝑏𝑐 − 𝐶𝑏𝑢 1 𝐾𝑢 − 𝐾𝑏 𝐶 − 𝐶𝑏𝑐 𝐶 𝐶 𝐵= ⏟ ( ) = 𝐶𝑏𝑢 ( 𝑏𝑢 ) = 𝐶𝑏𝑢 ( 𝑏𝑢 𝑏𝑐 ) = 𝐶𝑏𝑝 𝐶𝑏𝑝 𝐾𝑢 𝛼 ? 𝐶𝑏𝑐 𝐶𝑏𝑐 𝐶𝑏𝑝 𝐶𝑝𝑐 𝐶 − (𝐶 − ) 𝑏𝑐 𝑏𝑐 (𝐶𝑏𝑐 − 𝐶𝑏𝑢 )𝐶𝑏𝑐 𝐶𝑝𝑝 + 𝐶𝑓 𝐶𝑝𝑐 𝐶𝑏𝑢 ( )= = =𝐵 𝐶𝑏𝑝 𝐶𝑏𝑢 𝐶𝑏𝑐 𝐶𝑏𝑝 𝐶𝑝𝑝 + 𝐶𝑓 Terzaghi’s law (matrix form) with Biot’s coefficient: 𝑆𝑥𝑥 𝑆𝑥𝑦 𝑆𝑥𝑧 𝑃𝑝 0 𝜎𝑥𝑥 𝜎𝑥𝑦 𝜎𝑥𝑧 (𝜎𝑥𝑦 𝜎𝑦𝑦 𝜎𝑦𝑧 ) = (𝑆𝑥𝑦 𝑆𝑦𝑦 𝑆𝑦𝑧 ) − 𝛼 ( 0 𝑃𝑝 𝜎𝑥𝑧 𝜎𝑦𝑧 𝜎𝑧𝑧 0 0 𝑆𝑥𝑧 𝑆𝑦𝑧 𝑆𝑧𝑧 0 0) 𝑃𝑝 Pore pressure required for the intact rock failure: 𝑃𝑐𝑟𝑖𝑡 = 𝜎3 − 11. The Tectonic Stress Field and Methods of measuring in situ stress (𝜎1,𝑝 − 𝜎3 ) − 𝑞𝑢 𝜙 tan2 (45° + 2 ) − 1 𝜈 Recall that in a uniaxial strain 𝜎ℎ = 1−𝜈 𝜎𝑣 and that an effective stress is Pore pressure required for the reactivation of a fault: 𝑆𝑗 sin 𝜓 ⋅ cos 𝜓 𝑃𝑐𝑟𝑖𝑡 = + 𝜎3 + (𝜎1 − 𝜎3 ) (sin2 𝜓 − ) tan 𝜙𝑗 tan 𝜙𝑗 When 𝑆𝑗 and 𝜙𝑗 denote the joint’s cohesion and friction angle. 𝜎𝑣 = 𝑆𝑣 − 𝛼𝑃𝑝 . Vertical stresses in-situ onshore Vertical stresses in-situ offshore Recall that the rock’s failure criterion is 𝜏 = 𝑆0 + ⏟ tan 𝜙 𝜎𝑛 𝜇 The maximal deviatoric stress the rock can carry is: 2𝑆0 + 2𝜇𝜎3 Δ𝜎 = 𝜎1 − 𝜎3 = 𝜇 (1 − tan 𝛽 ) sin 2𝛽 𝑆𝜃𝜃 𝑆𝑟𝑟 𝜎𝜃𝜃 𝑎 2 0.5(𝑆𝐻,𝑚𝑎𝑥 + 𝑆ℎ,𝑚𝑖𝑛 ) (1 + ( ) ) 𝑟 𝑎 4 −0.5(𝑆𝐻,𝑚𝑎𝑥 − 𝑆ℎ,𝑚𝑖𝑛 ) (1 + 3 ( ) ) cos 2𝜃 𝑟 2 𝑎 0.5(𝑆𝐻,𝑚𝑎𝑥 + 𝑆ℎ,𝑚𝑖𝑛 ) (1 − ( ) ) 𝑟 𝑎 2 𝑎 4 +0.5(𝑆𝐻,𝑚𝑎𝑥 − 𝑆ℎ,𝑚𝑖𝑛 ) (1 − 4 ( ) + 3 ( ) ) cos 2𝜃 𝑟 𝑟 𝑎 2 ) 0.5(𝑆𝐻,𝑚𝑎𝑥 + 𝑆ℎ,𝑚𝑖𝑛 − 2𝑃𝑝 (1 + ( ) ) 𝑟 𝑎 4 𝑎 2 −0.5(𝑆𝐻,𝑚𝑎𝑥 − 𝑆ℎ,𝑚𝑖𝑛 ) (1 + 3 ( ) ) cos 2𝜃 − Δ𝑃 ( ) − 𝜎 Δ𝑇 𝑟 𝑟 𝑎 2 0.5(𝑆𝐻,𝑚𝑎𝑥 + 𝑆ℎ,𝑚𝑖𝑛 − 2𝑃𝑝 ) (1 − ( ) ) 𝑟 𝑎 2 𝑎 4 𝜎𝑟𝑟 +0.5(𝑆𝐻,𝑚𝑎𝑥 − 𝑆ℎ,𝑚𝑖𝑛 ) (1 − 4 ( ) + 3 ( ) ) cos 2𝜃 𝑟 𝑟 2 𝑎 +Δ𝑃 ( ) 𝑟 𝜃 is measures from the azimuth of 𝑆𝐻,𝑚𝑎𝑥 . Δ𝑃 = 𝑃𝑚𝑢𝑑 − 𝑃𝑝 . Δ𝑇 𝜎 is the thermal stress arising from the difference between the mud and formation temperatures. 𝑧 𝑧 𝜎𝑣 = ∫ 𝜌(𝑧)𝑔𝑑𝑧 = 𝜌̅ 𝑔𝑧 𝜎𝑣 = 𝜌𝑤 𝑔𝑧𝑤 + ∫ 𝜌(𝑧)𝑔𝑑𝑧 = 𝜌𝑤 𝑔𝑧𝑤 + 𝜌̅ 𝑔(𝑧 − 𝑧𝑤 ) 0 8. Deformation and Failure in Rocks 12. Stresses around Circular Holes 𝜎𝑣 = 𝑔Σ𝜌𝑖 𝑧𝑖 𝑧𝑤 𝜎𝑣 = 𝜌𝑤 𝑔𝑧𝑤 + 𝑔Σ𝜌𝑖 𝑧𝑖 Hydrostatic pore pressure gradient is approximately 10 𝑟 Variations of 𝑆/𝜎𝜃𝜃 by 𝑎 or 𝜃: 𝑀𝑃𝑎 𝑘𝑚 𝑃𝑝ℎ𝑦𝑑𝑟𝑜 = 𝜌𝑤 𝑔𝑧 Example of 𝑠𝑣 calculation: When we wish to draw Δ𝜎 as a function of 𝛽, 𝜎3 is a constant. Increasing 𝑆0 or 𝜎3 → less chance for a failure. Borehole breakout 𝜽 range determination Analytically: 𝑈𝐶𝑆 = 𝑆𝐻,𝑚𝑎𝑥 + 𝑆ℎ,𝑚𝑖𝑛 − 2(𝑆𝐻,𝑚𝑎𝑥 − 𝑆ℎ,𝑚𝑖𝑛 ) cos 2𝜃 Graphically: draw a horizontal line for the corresponding UCS value. Tensile fracture will occur if the minimum point of the stress curve is below its negative value of tensile strength. 9. Laboratory Testing of Rocks 13. In-situ stress estimation methods Hydrostatic test Volumetric Strain 𝜎1 = 𝜎2 = 𝜎3 = 𝑃 Δ𝑉 Δ = 𝑉 = 𝜀1 + 𝜀2 + 𝜀3 Bulk Modulus 𝐾=Δ Uniaxial test Triaxial test 𝜎1 > 𝜎2 = 𝜎3 = 0 𝜎1 > 𝜎2 = 𝜎3 > 0 The relation between the stresses’ magnitudes results in different types 1. Hydraulic fracturing (𝑷𝒄 , 𝑷𝒓 , 𝑷𝒔 → 𝝈𝒉 , 𝝈𝑯 , 𝑻𝟎 ) of faults: 𝑃 𝑃𝑐 = 3𝜎𝐻 − 𝜎𝐻 + 𝑇0 𝑃𝑟 = 3𝜎ℎ − 𝜎𝐻 𝑃𝑠 = 𝜎ℎ Δ𝜎 Δ𝜀 Δ𝜀𝑟𝑎𝑑𝑖𝑎𝑙,𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝜈=− Δ𝜀𝑎𝑥𝑖𝑎𝑙 2. Flat Jack 1 − 2 cos(2𝜃𝐴 ) 1 + 2 cos(2𝜃𝐴 ) 𝐴=[ ] 1 − 2 cos(2𝜃𝐵 ) 1 + 2 cos(2𝜃𝐵 ) We assume that 𝜃 = 0° horizontally and 𝜃 = 90° vertically (the principal stresses are 𝜎𝑣 and 𝜎ℎ ). 𝜎𝜃,𝐴 𝜎ℎ (𝜎 ) = 𝐴 (𝜎 ) 𝜃,𝐵 𝑣 𝜎𝜃,𝐴 𝜎ℎ (𝜎 ) = 𝐴−1 (𝜎 ) 𝜃,𝐵 𝑣 Rule of thumb: → 𝜎1 is in 60° to the normal of the fault plane → 𝜎3 is in 30° to the normal of the fault plane → 𝜎2 is parallel to the strike 𝐸= Minimal machine stiffness: |𝐸𝑠𝑠 |𝐴 𝑘𝑚 ≥ 𝑘𝑟 = 𝐿 𝑘𝑚 – machine stiffness 𝑘𝑟 – rock stiffness 𝐸𝑠𝑠 – the steepest slope of the rock after reaching ultimate strength. Normal fault → dip = 60‫ۥ‬ Reverse fault → dip = 30° 3. Over-coring – Strain Rosette First, find the strain tensor. cos 2 𝜃𝐴 sin2 𝜃𝐴 sin 2𝜃𝐴 𝐴 = [cos 2 𝜃𝐵 sin2 𝜃𝐵 sin 2𝜃𝐵 ] cos 2 𝜃𝐶 sin2 𝜃𝐶 sin 2𝜃𝐶 𝜀𝑥 𝜀𝑥 𝜀𝐴 𝜀𝐴 [𝜀𝐵 ] = 𝐴 [ 𝜀𝑦 ] → [ 𝜀𝑦 ] = 𝐴−1 [𝜀𝐵 ] 𝜀𝐶 Γxy Γxy 𝜀𝐶 Then, find the principal strains and their orientation. 1 1 𝜕𝑉 𝐶𝑏𝑐 = − 𝑉 𝑖 ( 𝜕𝑃𝑏) 𝑏 𝑐 1 𝑝 𝑐 𝜕𝑉 𝐶𝑏𝑝 = 𝑉 𝑖 (𝜕𝑃𝑏) 𝑃𝑝 𝑝 𝑏 𝜕𝑉 1 𝐶𝑝𝑐 = − 𝑉 𝑖 (𝜕𝑃𝑝) 𝑃𝑐 𝜕𝑉 1 𝐶𝑝𝑝 = 𝑉 𝑖 (𝜕𝑃𝑝 ) 𝑝 𝑃 𝑐 𝑝 𝑃𝑝 𝑑𝑉𝑏 𝜕𝑉𝑏 𝜕𝑉𝑏 𝑑𝜀𝑏 = − 𝑖 = − (( 𝑖 ) + ( 𝑖 ) ) = 𝐶𝑏𝑐 𝑑𝑃𝑐 − 𝐶𝑏𝑝 𝑑𝑃𝑝 𝑉𝑏 𝑉𝑏 𝑃 𝑉𝑏 𝑃 𝑝 𝑐 𝑑𝑉𝑝 𝜕𝑉𝑝 𝜕𝑉𝑝 𝑑𝜀𝑝 = − 𝑖 = − (( 𝑖 ) + ( 𝑖 ) ) = 𝐶𝑝𝑐 𝑑𝑃𝑐 − 𝐶𝑝𝑝 𝑑𝑃𝑝 𝑉𝑝 𝑉𝑝 𝑃 𝑉𝑝 𝑃 𝑝 𝐶𝑏𝑢 = ( 𝑐 𝐶𝑏𝑝 𝐶𝑝𝑐 𝜕𝑉𝑏 𝜕𝜀𝑏 ) = 𝐶𝑏𝑐 − =( ) 𝜕𝑃𝑐 𝑢𝑛𝑑𝑟𝑎𝑖𝑛𝑒𝑑 𝐶𝑝𝑝 + 𝐶𝑓 ⏟𝜕𝑃𝑐 𝑢𝑛𝑑𝑟𝑎𝑖𝑛𝑒𝑑 Calculation of h. stresses based on 𝝈𝒗 , fault type and location: Normal fault → 𝜎ℎ = 𝐾𝑎 𝜎𝑣 (tensional stress regime) Reverse fault → 𝜎ℎ = 𝐾𝑝 𝜎𝑣 (compressive stress regime) UCS is not included in the formulas near to a preexisting fault. Near a preexisting 𝐾𝑝 𝐾𝑎 fault? Yes No 1 𝜙 tan2 (45° + ) 2 𝑆𝑣 − 𝑈𝐶𝑆 𝜙 𝑆𝑣 tan2 (45° + ) 2 𝜙 tan2 (45° + ) 2 𝑈𝐶𝑆 𝜙 + tan2 (45° + ) 𝑆𝑣 2 The failure envelope will be tangent to the 2 Mohr circles based on 𝐾𝑎 𝜎𝑣 , 𝜎𝑣 and 𝐾𝑝 𝜎𝑣 . 𝑠𝑜𝑚𝑒𝑡𝑖𝑚𝑒𝑠 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑡ℎ𝑖𝑠 𝑤𝑎𝑦 𝐶𝑚 = 𝐶𝑏𝑐 − 𝐶𝑏𝑝 = 𝐶𝑝𝑐 − 𝐶𝑝𝑝 𝜕𝑃 𝐵 = ( 𝜕𝑃𝑝 ) 𝑐 = 𝑢𝑛𝑑𝑟𝑎𝑖𝑛𝑒𝑑 𝐾𝑏 𝐶𝑚 𝑚 𝑏𝑐 1 2 + (𝜀 − 𝜀 ) 𝜀1,2 = 2(𝜀𝑥 + 𝜀𝑦 ) ± √𝛤𝑥𝑦 𝑦 4 𝑥 10. Poroelasticity 𝐶𝑝𝑝 +𝐶𝑚 𝐶𝑝𝑝 +𝐶𝑓 = ⏟ 𝐶𝑚 ≪𝐶𝑝𝑝 𝐶𝑏𝑝 𝛼 = 1−𝐾 = 1−𝐶 →𝛼 = 𝐶𝑏𝑐 𝐶𝑝𝑝 𝐶𝑝𝑝 +𝐶𝑓 = 1 𝐶𝑓 ) 𝐶𝑝𝑝 1+( Higher Biot coefficient → higher effect of pore pressure. 1 𝐾𝑏 = 𝐾 = 𝐾𝑑 = 𝐶𝑏𝑐 1 𝐾𝑚 = 𝐶𝑚 1 𝐾𝑢 = 𝐶𝑏𝑢 𝐶𝑓 = 𝑆𝑤 𝐶𝑤 + (1 − 𝑆𝑊 )𝐶𝑎 Erosion effect on horizontal stress magnitude (at depth 𝒁 − 𝚫𝒁): 𝜈 Δ𝑍 𝐾(𝑍) = 𝐾0 + (𝐾0 − )⋅ 1−𝜈 𝑍 Thus, erosion of overlying rock will tend to increase the value of K, the horizontal stress becoming greater than the vertical stress at depths less than a certain value. 2Γ𝑥𝑦 tan 2𝜃 = 𝜀𝑥 − 𝜀𝑦 2

Formulas - Rock Mechanics

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