2. EXPLORATION GEOPHYSICS

3. THE EXPLORATION TASK PLAN EXPLORATION APPROACH FOR A MATURE TREND GATHER DATA FOR A MATURE TREND DEVELOP PLAY PROSPECT FRAMEWORK INITIAL DATA GATHANAL AND PROJECT PLANING FOR A FRONTIER TREND NEW DATA GATHERING FOR A FRONTIER TREND MAKE PLAY/PROSPECT ASSESSMENT COMMUNICATE ASSESSMENT TO MANAGEMENT PREPARE PRELOCATION REPORT DRILLING

4. EXPLORATION GEOPHYSICS

5. Elasticity Source Petroleum related rock mechanics Petroleum related rock mechanics Elsevier, 1992

6. ElasticityElasticity  Definition: The ability to resist and recover from deformations produced by forces.  It is the foundation for all aspects of Rock Mechanics  The simplest type of response is one where there is a linear relation between the external forces and the corresponding deformations.

7. StressStress defines a force field on a material Stress = Force / Area (pounds/sq. in. or psi)  F / A F Area: A

8. StressStress  In Rock Mechanics the sign convention states that the compressive stresses are positive.  In Rock Mechanics the sign convention states that the compressive stresses are positive.  Consider the cross section area at b, the force acting through this cross section area is F (neglecting weight of the column) and cross sectional area is A’. A’ is smaller than A, therefore stress  ’ = F/A’ acting at b is greater than  acting at a

9. StressStress W F F F a b c A A’ A’’ Area Load

10. StressStress  stress depends on the position within the stressed sample.  Consider the force acting through cross section area A’’. It is not normal to the cross section. We can decompose the force into one component F n normal to the cross section, and one component F p that is parallel to the section.

11. StressStress FpFp FnFn F Decomposition of forces

12. StressStress  The quantity  = F n /A’’ is called the normal stress, while the quantity   = F p / A’’ is called the shear stress.  Therefore, there are two types of stresses which may act through a surface, and the magnitude of each depend on the orientation of the surface.

13. General 3D State of Stress in a Reservoir  x ,  y ,  z  Normal stresses  xy ,  yz ,  zx  Shear stresses xx yy zz  yx  yz  zy  zx  xy

14. StressStress   =  x  xy  xz  yx  y  yz  yx  y  yz  zx  zy  z  zx  zy  z  Stress tensor

15. Principal Stresses Normal stresses on planes where shear stresses are zero       vv HH hh

16. Principal Stresses In case of a reservoir,   =  v Vertical stress,   =  h Minimum horizontal stresses   =  H Maximum horizontal stresses vv HH hh

17. Types of Stresses ä Tectonic Stresses: Due to relative displacement of lithospheric plates Based on the theory of earth’s tectonic plates ä Spreading ridge: plates move away from each other ä Subduction zone: plates move toward each other and one plate subducts under the other ä Transform fault: Plates slide past each other

18. Types of Stresses ä Gravitational Stresses: Due to the weight of the superincumbent rock mass ä Thermal Stresses: Due to temperature variation Induced, residual, regional, local, far-field, near-field, paleo...

19. Impact of In-situ Stress Important input during planning stage ä ä Fractures with larger apertures are oriented along the maximum horizontal stress

20. Natural fractures

21. Strain (x, y, z) (x ’, y ’, z ’ ) Initial Position Shifted Position

22. Strain   x’ = x – u   y’ = y – v   z’ = z – w   If the displacements u, v, and w are constants, i.e, they are the same for every particle within the sample, then the displacement is simply a translation of a rigid body.

23. Strain   Another simple form of displacement is the rotation of a rigid body.   If the relative positions within the sample are changed, so that the new positions cannot be obtained by a rigid translation and/or rotation of the sample, the sample is said to be strained. (figure 8)

24. Strain L O P L’ O’ P’ Initial Position Shifted Position

25. Strain   Elongation corresponding to point O and the direction OP is defined as    = (L – L’)/L   sign convention is that the elongation is positive for a contraction.   The other type of strain that may occur can be expressed by the change  of the angle between two initially orthogonal directions. (Figure 9)

26. Strain P O Q Initial Position P Q O  Shifted Position

27. Strain    = (1/2)tan  is called the shear strain corresponding to point O and the direction OP. We deal with infinitesimal strains.   The elongation (strain) in the x-direction at x can be written as    x =  u/  x

28. Strain   The shear strain corresponding to x- direction can be written as    xy = (  u/  y +  v/  x)/2   Strain tensor   Principal strains

29. Elastic Moduli D L L’ D’ F Y X Schematic showing deformation under load

30. Elastic Moduli   When force F is applied on its end surfaces, the length of the sample is reduced to L’.   The applied stress is then  x = F/A,   The corresponding elongation is  = (L – L’)/L   The linear relation between  x and  x, can be written as    x = E  x

31. Elastic Moduli   This equation is known as Hooke’s law   The coefficient E is called Young’s modulus.   Young’s modulus belongs to a group of coefficients called elastic moduli.   It is a measure of the stiffness of the sample, i.e., the sample’s resistance against being compressed by a uniaxial stress.

32. Elastic Moduli   Another consequence of the applied stress  x (Figure 10) is an increase in the width D of the sample. The lateral elongation is  y =  z = (D – D’)/D. In general D’ > D, thus  y and  z become negative.   The ratio defined as = -  y /  x is another elastic parameter known as Poisson’s ratio. It is a measure of lateral expansion relative to longitudinal contraction.

33. Elastic Moduli   Bulk modulus K is defined as the ratio of hydrostatic stress  p relative to the volumetric strain  v. For a hydrostatic stress state we have  p =  1 =  2 =  3 while  xy =  xz =  yz = 0. Therefore   K =  p /  v = + 2G/3 1   K is a measure of sample’s resistance against hydrostatic compression. The inverse of K, i.e., 1/K is known as compressibility

34. Elastic Moduli   Isotropic materials are materials whose response is independent of the orientation of the applied stress. For isotropic materials the general relations between stresses and strains may be written as:    x = ( + 2G)  x +  y +  z    y =  x + ( + 2G)  y +  z    z =  x +  y + ( + 2G)  z    xy = 2G  xy  xz = 2G  xz  yz = 2G  yz

35. Elastic Moduli   Expressing strains as function of stresses   E  x =  x - (  y +  z )   E  y =  y - (  x +  z )   E  z =  z - (  x +  y )   G  xy = (1/2)  xy   G  xz = (1/2)  xz   G  yz = (1/2)  yz

36. Elastic Moduli   In the definition of Young’s modulus and Poisson’s ratio, the stress is uniaxial, i.e.,  z =  y =  xy =  xz =  yz = 0. Therefore   E =  x /  x = G (3 + 2G)/ ( + G) 2   = -  y /  x = /(2( + G)) 3   Therefore from equations (1, 2, and 3 ), knowing any two of the moduli E,,, G and K, we can find other remaining moduli

37. Elastic Moduli   For rocks, is typically 0.15 – 0.25. For weak, porous rocks may approach zero or even become negative. For fluids, the rigidity G vanishes, which according to equation (3) implies  ½. Also for unconsolidated sand, is close to ½.