The cylindrical symmetry of a wellbore prompts the utilization of a cylindrical coordinate system rather than a rectangular cartesian coordinate system. The volume element of stresses in cylindrical coordinates is shown in Fig. 6.4. The distance is measured from the center axis of the wellbore. The angle is measured with respect to a predefined plane.
The normal stresses are radial stress , tangential or hoop stress , and axial stress . The shear stresses are , , and .
The Kirsch solution allows us to calculate normal and shear stresses around a circular cavity in a homogeneous linear elastic solid . The complete Kirsch solution assumes independent action of multiple factors, namely far-field isotropic stress, deviatoric stress, wellbore pressure and pore pressure.
Consider a vertical wellbore subjected to horizontal stresses and , both principal stresses, vertical stress , pore pressure , and wellbore pressure . The corresponding effective in-situ stresses are , , and . The Kirsch solution for a wellbore with radius within a linear elastic and isotropic solid is:
(6.2) |
where is the radial effective stress, is the tangential (hoop) effective stress, is the shear stress in a plane perpendicular to in tangential direction , and is the vertical effective stress in direction . The angle is the angle between the direction of and the point at which stress is considered. The distance is measured from the center of the wellbore. For example, at the wellbore wall .
An example of the solution of Kirsch equations for MPa, MPa, and MPa is available in Figure 6.8. The plots show radial and tangential effective stresses, as well as the calculated principal stresses and .
Let us obtain and at the wellbore wall . The radial stress for all is
(6.3) |
The hoop stress depends on ,
(6.4) |
and it is the minimum at and (azimuth of ) and the maximum at and (azimuth of ):
(6.5) |
These locations will be prone to develop tensile fractures ( and ) and shear fractures ( and ). The shear stress around the wellbore wall is . This makes sense because fluids (drilling mud) cannot apply steady shear stresses on the surface of a solid. Finally, the effective vertical stress is
(6.6) |