**3.4 Delta between plane strain and plane stress solutions with uniaxial far-field stress**

The residual displacements when subtracting Eqs. (9) from (5), and (10) from (6), respectively, are:

$$u\_r = \sigma\_{\rm xx-so} \left(\frac{\mathbf{1} + v}{2Er}\right) \left\{ \left[ r^2 (\mathbf{1} - 2v + \cos 2\theta) + a^2 \left[ \mathbf{1} + \left( 4(\mathbf{1} - v) - \frac{a^2}{r^2} \right) \cos 2\theta \right] \right] \right. $$

$$- \left[ \frac{\mathbf{1} - v}{\mathbf{1} + v} r^2 + a^2 + \left( \frac{4a^2}{\mathbf{1} + v} + r^2 - \frac{a^4}{r^2} \right) \cos 2\theta \right] \right\},$$

$$= -\sigma\_{\rm xx-so} \left( \frac{v^2}{Er} \right) \left\{ r^2 + 2a^2 \cos 2\theta \right\}. \tag{11}$$

$$u\_{\theta} = -\sigma\_{\text{xx} \to \infty} \left( \frac{\mathbf{1} + v}{2Er} \right) \left\{ \left[ r^2 + a^2 \left( 2(\mathbf{1} - 2v) + \frac{a^2}{r^2} \right) \right] \sin 2\theta - \left( \frac{\mathbf{1} - v}{\mathbf{1} + v} 2a^2 + r^2 + \frac{a^4}{r^2} \right) \sin 2\theta \right\}.$$

$$= \sigma\_{\text{xx} \to \infty} \left( \frac{v^2}{Er} \right) \left( 2a^2 \sin 2\theta \right). \tag{12}$$

#### **3.5 Plane strain and plane stress solutions for the hole with internal pressure**

From [23] general solution for a hollow cylinder (with infinite axial length) pressured inside and outside with different pressures, can be obtained a simple displacement solution for a hole in an infinite plate by letting the outer radius of the cylinder go to infinity such that only the term remains for the radial displacement due to the pressure inside the cylinder:

$$u\_r = -P\frac{a^2}{r} \frac{\mathbf{1} + v}{E} = -P\frac{a^2}{r} \frac{\mathbf{1}}{G} \tag{13}$$

$$
u\_{\theta} = \mathbf{0} \tag{14}$$

Above displacement field is due to a hole internally pressurized under plane strain boundary conditions and assumes *P* is given as a negative value when causing compression (as in mechanical engineering sign conventions); if we prefer to use *P* as a positive input, the minus sign in Eq. (13) must be dropped. Following ref. [3], we will consider positive *P* inputs (so minus sign will be dropped from Eq. (13) in the rest of this paper).

Plane strain solutions formulated with *G* can be converted to plane stress solutions by replacing *ν* with *<sup>ν</sup>* <sup>1</sup>þ*<sup>ν</sup>* (e.g., [17], page 115). Eq. (13) when formulated with *Asymptotic Solutions for Multi-Hole Problems: Plane Strain versus Plane Stress Boundary… DOI: http://dx.doi.org/10.5772/intechopen.105048*

*G* is identical to the plane strain solution, as applied in ref. [3], which means the plane stress solution is independent of *ν*. Therefore, the delta or residual displacement, in this case, will be zero. The displacement solution for an internally pressured hole in an infinite plate appears insensitive to the thickness of the plate, which can be understood via physical reasoning as follows. Adopting the definitions in [14] treatise on 3D stress systems in isotropic plates, for *<sup>a</sup> <sup>h</sup>* ! ∞ we have a thin-plate problem (plane stress) and for *<sup>a</sup> <sup>h</sup>* ! 0 we have a thick-plate problem (plane strain). In all 3D solutions for plane stress and plane strain cases, the solutions are identical in planes midway the plates at *h* ¼ 0. However, when the plates possess a finite thickness, differences in solutions for plates with holes subjected to a far-field stress under plane strain and plane stress arise when studying solutions, where *h* ! 0. The internally pressured hole solution is insensitive to plate thickness, because for both plane strain and plane stress cases, the pressure on the hole is assumed uniform along *h*, so essentially does not allow stresses to occur normal to the plate by strictly maintaining the pressure equal to *P* even at the rim of the hole near the surface (see Section 4).

#### **3.6 Conversion to Cartesian coordinates**

The conversion of the displacement solutions from polar to Cartesian coordinates is practical for single-hole and multi-hole analysis (which requires superposition) of practical borehole problems because the far-field (tectonic) stresses are assumed more or less constant in the three individual Cartesian directions. It is emphasized here that the solutions in Cartesian and polar coordinates only differ in coordinate transformation to facilitate the visualization of either polar or Cartesian vector displacements, each with their corresponding solutions for the stress and strain tensor components. However, the principal stresses remain invariant to the coordinate system used. The conversion from polar to Cartesian coordinates follows the same steps as in Eqs. (12–17) of ref. [3].

#### *3.6.1 Cartesian coordinates of plane strain solutions (uniaxial far-field stress)*

The displacement Eqs. (5) and (6) in polar vector coordinates *ur* ð Þ , *u<sup>θ</sup>* are converted to Cartesian displacements vector coordinates *ux*, *uy* � � using the appropriate coordinate transformation equations (see Appendix B for details):

$$u\_{\mathbf{x}} = u\_r \cos \theta - u\_\theta \sin \theta, \quad u\_\mathbf{y} = u\_r \sin \theta + u\_\theta \cos \theta$$

We get:

$$u\_x = \sigma\_{\text{xx}-\text{iso}} \left(\frac{1+\nu}{2E}\right) \left\{ \left[ (\mathbf{x}^2 + \mathbf{y}^2)(1-2\nu) + \mathbf{x}^2 - \mathbf{y}^2 + a^2 + 4a^2(1-\nu) \left(\frac{\mathbf{x}^2 - \mathbf{y}^2}{\mathbf{x}^2 + \mathbf{y}^2}\right) \right. \\ \left. + 2a^2 \mathbf{x} \frac{\mathbf{x} \cdot \mathbf{x}}{\mathbf{x}^2 + \mathbf{y}^2} \right] \\ \left. + \left[ 2\mathbf{x}\mathbf{y} + (\mathbf{1} - 2\nu) \left(\frac{4a^2 \mathbf{x} \mathbf{y}}{\mathbf{x}^2 + \mathbf{y}^2}\right) + \frac{2a^4 \mathbf{x} \mathbf{y}}{(\mathbf{x}^2 + \mathbf{y}^2)^2} \right] \left(\frac{\mathbf{y}}{\mathbf{x}^2 + \mathbf{y}^2}\right) \right\} \tag{15}$$

$$u\_{\gamma} = \sigma\_{\text{xx}-\text{so}} \left(\frac{1+\nu}{2E}\right) \left\{ \left[ \left(\mathbf{x}^2 + \mathbf{y}^2\right) (\mathbf{1} - 2\nu) + \mathbf{x}^2 - \mathbf{y}^2 + a^2 + 4a^2 (\mathbf{1} - \nu) \left(\frac{\mathbf{x}^2 - \mathbf{y}^2}{\mathbf{x}^2 + \mathbf{y}^2}\right) \right. \\ \left. \left. \left(\frac{\mathbf{x}^2 - \mathbf{y}^2}{\mathbf{x}^2 + \mathbf{y}^2}\right) \right] - \left[ 2\mathbf{x}\mathbf{y} + (\mathbf{1} - 2\nu) \left(\frac{4a^2 \mathbf{x}\mathbf{y}}{\mathbf{x}^2 + \mathbf{y}^2}\right) + \frac{2\mathbf{a}^4 \mathbf{x}\mathbf{y}}{(\mathbf{x}^2 + \mathbf{y}^2)^2} \right] \left(\frac{\mathbf{x}}{\mathbf{x}^2 + \mathbf{y}^2}\right) \right\} \right\} \tag{16}$$

$$u\_x = \sigma\_{\text{xx}-\text{so}}\left(\frac{1+\nu}{2E}\right)\left\{\left[\frac{1-\nu}{1+\nu}\left(\mathbf{x}^2+\mathbf{y}^2\right)+a^2+\left(\frac{4a^2}{1+\nu}+\mathbf{x}^2+\mathbf{y}^2-\frac{a^4}{\mathbf{x}^2+\mathbf{y}^2}\right)\left(\frac{\mathbf{x}^2-\mathbf{y}^2}{\mathbf{x}^2+\mathbf{y}^2}\right)\right]$$

$$\times \left(\frac{\mathbf{x}}{\mathbf{x}^2+\mathbf{y}^2}\right)+\left[\left(\frac{1-\nu}{1+\nu}2a^2+\mathbf{x}^2+\mathbf{y}^2+\frac{a^4}{\mathbf{x}^2+\mathbf{y}^2}\right)\left(\frac{2\mathbf{x}\mathbf{y}}{\mathbf{x}^2+\mathbf{y}^2}\right)\right]\left(\frac{\mathbf{y}}{\mathbf{x}^2+\mathbf{y}^2}\right)\right] \tag{17}$$

$$u\_y = \sigma\_{\text{xx}-\text{so}}\left(\frac{1+\nu}{2E}\right)\left\{\left[\frac{1-\nu}{1+\nu}(\mathbf{x}^2+\mathbf{y}^2)+a^2+\left(\frac{4a^2}{1+\nu}+\mathbf{x}^2+\mathbf{y}^2-\frac{a^4}{\mathbf{x}^2+\mathbf{y}^2}\right)\left(\frac{\mathbf{x}^2-\mathbf{y}^2}{\mathbf{x}^2+\mathbf{y}^2}\right)\right]\right\}$$

$$\times \left(\frac{\mathbf{y}}{\mathbf{x}^2+\mathbf{y}^2}\right)-\left[\left(\frac{1-\nu}{1+\nu}2a^2+\mathbf{x}^2+\mathbf{y}^2+\frac{a^4}{\mathbf{x}^2+\mathbf{y}^2}\right)\left(\frac{2\mathbf{y}}{\mathbf{x}^2+\mathbf{y}^2}\right)\right]\left(\frac{\mathbf{x}}{\mathbf{x}^2+\mathbf{y}^2}\right)\right]\tag{18}$$

$$
\mu\_{\mathbf{x}} = \frac{-(n-1)}{n} \frac{\sigma\_{\mathbf{x}\mathbf{x}-\mathbf{e}\mathbf{e}}(\mathbf{1}-\mathbf{v}^{2})}{E} \mathbf{x} \tag{19}
$$

$$
\mu\_{\mathcal{V}} = \frac{(n-1)}{n} \frac{\sigma\_{\text{xx}-\text{ss}} v (1 - v^{\omega})}{E(1 - v)} \mathcal{Y} \tag{20}
$$

$$
\mu\_{\rm x} = \frac{-(n-1)}{n} \frac{\sigma\_{\rm xx-so}}{E} \text{x} \tag{21}
$$

$$
\mu\_{\mathcal{Y}} = \frac{(n-1)}{n} \frac{\sigma\_{\text{xx}-\text{os}} v}{E} \mathcal{Y} \tag{22}
$$

*Asymptotic Solutions for Multi-Hole Problems: Plane Strain versus Plane Stress Boundary… DOI: http://dx.doi.org/10.5772/intechopen.105048*

where *n* denotes the number of boreholes. The total displacement vectors (*ux*,total and *uy*,total) due to all holes combined can be computed by the following summations:

$$u\_{\mathbf{x}, \text{total}} = \sum\_{i=1}^{n} u\_{\mathbf{x},i} \tag{23}$$

$$u\_{\rm y,total} = \sum\_{i=1}^{n} u\_{\rm y,i} \tag{24}$$
