**A. Validate the computational integrity of the plane strain displacement solutions of Section 2.2 using the plane stress solutions of Section 2.3**

The plane strain solution, Eqs. (5) and (6), can be obtained directly from the plane stress solution, Eqs. (9) and (10), by replacing E with *<sup>E</sup><sup>=</sup>* <sup>1</sup> � *<sup>v</sup>*<sup>2</sup> ð Þ, and v with *<sup>v</sup>=*ð Þ <sup>1</sup> � *<sup>v</sup>* as follows:

Starting from the stress solution given by Eq. (9)

$$u\_r = \sigma\_{\text{xx}-\infty} \left( \frac{\mathbf{1} + \boldsymbol{\nu}}{2Er} \right) \left\{ \frac{\mathbf{1} - \boldsymbol{\nu}}{\mathbf{1} + \boldsymbol{\nu}} r^2 + a^2 + \left( \frac{4a^2}{\mathbf{1} + \boldsymbol{\nu}} + r^2 - \frac{a^4}{r^2} \right) \cos 2\theta \right\} \tag{A1}$$

and applying the above replacements we get

$$\begin{split} u\_r &= \sigma\_{\text{xx}-\text{so}} \left( \frac{1 + \left(\frac{v}{1-v}\right)}{2\left(\frac{E}{1-v^2}\right)r} \right) \left\{ \frac{1 - \left(\frac{v}{1-v}\right)}{1 + \left(\frac{v}{1-v}\right)} r^2 + a^2 + \left(\frac{4a^2}{1 + \left(\frac{v}{1-v}\right)} + r^2 - \frac{a^4}{r^2}\right) \cos 2\theta \right\} \\ &= \sigma\_{\text{xx}-\text{so}} \left( \frac{1+v}{2Er} \right) \left\{ (1-2v)r^2 + a^2 + \left(4a^2(1-v) + r^2 - \frac{a^4}{r^2}\right) \cos 2\theta \right\} \\ &= \sigma\_{\text{xx}-\text{so}} \left( \frac{1+v}{2Er} \right) \left\{ r^2(1-2v + \cos 2\theta) + a^2 \left[ 1 + \left(4(1-v) - \frac{a^2}{r^2}\right) \cos 2\theta \right] \right\} \end{split} \tag{A2}$$

Eq. (A2) represents the plan strain solution as given by Eq. (5). Similarly, one can obtain Eq. (6) from (10) using the same modification.
