Shear failure and wellbore breakouts

Wellbore breakouts occur when the stress anisotropy $\sigma_1/\sigma_3$ surpasses the shear strength limit of the wellbore rock. Maximum anisotropy is found at $\theta = \pi/2$ and $3\pi/2$, for which

$$\left\lbrace \begin{array}{rcl} \sigma_1 = \sigma_{\theta \th... ...\\ \sigma_3 = \sigma_{rr} & = & (P_W - P_p) \end{array}\right.$$ (6.7)

Figure 6.8: Initiation of wellbore breakouts

Hence, replacing $\sigma _1$ and $\sigma _3$ into a shear failure equation (Eq. ) permits finding the mud pressure $P_{Wshear}$ that would produce a tiny shear failure (or breakout) at $\theta = \pi/2$ and $3\pi/2$:

$$\left[ -(P_W - P_p) +3\sigma_{Hmax} -\sigma_{hmin} \right] = UCS + q \left[ P_W - P_p \right]$$ (6.8)

Hence,

$$P_{Wshear} = P_p + \frac{3\sigma_{Hmax} -\sigma_{hmin} - UCS}{1+q}$$ (6.9)

Mud pressure $P_W < P_{Wshear}$ would extend the breakout further in the neighbohood of $\theta = \pi/2$ and $3\pi/2$ (See Fig. 6.9). Thus, $P_{Wshear}$ is the lowest mud pressure before initiation of breakouts.

Figure 6.9: Example of wellbore breakout [Zoback 2013 - Fig. 6.15]. Observe the inclination of shear fractures as the propagate into the rock. The material that falls into the wellbore is taken out as drilling cuttings.

PROBLEM 6.1: Calculate the minimum mud weight (ppg) in a vertical wellbore for avoiding shear failure (breakouts) in a site onshore at 7000 ft of depth where $S_{hmin} =$ 4300 psi and $S_{Hmax} =$ 6300 psi and with hydrostatic pore pressure. The rock mechanical properties are $UCS =$ 3500 psi, $\mu_i=$ 0.6, and $T_s$ = 800 psi.

SOLUTION
Hydrostatic pore pressure results in:

$$P_p = 7000$$    ft$$\times 0.44$$    psi/ft$$= 3080$$    psi

The effective horizontal stresses are:

$$\sigma_{Hmax} = 6300$$    psi$$- 3080$$    psi$$= 3220$$    psi$$$$

and

$$\sigma_{hmin} = 4300$$    psi$$- 3080$$    psi$$= 1220$$    psi$$$$

The friction angle is $\arctan(0.6)=30.96^{\circ}$, and therefore, the friction stress anisotropy ratio is

$$q = \frac{1+ \sin 30.96^{\circ} }{1- \sin 30.96^{\circ} } = 3.12$$

Thus, the minimum mud pressure for avoiding shear failure (breakouts) is

$$P_{Wshear} = 3080$$    psi$$+ \frac{3 \times 3220 \text{ psi} - 1220 \text{ psi} - 3500 \text{ psi} }{1+3.12} = 4279 \text{ psi}$$

This pressure can be achieved with an equivalent circulation density of

$$\frac{4279 \text{ psi}} {7000 \text{ ft}} \times \frac{8.3 \text{ ppg}} {0.44 \text{ psi/ft}} = 11.57 \text{ ppg} \: \: \blacksquare$$