6.3 Shear failure and wellbore breakouts

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

\sigma_1 = \sigma_{\theta \th...
\sigma_3 = \sigma_{rr} & = & (P_W - P_p)
\end{array}\right.\end{displaymath} (6.7)

Figure 6.9: Wellbore pressure for initiation of wellbore breakouts. Shear fractures will develop if $P_W < P_{Wshear}$.

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 $3\pi/2$:

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


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

Mud pressure $P_W < P_{Wshear}$ would extend rock failure and breakouts further in the neighborhood of $\theta = \pi/2$ and $3\pi/2$ (See Fig. 6.10). Thus, $P_{Wshear}$ is the lowest mud pressure before initiation of breakouts.

Figure 6.10: 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 7,000 ft of depth where $S_{hmin} =$ 4,300 psi and $S_{Hmax} =$ 6,300 psi and with hydrostatic pore pressure. The rock mechanical properties are $UCS =$ 3,500 psi, $\mu_i=$ 0.6, and $T_s$ = 800 psi.

Hydrostatic pore pressure results in:

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

The effective horizontal stresses are:

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


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

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

$\displaystyle 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

$\displaystyle P_{Wshear} = 3080$    psi$\displaystyle + \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

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

6.3.1 Breakout angle determination

For a given set of problem variables (far field stress, pore pressure, and mud pressure), we can calculate the required strength of the rock to have a stable wellbore. Let us consider the example of Fig. 6.11 that shows the required $UCS$ to resist shear failure assuming the friction angle is $\varphi = 30^{\circ}$. For example, if the rock had a $UCS \sim 25$ MPa, one may expect a $\sim 90^{\circ}$ wide breakout in Fig. 6.11.

Figure 6.11: Example of required $UCS$ with Mohr-Coulomb analysis to verify likelihood of wellbore breakouts.

Alternatively, you could solve the previous problem analytically. The procedure consists in setting shear failure at the point in the wellbore at an angle $(w_{BO}/2)$ from $\pi/2$ or $3\pi/2$. Hence, at a point on the wellbore wall at $\theta_B = \pi/2 - w_{BO}/2$:

\sigma_{\theta \theta} & = &
...\theta_B) \\
\sigma_{rr} & = & +(P_W - P_p)
\end{array}\right.\end{displaymath} (6.10)

Figure 6.12: Determination of breakout angle with Mohr-Coulomb failure criterion.

Say hoop stress reaches the maximum principal stress anisotropy allowed by the Mohr-Coulomb shear failure criterion ( $\sigma_1 = UCS + q \: \sigma_3$) where the breakout begins (rock about to fail - Fig. 6.12), then

$\displaystyle \left[ -(P_W - P_p) + (\sigma_{Hmax} + \sigma_{hmin})
- 2(\sigma_{Hmax} - \sigma_{hmin}) \cos (2 \theta_B) \right]
= UCS + q (P_W - P_p)$ (6.11)

which after some algebraic manipulations results in:

$\displaystyle 2 \theta_B = \arccos \left[ \frac{ \sigma_{Hmax} + \sigma_{hmin} - UCS - (1+q)(P_W - P_p)}{2(\sigma_{Hmax} - \sigma_{hmin})} \right]$ (6.12)

The breakout angle is

$\displaystyle w_{BO} = \pi - 2 \theta_B$ (6.13)

The procedure assumes the rock in the breakout (likely already gone) is still resisting hoop stresses and therefore it is not accurate for large breakouts ( $w_{BO} \gtrsim 60^{\circ}$).

You could also calculate the wellbore pressure for a predetermined breakout angle by rearranging Eq. 6.12 to

$\displaystyle P_{WBO} = P_p + \frac{ (\sigma_{Hmax} + \sigma_{hmin})
- 2(\sigma_{Hmax} - \sigma_{hmin}) \cos (\pi - w_{BO}) - UCS}{1+q}$ (6.14)

PROBLEM 6.2: Calculate the breakout angle in a vertical wellbore for a mud weight of 10 ppg in a site onshore at 7,000 ft of depth where $S_{hmin} =$ 4,300 psi and $S_{Hmax} =$ 6,300 psi and with hydrostatic pore pressure. The rock mechanical properties are $UCS =$ 3,500 psi, $\mu_i=$ 0.6, and $T_s$ = 800 psi.

The problem variables are the same of Problem 6.1. For a 10 ppg mud, the resulting mud pressure is

$\displaystyle 10$    ppg$\displaystyle \times
\frac{0.44 \text{ psi/ft}} {8.3 \text{ ppg}} \times
7000 \text{ ft} = 3710 \text{ psi}

Hence, the expected wellbore breakout angle is

$\displaystyle w_{BO} = 180^{\circ} - \arccos \left[ \frac{ 3220 \text{ psi} + 1...
...3220 \text{ psi} - 1220 \text{ psi})} \right] = 66 ^{\circ} \: \: \blacksquare

6.3.2 Breakout measurement

Breakouts and tensile induced fractures (Section 6.4) can be identified and measured with borehole imaging tools (Fig. 6.13). Breakouts appear as wide bands of longer travel time or higher electrical resistivity in borehole images. Tensile fractures appear as narrow stripes of longer travel time or higher electrical resistivity. Borehole images also permit identifying the direction of the stresses that caused such breakouts or tensile induced fractures. For example, the azimuth of breakouts coincides with the direction of $S_{hmin}$ in vertical wells.

Figure 6.13: Examples of wellbore breakouts and tensile fractures in borehole images. Learn more about wellbore imaging tools here:

Breakouts can also be detected from caliper measurements. Caliper tools permit measuring directly the size and shape of the borehole ( Thus, the caliper log is extremely useful to measure breakouts and extended breakouts (washouts). For the same mud density, the caliper log reflects changes of rock properties along the well and correlate with other well logging measurements.

6.3.3 Maximum horizontal stress determination from breakout angle

Breakouts are a consequence of stress anisotropy in the plane perpendicular to the wellbore. Hence, knowing the size and orientation of breakouts permits measuring and calculating the direction and magnitude of stresses that caused such breakouts. This technique is very useful for measuring orientation of horizontal stresses. In addition, if we know the rock properties $UCS$ and $q$, then it is possible to calculate the maximum principal stress in the plane perpendicular to the wellbore. For example, for a vertical wellbore the total maximum horizontal stress would be

$\displaystyle S_{Hmax} = P_p + \frac{UCS + (1+q)(P_W - P_p)
- \sigma_{hmin} \left[ 1 + 2 \cos (\pi - w_{BO}) \right]}
{1 - 2 \cos (\pi - w_{BO})}$ (6.15)