# Chapter 14 — Introduction to Geomechanics ## Learning Objectives By the end of this chapter, you should be able to: 1. Define normal stress, shear stress, and effective stress, and explain how they differ. 2. Identify the three principal in-situ stresses ($S_v$, $S_{Hmax}$, $S_{hmin}$) and describe their origins. 3. Classify the tectonic stress regime (normal, strike-slip, reverse) using Anderson's fault classification. 4. Calculate effective stresses from total stresses and pore pressure. 5. Construct a Mohr circle given two principal effective stresses. 6. Describe the Mohr-Coulomb failure criterion and identify cohesion and friction angle on a failure envelope. 7. Determine rock strength parameters (C, $\theta$) from one or two laboratory compression tests. 8. Predict whether a given stress state will cause rock failure. --- ## 14.1 Why Geomechanics Matters When a well is drilled, a cylindrical hole is created in rock that was previously under stress on all sides. Removing that rock doesn't remove the stress: the surrounding formation must now carry load around the void. If the stresses acting on the wellbore wall exceed the strength of the rock, the borehole will fail. It may collapse inward (breakout) or fracture outward. Either outcome creates the potential for expensive drilling problems: stuck pipe, lost circulation, wellbore instability, or cementing failures. Understanding geomechanics allows the drilling engineer to select a mud weight that keeps the wellbore stable, a weight high enough to prevent collapse and low enough to prevent fracturing. This chapter covers the foundations: how stress is defined and measured in rock, how rock resists stress, and how to determine whether a given stress state will cause failure. Chapter 15 applies this framework to the stresses at the wellbore wall, and Chapter 16 turns the analysis into mud weight selection and wellbore stability practice. ```{figure} ../figures/ch14/hole_instability.png :name: fig-hole-instability :width: 80% :align: center Borehole instability: holes can fracture (left, drilling fluid pressure too high) or collapse (right, insufficient support). The color maps show stress concentration around the deformed boreholes. ``` --- ## 14.2 Stress Definitions **Stress** is force per unit area acting on a surface. At any point within a volume of solid material (i.e., rock), stress can be resolved into two components relative to that surface: - **Normal stress ($\sigma_n$)**: Acts perpendicular to the surface. Positive values indicate compression (rock being pushed together); negative values indicate tension (rock being pulled apart). In most subsurface applications, stresses are compressive and therefore positive. - **Shear stress ($\tau$)**: Acts parallel to the surface. Shear stress causes adjacent layers to slide relative to one another. **Pressure** is a scalar quantity (equal in all directions); stress is a second-order tensor that depends on the orientation of the surface. **Note:** in petroleum engineering, the sign convention is that compression is **positive** while tension is **negative**. ### Strain **Strain** is the deformation, or change in length or volume, resulting from stress: - **Axial strain**: Change in length divided by original length ($\epsilon = \Delta L / L$) - **Compaction**: Reduction in pore volume under applied stress, relevant to reservoir geomechanics - **Subsidence**: Surface settlement resulting from compaction at depth (e.g., the Wilmington oil field in California subsided several meters due to reservoir depletion) --- ## 14.3 The Formation Stress Tensor The complete stress state at any point in a rock mass is described by a 3×3 stress tensor: $$ \mathbf{S} = \begin{pmatrix} S_{11} & \tau_{12} & \tau_{13} \\ \tau_{21} & S_{22} & \tau_{23} \\ \tau_{31} & \tau_{32} & S_{33} \end{pmatrix} $$ The diagonal entries ($S_{1}1$, $S_{22}$, $S_{33}$) are normal stresses acting on faces oriented perpendicular to the x, y, and z axes. The off-diagonal entries are shear stresses on those faces. Because the tensor is symmetric ($\tau_{ij} = \tau_{ji}$), there are six independent components. For any stress state, it is always possible to find a set of three orthogonal axes — the **principal stress axes** — on which all shear stresses vanish. The normal stresses on these planes are the **principal stresses**. These principle stresses are labeled $S_1 \ge S_2 \ge S_3$ (maximum, intermediate, and minimum, respectively). --- ## 14.4 Principal Stresses in the Earth In the subsurface, the three principal stresses are conventionally aligned with: - **Vertical stress ($S_v$ or Sv)**: Acts vertically downward. This is the overburden stress — the weight of all rock and fluid above the point of interest. - **Maximum horizontal stress ($S_{Hmax}$ or $S_H$ or SHmax)**: The larger of the two horizontal principal stresses. - **Minimum horizontal stress ($S_{hmin}$ or $S_h$ or Shmin)**: The smaller of the two horizontal principal stresses. These three stresses are orthogonal and define the in-situ stress state. ### Vertical Stress The vertical stress is calculated by integrating the density of all overlying rock: $$ S_v = \int_0^z \rho(z) \cdot g \cdot dz $$ In oilfield units with a typical rock density gradient of approximately 1.0 psi/ft: $$ S_v \approx 1.0 \text{ psi/ft} \times z \text{ (ft)} $$ This gradient varies with lithology and compaction: shallow unconsolidated sediments have lower densities (lower gradient), while deeper compacted formations and igneous rock may exceed 1.0 psi/ft. ### Horizontal Stresses Horizontal stresses originate from tectonic activity (plate tectonics, mountain building, basin extension) and the elastic response of rock to vertical loading. They are generally not equal to each other or to Sv. Estimating horizontal stresses requires a combination of: - Understanding of the regional tectonic setting - Well log data (density, sonic) - In-situ measurements (leak-off tests, extended leak-off tests, mini-fracs) - Analysis of borehole failures (breakout orientation, induced fractures) ### An Important Exception: Salt Salt domes are a notable exception to the assumption that principal stresses align with the vertical and horizontal directions. Salt is a ductile material that flows on geologic timescales. Because it cannot sustain differential stress over millions of years, the stress state inside a salt body approaches isotropic: S1 ≈ S2 ≈ S3. Additionally, near-salt formations experience stress perturbations because the salt redistributes the load around it. Drilling near salt structures requires special geomechanical analysis. --- ## 14.5 Mean Stress and Maximum Shear Stress Two useful scalar measures summarize the 3D stress state: **Mean stress** (sometimes called hydrostatic stress): $$ S_{mean} = \frac{S_1 + S_2 + S_3}{3} $$ **Maximum shear stress** (acting on a plane at 45° to $S_1$ and $S_3$): $$ \tau_{max} = \frac{S_1 - S_3}{2} $$ The maximum shear stress drives rock failure: a large difference between the maximum and minimum principal stresses creates a large shear load on favorably oriented planes. --- ## 14.6 Anderson's Fault Classification Anderson (1951) recognized that the three principal stresses determine which type of faulting will occur in a region, and classified tectonic regimes accordingly: ```{figure} ../figures/ch14/anderson_faults.png :name: fig-anderson-faults :width: 85% :align: center Anderson's (1951) three fault types and their associated stress orderings: normal faulting ($S_v \ge S_{Hmax} \ge S_{hmin}$), strike-slip faulting ($S_{Hmax} \ge S_v \ge S_{hmin}$), and reverse faulting ($S_{Hmax} \ge S_{hmin} \ge S_v$). ``` **Normal Faulting Regime (extensional)** $$S_v = S_1 > S_{Hmax} = S_2 > S_{hmin} = S_3$$ The overburden (vertical stress) is the largest stress. The crust is extending horizontally, and faults dip steeply, with hanging-wall blocks dropping relative to footwall blocks. This regime is typical of rift basins (e.g., the Gulf of Mexico, the North Sea), and is common in Texas outside of tectonically active areas. **Strike-Slip Faulting Regime** $$S_{Hmax} = S_1 > S_v = S_2 > S_{hmin} = S_3$$ One horizontal stress dominates. Faults are sub-vertical, and motion is horizontal (lateral). Strike-slip faults are common in California (e.g., the San Andreas fault system). **Reverse Faulting Regime (compressional)** $$S_{Hmax} = S_1 > S_{hmin} = S_2 > S_v = S_3$$ Both horizontal stresses exceed the vertical stress. The crust is being compressed, and blocks are thrust upward over one another. Reverse faulting occurs in mountain-building regions (e.g., the Rocky Mountains, the Canadian Rockies, the Zagros Mountains). In these settings, even vertical wells may require significant mud weight to prevent wellbore collapse because the horizontal stresses are very high. ```{admonition} Connecting Geology to Drilling :class: note The stress regime at a drilling location is not always obvious. Geologists use the regional tectonic history, fault orientations, and seismic data to estimate which regime applies. As the drilling engineer, you translate that estimate into a mud weight window. A reverse faulting environment generally requires heavier mud than a normal faulting environment at the same depth. ``` --- ## 14.7 Normal Stress and Shear Stress on Inclined Planes Consider a block of rock with two principal stresses $S_1$ (maximum) and $S_3$ (minimum) acting on it. Imagine standing inside the block, rotating to face different directions. The stress that "pushes into your face" changes depending on which direction you face — it is the **normal stress** on that plane. The stress that tries to rotate you is the **shear stress**. For a plane oriented at angle θ measured from the direction of S1: $$ S_n(\theta) = S_1 \cos^2\theta + S_3 \sin^2\theta $$ $$ \tau(\theta) = \frac{S_1 - S_3}{2} \sin(2\theta) $$ The normal stress varies between $S_3$ (at $\theta = 90^o$, facing the $S_3$ direction) and $S_1$ (at $\theta = 0^o$, facing the $S_1$ direction). The shear stress is maximum at $\theta = 45^o$ and zero at $\theta = 0^o$ and $90^o$. ```{figure} ../figures/ch14/stress_orientation.png :name: fig-stress-orientation :width: 80% :align: center Normal stress versus plane orientation for $S_1 = 22$ and $S_3 = 12$: the normal stress oscillates sinusoidally between the two principal values as the plane rotates through 360°. ``` --- ## 14.8 Effective Stress Rock failure is controlled not by the total stress but by the **effective stress**: the portion of stress carried by the rock matrix rather than the pore fluid. Terzaghi's effective stress principle states: $$\sigma = S - P_p$$ where: - $\sigma$ = effective stress (carried by rock matrix) - $S$ = total stress (total load applied) - $P_p$ = pore pressure (fluid pressure in pores) The pore fluid acts like a hydraulic cushion. High pore pressure "pushes back" on the rock grains, reducing the net stress they experience. This is why elevated pore pressure (overpressure) can destabilize a rock that would otherwise be stable: increasing or decreasing pore pressure shifts the stress state toward failure without any change in the applied total stresses. **Convention**: In this text, capital $S$ denotes total stress; $\sigma$ denotes effective stress. | Quantity | Symbol | Definition | |----------|--------|------------| | Total maximum principal stress | $S_1$ | Applied load | | Effective maximum principal stress | $\sigma_1$ | $S_1 − P_p$ | | Total minimum principal stress | $S_3$ | Applied load | | Effective minimum principal stress | $\sigma_3$ | $S_3 − P_p$ | --- ## 14.9 Rock Strength Properties To predict whether applied stresses will cause failure, we need to quantify how strong the rock is. The key mechanical properties are: **Elastic (pre-failure) properties:** - **Young's modulus (E)**: Ratio of axial stress to axial strain. Measures the stiffness of the rock. Units: psi or GPa. High E → stiff rock (granite); low E → soft rock (unconsolidated sand). - **Poisson's ratio (ν)**: Ratio of lateral strain to axial strain under uniaxial compression. Dimensionless, typically 0.1–0.4 for rocks. Governs how horizontal stress develops from overburden. **Strength (failure) properties:** - **Unconfined Compressive Strength (UCS or $C_o$)**: The axial stress at which a rock fails with zero confining pressure ($S_3 = 0$). This is the simplest direct measure of rock strength. Typical values range from ~500 psi for weak shales to over 30,000 psi for hard granite. - **Cohesion ($S_o$ or C)**: The shear strength a rock has even at zero normal stress, i.e. the "stickiness" holding rock grains together through cementation. - **Internal friction coefficient ($\mu_i$)**: Relates how much additional shear strength the rock gains as normal stress increases. Related to the **friction angle (θ)** by $\mu_i = \tan(\theta)$. These properties are measured in the laboratory and are essential inputs to any wellbore stability analysis. --- ## 14.10 The Mohr Circle The Mohr Circle is a tool for understandinng rock strength, and by extension, rock failure. It serves both as a graphical tool for visualizing the stress state on all possible planes through a point, and a mathematical framework to estimate rock failure. The x-axis represents effective normal stress ($\sigma_n$) and the y-axis represents shear stress ($\tau$). For two principal effective stresses $\sigma_1$ and $\sigma_3$: - The center of the Mohr circle is at $\left(\frac{\sigma_1-\sigma_3}{2}+\sigma_3, 0\right)$ - The radius is $\left(\frac{\sigma_1-\sigma_3}{2}\right)$ - Each point on the circle represents the $(\sigma_n, \tau)$ combination on a plane at some angle to the principal stress axes The specific equations for the normal stress and shear stress on a plane at angle $\phi$ from the $\sigma_1$ direction are: $$ \sigma_n = \frac{\sigma_1 + \sigma_3}{2} - \frac{\sigma_1 - \sigma_3}{2}\cos(2\phi) $$ $$ \tau = \frac{\sigma_1 - \sigma_3}{2}\sin(2\phi) $$ ```{figure} ../figures/ch14/mohr_circle.png :name: fig-mohr-circle :width: 85% :align: center The Mohr circle represents the complete 2D stress state in effective stress space. The endpoints on the σn axis are $\sigma_3$ (left) and $\sigma_1$ (right); the maximum shear stress τmax occurs at the top of the circle. ``` ### Effect of Pore Pressure on the Mohr Circle Because we work with effective stresses, the Mohr circle is plotted using ($S-P_p$) values. Increasing pore pressure shifts the entire circle to the left along the σn axis (toward lower effective stresses), while decreasing pore pressure shifts it to the right. This has a critical implication: **a formation that is stable at its original pore pressure can become unstable if pore pressure increases** (e.g., due to injection of water or drilling with insufficient overbalance). ```{figure} ../figures/ch14/mohr_circle_pp.png :name: fig-mohr-circle-pp :width: 90% :align: center The effect of pore pressure on the Mohr circle. For the same total stresses ($S_1$ = 22, $S_3$ = 15), raising the pore pressure from 0 (right) to 10 (left) shifts the circle toward the origin — and toward the failure envelope. The circle's size is unchanged; only its position moves. ``` Conversely, pressure depletion during production (lower $P_p$) shifts the circle to the right, moving the stress state away from the failure envelope and generally making the rock more stable, though other effects such as increased compaction can complicate this. ### 3D Mohr Circles For a fully 3D stress state $(\sigma_1, \sigma_2, \sigma_3)$, there are three Mohr circles: - Large circle between $\sigma_3$ and $\sigma_1$ - Medium circle between $\sigma_3$ and $\sigma_2$ - Small circle between $\sigma_2$ and $\sigma_1$ The accessible stress states are the shaded region between the three circles. The intermediate stress σ2 does influence rock strength in a 3D sense, but the Mohr-Coulomb criterion (discussed below) captures the dominant behavior using only $\sigma_1$ and $\sigma_3$. --- ## 14.11 Rock Deformation and Failure Modes Laboratory rock mechanics experiments compress cylindrical core samples to measure how rocks respond to stress. Three standard test types are used: ```{figure} ../figures/ch14/rock_experiments.png :name: fig-rock-experiments :width: 85% :align: center Rock deformation experiments: cylindrical core samples are loaded axially (with or without confining pressure) until failure, mapping the stress-strain response and the failure stress. ``` **Uniaxial (Unconfined) Compressive Test (UCS test)** The core is loaded axially with no lateral confinement ($S_2$ = $S_3$ = 0). The axial stress at which the core fails is the **Unconfined Compressive Strength ($C_o$)**. In the stress-strain curve, the rock deforms nearly linearly until failure, then the stress drops abruptly as the failure plane forms. Loading occurs over minutes; failure itself occurs in milliseconds. **Triaxial Test** A confining pressure ($S_2$ = $S_3$ = $S_c$) is applied to the sides of the core (via a pressurized jacket), and then the axial stress is increased until failure. Triaxial tests at multiple confining pressures are used to map out the full Mohr-Coulomb failure envelope. **Rock Failure Modes** The confining stress dramatically changes how a rock fails: - **Brittle failure** (low confining stress, high rock strength): Failure occurs abruptly along a well-defined shear plane. This is the dominant mode for most rocks at the depths and stresses typical in drilling. The failure angle from the axis of the core is related to the friction angle. - **Semi-brittle failure** (moderate confining stress): Some plastic deformation before failure; fracture still forms but with more ductile character. - **Ductile failure** (high confining stress, low rock strength): Rock deforms plastically without a distinct fracture. Unconsolidated sands often fail in a ductile manner. Salt behaves ductilely at all practical downhole conditions. ```{figure} ../figures/ch14/deformation_confining.png :name: fig-deformation-confining :width: 85% :align: center The transition from brittle to ductile failure with increasing confining pressure. At low confinement a sharp shear plane forms; at high confinement the rock deforms plastically without a discrete fracture. ``` --- ## 14.12 The Mohr-Coulomb Failure Criterion The **Mohr-Coulomb failure criterion** defines the stress combinations that will cause shear failure in a rock. The linearized failure envelope is a straight line in $\tau$ vs. $\sigma_n$ space: $$ \tau = C + \sigma_n \tan\theta $$ where: - $C$ = cohesion (y-intercept of the failure line; units of psi or MPa) - $\theta$ = angle of internal friction (slope angle of the failure line) - $\sigma_n$ = effective normal stress on the failure plane The Mohr circle touches the failure envelope at the point representing the failure plane. At that point, the angle $2\phi$ between $\sigma_1$ and the normal to the failure plane satisfies: $$ 90° = \theta + 2\phi $$ ```{figure} ../figures/ch14/mohr_coulomb.png :name: fig-mohr-coulomb :width: 85% :align: center The Mohr-Coulomb failure envelope $\tau = \pm (C+ \sigma_n tan \theta)$ and Mohr circles at failure. The circle touches the envelope at the point ($\sigma_n, \tau$) corresponding to the failure plane orientation, where $90^o = \theta + 2\phi$. ``` ### Failure Condition For a given stress state ($\sigma_1, \sigma_3$), the rock: - **Remains stable** if the Mohr circle lies entirely below the failure envelope - **Fails** if the Mohr circle intersects or touches the failure envelope --- ## 14.13 Determining Rock Strength Parameters ### Four-Step Procedure Given any problem involving the Mohr-Coulomb failure criterion: 1. **Identify the total stresses and pore pressure** at the location of interest 2. **Calculate effective stresses**: $\sigma_1 = S_1 - P_p$ and $\sigma_3 = S_3 - P_p$ 3. **Plot (or calculate) the Mohr circle** using $\sigma_1$ and $\sigma_3$ as endpoints on the $\sigma_n$ axis 4. **Determine the failure point** by finding ($\sigma_n, \tau$) on the circle and comparing to the failure envelope ### From a Single UCS Test If only one compression test is available, the friction angle must be assumed. For sedimentary rocks, $θ ≈ 30^o$ is a reasonable default. For example, if the following is given: - Test failure stress: $S = 4,500$ psi (at $P_p$ = 0, so $\sigma_1$ = 4,500 psi, $\sigma_3$ = 0) - If we assume $\theta = 30^o$, then $\phi = (90^o − \theta)/2 = 30^o$, and $2 \phi = 60^o$. Data from prior lab tests of similar cores can be used to improve the estimate of $\theta$ The stress state at the failure point on the Mohr circle: $$ \sigma_n = \frac{\sigma_1 + \sigma_3}{2} - \frac{\sigma_1 - \sigma_3}{2}\cos(2\phi) = \frac{4500 + 0}{2} - \frac{4500 - 0}{2}\cos(60°) = 2250 - 1125 = 1125 \text{ psi} $$ $$ \tau = \frac{\sigma_1 - \sigma_3}{2}\sin(2\phi) = \frac{4500 - 0}{2}\sin(60°) = 2250 \times 0.866 = 1948.6 \text{ psi} $$ Solving for cohesion using the failure criterion: $$ C = \tau - \sigma_n\tan\theta = 1948.6 - 1125 \times \tan(30°) = 1948.6 - 649.5 \approx 1300 \text{ psi} $$ The Mohr-Coulomb failure envelope for this rock: $$ \tau = 1300 + \sigma_n \tan(30°) $$ ### From Two Tests (System of Equations) When two compression tests are available (one uniaxial test and one triaxial test), both cohesion $C$ and friction angle $\theta$ can be solved simultaneously without assuming $\theta$. **Test 1** (uniaxial): $S_1$ = S (failure stress), $S_3$ = 0, $P_p$ = 0 or given - Calculate $\sigma_1$ and $\sigma_3$ - Find $\sigma_{n,1}$ and $\tau_1$ from the Mohr failure criteria equations **Test 2** (Triaxial): $S_1 = S_{fail}$, $S_3 = S_{confine}$, $P_p = 0$ or given - Calculate $\sigma_1$ and $\sigma_3$ - Find $\sigma_{n,2}$ and $\tau_2$ from the Mohr failure criteria equations Since both tests use the same rock: $$ \tau_1 = C + \sigma_{n,1}\tan\theta \quad \text{(Equation A)} $$ $$ \tau_2 = C + \sigma_{n,2}\tan\theta \quad \text{(Equation B)} $$ Subtract Equation A from Equation B to eliminate $C$ and solve for $\theta$, then substitute back to find $C$. ```{admonition} Units and Calculator Tips :class: tip When solving the two-equation system, the angle $\theta$ will typically be returned in radians by a calculator or solver. Convert to degrees: $\theta$ (degrees) = $\theta$ (radians) × 180/π. The answer should be between about 15° and 45° for typical rocks; values near 30° are common for sedimentary rocks. ``` ### Using Rock Strength to Predict Stability With $C$ and $\theta$ known, the failure envelope can be used to assess any new stress combination. Given $\sigma_1$ and $\sigma_3$ at a depth of interest: 1. Calculate $\sigma_n$ and $\tau$ from the Mohr circle 2. Calculate the failure criterion value: $C + \sigma_n·tan(\theta)$ 3. If $\tau > C + \sigma_n·tan(\theta)$: **failure occurs** 4. If $\tau < C + \sigma_n·tan(\theta)$: **rock remains stable** --- ## 14.14 Worked Example: Geomechanical Characterization **Problem**: A core sample is obtained from a sandstone formation during drilling. Two compression tests are performed: - **Test 1** (UCS): Failure occurs at S = 6,000 psi. No confining pressure, no pore pressure. - **Test 2** (Triaxial): Failure occurs at $S_1$ = 9,200 psi with $S_3$ = 1,500 psi and $P_p$ = 500 psi. Determine the Mohr-Coulomb failure envelope ($C$ and $\theta$) for this rock. **Solution** **Test 1:** - σ1 = S − Pp = 6,000 − 0 = 6,000 psi - σ3 = 0 − 0 = 0 psi **Test 2:** - σ1 = 9,200 − 500 = 8,700 psi - σ3 = 1,500 − 500 = 1,000 psi Set up the Mohr circle equations for both tests (leaving φ unknown): $$\sigma_{n,1} = \frac{6000 + 0}{2} - \frac{6000 - 0}{2}\cos(2\phi) = 3000 - 3000\cos(2\phi)$$ $$\tau_1 = \frac{6000 - 0}{2}\sin(2\phi) = 3000\sin(2\phi)$$ $$\sigma_{n,2} = \frac{8700 + 1000}{2} - \frac{8700 - 1000}{2}\cos(2\phi) = 4850 - 3850\cos(2\phi)$$ $$\tau_2 = \frac{8700 - 1000}{2}\sin(2\phi) = 3850\sin(2\phi)$$ Writing the failure criterion for each: - Equation A: $3000\sin(2\phi) = C + [3000 - 3000\cos(2\phi)]\tan\theta$ - Equation B: $3850\sin(2\phi) = C + [4850 - 3850\cos(2\phi)]\tan\theta$ Subtracting A from B: $$850\sin(2\phi) = 1850\tan\theta - 850\cos(2\phi)\tan\theta$$ Note that φ and θ are related by 90° = θ + 2φ, so 2φ = 90° − θ. Substituting this relationship and solving (numerically or symbolically) gives θ ≈ 30° and C ≈ 1,200 psi. The failure envelope is: $$\tau = 1200 + \sigma_n\tan(30°)$$ --- ## Practice Problems **Problem 14.1** — *Stress Regime Identification* At a proposed drilling location at 9,000 ft TVD, the geomechanics study reports: Sv gradient = 1.0 psi/ft, SHmax = 10,400 psi, Shmin = 7,800 psi. (a) Calculate Sv at 9,000 ft. (b) Order the three principal stresses and identify the Anderson fault regime. (c) Calculate the mean stress and the maximum shear stress. (d) If the pore pressure is 4,200 psi, calculate the three effective principal stresses. ```{dropdown} Solution **(a)** $S_v = 1.0 \times 9{,}000 = 9{,}000$ psi **(b)** $S_{Hmax} = 10{,}400 > S_v = 9{,}000 > S_{hmin} = 7{,}800$ psi → S1 = SHmax, S2 = Sv, S3 = Shmin → **strike-slip regime**. **(c)** $S_{mean} = (10{,}400 + 9{,}000 + 7{,}800)/3 = 9{,}067$ psi; $\tau_{max} = (10{,}400 - 7{,}800)/2 = 1{,}300$ psi **(d)** $\sigma_1 = 10{,}400 - 4{,}200 = 6{,}200$ psi; $\sigma_2 = 9{,}000 - 4{,}200 = 4{,}800$ psi; $\sigma_3 = 7{,}800 - 4{,}200 = 3{,}600$ psi ``` **Problem 14.2** — *Failure Envelope from a Single Test* A UCS test on a shale core fails at 3,800 psi (no confining pressure, no pore pressure). Assuming θ = 30°: (a) Find the normal and shear stress at the failure point on the Mohr circle. (b) Determine the cohesion C. (c) Write the failure envelope. (d) Using this envelope, will the rock fail at σ1 = 7,000 psi, σ3 = 1,500 psi? ```{dropdown} Solution **(a)** With 2φ = 60°: $\sigma_n = 1{,}900 - 1{,}900\cos 60° = 950$ psi; $\tau = 1{,}900 \sin 60° = 1{,}645$ psi **(b)** $C = 1{,}645 - 950\tan 30° = 1{,}645 - 548 = 1{,}097 \approx 1{,}100$ psi **(c)** $\tau = 1{,}100 + \sigma_n \tan 30°$ **(d)** At σ1 = 7,000, σ3 = 1,500 psi: $\sigma_n = 4{,}250 - 2{,}750(0.5) = 2{,}875$ psi; $\tau = 2{,}750(0.866) = 2{,}382$ psi. Criterion value: $1{,}100 + 2{,}875(0.577) = 2{,}759$ psi. Since $\tau = 2{,}382 < 2{,}759$ → **stable, no failure**. ``` **Problem 14.3** — *Pore Pressure and Stability* A formation at depth has total stresses S1 = 12,000 psi and S3 = 7,000 psi, with rock strength C = 1,500 psi and θ = 30°. (a) At the original pore pressure of 4,000 psi, does the rock fail? (b) Water injection raises the pore pressure to 5,500 psi. Does the rock fail now? (c) Explain the result using the Mohr circle picture. ```{dropdown} Solution **(a)** σ1 = 8,000, σ3 = 3,000 psi. With 2φ = 60°: $\sigma_n = 5{,}500 - 2{,}500(0.5) = 4{,}250$ psi; $\tau = 2{,}500(0.866) = 2{,}165$ psi. Criterion: $1{,}500 + 4{,}250(0.577) = 3{,}952$ psi. $\tau < 3{,}952$ → **stable**. **(b)** σ1 = 6,500, σ3 = 1,500 psi. $\sigma_n = 4{,}000 - 2{,}500(0.5) = 2{,}750$ psi; $\tau = 2{,}165$ psi. Criterion: $1{,}500 + 2{,}750(0.577) = 3{,}087$ psi. $\tau = 2{,}165 < 3{,}087$ → **still stable**, but the margin has dropped from 1,787 psi to 922 psi — roughly half the safety margin has been consumed. **(c)** Raising Pp slides the Mohr circle left (toward the origin) without changing its size. Because the Mohr-Coulomb envelope has a positive intercept and slope, a circle closer to the origin is closer to the envelope — continued injection would eventually bring the circle into contact with the envelope and trigger shear failure (this is the mechanism of injection-induced fault reactivation). ``` --- ## Chapter Summary Geomechanics provides the physical framework for understanding how in-situ stresses drive wellbore failure. The key concepts are: - Rock is under stress in three dimensions: vertical overburden stress $S_v$ and two horizontal stresses $S_{Hmax}$ and $S_{hmin}$. - The tectonic regime (normal, strike-slip, or reverse faulting) determines the relative magnitudes of these stresses. - Effective stress equals total stress minus pore pressure. Rock failure is controlled by effective stress. - The Mohr circle graphically represents the stress state on all planes. Higher pore pressure shifts the circle toward failure. - The Mohr-Coulomb failure envelope defines the boundary between stable and failed rock. It is described by cohesion $C$ and friction angle $\theta$. - Rock strength parameters are determined from laboratory compression tests (UCS and triaxial). One test allows estimation if $\theta$ is assumed; two tests fully determine both $C$ and $\theta$. Chapter 15 applies this framework to the wellbore, calculating the stresses that develop on the borehole wall. Chapter 16 uses those results to select the mud weight window that keeps the hole stable. --- ## Key Equations | Equation | Description | |----------|-------------| | $\sigma = S - P_p$ | Effective stress | | $S_v = \int_0^z \rho(z) g dz$ | Vertical (overburden) stress | | $S_{mean} = \frac{S_1 + S_2 + S_3}{3}$ | Mean stress | | $\tau_{max} = \frac{S_1-S_3}{2}$ | Maximum shear stress | | $\sigma_n = \frac{\sigma_1+\sigma_3}{2} − \frac{\sigma_1−\sigma_3}{2} \cdot cos(2 \phi)$ | Normal stress given $\sigma_1$ and $\sigma_3$ | | $\tau = \frac{\sigma_1 - \sigma_3}{2} \cdot sin (2\phi)$ | Shear stress given $\sigma_1$ and $\sigma_3$ | | $\tau = C + \sigma_n \cdot tan (\theta)$ | Linear Mohr-Coulomb failure criterion | | $90^o = \theta + 2\phi$ | Failure plane angle relationship | ## Key Terms **Anderson's fault classification**: Framework relating stress ordering to fault type (normal, strike-slip, reverse). **Cohesion ($C$ or $C_o$)**: Shear strength at zero normal stress; represents cementation between grains. **Effective stress ($\sigma$)**: Total stress minus pore pressure; governs rock mechanical behavior. **Friction angle ($\theta$)**: Angle of the Mohr-Coulomb failure envelope; determines additional strength gained under confinement. **Mohr circle**: Graphical representation of the 2D stress state showing normal and shear stress on all planes. **Mohr-Coulomb failure criterion**: Linear failure envelope in shear vs. normal stress space: $\tau = C + \sigma_n \cdot tan (\theta)$. **Overburden stress ($S_v$)**: Vertical stress due to the weight of rock and fluid above a given depth. **Pore pressure ($P_p$)**: Fluid pressure in the connected pore space of the rock. **Principal stresses**: Three orthogonal stresses with no shear components: $S_1 \ge S_2 \ge S_3$. **Unconfined Compressive Strength (UCS)**: Axial stress to failure with no lateral confinement.