Tensor method

The following subsection describes the procedure to calculate stresses $(\sigma_n,\tau)$ on an arbitrary plane given its orientation respect to the geographical coordinate system $(strike,dip)$ and the in-situ stress tensor of principal stresses $\uuline{S}_P$ (given its principal values and principal directions).

The first step consists on defining the principal stress coordinate system and the geographical coordinate system (both right-handed coordinate systems).

• The principal stress coordinate system has the three principal directions as bases of the system in the order of most compressive to least compressive: $\uline{e}_1$ for $S_1$, $\uline{e}_2$ for $S_2$, and $\uline{e}_3$ for $S_3$ (Fig. 5.19). The stress tensor is termed $\uuline{S}_P$ in this coordinate system.
• The geographical coordinate system has bases $\uline{e}_1$ pointing in North direction, $\uline{e}_2$ pointing in East direction, and $\uline{e}_3$ pointing down in direction of increasing depth (Fig. 5.19). We will refer this basis as the “NED” basis. The stress tensor is termed $\uuline{S}_G$ in this coordinate system.

Figure 5.19: The stress tensor in principal directions and geographical coordinate systems.

The second step involves constructing a change of basis matrix $R_{PG}$ from the Principal Stress to the Geographical Coordinate system. This matrix depends on the projections of the elements of the new base on the old base according to the cosines of the director angles $\alpha$, $\beta$, and $\gamma$. Table 5.3 summarizes the meaning of $\alpha$, $\beta$, and $\gamma$ for cases in which vertical stress is a principal stress.

Table 5.3: Summary of possible values of $\alpha$, $\beta$, and $\gamma$ for vertical stress being a principal stress. Notes: (1) these angles indicate solely the orientation of principal stresses with respect to the geographical coordinate system, (2) this $\alpha$ has nothing to do with the angle used in the Mohr's circle method to solve for stresses on a fault.

	Normal faulting	Strike slip	Reverse faulting
$\alpha$	Azimuth of $S_{hmin}$	Azimuth of $S_{Hmax}$	Azimuth of $S_{Hmax}$
$\beta$	$90^{\circ}$	$0^{\circ}$	$0^{\circ}$
$\gamma$	$0^{\circ}$	$90^{\circ}$	$0^{\circ}$

Check out this link https://mybinder.org/v2/gh/johntfoster/rotation_widget/master?filepath=rotation_widget-rise.ipynb for an animation of $\alpha$, $\beta$, and $\gamma$ in arbitrary directions.

Figure 5.20: Transformation matrix from the principal directions to geographical coordinate system and corresponding angles.

With the matrix $R_{PG}$, we can calculate the stress tensor $\uuline{S}_P$ as a function of $\uuline{S}_G$,

$$\uuline{S}_P = R_{PG} \uuline{S}_G R_{PG}^T$$ (5.6)

and therefore:

$$\uuline{S}_G = R_{PG}^T \uuline{S}_P R_{PG}$$ (5.7)

where $T$ stands for “transpose”.

PROBLEM 5.5: Calculate $\uuline{S}_G$ in a normal faulting stress regime case ($S_1 = 30$ MPa, $S_2 = 25$ MPa, $S_3 = 20$ MPa) with azimuth of $S_{hmin}$ N-S. $S_v$ is a principal stress.

SOLUTION

$\: \: \blacksquare$

PROBLEM 5.6: Calculate $\uuline{S}_G$ in a strike-slip faulting stress regime case ($S_1 = 30$ MPa, $S_2 = 25$ MPa, $S_3 = 20$ MPa) with azimuth of $S_{Hmax}$ N-S. $S_v$ is a principal stress.

SOLUTION

$\: \: \blacksquare$

PROBLEM 5.7: Calculate $\uuline{S}_G$ in a reverse faulting stress regime case ($S_1 = 30$ MPa, $S_2 = 25$ MPa, $S_3 = 20$ MPa) with azimuth of $S_{Hmax}$ E-W. $S_v$ is a principal stress.

SOLUTION

$\: \: \blacksquare$

PROBLEM 5.8: Calculate $\uuline{S}_G$ in a strike-slip faulting stress regime case ($S_1 = 60$ MPa, $S_2 = 40$ MPa, $S_3 = 35$ MPa) with azimuth of $S_{Hmax}$ being 135$^{\circ }$. $S_v$ is a principal stress.

SOLUTION

$\: \: \blacksquare$

The third step consists in defining the fault plane coordinate system. The coordinate system basis is comprised of $n_d$ (dip), $n_s$ (strike), and $n_n$ (normal) vectors: d-s-n right-handed basis. The three vectors depend solely in two variables: $strike$ and $dip$.

Figure 5.21: Fault coordinate system as a function of strike and dip.

The fourth (and last) step consists in projecting the stress tensor based on the geographical coordinate system onto the fault base vectors. The stress vector acting on the plane of the fault is $\uline{t}$ (note that $\uline{t}$ is not necessarily aligned with $\uline{n}_d$, $\uline{n}_s$ or $\uline{n}_n$) and is calculated according to:

$$\uline{t} = \uuline{S}_G \uline{n}_n$$ (5.8)

The total normal stress on the plane of the fault is $S_n$ (aligned with $\uline{n}_n$):

$$S_n = \uline{t} \cdot \uline{n}_n$$ (5.9)

The effective normal stress on the fault plane is $\sigma_n = S_n - P_p$. The shear stresses on the plane of the fault is aligned with $\uline{n}_d$ and $\uline{n}_s$ are:

$$\left\lbrace \begin{array}{l} \tau_d = \uline{t} \cdot \uline{n}_d \\ \tau_s = \uline{t} \cdot \uline{n}_s \end{array}\right.$$ (5.10)

The dot product is used in all these vector to vector multiplications. The geometrical meaning is the projection of one vector onto the other.

The effective normal stress $\sigma_n$ and absolute shear $\tau = \sqrt{\tau_d^2 + \tau_s^2}$ can also be calculated with the following equations:

$$\sigma_n = \uline{t} \cdot \uline{n}_n - P_p$$ (5.11)

$$\tau^2 = \vert\vert\uline{t}\vert\vert^2 - \vert\vert\uline{t} \cdot \uline{n}_n\vert\vert^2$$ (5.12)

The $rake$ is the angle of the shear stress $\uline{\tau}_d + \uline{\tau}_s$ with respect to $\uline{n}_s$ (horizontal line) and quantifies the direction of expected fault movement in the fault plane.

$$rake = \arctan \left( \frac{\tau_d}{\tau_s} \right)$$ (5.13)

PROBLEM 5.9: Calculate $t$, $S_n$, $\tau_d$, $\tau_s$, and $rake$ for a fault with strike 000$^{\circ }$ and dip 60$^{\circ }$E in a place with normal faulting stress regime ($S_v = 23$ MPa, $S_{Hmax} = 15$ MPa, $S_{hmin} = 13.8$ MPa) with azimuth of $S_{hmin}$ equal to 90$^{\circ }$. $S_v$ is a principal stress.

SOLUTION

$\: \: \blacksquare$

PROBLEM 5.10: Calculate $t$, $S_n$, $\tau_d$, $\tau_s$, and $rake$ for a fault with strike 060$^{\circ }$ and dip 90$^{\circ }$ in a place with strike-slip stress regime ($S_v = 30$ MPa, $S_{Hmax} = 45$ MPa, $S_{hmin} = 25$ MPa) with azimuth of $S_{Hmax}$ equal to 120$^{\circ }$. $S_v$ is a principal stress.

SOLUTION

$\: \: \blacksquare$

PROBLEM 5.11: Calculate $t$, $S_n$, $\tau_d$, $\tau_s$, and $rake$ for conjugate faults with strike 045$^{\circ }$ and 225$^{\circ }$ both with dip 60$^{\circ }$ in a place with normal faulting stress regime ( $S_v = 5000$ psi, $S_{Hmax} = 4000$ psi, $S_{hmin} = 3000$ psi) with azimuth of $S_{hmin}$ equal to 90$^{\circ }$. $S_v$ is a principal stress.

SOLUTION

$\: \: \blacksquare$

PROBLEM 5.12: Calculate $t$, $S_n$, $\tau_d$, $\tau_s$, $rake$, and $\tau /\sigma _n$ for a fault with strike 120$^{\circ }$ and 70$^{\circ }$ dip in a place with reverse faulting stress regime ( $S_v = 1000$ psi, $S_{Hmax} = 2400$ psi, $S_{hmin} = 1200$ psi) with azimuth of $S_{Hmax}$ equal to 150$^{\circ }$ and pore pressure $P_p = 440$ psi. $S_v$ is a principal stress.

SOLUTION

$\: \: \blacksquare$