5.4 Determination of normal and shear stresses on the fault plane

In this section we will review two methods to calculate normal and shear stresses on fractures and faults. The first part reviews the Mohr circle method in order to have a conceptual understanding of stress projection on faults and maximum ratio between shear stress and effective normal stress. The second part discusses the tensor method, which requires the definition of three coordinate systems and matrix multiplication. The tensor method can be easily implemented in a computer script but is laborious to work out manually.

5.4.1 Mohr's circle method

The 3D Mohr circle is a graphical representation of the stress tensor and all its projections (or possibles values of normal effective stress $\sigma_n$ and shear stress $\tau$) on a given plane. Consider a horizontal plane in Fig. 5.20, the normal stress is the vertical stress $S_v$ and there is no shear stress. Consider a vertical plane with strike East-West in Fig. 5.20, you get the minimum principal stress $S_{hmin}$. Consider a vertical plane with strike North-South in Fig. 5.20, you get the maximum principal stress $S_{Hmax}$.

Likewise, non-trivial solutions of stress projection at an arbitrary plane angle include all the points delimited by the three Mohr circles. Let's consider solutions along each circle in Fig. 5.20.

• The blue circle (center C$_1$) represents the possible state of stresses that result as a combination of $\sigma _v$ and $\sigma_{hmin}$. The shear stress $\tau$ and normal effective stress $\sigma_n$ of any plane in between the planes of $\sigma _v$ and $\sigma_{hmin}$ and colinear with $\sigma_{Hmax}$ can be found through the angle $\beta_1$ measured from $\sigma _v$.
• The green circle (center C$_2$) represents the possible state of stresses that result as a combination of $\sigma_{v}$ and $\sigma_{Hmax}$. The shear stress $\tau$ and normal effective stress $\sigma_n$ of any plane in between the planes of $\sigma_{v}$ and $\sigma_{Hmax}$ and colinear with $\sigma_{hmin}$ can be found through the angle $\beta_2$ measured from $\sigma_{Hmax}$
• The red circle (center C$_3$) represents the possible state of stresses that result as a combination of $\sigma_{Hmax}$ and $\sigma_{hmin}$. The shear stress $\tau$ and normal effective stress $\sigma_n$ of any plane in between the planes of $\sigma_{Hmax}$ and $\sigma_{hmin}$ and colinear with $\sigma_{v}$ can be found through the angle $\beta_3$ measured from $\sigma_{Hmax}$

Figure 5.20: The 3D Mohr circle

For this example (normal faulting, $S_{Hmax}$ azimuth E-W), an ideal fault would occur with a strike E-W and dip 60$^{\circ }$ (assuming $\varphi = 30^{\circ}$). This is the orientation of a plane with maximum $\tau /\sigma _n$.

PROBLEM 5.3: Find the shear and normal effective stresses on a fault plane within the following state of stress and conditions:

• Fault: strike azimuth = 000$^{\circ }$, dip = 60$^{\circ }$.
• State of stress: $S_v =$ 23 MPa (principal), $S_{Hmax} =$ 20 MPa, $S_{hmin} =$ 13.8 MPa (azimuth: 090$^{\circ }$), and $P_p =$ 10 MPa.

SOLUTION

The effective stresses are: $\sigma_v =$ 13 MPa, $\sigma_{Hmax} =$ 10 MPa, $\sigma_{hmin} =$ 3.8 MPa. Based on the Mohr circle of $\sigma _v$ with $\sigma_{hmin}$ and trigonometry:

$$\sigma_n = \left( \frac{13 \text{ MPa} + 3.8 \text{ MPa}}{2} \rig... ... MPa} - 3.8 \text{ MPa}}{2} \right) \cos(2 \cdot 60^{\circ}) = 6.1 \text{ MPa}$$

$$\tau = \left( \frac{13 \text{ MPa} - 3.8 \text{ MPa}}{2} \right) \sin(2 \cdot 60^{\circ}) = 4.0 \text{ MPa} \: \: \blacksquare$$

PROBLEM 5.4: Find the shear and normal effective stresses on a fault plane within the following state of stress and conditions:

• Fault: strike azimuth = N60$^{\circ }$E, dip: 90$^{\circ }$.
• State of stress: $S_v =$ 30 MPa (principal), $S_{Hmax} =$ 45 MPa, $S_{hmin} =$ 25 MPa (azimuth: N30$^{\circ }$E), and $P_p =$ 15 MPa.

SOLUTION

The effective stresses are: $\sigma_v =$ 15 MPa, $\sigma_{Hmax} =$ 30 MPa, $\sigma_{hmin} =$ 10 MPa. Based on the Mohr circle of $\sigma_{Hmax}$ with $\sigma_{hmin}$ and trigonometry:

$$\sigma_n = \left( \frac{30 \text{ MPa} + 10 \text{ MPa}}{2} \righ... ...t{ MPa} - 10 \text{ MPa}}{2} \right) \cos(2 \cdot 30^{\circ}) = 25 \text{ MPa}$$

$$\tau = \left( \frac{30 \text{ MPa} - 10 \text{ MPa}}{2} \right) \sin(2 \cdot 30^{\circ}) = 8.7 \text{ MPa} \: \: \blacksquare$$

5.4.2 Tensor method

This 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. 5.21). 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. 5.21). We will refer this basis as the “NED” basis. The stress tensor is termed $\uuline{S}{}_G$ in this coordinate system.

Figure 5.21: 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 $\gamma$ (Fig. 5.22). Table 5.3 summarizes the meaning of $\alpha$, $\beta$, and $\gamma$ for cases in which vertical stress is a principal stress.

Figure 5.22: Transformation matrix from the principal directions to geographical coordinate system and corresponding angles.

$$R_{PG} = \left[ \begin{array}{ccc} \cos \alpha \cos \beta & \... ... \alpha \sin \gamma & \cos \beta \cos \gamma \end{array}\right]$$ (5.7)

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 $\beta$ 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.

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

$$\uuline{S}_P = R_{PG} \uuline{S}{}_G R_{PG}^T$$ (5.8)

and therefore:

$$\uuline{S}_G = R_{PG}^T \uuline{S}{}_P R_{PG}$$ (5.9)

where the superscript $T$ 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
The principal stress tensor is

$$\uuline{S}{}_P = \left[ \begin{array}{ccc} 30 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 20 \end{array}\right]$$

Using Table 5.3 and considering that $S_1 = S_v$, the angles of the principal stress coordinate result $\alpha = 0^{\circ}$, $\beta = 90^{\circ}$, and $\gamma = 0^{\circ}$. The change of coordinate system matrix results

$$R_{PG} = \left[ \begin{array}{ccc} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right]$$

Finally, using equation 5.9

$$\uuline{S}{}_G = \left[ \begin{array}{ccc} 0 & 0 & -1 \\ 0 &... ...0 & 25 & 0 \\ 0 & 0 & 30 \end{array}\right] \: \: \blacksquare$$

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
The principal stress tensor is

$$\uuline{S}{}_P = \left[ \begin{array}{ccc} 30 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 20 \end{array}\right]$$

Using Table 5.3 and considering that $S_1 = S_{Hmax}$ and $S_2 = S_v$, the angles of the principal stress coordinate result $\alpha = 0^{\circ}$, $\beta = 0^{\circ}$, and $\gamma = 90^{\circ}$. The change of coordinate system matrix results

$$R_{PG} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array}\right]$$

Finally, using equation 5.9

$$\uuline{S}{}_G = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & ... ...0 & 20 & 0 \\ 0 & 0 & 25 \end{array}\right] \: \: \blacksquare$$

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

The principal stress tensor is

$$\uuline{S}{}_P = \left[ \begin{array}{ccc} 30 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 20 \end{array}\right]$$

Using Table 5.3 and considering that $S_1 = S_{Hmax}$ and $S_2 = S_{hmin}$, the angles of the principal stress coordinate result $\alpha = 90^{\circ}$, $\beta = 0^{\circ}$, and $\gamma = 0^{\circ}$. The change of coordinate system matrix results

$$R_{PG} = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right]$$

Finally, using equation 5.9

$$\uuline{S}{}_G = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ -1 &... ...0 & 30 & 0 \\ 0 & 0 & 20 \end{array}\right] \: \: \blacksquare$$

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

The principal stress tensor is

$$\uuline{S}{}_P = \left[ \begin{array}{ccc} 60 & 0 & 0 \\ 0 & 40 & 0 \\ 0 & 0 & 35 \end{array}\right]$$

Using Table 5.3 and considering that $S_1 = S_{Hmax}$ and $S_2 = S_v$, the angles of the principal stress coordinate result $\alpha = 135^{\circ}$, $\beta = 0^{\circ}$, and $\gamma = 90^{\circ}$. The change of coordinate system matrix results

$$R_{PG} = \left[ \begin{array}{ccc} -0.707 & 0.707 & 0 \\ 0 & 0 & 1 \\ 0.707 & 0.707 & 0 \end{array}\right]$$

Finally, using equation 5.9

$$\uuline{S}{}_G = \left[ \begin{array}{ccc} -0.707 & 0.707 & 0... ...& 47.5 & 0 \\ 0 & 0 & 40 \end{array}\right] \: \: \blacksquare$$

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 $dip$ of the fault.

Figure 5.23: Fault coordinate system as a function of strike and dip.

The fourth step (and last) 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:

$$\uline{t} = \uuline{S}{}_G \uline{n}_n$$ (5.10)

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

$$S_n = \uline{t} \cdot \uline{n}_n$$ (5.11)

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:

$$\left\lbrace \begin{array}{l} \tau_d = \uline{t} \cdot \uline{n}_d \\ \tau_s = \uline{t} \cdot \uline{n}_s \end{array}\right.$$ (5.12)

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:

$$\sigma_n = \uline{t} \cdot \uline{n}_n - P_p$$ (5.13)

$$\tau^2 = \vert\vert\uline{t}\vert\vert^2 - \vert\vert\uline{t} \cdot \uline{n}_n\vert\vert^2$$ (5.14)

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.

$$rake = \arctan \left( \frac{\tau_d}{\tau_s} \right)$$ (5.15)

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

The tensor of principal stresses is

$$\uuline{S}{}_P = \left[ \begin{array}{ccc} 23 & 0 & 0 \\ 0 & 15 & 0 \\ 0 & 0 & 13.8 \end{array}\right]$$

Using Table 5.3 and considering that $S_1 = S_v$ and the azimuth of $S_{hmin}$ is 90$^{\circ }$, the angles of the principal stress coordinate result $\alpha = 90^{\circ}$, $\beta = 90^{\circ}$, and $\gamma = 0^{\circ}$. The change of coordinate system matrix results

$$R_{PG} = \left[ \begin{array}{ccc} 0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right]$$

and the total stress in the geographical coordinate system results

$$\uuline{S}{}_G = \left[ \begin{array}{ccc} 15 & 0 & 0 \\ 0 & 13.8 & 0 \\ 0 & 0 & 23 \end{array}\right]$$

Given the orientation of the fault, the vector normal to the fault is

$$\uline{n}{}_n = \left[ \begin{array}{c} 0 \\ 0.867 \\ -0.5 \end{array}\right]$$

Finally, the stresses on the fault are

$\uline{t}= [0,11.95,-11.50]$ MPa.
$S_n = 16.1$ MPa
$\tau_d = -3.98$ MPa
$\tau_s = 0$ MPa
$rake$ = 90 $^{\circ} \: \: \blacksquare$

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
The tensor of principal stresses is

$$\uuline{S}{}_P = \left[ \begin{array}{ccc} 45 & 0 & 0 \\ 0 & 30 & 0 \\ 0 & 0 & 25 \end{array}\right] \text{ MPa}$$

Using Table 5.3 and considering that $S_1 = S_v$ and the azimuth of $S_{Hmax}$ is 120$^{\circ }$, the angles of the principal stress coordinate result $\alpha = 120^{\circ}$, $\beta = 0^{\circ}$, and $\gamma = 90^{\circ}$. The change of coordinate system matrix results

$$R_{PG} = \left[ \begin{array}{ccc} -0.5 & 0.866 & 0 \\ 0 & 0 & 1 \\ 0.866 & 0.5 & 0 \end{array}\right]$$

and the total stress in the geographical coordinate system results

$$\uuline{S}{}_G = \left[ \begin{array}{ccc} 30 & -8.66 & 0 \\ -8.66 & 40 & 0 \\ 0 & 0 & 30 \end{array}\right] \text{ MPa}$$

Given the orientation of the fault, the vector normal to the fault is

$$\uline{n}{}_n = \left[ \begin{array}{c} -0.866 \\ 0.5 \\ 0 \end{array}\right]$$

Finally, the stresses on the fault are

$\uline{t}= [-30.31,27.5,0]$ MPa.
$S_n = 40$ MPa
$\tau_d = 0$ MPa
$\tau_s = 8.66$ MPa
$rake$ = 0 $^{\circ} \: \: \blacksquare$

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
The tensor of principal stresses is

$$\uuline{S}{}_P = \left[ \begin{array}{ccc} 5000 & 0 & 0 \\ 0 & 4000 & 0 \\ 0 & 0 & 3000 \end{array}\right] \text{ psi}$$

Using Table 5.3 and considering that $S_1 = S_v$ and the azimuth of $S_{hmin}$ is 90$^{\circ }$, the angles of the principal stress coordinate result $\alpha = 90^{\circ}$, $\beta = 90^{\circ}$, and $\gamma = 0^{\circ}$. The change of coordinate system matrix results

$$R_{PG} = \left[ \begin{array}{ccc} 0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right]$$

and the total stress tensor in the geographical coordinate system results

$$\uuline{S}{}_G = \left[ \begin{array}{ccc} 4000 & 0 & 0 \\ 0 & 3000 & 0 \\ 0 & 0 & 5000 \end{array}\right] \text{ psi}$$

Let us consider the first fault with strike of 045$^{\circ }$ and dip of 60$^{\circ }$, the vector normal to the faults is

$$\uline{n}{}_{n} = \left[ \begin{array}{c} -0.612 \\ 0.612 \\ -0.5 \end{array}\right]$$

The stresses on this fault are

$\uline{t}= [-2450,1840,-2500]$ psi.
$S_n = 3870$ psi
$\tau_d = -649$ psi
$\tau_s = -433$ psi
$rake$ = 56.3$^{\circ }$

Let us consider the fault with strike of 225$^{\circ }$ and dip of 60$^{\circ }$, the vector normal to the faults is

$$\uline{n}{}_{n} = \left[ \begin{array}{c} 0.612 \\ -0.612 \\ -0.5 \end{array}\right]$$

The stresses on this fault are

$\uline{t}= [2450,-1840,-2500]$ psi.
$S_n = 3870$ psi
$\tau_d = -649$ psi
$\tau_s = -433$ psi
$rake$ = 56.3 $^{\circ} \: \: \blacksquare$

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
The tensor of principal stresses is

$$\uuline{S}{}_P = \left[ \begin{array}{ccc} 2400 & 0 & 0 \\ 0 & 1200 & 0 \\ 0 & 0 & 1000 \end{array}\right] \text{ psi}$$

and the pore pressure is $P_p = 440$ psi.

Using Table 5.3 and considering that $S_3 = S_v$ and the azimuth of $S_{Hmax}$ is 150$^{\circ }$, the angles of the principal stress coordinate result $\alpha = 150^{\circ}$, $\beta = 0^{\circ}$, and $\gamma = 0^{\circ}$. The change of coordinate system matrix results

$$R_{PG} = \left[ \begin{array}{ccc} -0.866 & 0.5 & 0 \\ -0.5 & -0.866 & 0 \\ 0 & 0 & 1 \end{array}\right]$$

and the total stress in the geographical coordinate system results

$$\uuline{S}{}_G = \left[ \begin{array}{ccc} 2100 & -520 & 0 \\ -520 & 1500 & 0 \\ 0 & 0 & 1000 \end{array}\right] \text{ psi}$$

Given the orientation of the fault, the vector normal to the fault is

$$\uline{n}{}_n = \left[ \begin{array}{c} -0.814 \\ -0.470 \\ -0.342 \end{array}\right]$$

Finally, the stresses on the fault are

$\uline{t}= [-1465,-281,-342]$ psi.
$S_n = 1441$ psi, $\sigma_n = 1001$ psi
$\tau_d = 160$ psi, $\tau_s = 488$ psi, $\tau = 514$ psi
$rake$ = 18.21$^{\circ }$

The ratio of shear to normal effective stress is $\tau / \sigma_n = 0.51 \: \: \blacksquare$

Example: make 3D Mohr-Circle filled with color corresponding to $\tau /\sigma _n$ value, stereo net, and fractures in 3D.