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.

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

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) represents the possible state of stresses that result as a combination of and . The shear stress and normal effective stress of any plane in between the planes of and and colinear with can be found through the angle measured from .
- The green circle (center C) represents the possible state of stresses that result as a combination of and . The shear stress and normal effective stress of any plane in between the planes of and and colinear with can be found through the angle measured from
- The red circle (center C) represents the possible state of stresses that result as a combination of and . The shear stress and normal effective stress of any plane in between the planes of and and colinear with can be found through the angle measured from

For this example (normal faulting, azimuth E-W), an ideal fault would occur with a strike E-W and dip 60 (assuming ). This is the orientation of a plane with maximum .

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, dip = 60.
- State of stress: 23 MPa (principal), 20 MPa, 13.8 MPa (azimuth: 090), and 10 MPa.

SOLUTION

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 = N60E, dip: 90.
- State of stress: 30 MPa (principal), 45 MPa, 25 MPa (azimuth: N30E), and 15 MPa.

SOLUTION

The effective stresses are: 15 MPa, 30 MPa, 10 MPa. Based on the Mohr circle of with and trigonometry:

This subsection describes the procedure to calculate stresses on an arbitrary plane given its orientation respect to the geographical coordinate system and the in-situ stress tensor of principal stresses (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: for , for , and for (Fig. 5.21). The stress tensor is termed in this coordinate system.
- The geographical coordinate system has bases pointing in North direction, pointing in East direction, and pointing down in direction of increasing depth (Fig. 5.21). We will refer this basis as the “NED” basis. The stress tensor is termed in this coordinate system.

The second step involves constructing a change of basis matrix 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 , , and (Fig. 5.22). Table 5.3 summarizes the meaning of , , and for cases in which vertical stress is a principal stress.

Normal faulting | Strike slip | Reverse faulting | |

Azimuth of | Azimuth of | Azimuth of | |

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

With the matrix , we can calculate the stress tensor as a function of ,

(5.8) |

and therefore:

where the superscript stands for “transpose”.

PROBLEM 5.5: Calculate
in a normal faulting stress regime case ( MPa, MPa, MPa) with azimuth of N-S. is a principal stress.

SOLUTION

The principal stress tensor is

Using Table 5.3 and considering that , the angles of the principal stress coordinate result , , and . The change of coordinate system matrix results

Finally, using equation 5.9

PROBLEM 5.6: Calculate
in a strike-slip faulting stress regime case ( MPa, MPa, MPa) with azimuth of N-S. is a principal stress.

SOLUTION

The principal stress tensor is

Using Table 5.3 and considering that and , the angles of the principal stress coordinate result , , and . The change of coordinate system matrix results

Finally, using equation 5.9

PROBLEM 5.7: Calculate
in a reverse faulting stress regime case ( MPa, MPa, MPa) with azimuth of E-W. is a principal stress.

SOLUTION

The principal stress tensor is

Using Table 5.3 and considering that and , the angles of the principal stress coordinate result , , and . The change of coordinate system matrix results

Finally, using equation 5.9

PROBLEM 5.8: Calculate
in a strike-slip faulting stress regime case ( MPa, MPa, MPa) with azimuth of being 135. is a principal stress.

SOLUTION

The principal stress tensor is

Using Table 5.3 and considering that and , the angles of the principal stress coordinate result , , and . The change of coordinate system matrix results

Finally, using equation 5.9

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

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 (note that is not necessarily aligned with , or ) and is calculated according to:

(5.10) |

The total normal stress on the plane of the fault is (aligned with ):

(5.11) |

The effective normal stress on the fault plane is . The shear stresses on the plane of the fault is aligned with and are:

(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 and absolute shear can also be calculated with the following equations:

(5.13) |

(5.14) |

The is the angle of the shear stress with respect to (horizontal line) and quantifies the direction of expected fault movement in the fault plane.

(5.15) |

PROBLEM 5.9: Calculate , , , , and for a fault with strike 000 and dip 60E in a place with normal faulting stress regime ( MPa,
MPa,
MPa) with azimuth of equal to 90. is a principal stress.

SOLUTION

The tensor of principal stresses is

Using Table 5.3 and considering that and the azimuth of is 90, the angles of the principal stress coordinate result , , and . The change of coordinate system matrix results

and the total stress in the geographical coordinate system results

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

Finally, the stresses on the fault are

MPa.

MPa

MPa

MPa

= 90

PROBLEM 5.10: Calculate , , , , and for a fault with strike 060 and dip 90 in a place with strike-slip stress regime ( MPa,
MPa,
MPa) with azimuth of equal to 120. is a principal stress.

SOLUTION

The tensor of principal stresses is

Using Table 5.3 and considering that and the azimuth of is 120, the angles of the principal stress coordinate result , , and . The change of coordinate system matrix results

and the total stress in the geographical coordinate system results

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

Finally, the stresses on the fault are

MPa.

MPa

MPa

MPa

= 0

PROBLEM 5.11: Calculate , , , , and for conjugate faults with strike 045 and 225 both with dip 60 in a place with normal faulting stress regime (
psi,
psi,
psi) with azimuth of equal to 90. is a principal stress.

SOLUTION

The tensor of principal stresses is

Using Table 5.3 and considering that and the azimuth of is 90, the angles of the principal stress coordinate result , , and . The change of coordinate system matrix results

and the total stress tensor in the geographical coordinate system results

Let us consider the first fault with strike of 045 and dip of 60, the vector normal to the faults is

The stresses on this fault are

psi.

psi

psi

psi

= 56.3

Let us consider the fault with strike of 225 and dip of 60, the vector normal to the faults is

The stresses on this fault are

psi.

psi

psi

psi

= 56.3

PROBLEM 5.12: Calculate , , , , , and
for a fault with strike 120 and 70 dip in a place with reverse faulting stress regime (
psi,
psi,
psi) with azimuth of equal to 150 and pore pressure psi. is a principal stress.

SOLUTION

The tensor of principal stresses is

and the pore pressure is psi.

Using Table 5.3 and considering that and the azimuth of is 150, the angles of the principal stress coordinate result , , and . The change of coordinate system matrix results

and the total stress in the geographical coordinate system results

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

Finally, the stresses on the fault are

psi.

psi,
psi

psi,
psi,
psi

= 18.21

The ratio of shear to normal effective stress is

Example: make 3D Mohr-Circle filled with color corresponding to value, stereo net, and fractures in 3D.