Question

The state of stress at a point on a fault plane is \(\sigma_{y y}=150 \mathrm{MPa}, \sigma_{x x}=200 \mathrm{MPa}\), and \(\sigma_{x y}=0\) ( \(y\) is depth and the \(x\) axis points westward). What are the normal stress and the tangential stress on the fault plane if the fault strikes \(\mathrm{N}-\mathrm{S}\) and dips \(35^{\circ}\) to the west?

Short answer

Normal stress: 166.35 MPa; Tangential stress: 23.49 MPa.

Step-by-step solution

Understanding the Problem

We need to find both the normal and tangential stresses on a fault plane given its orientation and initial stress conditions.

Define Components of Stress

The initial state of stress is given by \[ \sigma_{yy} = 150 \, \text{MPa}, \quad \sigma_{xx} = 200 \, \text{MPa}, \quad \sigma_{xy} = 0 \]These are principal stresses along the vertical and horizontal directions.

Setup Coordinate Alignment

Since the fault strikes north-south (N-S) and dips at an angle of 35 degrees to the west, we align our coordinates with the fault plane to analyze stresses. The fault orientation means the x-axis is parallel to the strike and the y-axis is normal to the strike.

Calculate Normal Stress

The normal stress \(\sigma_n\) on the fault plane is given by the formula \[ \sigma_n = \sigma_{xx}\sin^2(\theta) + \sigma_{yy}\cos^2(\theta) \]where \(\theta\) is the dip angle (35 degrees). Substituting the given values:\[ \sigma_n = 200 \times \sin^2(35^\circ) + 150 \times \cos^2(35^\circ) \]

Solve Normal Stress Equation

Calculating each term:\[ \sin^2(35^\circ) \approx 0.327, \quad \cos^2(35^\circ) \approx 0.673 \]Substitute these into the equation:\[ \sigma_n = 200 \times 0.327 + 150 \times 0.673 = 65.4 + 100.95 = 166.35 \, \text{MPa} \]

Calculate Tangential Stress

The tangential stress \(\tau\) on the fault plane is determined by the formula:\[ \tau = (\sigma_{xx} - \sigma_{yy})\sin(\theta)\cos(\theta) \]Substitute the given values:\[ \tau = (200 - 150) \times \sin(35^\circ) \times \cos(35^\circ) \]

Solve Tangential Stress Equation

Using \(\sin(35^\circ) \approx 0.574\) and \(\cos(35^\circ) \approx 0.819\), we find:\[ \tau = 50 \times 0.574 \times 0.819 = 23.49 \, \text{MPa} \]

Conclusion: Normal and Tangential Stresses

The calculated normal stress is approximately 166.35 MPa, and the tangential stress is approximately 23.49 MPa.

Key concepts

Normal Stress Calculation

Understanding how to calculate normal stress on a fault plane is crucial in geomechanics. The normal stress, denoted as \( \sigma_n \), applies perpendicularly to the fault plane.

The formula used to calculate normal stress is:

• \( \sigma_n = \sigma_{xx}\sin^2(\theta) + \sigma_{yy}\cos^2(\theta) \)

Where:

• \( \sigma_{xx} \) is the horizontal principal stress.
• \( \sigma_{yy} \) is the vertical principal stress.
• \( \theta \) is the dip angle of the fault plane.

For a given system with \( \sigma_{xx} = 200 \, \text{MPa} \) and \( \sigma_{yy} = 150 \, \text{MPa} \), and a dip angle 35°, you substitute these into the normal stress formula, calculate \( \sin^2(35°) \) and \( \cos^2(35°) \), and proceed to find the normal stress.
This approach ensures the stress perpendicular to the fault plane is accurately determined. Knowing this helps in assessing how faults may slip under certain conditions.

Tangential Stress Calculation

Tangential stress is associated with how forces are aligned parallel to the fault plane, impacting shear movement. It's symbolized as \( \tau \).
The equation for tangential stress is:

• \( \tau = (\sigma_{xx} - \sigma_{yy})\sin(\theta)\cos(\theta) \)

Here, the parameters have these interpretations:

• \( \sigma_{xx} \) and \( \sigma_{yy} \) are the principal stresses, as earlier described.
• \( \theta \) is the angle of the dip of the fault plane.

By substituting \( \sigma_{xx} = 200 \, \text{MPa} \) and \( \sigma_{yy} = 150 \, \text{MPa} \), along with \( \sin(35°) \) and \( \cos(35°) \) values, you evaluate the tangential stress.

This computation is essential, as it provides insight into the shear forces acting on a fault, which play a critical role in fault slip potential.

Fault Plane Orientation

The fault plane orientation is defined by two attributes: strike and dip.

• Strike describes the direction of the intersection line between the fault plane and the horizontal plane.
• Dip is the angle between the fault plane and the horizontal plane.

In this scenario, the fault plane strikes north-south (N-S) and dips 35° to the west. This means:

• The orientation of the fault is longitudinally north-south.
• It inclines downward towards the west.

Correctly orienting the fault plane is vital in analyzing stress components, as it governs how you align the coordinate system to apply stress transformation equations.
Such detailed orientation ensures a precise analysis of stress distribution and behaviour.

Principal Stresses

Principal stresses are crucial in stress analysis as they represent the normal stresses acting on particular planes where shear stress is zero. Understanding this helps in simplifying complex stress conditions.
For the given fault plane problem, we identified:

• \( \sigma_{yy} = 150 \, \text{MPa} \) represents the vertical stress.
• \( \sigma_{xx} = 200 \, \text{MPa} \) represents the horizontal stress.
• \( \sigma_{xy} = 0 \) indicating no shear stress initially at that plane alignment.

Interpreting principal stresses allows you to understand how each influences the potential slip along the fault plane. As a foundational concept, principal stresses help predict structural behavior and failure modes.
Correctly identifying and using principal stresses ensures comprehensive stress analysis on geological structures.

Geomechanics

Geomechanics involves studying how geological materials deform under various stress conditions. This study covers the behavior of both solid rocks and faults, such as the one in the example with normal and tangential stresses.

• Understanding geomechanics helps in predicting fault movements.
• It allows for the assessment of stability in the earth's subsurface structures.

In the context of the problem:

• It explains how stresses on a fault plane are derived from larger-scale stress fields in the crust.

Incorporating these calculations into geomechanics enables geologists and engineers to design more robust infrastructures and manage geological hazards effectively.
Mastering geomechanics is essential for those involved in fields such as earthquake engineering, oil extraction, and any area where understanding earth stresses is crucial.