Question

On a normal fault plane dipping 60° towards east, the measured heave and throw are 5 m and 12 m, respectively. If the strike-slip component of the fault is 13 m, the magnitude of true displacement of the fault is ________m.

Hint:

When calculating the true displacement of a fault, use the Pythagorean theorem to combine the horizontal components (heave and strike-slip) and the vertical component (throw).

Answer 1

Step 1: Understanding the components of displacement.
In fault mechanics, the total displacement is the vector sum of the heave (horizontal component), throw (vertical component), and strike-slip (horizontal shear component). The true displacement (D) is calculated using the Pythagorean theorem, which accounts for all components of displacement. The formula for the magnitude of true displacement is: \[ D = \sqrt{({heave})^2 + ({throw})^2 + ({strike-slip})^2} \] Step 2: Substituting the given values.
Given:
Heave = 5 m
Throw = 12 m
Strike-slip = 13 m
Now, calculate the true displacement: \[ D = \sqrt{(5)^2 + (12)^2 + (13)^2} \] \[ D = \sqrt{25 + 144 + 169} = \sqrt{338} \] \[ D \approx 18.4 \, {m} \] Therefore, the magnitude of the true displacement of the fault is approximately \(18.4\) m.