Question

In an oblique slip fault with an attitude 000°, 30° E, the net slip vector has a length of 20 m and a rake of 30° S on the fault plane. The displacement of a horizontal bed along the fault trace in a plane perpendicular to the strike of the fault is ________ m. (Answer in integer.)

Hint:

In oblique faults: Total slip = \(\sqrt{(strike-slip)^2 + (dip-slip)^2}\); Dip-slip = \(S \sin(\text{rake})\).

Answer 1

Correct Answer: 10

Step 1: Understanding the geometry.
The net slip vector (20 m) has a rake of 30° on the fault plane. The rake measures the angle between the slip direction and the horizontal line (strike line) on the fault plane.
Step 2: Components of motion.
The horizontal (strike-slip) component \( = 20 \cos(30°) = 17.32 \, \text{m} \).
The vertical (dip-slip) component \( = 20 \sin(30°) = 10 \, \text{m} \).
Step 3: Interpretation.
The displacement of a horizontal bed along a section perpendicular to the strike corresponds to the dip-slip component. Step 4: Conclusion.
Hence, displacement = 10 m.