Question

As per the Rankine's earth pressure theory, which of the following statements is/are FALSE?

• A. For the active earth pressure, the inclination of failure plane is \((45^{\circ} + \phi/2)\) with respect to the major principal plane.
• B. For the active earth pressure, the inclination of failure plane is \((45^{\circ} - \phi/2)\) with respect to the major principal plane.
• C. For the passive earth pressure, the inclination of failure plane is \((45^{\circ} + \phi/2)\) with respect to the major principal plane.
• D. For the passive earth pressure, the inclination of failure plane is \((45^{\circ} - \phi/2)\) with respect to the major principal plane.

Hint:

The failure plane always makes \(45^{\circ}+\phi/2\) with the major principal plane in Rankine's theory, for both the active and passive cases.

Answer 1

The Correct Option is B, D

Step 1: Understanding the Concept.
Rankine's earth pressure theory treats the soil mass as being on the verge of failure everywhere (a state of plastic equilibrium), and uses the Mohr-Coulomb failure criterion to find the orientation of the failure planes. This orientation is a general result of the Mohr circle construction and does not change in form between the active and passive cases; only which stress is major and which is minor changes.

Step 2: Key Formula or Approach.
From the Mohr-Coulomb failure envelope, the failure plane always makes an angle of \(\left(45^{\circ} + \frac{\phi}{2}\right)\) with the plane on which the major principal stress acts, where \(\phi\) is the angle of internal friction. This holds true in both the active and the passive case, because it comes from the geometry of the Mohr circle touching the failure envelope, not from which case we are in.

Step 3: Detailed Explanation.
In the active case, the vertical overburden stress is the major principal stress, and it acts on the horizontal plane, so the major principal plane here is horizontal. The failure plane is inclined at \(\left(45^{\circ}+\frac{\phi}{2}\right)\) to this horizontal plane.
(A) states exactly this for the active case, so (A) is TRUE.
(B) claims \(\left(45^{\circ}-\frac{\phi}{2}\right)\) for the active case, which contradicts the correct result, so (B) is FALSE.
In the passive case, the horizontal pressure becomes the major principal stress (it now exceeds the vertical overburden stress), acting on the vertical plane, so the major principal plane here is vertical. The failure plane is still inclined at \(\left(45^{\circ}+\frac{\phi}{2}\right)\) to this vertical major principal plane, by the same Mohr-Coulomb geometry.
(C) states this correctly for the passive case, so (C) is TRUE.
(D) claims \(\left(45^{\circ}-\frac{\phi}{2}\right)\) for the passive case, which is wrong for the same reason as (B), so (D) is FALSE.
A common mix-up is remembering that active failure planes look steeper and passive ones look flatter when measured from the horizontal ground surface, and wrongly carrying that difference into the angle measured from the major principal plane itself. But because the major principal plane rotates by 90 degrees between the two cases (horizontal for active, vertical for passive), the angle to the major principal plane stays \(\left(45^{\circ}+\frac{\phi}{2}\right)\) in both cases.

Step 4: Final Answer.
Statements (B) and (D) are false.
\[ \boxed{\text{Options (B) and (D)}} \]