Question

The intensity of active earth pressure on retaining wall at a depth of 4 m with backfill having an angle of shearing resistance of 30º and unit weight of 18 kN/m$^3$ is

• A. 12 kN/m$^2$
• B. 24 kN/m$^2$
• C. 48 kN/m$^2$
• D. 60 kN/m$^2$

Hint:

Memorize the Rankine coefficients for the common case of $\phi = 30^\circ$:
- Active coefficient: $K_a = \frac{1-\sin(30)}{1+\sin(30)} = \frac{1-0.5}{1+0.5} = 1/3$.
- Passive coefficient: $K_p = \frac{1+\sin(30)}{1-\sin(30)} = \frac{1+0.5}{1-0.5} = 3$.
Note that $K_p = 1/K_a$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the active earth pressure intensity ($p_a$) at a specific depth behind a retaining wall, given the soil properties.

Step 2: Key Formula or Approach:
According to Rankine's theory for active earth pressure in a cohesionless soil with a horizontal backfill, the pressure at a depth $h$ is given by:
\[ p_a = K_a \gamma h \] where:
$\gamma$ = unit weight of the soil
$h$ = depth
$K_a$ = coefficient of active earth pressure
The coefficient $K_a$ is calculated as:
\[ K_a = \frac{1 - \sin(\phi)}{1 + \sin(\phi)} \] where $\phi$ is the angle of shearing resistance (or angle of internal friction).

Step 3: Detailed Explanation:
First, calculate the coefficient of active earth pressure, $K_a$.
- Angle of shearing resistance ($\phi$) = $30^\circ$
- We know that $\sin(30^\circ) = 0.5$
\[ K_a = \frac{1 - \sin(30^\circ)}{1 + \sin(30^\circ)} = \frac{1 - 0.5}{1 + 0.5} = \frac{0.5}{1.5} = \frac{1}{3} \] Now, calculate the active earth pressure intensity, $p_a$.
- Unit weight ($\gamma$) = 18 kN/m$^3$
- Depth ($h$) = 4 m
\[ p_a = K_a \gamma h = \left(\frac{1}{3}\right) \times (18 \text{ kN/m}^3) \times (4 \text{ m}) \] \[ p_a = 6 \times 4 \text{ kN/m}^2 \] \[ p_a = 24 \text{ kN/m}^2 \]

Step 4: Final Answer:
The intensity of active earth pressure at a depth of 4 m is 24 kN/m$^2$.