Question

The intensity of active earth pressure at a depth of 10 metres in dry cohesionless sand with an angle of internal friction of \( 30^\circ \) and with a weight of \( 1.8 \, \text{t/m}^3 \) is

• A. \( 4 \, \text{t/m}^2 \)
• B. \( 5 \, \text{t/m}^2 \)
• C. \( 6 \, \text{t/m}^2 \)
• D. \( 7 \, \text{t/m}^2 \)

Hint:

The intensity of active earth pressure increases linearly with depth in a homogeneous soil. It is calculated by multiplying the effective vertical stress ($\gamma z$) by the coefficient of active earth pressure ($K_a$). For dry cohesionless soils, cohesion ($c$) is zero, simplifying the pressure calculation.

Answer 1

The Correct Option is C

Step 1: Identify the given data.
Depth \( z = 10 \, \text{m} \)
Angle of internal friction \( \phi = 30^\circ \)
Unit weight of soil \( \gamma = 1.8 \, \text{t/m}^3 \)
The soil is dry and cohesionless, which means cohesion \( c = 0 \).
Step 2: Calculate the coefficient of active earth pressure (\( K_a \)).
For a cohesionless soil, the coefficient of active earth pressure is given by:
$$K_a = \tan^2 \left( 45^\circ - \frac{\phi}{2} \right)$$
Substitute \( \phi = 30^\circ \):
$$K_a = \tan^2 \left( 45^\circ - \frac{30^\circ}{2} \right)$$
$$K_a = \tan^2 (45^\circ - 15^\circ)$$
$$K_a = \tan^2 (30^\circ)$$
We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).
$$K_a = \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3}$$
Step 3: Calculate the intensity of active earth pressure (\( p_a \)).
For a dry cohesionless soil, the intensity of active earth pressure at a depth \( z \) is given by:
$$p_a = \gamma z K_a$$
Substitute the values:
$$p_a = (1.8 \, \text{t/m}^3) \times (10 \, \text{m}) \times \left( \frac{1}{3} \right)$$
$$p_a = 18 \, \text{t/m}^2 \times \frac{1}{3}$$
$$p_a = 6 \, \text{t/m}^2$$
Step 4: Select the correct option.
Based on the calculation, the intensity of active earth pressure is \( 6 \, \text{t/m}^2 \).
$$\boxed{6 \, \text{t/m}^2}$$