Question

The active earth pressure of a soil is proportional to (where \( \phi \) is the angle of internal friction of soil)

• A. \( \tan (45^\circ - \phi) \)
• B. \( \tan^2 (45^\circ + \phi / 2) \)
• C. \( \tan^2 (45^\circ - \phi / 2) \)
• D. \( \tan (45^\circ + \phi) \)

Hint:

The coefficient of earth pressure ($K$) is a critical parameter in geotechnical engineering, particularly for the design of retaining structures. It represents the ratio of effective horizontal stress to effective vertical stress. $K_a$ (active), $K_p$ (passive), and $K_0$ (at-rest) are key values that describe the soil's lateral pressure behavior under different deformation conditions.

Answer 1

The Correct Option is C

Step 1: Understand active earth pressure and its coefficient.
Active earth pressure is the lateral pressure exerted by a soil mass on a retaining structure when the soil tends to move away from the structure, reaching a state of plastic equilibrium (active state). This pressure is quantified by the coefficient of active earth pressure, \( K_a \).
Step 2: Recall the formula for the coefficient of active earth pressure (\( K_a \)).
For a cohesionless soil, according to Rankine's theory, the coefficient of active earth pressure \( K_a \) is given by:
$$K_a = \frac{1 - \sin \phi}{1 + \sin \phi}$$ where \( \phi \) is the angle of internal friction of the soil.
Step 3: Transform the expression for \( K_a \) using trigonometric identities.
We can express \( K_a \) in terms of tangent functions using trigonometric identities.
Recall the identity \( \tan \left( 45^\circ - \frac{\theta}{2} \right) = \frac{1 - \tan (\theta/2)}{1 + \tan (\theta/2)} \).
We also know that \( \sin \phi = \frac{2 \tan (\phi/2)}{1 + \tan^2 (\phi/2)} \).
Substitute this into the expression for \( K_a \):
$$K_a = \frac{1 - \frac{2 \tan (\phi/2)}{1 + \tan^2 (\phi/2)}}{1 + \frac{2 \tan (\phi/2)}{1 + \tan^2 (\phi/2)}} = \frac{(1 + \tan^2 (\phi/2)) - 2 \tan (\phi/2)}{(1 + \tan^2 (\phi/2)) + 2 \tan (\phi/2)}$$ $$K_a = \frac{1 - 2 \tan (\phi/2) + \tan^2 (\phi/2)}{1 + 2 \tan (\phi/2) + \tan^2 (\phi/2)} = \frac{(1 - \tan (\phi/2))^2}{(1 + \tan (\phi/2))^2}$$
$$K_a = \left( \frac{1 - \tan (\phi/2)}{1 + \tan (\phi/2)} \right)^2$$ By comparison with the tangent identity, we get:
$$K_a = \tan^2 \left( 45^\circ - \frac{\phi}{2} \right)$$
Step 4: Conclude the proportionality.
The active earth pressure \( P_a \) on a retaining wall is directly proportional to \( K_a \). For example, for a dry cohesionless soil with a horizontal backfill, the pressure is given by \( p_a = \gamma z K_a \), and the total active force is \( P_a = \frac{1}{2} \gamma H^2 K_a \). Thus, the active earth pressure is proportional to \( \tan^2 (45^\circ - \phi / 2) \).
Step 5: Select the correct option.
Based on the derivation, the active earth pressure of a soil is proportional to \( \tan^2 (45^\circ - \phi / 2) \).
$$\boxed{\tan^2 (45^\circ - \phi / 2)}$$