Rankine's Formula: Minimum Foundation Depth Calculation

This solution explains how to determine the minimum depth required for a foundation using Rankine's formula. It involves understanding the relationship between the soil's properties, its safe bearing capacity, and the depth needed to ensure stability.

Soil Properties and Safe Bearing Capacity

The problem provides the following essential information about the soil and the foundation requirements:

• Safe bearing capacity of the soil, q_a = 90 kN/m²
• Unit weight of the soil, γ = 18 kN/m³
• Sine of the angle of repose (friction angle), sin φ = 1/3

Understanding Rankine's Formula for Minimum Depth

Rankine's theory provides a formula to estimate the minimum depth of foundation (D_f) needed. This depth ensures that the soil pressure at that depth does not exceed the safe bearing capacity. The formula is:

$$D_f = \frac{q_a}{\gamma} \left( \frac{1 - \sin \phi}{1 + \sin \phi} \right)^2$$

Key terms in the formula:

• D_f represents the minimum depth of the foundation.
• q_a is the safe bearing capacity of the soil.
• γ is the unit weight of the soil.
• φ is the angle of repose, also known as the internal friction angle of the soil.

Step-by-Step Calculation

Let's calculate the minimum depth using the provided values:

1. Calculate the factor related to the angle of repose:
   We need to compute the value of the expression $\left( \frac{1 - \sin \phi}{1 + \sin \phi} \right)$.
   Given that $\sin \phi = \frac{1}{3}$:
   $$ \frac{1 - \sin \phi}{1 + \sin \phi} = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} $$
   To simplify the fraction:
   $$ = \frac{\frac{3}{3} - \frac{1}{3}}{\frac{3}{3} + \frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} $$
   $$ = \frac{2}{3} \times \frac{3}{4} = \frac{2}{4} = \frac{1}{2} $$
   Now, we square this result:
   $$ \left( \frac{1 - \sin \phi}{1 + \sin \phi} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4} $$
2. Calculate the minimum depth (D_f):
   Substitute the calculated factor and the given soil parameters into Rankine's formula:
   $$ D_f = \frac{q_a}{\gamma} \times \left( \frac{1 - \sin \phi}{1 + \sin \phi} \right)^2 $$
   Plug in the values for q_a and γ:
   $$ D_f = \frac{90 \text{ kN/m}^2}{18 \text{ kN/m}^3} \times \frac{1}{4} $$
   First, calculate the ratio $\frac{q_a}{\gamma}$:
   $$ \frac{90}{18} \text{ m} = 5 \text{ m} $$
   Now, multiply by the squared factor:
   $$ D_f = 5 \text{ m} \times \frac{1}{4} $$
   $$ D_f = \frac{5}{4} \text{ m} $$
   $$ D_f = 1.25 \text{ m} $$

Final Result and Conclusion

Based on the calculations using Rankine's formula, the minimum depth required for the foundation is 1.25 meters. This result aligns with the first option presented in the question.