Question

A circular pile of diameter 0.6 m and length 8 m was constructed in a cohesive soil stratum having the following properties: bulk unit weight = 19 kN/m$^3$, angle of internal friction = 0$^\circ$ and cohesion = 25 kPa. The allowable load the pile can carry with a factor of safety of 3 is  kN (round off to one decimal place).}

Hint:

The ultimate bearing capacity for cohesive soils is based on the cohesion and the area of the pile shaft. Always apply the factor of safety to get the allowable load.

Answer 1

Correct Answer: 21.2

- Diameter of pile, \( D = 0.6 \, \text{m} \)
- Length of pile, \( L = 8 \, \text{m} \)
- Bulk unit weight, \( \gamma = 19 \, \text{kN/m}^3 \)
- Cohesion, \( c = 25 \, \text{kPa} \)
- Adhesion factor, \( \alpha = 1.0 \)
- Bearing capacity factor, \( N_c = 9.0 \)
- Factor of safety, \( F_s = 3 \)

Step 1: Calculate the ultimate bearing capacity \( Q_u \).
The ultimate bearing capacity \( Q_u \) for a pile in cohesive soil is given by the formula:
\[ Q_u = N_c \cdot c \cdot A \] Where:
- \( N_c = 9.0 \) (bearing capacity factor)
- \( c = 25 \, \text{kPa} \) (cohesion)
- \( A = \text{Area of pile tip} = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.6}{2}\right)^2 = 0.283 \, \text{m}^2 \)
Thus,
\[ Q_u = 9.0 \cdot 25 \, \text{kPa} \cdot 0.283 \, \text{m}^2 = 63.675 \, \text{kN} \]

Step 2: Calculate the allowable load.
The allowable load \( Q_a \) is given by:
\[ Q_a = \frac{Q_u}{F_s} \] Where \( F_s = 3 \) (factor of safety).
Substituting the values:
\[ Q_a = \frac{63.675 \, \text{kN}}{3} = 21.225 \, \text{kN} \]

Step 3: Conclusion.
Thus, the allowable load the pile can carry is \( \boxed{21.2 \, \text{kN}} \).