Question

A single pile with 450 mm diameter has been driven into a homogeneous clay layer, which has an undrained cohesion (\(c_u\)) of 20 kPa and unit weight of 18 kN/m³. The ground water table is found to be at the surface of the clay layer. The adhesion factor (\(\alpha\)) of the soil is 0.95 and bearing capacity factor (\(N_c\)) is 9. The pile is supporting a column load of 144 kN with a factor of safety of 3.0 against ultimate axial pile capacity in compression. The required embedment depth of the pile (in m) is ......... (rounded off to the nearest integer).

Hint:

When calculating the embedment depth of a pile, use the factors of safety and adhesion factor to calculate the ultimate load and then divide by the factor of safety to get the required depth.

Answer 1

The ultimate load capacity of the pile is given by: \[ Q_{{up}} = 9 c' A_s + \alpha C_u A_s \] Substituting the values: \[ Q_{{up}} = 9 \times 20 \times \left( \frac{\pi \times 0.45^2}{4} \right) + 0.95 \times 20 \times \left( \pi \times 0.45 \times L \right) \] Now, the factor of safety is given by: \[ Q_{{safe}} = \frac{Q_{{up}}}{FOS} = \frac{144}{3} = 48 \, {kN}. \] By equating the ultimate load capacity and the safe load: \[ 432 = 28.627 + 26.861 \quad \Rightarrow L = 15.01 \, {m} \quad {or} \quad L \approx 15 \, {m}. \] Thus, the required embedment depth is \( \boxed{15} \, {m} \) (rounded to the nearest integer).