Question

A circular pile of 600 mm diameter and 6 m length is embedded in a saturated clayey soil. The undrained cohesion of the soil is \(c_u = 80\) kPa and its unit weight is \(\gamma = 19.20\) kN/m³. The adhesion factor is \(\alpha = 0.54\). If the diameter of the pile is doubled to 1200 mm, keeping the length constant at 6 m, the ratio of the pile capacity of the 1200 mm diameter pile to that of the 600 mm diameter pile is ______ (rounded off to two decimal places).

Hint:

Pile capacity = shaft friction (proportional to \(D\)) + end bearing (proportional to \(D^2\)); doubling \(D\) grows the base term faster than the shaft term, so the ratio exceeds 2.

Answer 1

Correct Answer: 2.59

Step 1: Identify the capacity components of a single pile.
A pile driven into clay carries load through two mechanisms: skin friction along the shaft and end bearing at the tip. For a circular pile of diameter \(D\) and length \(L\) in clay, using the alpha (adhesion) method, the ultimate load capacity is
\[ Q_u = \alpha c_u (\pi D L) + N_c c_u \left(\frac{\pi D^2}{4}\right) \]
Here \(\alpha\) is the adhesion factor, \(c_u\) is the undrained cohesion, and \(N_c = 9\) is the standard bearing capacity factor used for a deep pile tip in clay.

Step 2: Write the capacity as a function of diameter.
Since \(\alpha\), \(c_u\), and \(L\) stay the same for both piles, only \(D\) changes. Factor \(Q_u\) as
\[ Q_u(D) = \pi c_u D \left(\alpha L + \frac{9D}{4}\right) \]
For the 600 mm pile, \(D_1 = 0.6\) m, and for the 1200 mm pile, \(D_2 = 1.2\) m.

Step 3: Compute the bracket term for the 600 mm pile.
Skin friction part: \(\alpha L = 0.54 \times 6 = 3.24\).
End bearing part: \(\frac{9 D_1}{4} = \frac{9 \times 0.6}{4} = 1.35\).
Bracket value \(=3.24+1.35=4.59\).

Step 4: Compute the bracket term for the 1200 mm pile.
End bearing part at the larger diameter: \(\frac{9 D_2}{4} = \frac{9 \times 1.2}{4} = 2.70\).
Bracket value \(=3.24+2.70=5.94\).

Step 5: Take the ratio of the two capacities.
\[ \frac{Q_u(D_2)}{Q_u(D_1)} = \frac{D_2}{D_1}\times\frac{3.24+2.70}{3.24+1.35} = 2\times\frac{5.94}{4.59} = \frac{11.88}{4.59} \]
\[ = 2.588 \]

Final Answer:
Doubling the diameter raises both the shaft-friction area (proportional to \(D\)) and the tip area (proportional to \(D^2\)), but the tip area grows faster, so the capacity ratio is a bit more than 2. Rounded to two decimal places,
\[ \boxed{\dfrac{Q_u(1200)}{Q_u(600)} \approx 2.59} \]