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: List the known values.
$c_u = 80$ kPa, $\alpha = 0.54$, $L = 6$ m, $N_c = 9$ (standard deep-pile factor for clay), pile 1 diameter $D_1=0.6$ m, pile 2 diameter $D_2=1.2$ m.

Step 2: Find the shaft (skin friction) resistance of each pile.
Shaft resistance $= \alpha c_u \times (\pi D L)$.
For $D_1=0.6$ m: shaft area $= \pi \times 0.6 \times 6 = 11.310$ m$^2$, so $Q_{s1} = 0.54 \times 80 \times 11.310 = 488.6$ kN.
For $D_2=1.2$ m: shaft area $= \pi \times 1.2 \times 6 = 22.619$ m$^2$, so $Q_{s2} = 0.54 \times 80 \times 22.619 = 977.2$ kN.

Step 3: Find the base (end bearing) resistance of each pile.
Base resistance $= N_c c_u \times \left(\frac{\pi D^2}{4}\right)$.
For $D_1=0.6$ m: base area $=\frac{\pi (0.6)^2}{4}=0.2827$ m$^2$, so $Q_{p1}=9\times80\times0.2827=203.6$ kN.
For $D_2=1.2$ m: base area $=\frac{\pi (1.2)^2}{4}=1.1310$ m$^2$, so $Q_{p2}=9\times80\times1.1310=814.3$ kN.

Step 4: Add the shaft and base resistance to get the total capacity of each pile.
$Q_{u1} = Q_{s1}+Q_{p1} = 488.6+203.6=692.2$ kN (600 mm pile).
$Q_{u2} = Q_{s2}+Q_{p2} = 977.2+814.3=1791.5$ kN (1200 mm pile).

Step 5: Divide the two capacities to get the required ratio.
\[ \frac{Q_{u2}}{Q_{u1}} = \frac{1791.5}{692.2} = 2.588 \]
This full numeric route confirms the same value reached by scaling the terms algebraically: the base area grows with $D^2$ while the shaft area grows only with $D$, so doubling $D$ more than doubles the total capacity. \[ \boxed{2.59} \]