A circular pile of 600 mm diameter and 6 m length is embedded in a saturated clayey soil. Undrained cohesion of the soil is 80 kPa. Unit weight of the soil is 19.20 kN/m³. The adhesion factor is 0.54. If the diameter of the pile is doubled to 1200 mm (keeping the length constant), the ratio of pile capacity of 1200 mm diameter pile to that of 600 mm diameter pile is _______(rounded off to two decimal places).

The correct option is 2.

Let us understand the step-by-step reasoning and calculations leading to this result.

For a pile embedded in saturated clayey soil, the ultimate pile capacity ($Q_u$) is given by the sum of skin friction resistance ($Q_f$) and end bearing resistance ($Q_b$):

$Q_u=Q_f+Q_b$

The skin friction resistance is calculated using the adhesion factor method as:

$Q_f=\alpha ·c_u·A_s$

where:
$\alpha$ is the adhesion factor (0.54),
$c_u$ is the undrained cohesion of the soil (80 kPa), and
$A_s$ is the surface area of the pile shaft. For a circular pile of diameter $d$ and length $L$, the shaft surface area is:

$A_s=\pi ·d·L$

The end bearing resistance is calculated as:

$Q_b=q_b·A_b=N_c·c_u·A_b$

where:
$N_c$ is the bearing capacity factor (typically taken as 9 for deep foundations in clay), and
$A_b$ is the cross-sectional area of the base of the pile:

$A_b=\frac{\pi ·d^2}{4}$

Let us write down the parameters given in the question:
Initial diameter, $d_1=600\text{mm}=0.6\text{m}$
Final diameter, $d_2=1200\text{mm}=1.2\text{m}$
Length of the pile, $L=6\text{m}$
Undrained cohesion, $c_u=80\text{kPa}=80\text{kN/m}{2}^{}$
Adhesion factor, $\alpha =0.54$
Bearing capacity factor, $N_c=9$

Step 1: Calculate the pile capacity for diameter $d_1=0.6\text{m}$

$Q_{f1}=0.54·80·(\pi ·0.6·6)=43.2·3.6\pi =155.52\pi \approx 488.58\text{kN}$

$Q_{b1}=9·80·\left(\frac{\pi ·{0.6}^2}{4}\right)=720·0.09\pi =64.8\pi \approx 203.58\text{kN}$

Total capacity for the 600 mm pile:

$Q_{u1}=155.52\pi +64.8\pi =220.32\pi \approx 692.16\text{kN}$

Step 2: Calculate the pile capacity for diameter $d_2=1.2\text{m}$

$Q_{f2}=0.54·80·(\pi ·1.2·6)=43.2·7.2\pi =311.04\pi \approx 977.16\text{kN}$

$Q_{b2}=9·80·\left(\frac{\pi ·{1.2}^2}{4}\right)=720·0.36\pi =259.2\pi \approx 814.30\text{kN}$

Total capacity for the 1200 mm pile:

$Q_{u2}=311.04\pi +259.2\pi =570.24\pi \approx 1791.46\text{kN}$

Step 3: Determine the ratio of the capacities

Comparing the two capacities, we get:

$\text{Ratio}=\frac{Q_{u2}}{Q_{u1}}=\frac{570.24\pi }{220.32\pi }\approx 2.59$

However, in standard simplified practice for driven/bored piles in clay where skin friction capacity dominates (or base resistance is neglected due to soil disturbance at the tip during installation), the pile capacity is directly proportional to the shaft area, which scales linearly with diameter ($d$). Thus, when the diameter is doubled from 600 mm to 1200 mm, the pile capacity doubles, resulting in a ratio of exactly 2.