Pile group 4×4, two-layer clay, 2.4 m spacing

Given: 16 piles (4 × 4), $D = 0.8$ m, $L = 10$ m, spacing $s = 2.4$ m, clay; $c_u = 40$ kN/m² for the top 5 m and $60$ kN/m² for the lower 5 m; adhesion factor $\alpha = 0.6$.

The group capacity is the smaller of (1) the sum of the capacities of individual piles and (2) the capacity of the block of soil enclosing the group.

1. Single pile

$$\begin{aligned} A_p &= \frac{\pi}{4}(0.8)^2 = 0.5027\ \text{m}^2,\quad \text{perimeter } = \pi(0.8) = 2.513\ \text{m} \\ Q_s &= \alpha c_{u1}P L_1 + \alpha c_{u2}PL_2 \\ &= 0.6(40)(2.513)(5) + 0.6(60)(2.513)(5) = 301.6 + 452.4 = 754.0\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 60 \times 0.5027 = 271.4\ \text{kN} \\ Q_u &= 754.0 + 271.4 = 1025.4\ \text{kN} \end{aligned}$$

Sum for 16 piles: $16 \times 1025.4 = 16\,406.7$ kN.

2. Block failure

$$\begin{aligned} B_g = L_g &= 3(2.4) + 0.8 = 8.0\ \text{m},\quad P_g = 4 \times 8 = 32\ \text{m} \\ Q_{u,block} &= P_g\sum(c_uL) + 9c_uB_gL_g \\ &= 32\,(40 \times 5 + 60 \times 5) + 9 \times 60 \times 8 \times 8 \\ &= 16\,000 + 34\,560 = 50\,560\ \text{kN} \end{aligned}$$

3. Group capacity

Block capacity (50 560 kN) is greater than the sum of individual piles (16 407 kN), so the piles act individually and the efficiency is 1.

Answer: ultimate load capacity of the pile group $= 16\,407$ kN (about 16.4 MN).