12 piles, spacing and diameter, 100% efficiency

Given: 12 circular piles in a rectangular (4 × 3) pattern, $s/d = 3.0$, $q_u = 80$ kN/m² so $c_u = 40$ kN/m², $L = 15$ m, $\alpha = 0.85$. Efficiency $= 100\%$ means block capacity $\ge$ sum of individual pile capacities.

Individual pile (diameter $d$)

$$Q_u = \alpha c_u\pi dL + 9c_u\frac{\pi d^2}{4} = 40\left(40.06\,d + 7.07\,d^2\right)$$

Sum for 12 piles $= 40(480.7\,d + 84.8\,d^2)$.

Block (4 piles by 3 piles, $s = 3d$)

$$\begin{aligned} B_g &= 2s + d = 7d,\quad L_g = 3s + d = 10d \\ Q_{block} &= 2(B_g + L_g)Lc_u + 9c_uB_gL_g = 40\left(510\,d + 630\,d^2\right) \end{aligned}$$

The block capacity ($510d + 630d^2$) is larger than the sum of individual piles ($480.7d + 84.8d^2$) for every diameter, so the group fails as individual piles and the efficiency is 100% for any $d$ when $s = 3d$. The ratio $s/d$ governs, not the diameter.

Minimum spacing

For a general ratio $r = s/d$, $B_g = (2r+1)d$ and $L_g = (3r+1)d$. Setting block $=$ sum gives

$$(150r + 60) + 9d\,(6r^2 + 5r + 1) = 480.7 + 84.8\,d$$

The smallest ratio for 100% efficiency therefore depends on $d$:

So $s/d = 3$ is satisfactory for all practical diameters.

The diameter is then fixed by the load to be carried or practical size. For example, with $d = 0.40$ m:

$$s = 3 \times 0.40 = 1.2\ \text{m},\quad Q_u(\text{one pile}) = 686.2\ \text{kN},\quad Q_{u,group} = 8234\ \text{kN}$$

Answer: spacing $= 3d$ (for example $s = 1.2$ m with a pile diameter of 0.40 m, centre to centre), which gives an efficiency of 100%. The smallest spacing that still gives 100% efficiency is about $2.1d$ for $d = 0.40$ m (it falls as the diameter increases). The data given do not fix the diameter, so a practical value of 0.3 to 0.5 m is adopted.