Spacing for 4×5 pile group, block failure

Given: 20 piles in a 4 × 5 pattern, $d = 0.3$ m, $L = 20$ m, $q_u = 80$ kN/m² so $c_u = 40$ kN/m², $\alpha = 0.5$. "The same possibility of individual and block failure" means the block capacity equals the sum of the individual pile capacities (efficiency $= 1$). Point bearing is included.

Individual piles

$$\begin{aligned} Q_s &= \alpha c_u\pi dL = 0.5 \times 40 \times \pi \times 0.3 \times 20 = 377.0\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 40 \times 0.0707 = 25.4\ \text{kN} \\ Q_u &= 402.4\ \text{kN} \\ \sum Q_u &= 20 \times 402.4 = 8048.8\ \text{kN} \end{aligned}$$

Block (4 piles in one direction, 5 in the other, spacing $s$)

$$\begin{aligned} B_g &= 3s + 0.3,\quad L_g = 4s + 0.3,\quad P_g = 2(B_g + L_g) = 14s + 1.2 \\ Q_{block} &= P_g Lc_u + 9c_uB_gL_g \\ &= (14s + 1.2)(20)(40) + 9(40)(3s + 0.3)(4s + 0.3) \end{aligned}$$

Equate

$$\begin{aligned} 800(14s + 1.2) + 360(12s^2 + 2.1s + 0.09) &= 8048.8 \\ 4320\,s^2 + 11\,956\,s + 992.4 &= 8048.8 \\ s &= 0.50\ \text{m} \end{aligned}$$

Answer: centre-to-centre spacing $\approx 0.50$ m ($\approx 1.7d$) for equal chances of individual and block failure. This is smaller than the practical minimum of $2.5d$ to $3d$ (0.75 to 0.9 m) needed for driving, so at a practical spacing the block capacity is higher and the group fails as individual piles (efficiency 1).