Question

A group of 25 circular piles is arranged in a \(5 \times 5\) uniform pattern in a soft clay soil, with equal spacing in both directions. These are friction piles with negligible end bearing. Diameter of each pile = 1 m. Length of each pile = 15 m. Cohesion of the soil = 20 kN/m². Unit weight of the soil = 16 kN/m³. Adhesion factor = 0.75. Considering the efficiency of the pile group as unity, the optimum value of the ratio of the pile spacing to pile diameter is ______ (rounded off to one decimal place).

Hint:

Set the sum of the individual (friction-only) pile capacities equal to the block's perimeter-friction capacity and solve for the spacing-to-diameter ratio.

Answer 1

Correct Answer: 3.4

Step 1: Set the target condition.
Group efficiency $\eta = \dfrac{Q_u(\text{block})}{n \times Q_u(\text{single pile})}$. We want the spacing where $\eta = 1$, so the block capacity equals $n$ times the single-pile capacity. Since these are friction piles with negligible end bearing, both sides use shaft/perimeter friction only.

Step 2: Capacity of one single pile.
$Q_u(\text{single}) = \alpha c (\pi d) L = 0.75 \times 20 \times \pi \times 1 \times 15 = 225\pi = 706.86$ kN.
For 25 piles, $n \times Q_u(\text{single}) = 25 \times 706.86 = 17671.5$ kN.

Step 3: Express the block side directly in terms of the ratio $r = s/d$.
Let $r = s/d$, so $s = r$ m (since $d=1$ m). The block plan dimension is $B_g = 4s+d = 4r+1$.
Block capacity (perimeter friction only, base bearing negligible):
$Q_u(\text{block}) = c(4B_g)L = 20\times4(4r+1)\times15 = 1200(4r+1)$

Step 4: Equate and solve for $r$.
$1200(4r+1) = 17671.5$
$4r+1 = \dfrac{17671.5}{1200} = 14.726$
$4r = 13.726 \implies r = 3.4316$

Step 5: Round and state the answer.
The optimum spacing-to-diameter ratio is $r \approx 3.4$. Below this ratio the block behaves as one unit and the group underperforms the sum of the single piles (efficiency below 1); at or above it, the piles act independently at full single-pile capacity, which is the economical target of "efficiency = unity". \[ \boxed{s/d \approx 3.4} \]