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).
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).
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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.
Updated On: Jul 17, 2026
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Correct Answer: 3.4
Solution and Explanation
Step 1: Understand what "efficiency of the pile group as unity" means.
A pile group in soft clay can fail in two ways: each pile fails individually (shaft friction only, since end bearing is negligible here), or the whole group fails together as one large "block" pile (perimeter friction of the block, again with negligible base resistance). Group efficiency \(\eta\) is the ratio of the block capacity to the sum of the individual pile capacities. \(\eta = 1\) marks the optimum (most economical) spacing at which the block capacity exactly equals the sum of the individual pile capacities.
Step 2: Write the sum of the individual pile capacities.
With negligible end bearing, each pile's capacity is pure shaft friction: \(\alpha c \pi d L\). For \(n = 25\) piles,
\[ \sum Q_u(\text{individual}) = n\,\alpha\, c\, \pi\, d\, L = 25 \times 0.75 \times 20 \times \pi \times 1 \times 15 = 17671.5 \text{ kN} \]
Step 3: Write the block (group) capacity.
The 25 piles form a \(5 \times 5\) grid (\(m = 5\) piles per side). The plan size of the equivalent block is
\[ B_g = (m-1)s + d = 4s + 1 \]
where \(s\) is the center-to-center spacing. With negligible end bearing at the block base too, the block capacity is pure perimeter friction of the whole block:
\[ Q_u(\text{block}) = c \times (4B_g) \times L \]
Step 4: Set the two capacities equal (efficiency = 1) and solve for \(B_g\).
\[ n\,\alpha\,\pi\, d\, L = 4\, c\, B_g\, L \]
The cohesion \(c\) and length \(L\) cancel, leaving
\[ n\,\alpha\,\pi\, d = 4 B_g \implies 25\times0.75\times\pi\times1 = 4B_g \]
\[ 58.905 = 4B_g \implies B_g = 14.726 \text{ m} \]
Step 5: Back out the spacing and the ratio.
\[ B_g = 4s+d \implies 14.726 = 4s+1 \implies s = \frac{13.726}{4}=3.4316 \text{ m} \]
\[ \frac{s}{d} = \frac{3.4316}{1} = 3.43 \]
Final Answer:
The soil unit weight is not needed because end bearing is negligible for both the single pile and the block, so no weight-dependent bearing term enters the calculation. Rounded to one decimal place,
\[ \boxed{\dfrac{s}{d} \approx 3.4} \]
A pile group in soft clay can fail in two ways: each pile fails individually (shaft friction only, since end bearing is negligible here), or the whole group fails together as one large "block" pile (perimeter friction of the block, again with negligible base resistance). Group efficiency \(\eta\) is the ratio of the block capacity to the sum of the individual pile capacities. \(\eta = 1\) marks the optimum (most economical) spacing at which the block capacity exactly equals the sum of the individual pile capacities.
Step 2: Write the sum of the individual pile capacities.
With negligible end bearing, each pile's capacity is pure shaft friction: \(\alpha c \pi d L\). For \(n = 25\) piles,
\[ \sum Q_u(\text{individual}) = n\,\alpha\, c\, \pi\, d\, L = 25 \times 0.75 \times 20 \times \pi \times 1 \times 15 = 17671.5 \text{ kN} \]
Step 3: Write the block (group) capacity.
The 25 piles form a \(5 \times 5\) grid (\(m = 5\) piles per side). The plan size of the equivalent block is
\[ B_g = (m-1)s + d = 4s + 1 \]
where \(s\) is the center-to-center spacing. With negligible end bearing at the block base too, the block capacity is pure perimeter friction of the whole block:
\[ Q_u(\text{block}) = c \times (4B_g) \times L \]
Step 4: Set the two capacities equal (efficiency = 1) and solve for \(B_g\).
\[ n\,\alpha\,\pi\, d\, L = 4\, c\, B_g\, L \]
The cohesion \(c\) and length \(L\) cancel, leaving
\[ n\,\alpha\,\pi\, d = 4 B_g \implies 25\times0.75\times\pi\times1 = 4B_g \]
\[ 58.905 = 4B_g \implies B_g = 14.726 \text{ m} \]
Step 5: Back out the spacing and the ratio.
\[ B_g = 4s+d \implies 14.726 = 4s+1 \implies s = \frac{13.726}{4}=3.4316 \text{ m} \]
\[ \frac{s}{d} = \frac{3.4316}{1} = 3.43 \]
Final Answer:
The soil unit weight is not needed because end bearing is negligible for both the single pile and the block, so no weight-dependent bearing term enters the calculation. Rounded to one decimal place,
\[ \boxed{\dfrac{s}{d} \approx 3.4} \]
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