A group of 25 circular piles is arranged in 5×5 uniform pattern in a soft clay soil with equal spacing in both the directions. These are friction piles with negligible end bearing.
Consider the following details:
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).

The correct option is 3.4.

To find the optimum spacing-to-diameter ratio, we need to compare the load-carrying capacity of the piles acting individually versus the capacity of the pile group acting as a single block. The optimum spacing occurs when the group efficiency is unity ($\eta =1$), meaning the ultimate load capacity of the pile group acting as a block ($Q_g$) is equal to the sum of the individual capacities of the piles ($n·Q_i$).

Let's first define the parameters given in the problem:
Number of piles, $n=25$ arranged in a $5\times 5$ grid.
Diameter of each pile, $d=1\text{m}$.
Length of each pile, $L=15\text{m}$.
Cohesion of the clay, $c=20\text{kN/m}{\sup }^2$.
Adhesion factor, $\alpha =0.75$.
Spacing between the piles center-to-center is $s$.

Since the piles are friction piles with negligible end bearing, we only consider the skin friction resistance for both individual pile action and group block action.

1. Ultimate capacity of a single pile ($Q_i$):
The ultimate skin friction capacity of a single circular pile is given by:
$Q_i=\alpha ·c·A_s$
where $A_s$ is the surface area of a single pile:
$A_s=\pi ·d·L$
Thus, the individual capacity is:
$Q_i=\alpha ·c·(\pi ·d·L)$

The total capacity of $n=25$ piles acting individually is:
$n·Q_i=25·\alpha ·c·\pi ·d·L$

2. Ultimate capacity of the pile group acting as a block ($Q_g$):
For a $5\times 5$ pile group, the width (or side dimension) of the block $B$ is:
$B=4s+d$
The perimeter of the block enclosing the pile group is:
$P=4·B=4·(4s+d)$
For block failure, the adhesion factor is taken as $1.0$ because the soil shears against soil along the perimeter of the block. Therefore, the group block skin friction capacity is:
$Q_g=1.0·c·(P·L)=c·4(4s+d)·L$

3. Setting the group efficiency to unity ($\eta =1$):
For optimum spacing, we equate the block capacity to the sum of individual pile capacities:
$Q_g=n·Q_i$
$c·4(4s+d)·L=25·\alpha ·c·\pi ·d·L$

We can cancel out $c$ and $L$ from both sides of the equation:
$4(4s+d)=25·\alpha ·\pi ·d$

Divide both sides by $d$ to express the equation in terms of the ratio $s/d$:
$4·(4\frac{s}{d}+1)=25·\alpha ·\pi$

Substitute $\alpha =0.75$ into the equation:
$16\frac{s}{d}+4=25·0.75·\pi$
$16\frac{s}{d}+4=18.75·\pi$
$16\frac{s}{d}+4\approx 18.75·3.14159\approx 58.905$

Solving for $s/d$:
$16\frac{s}{d}=58.905-4=54.905$
$\frac{s}{d}=\frac{54.905}{16}\approx 3.43$

Rounding to one decimal place, the optimum ratio of pile spacing to pile diameter is 3.4.