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Pile spacing for 100% efficiency, 16 piles

Short answer

Equate block and sum of individual capacities (both proportional to c_u). Solve 9B_g^2+48B_g–197.28=0, giving B_g=2.721 m. For a 4×4 square group, centre-to-centre spacing s=(B_g–b)/3=0.807 m≈0.81 m.

A group of 16 piles arranged in a square pattern is to be proportioned in a deposit of soft saturated clay. Assuming the piles to be square with side 30 cm and 12 m long, determine the centre to centre spacing of piles for 100% efficiency of the pile group. Take adhesion factor = 0.8 and consider both point bearing and skin friction. (Cu not legible in the 2080 Bhadra scan.)

Given: 16 square piles in a 4 × 4 group, side b=0.30b = 0.30 m, L=12L = 12 m, soft clay, α=0.8\alpha = 0.8. The value of cuc_u is not legible, but it cancels out, as both the individual and block capacities are proportional to cuc_u. Point bearing and skin friction are both considered.

For 100% efficiency, block capacity = sum of individual pile capacities.

Individual pile

Qu=αcu(4bL)+9cub2=cu[0.8(4)(0.3)(12)+9(0.3)2]=cu(11.52+0.81)=12.33 cu\begin{aligned} Q_u &= \alpha c_u(4bL) + 9c_ub^2 \\ &= c_u\left[0.8(4)(0.3)(12) + 9(0.3)^2\right] \\ &= c_u(11.52 + 0.81) = 12.33\,c_u \end{aligned}

Sum for 16 piles =16×12.33 cu=197.28 cu= 16 \times 12.33\,c_u = 197.28\,c_u.

Block of side BgB_g (square)

Qblock=cu(4Bg)L+9cuBg2=cu(48Bg+9Bg2)Q_{block} = c_u(4B_g)L + 9c_uB_g^2 = c_u\left(48B_g + 9B_g^2\right)

Equate

9Bg2+48Bg−197.28=0Bg=−48+482+4(9)(197.28)2(9)=2.721 m\begin{aligned} 9B_g^2 + 48B_g - 197.28 &= 0 \\ B_g &= \frac{-48 + \sqrt{48^2 + 4(9)(197.28)}}{2(9)} = 2.721\ \text{m} \end{aligned}

For a 4 × 4 group Bg=3s+bB_g = 3s + b:

s=2.721−0.303=0.807 ms = \frac{2.721 - 0.30}{3} = 0.807\ \text{m}

Answer: centre-to-centre spacing of piles ≈0.81\approx 0.81 m (≈2.7b\approx 2.7b) for 100% efficiency. A larger spacing also gives an efficiency of 100%.

✓ Answer checked by Claude Sonnet 5.5

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