---
title: "Pile group 16 piles of 50 cm, efficiency"
subject: "Foundation Engineering"
chapter: 8
key: "semester_5/foundation_engineering:pile-group-16-piles-50cm"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-16-piles-50cm
site: BCT Engineering
---

# Pile group 16 piles of 50 cm, efficiency

Foundation Engineering · Chapter 8: Pile Foundations

- Asked in: 2073 Shrawan, 6 marks
- Video: [Efficiency of Pile group | Deep Foundation | Geotechnical Engineering - GATE](https://www.youtube.com/watch?v=gix87aYtLEc) — Ekeeda GATE & ESE
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-16-piles-50cm
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-8

## Question

A group of 16 piles of 50 cm diameter is arranged with a center to center spacing of 1.0 m. The piles are 90 m long (as printed) and are embedded in soft clay with cohesion of $30\ \text{kN/m}^2$. Bearing resistance may be neglected for the piles. Adhesion factor is 0.6. Determine the ultimate load capacity of the pile group. Also check the efficiency of the group of pile.

## Short answer

Assuming pile length 9 m, single pile skin friction 254.5 kN, sum 4071.5 kN. Block capacity 3780 kN governs. Group ultimate capacity 3780 kN, efficiency 0.93.

## Answer

**Given:** 16 piles (4 × 4), $d = 0.50$ m, spacing $s = 1.0$ m, soft clay $c_u = 30$ kN/m², $\alpha = 0.6$, point bearing neglected. The length is printed as 90 m; **9 m** is taken as the intended value (the efficiency does not depend on $L$, and for 90 m the capacities are 10 times larger).

#### Sum of individual piles
$$
Q_u = \alpha c_u\pi dL = 0.6 \times 30 \times \pi \times 0.5 \times 9 = 254.5\ \text{kN},\quad 16Q_u = 4071.5\ \text{kN}
$$

#### Block
$$
\begin{aligned} B_g &= L_g = 3(1.0) + 0.5 = 3.5\ \text{m},\quad P_g = 14\ \text{m} \\ Q_{block} &= P_gLc_u = 14 \times 9 \times 30 = 3780\ \text{kN} \end{aligned}
$$

#### Group capacity and efficiency
The smaller value governs: $Q_{g} = 3780$ kN.

$$
\eta = \frac{3780}{4071.5} = 0.928\ (92.8\%)
$$

**Answer: ultimate load capacity of the group $= 3780$ kN and efficiency $= 0.93$** (block failure governs). For $L = 90$ m as printed: $Q_g = 37\,800$ kN, $\eta = 0.93$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Block capacity (skin only) is:
   A. B_g L c_u
   B. 9 c_u B_g^2
   C. 4071 kN
   D. P_g L c_u = 14 x 9 x 30 = 3780 kN
   Answer: D. Block perimeter 4 x 3.5 = 14 m times length times c_u. (Key point: Block capacity)

2. Efficiency of the group is:
   A. 1.08
   B. 3780/4071.5 = 0.93
   C. 0.5
   D. 1.0
   Answer: B. Efficiency is group capacity divided by sum of individual capacities. (Key point: Efficiency)

## Related in Foundation Engineering

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- [Pile spacing for 100% efficiency, 16 piles](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Apile-spacing-100-efficiency-square-30cm) — Foundation Engineering, chapter 8, asked 2×
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