---
title: "Spacing for 4×5 pile group, block failure"
subject: "Foundation Engineering"
chapter: 8
key: "semester_5/foundation_engineering:pile-group-20-piles-block-failure"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-20-piles-block-failure
site: BCT Engineering
---

# Spacing for 4×5 pile group, block failure

Foundation Engineering · Chapter 8: Pile Foundations

- Asked in: 2075 Chaitra, 8 marks
- Video: [Group Action of Pile। Block Failure in Piles। Lecture 7। Pile Foundation। Foundation Engineering।](https://www.youtube.com/watch?v=wGWZWO8cIC8) — Civil Gyani
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-20-piles-block-failure
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-8

## Question

The 20 numbers of concrete pile of 0.3 m diameter and 20 meter depth are designed to construct 4 × 5 layout pattern. The site consists of clay with unconfined compressive strength $80\ \text{kN/m}^2$. Design center to center spacing of piles so that the group pile has the same possibility of individual pile failure and block failure (printed "some"). Take adhesion factor $\alpha = 0.5$.

## Short answer

Equate block capacity (skin + end bearing) with sum of individual pile capacities (8049 kN). Solve for spacing s yielding s=0.50 m, at which both failure modes are equally likely.

## Answer

**Given:** 20 piles in a 4 × 5 pattern, $d = 0.3$ m, $L = 20$ m, $q_u = 80$ kN/m² so $c_u = 40$ kN/m², $\alpha = 0.5$. "The same possibility of individual and block failure" means the **block capacity equals the sum of the individual pile capacities** (efficiency $= 1$). Point bearing is included.

#### Individual piles
$$
\begin{aligned} Q_s &= \alpha c_u\pi dL = 0.5 \times 40 \times \pi \times 0.3 \times 20 = 377.0\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 40 \times 0.0707 = 25.4\ \text{kN} \\ Q_u &= 402.4\ \text{kN} \\ \sum Q_u &= 20 \times 402.4 = 8048.8\ \text{kN} \end{aligned}
$$

#### Block (4 piles in one direction, 5 in the other, spacing $s$)
$$
\begin{aligned} B_g &= 3s + 0.3,\quad L_g = 4s + 0.3,\quad P_g = 2(B_g + L_g) = 14s + 1.2 \\ Q_{block} &= P_g Lc_u + 9c_uB_gL_g \\ &= (14s + 1.2)(20)(40) + 9(40)(3s + 0.3)(4s + 0.3) \end{aligned}
$$

#### Equate
$$
\begin{aligned} 800(14s + 1.2) + 360(12s^2 + 2.1s + 0.09) &= 8048.8 \\ 4320\,s^2 + 11\,956\,s + 992.4 &= 8048.8 \\ s &= 0.50\ \text{m} \end{aligned}
$$

**Answer: centre-to-centre spacing $\approx 0.50$ m ($\approx 1.7d$)** for equal chances of individual and block failure. This is smaller than the practical minimum of $2.5d$ to $3d$ (0.75 to 0.9 m) needed for driving, so at a practical spacing the block capacity is higher and the group fails as individual piles (efficiency 1).

## Check yourself

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

1. 'Same possibility of individual and block failure' means:
   A. Spacing equals 3D
   B. Block capacity equals the sum of individual capacities
   C. Tip bearing is ignored
   D. Efficiency is zero
   Answer: B. Setting Q_block = sum Q_u (efficiency 1) gives the spacing. (Key point: Meaning of the condition)

2. For a 4 x 5 layout the block dimensions are:
   A. B_g = 4s, L_g = 5s
   B. B_g = 4s + 0.3, L_g = 5s + 0.3
   C. B_g = 3s + 0.3, L_g = 4s + 0.3
   D. B_g = 2s, L_g = 3s
   Answer: C. Gaps are one less than the number of piles in each direction. (Key point: Block geometry)

3. The sum of individual pile capacities used is:
   A. 402.4 kN
   B. 8049 kN
   C. 377 kN
   D. 20 x 377 kN
   Answer: B. 20 x (377.0 + 25.4) = 8048.8 kN. (Key point: Individual sum)

4. The result s = 0.50 m (1.7d) is:
   A. Larger than the practical minimum
   B. Below the practical minimum of 2.5D to 3D, so at practical spacing the group fails as individual piles
   C. Exactly 3D
   D. Needed for design
   Answer: B. At practical spacing the block capacity is higher, so efficiency is 1. (Key point: Practical meaning)

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