---
title: "Pile group 16 piles, 0.40 m, soft clay"
subject: "Foundation Engineering"
chapter: 8
key: "semester_5/foundation_engineering:pile-group-16-soft-clay-040"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-16-soft-clay-040
site: BCT Engineering
---

# Pile group 16 piles, 0.40 m, soft clay

Foundation Engineering · Chapter 8: Pile Foundations

- Asked in: 2081 Bhadra, 8 marks
- Video: [Numerical on Group Capacity of Pile in Clay](https://www.youtube.com/watch?v=UHjqId4wkvc) — Curious Civil Engineer
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-16-soft-clay-040
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-8

## Question

A group of 16 piles with 4 piles in a row was driven into soft clay. The diameter and length of piles were 0.40 m and 12 m, respectively. The undrained cohesion is 35 kPa. The piles were placed 80 cm centre to centre. Calculate the allowable load carrying capacity of foundation. Take adhesion factor = 0.8 and FoS = 2.5.

## Short answer

Individual pile capacity = 461.8 kN; sum for 16 piles = 7389 kN. Block capacity = 7173.6 kN (governs). Allowable load = 7173.6 / 2.5 = 2869.4 kN ≈ 2869 kN.

## Answer

**Given:** 16 piles (4 × 4), $D = 0.40$ m, $L = 12$ m, $c_u = 35$ kPa, spacing $s = 0.80$ m (c/c, $= 2D$), $\alpha = 0.8$, FoS $= 2.5$.

#### 1. Individual pile
$$
\begin{aligned} A_p &= \frac{\pi}{4}(0.4)^2 = 0.1257\ \text{m}^2,\quad P = \pi(0.4) = 1.257\ \text{m} \\ Q_s &= \alpha c_uPL = 0.8 \times 35 \times 1.257 \times 12 = 422.2\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 35 \times 0.1257 = 39.6\ \text{kN} \\ Q_u &= 461.8\ \text{kN} \end{aligned}
$$

Sum of 16 piles $= 16 \times 461.8 = 7389.0$ kN.

#### 2. Block failure
$$
\begin{aligned} B_g = L_g &= 3(0.8) + 0.4 = 2.8\ \text{m},\quad P_g = 4(2.8) = 11.2\ \text{m} \\ Q_{block} &= P_gLc_u + 9c_uB_gL_g \\ &= 11.2 \times 12 \times 35 + 9 \times 35 \times 2.8^2 \\ &= 4704.0 + 2469.6 = 7173.6\ \text{kN} \end{aligned}
$$

#### 3. Group capacity
The block capacity (7173.6 kN) is smaller than the sum of individual piles (7389.0 kN), so the **block failure governs**. Efficiency $= 7173.6/7389.0 = 0.97$.

$$
Q_{all} = \frac{7173.6}{2.5} = 2869.4\ \text{kN}
$$

**Answer: allowable load of the pile group $\approx 2869$ kN.**

## Check yourself

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

1. Which expression gives the ultimate capacity of one pile here?
   A. alpha c_u A_p L + 9 c_u P
   B. alpha c_u P L + 9 c_u A_p
   C. c_u P L only
   D. 9 c_u P L
   Answer: B. Shaft adhesion alpha c_u P L plus end bearing 9 c_u A_p. (Key point: Single pile capacity)

2. With 4 piles per row at 0.8 m and D = 0.4 m, the block width B_g is:
   A. 3.2 m
   B. 2.4 m
   C. 2.8 m
   D. 3.6 m
   Answer: C. B_g = 3(0.8) + 0.4 = 2.8 m. (Key point: Block dimensions)

3. Which value governs the group capacity and why?
   A. Sum of piles, 7389 kN, because it is larger
   B. Block, 7173.6 kN, because it is smaller
   C. Sum of piles, because efficiency is 1
   D. Neither; FoS decides
   Answer: B. The smaller of the two governs, and the block (7173.6 kN) is smaller than 7389 kN. (Key point: Governing mode)

4. The allowable load of the group is:
   A. 7173.6 x 2.5 kN
   B. 7173.6 / 2.5 = 2869 kN
   C. 7389 / 2.5 = 2956 kN
   D. 16 x 461.8 / 2.5 kN
   Answer: B. Governing capacity divided by FoS 2.5 gives about 2869 kN. (Key point: Allowable load)

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