---
title: "Steel pipe pile group 3×4, layered clay"
subject: "Foundation Engineering"
chapter: 8
key: "semester_5/foundation_engineering:pile-group-steel-pipe-layered-clay"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-steel-pipe-layered-clay
site: BCT Engineering
---

# Steel pipe pile group 3×4, layered clay

Foundation Engineering · Chapter 8: Pile Foundations

- Asked in: 2076 Asoj, 8 marks
- Video: [CIVE3315 RL  Example 1 Pile Foundation Group pile](https://www.youtube.com/watch?v=RK0GTMpMsz0) — Ir. Ts. Dr. Ayu Kasim
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-steel-pipe-layered-clay
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-8

## Question

A 15 m long closed end steel pipe pile group (3×4) consists of 12 piles of a 300 mm diameter and evenly spaced at 900 mm, center-to-center, is driven into layered undrained clay. The top 6 m consists of clay with undrained cohesion of 50 kPa and adhesion factor of 0.74, followed by 6 m of clay with undrained cohesion of 65 kPa and adhesion factor of 0.62, which was underlain by stiff clay with undrained cohesion of 90 kPa and adhesion factor of 0.50. Estimate the allowable load carrying capacity of pile group. Take factor of safety = 2.5.

## Short answer

Single pile skin friction by α-method: 564.4 kN, end bearing 57.3 kN, ultimate 621.6 kN. Sum for 12 piles 7459 kN; block capacity 14895 kN. Group capacity governed by sum, allowable 2984 kN with FS 2.5.

## Answer

**Given:** closed-end steel pipe piles, 3 × 4 group (12 piles), $D = 0.30$ m, $L = 15$ m, spacing $0.9$ m. Layers: 0–6 m ($c_u = 50$, $\alpha = 0.74$); 6–12 m ($c_u = 65$, $\alpha = 0.62$); 12–15 m ($c_u = 90$, $\alpha = 0.50$). FS $= 2.5$.

#### 1. Single pile
$$
\begin{aligned} P &= \pi(0.3) = 0.9425\ \text{m},\quad A_p = 0.07069\ \text{m}^2 \\ \sum\alpha c_uL &= 0.74(50)(6) + 0.62(65)(6) + 0.50(90)(3) = 222.0 + 241.8 + 135.0 = 598.8\ \text{kN/m} \\ Q_s &= 0.9425 \times 598.8 = 564.4\ \text{kN} \\ Q_p &= 9c_uA_p = 9 \times 90 \times 0.07069 = 57.3\ \text{kN} \\ Q_u &= 621.6\ \text{kN} \end{aligned}
$$

Sum for 12 piles $= 12 \times 621.6 = 7459.3$ kN.

#### 2. Block failure
$$
\begin{aligned} B_g &= 2(0.9) + 0.3 = 2.1\ \text{m},\quad L_g = 3(0.9) + 0.3 = 3.0\ \text{m},\quad P_g = 2(2.1 + 3.0) = 10.2\ \text{m} \\ Q_{block} &= P_g\sum c_uL + 9c_uB_gL_g \\ &= 10.2(50 \times 6 + 65 \times 6 + 90 \times 3) + 9(90)(2.1)(3.0) \\ &= 9792 + 5103 = 14\,895\ \text{kN} \end{aligned}
$$

#### 3. Group capacity
Group capacity $=$ smaller of $7459.3$ kN and $14\,895$ kN $= 7459.3$ kN (efficiency 1).

$$
Q_{all} = \frac{7459.3}{2.5} = 2983.7\ \text{kN}
$$

**Answer: allowable load of the pile group $\approx 2984$ kN.**

## Check yourself

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

1. In layered clay the shaft resistance is:
   A. P x average c_u x L
   B. P x sum of alpha c_u L for each layer
   C. Only the top layer
   D. P x largest layer
   Answer: B. Each layer has its own alpha and c_u: sum = 598.8 kN/m, Q_s = 564.4 kN. (Key point: Layered shaft friction)

2. End bearing uses c_u of:
   A. Top layer, 50 kPa
   B. Average, 68 kPa
   C. The stiff clay at the tip, 90 kPa
   D. Layer 2, 65 kPa
   Answer: C. Q_p = 9 c_u A_p with c_u at the tip = 90 kPa gives 57.3 kN. (Key point: End bearing)

3. For a 3 x 4 group at 0.9 m, D = 0.3 m, block dimensions are:
   A. 2.7 m x 3.6 m
   B. 1.8 m x 2.7 m
   C. 2.1 m x 3.0 m
   D. 3.0 m x 4.0 m
   Answer: C. B_g = 2(0.9)+0.3 = 2.1 m, L_g = 3(0.9)+0.3 = 3.0 m. (Key point: Block dimensions)

4. Group allowable load is 2984 kN because:
   A. Block (14895) governs
   B. Sum of piles (7459 kN) is smaller, then /2.5
   C. Efficiency is 0.5
   D. Tip only counts
   Answer: B. 7459.3 < 14895, so individual failure governs; 7459.3/2.5 = 2984 kN. (Key point: Governing capacity)

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