---
title: "Friction pile length, 350 mm, 1200 kN"
subject: "Foundation Engineering"
chapter: 8
key: "semester_5/foundation_engineering:friction-pile-350mm-1200kn"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/friction-pile-350mm-1200kn
site: BCT Engineering
---

# Friction pile length, 350 mm, 1200 kN

Foundation Engineering · Chapter 8: Pile Foundations

- Asked in: 2078 Kartik, 7 marks
- Video: [Geotechnical Engineering: Piles: undrained side friction resistance.](https://www.youtube.com/watch?v=Q0n8BBLgGSU) — GEOTECH VIDEOS
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/friction-pile-350mm-1200kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-8

## Question

A friction pile 350 mm diameter is proposed to be driven in a layer of uniform cohesive soil with unconfined compressive strength of 80 kPa. Considering the pile end carry only 30% of the total load, determine the length of the pile required for the ultimate load of 1200 kN. Assume adhesion factor = 0.80.

## Short answer

Shaft load =840 kN (70% of ultimate). Using Q_s = α c_u π D L, L =840/(0.8×40×π×0.35) =23.87 m. Required length ≈24 m.

## Answer

**Given:** $D = 0.35$ m, $q_u = 80$ kPa so $c_u = 40$ kPa, $\alpha = 0.80$, ultimate load $Q_u = 1200$ kN, end bearing is 30% of the total load.

$$
\begin{aligned} Q_p &= 0.30 \times 1200 = 360\ \text{kN} \\ Q_s &= 1200 - 360 = 840\ \text{kN} \end{aligned}
$$

Shaft friction: $Q_s = \alpha c_u\pi DL$

$$
\begin{aligned} 840 &= 0.80 \times 40 \times \pi \times 0.35 \times L = 35.19\,L \\ L &= \frac{840}{35.19} = 23.87\ \text{m} \end{aligned}
$$

**Answer: length of pile required $\approx 24$ m.**

## Check yourself

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

1. With the tip carrying 30% of 1200 kN, the shaft must carry:
   A. 360 kN
   B. 840 kN
   C. 1200 kN
   D. 480 kN
   Answer: B. Q_s = 1200 - 360 = 840 kN. (Key point: Load split)

2. For q_u = 80 kPa, c_u is:
   A. 80 kPa
   B. 160 kPa
   C. 40 kPa
   D. 26.7 kPa
   Answer: C. c_u = q_u / 2 = 40 kPa. (Key point: Cohesion)

3. Which equation gives L?
   A. 1200 = alpha c_u pi D L
   B. 360 = alpha c_u pi D L
   C. 840 = alpha c_u pi D L
   D. 840 = c_u pi D^2 L
   Answer: C. Shaft friction Q_s = alpha c_u pi D L = 35.19 L equals 840 kN. (Key point: Length equation)

4. The required length is about:
   A. 12 m
   B. 34 m
   C. 24 m
   D. 8 m
   Answer: C. L = 840 / 35.19 = 23.9 m, about 24 m. (Key point: Result)

## Related in Foundation Engineering

- [Friction pile length, 300 mm, 250 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Afriction-pile-300mm-250kn) — Foundation Engineering, chapter 8
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- [Active pressure on wall, cohesive soil](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Acohesive-wall-pressure-diagram) — Foundation Engineering, chapter 3

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