---
title: "Friction pile length, 300 mm, 250 kN"
subject: "Foundation Engineering"
chapter: 8
key: "semester_5/foundation_engineering:friction-pile-300mm-250kn"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/friction-pile-300mm-250kn
site: BCT Engineering
---

# Friction pile length, 300 mm, 250 kN

Foundation Engineering · Chapter 8: Pile Foundations

- Asked in: 2076 Chaitra, 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-300mm-250kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-8

## Question

A friction pile 300 mm in diameter is proposed to be driven in a layer of uniform soil having unit skin friction between pile surface and soil as $60\ \text{kN/m}^2$. Determine the length of the pile required to carry allowable load of 250 kN assuming the pile tip carries 20% of the total load.

## Short answer

Assuming FoS=2.5, ultimate load =625 kN, shaft carries80% =500 kN. Length L =500/(60×π×0.30) =8.84 m ≈8.9 m.

## Answer

**Given:** $D = 0.30$ m, unit skin friction $f_s = 60$ kN/m² (ultimate), allowable load $250$ kN, tip carries 20% of the total load.

**Assumption:** a factor of safety of 2.5 is applied to the allowable load to obtain the ultimate load.

$$
\begin{aligned} Q_u &= 2.5 \times 250 = 625\ \text{kN} \\ Q_s &= 0.80 \times 625 = 500\ \text{kN} \quad (\text{shaft share}) \end{aligned}
$$

$$
\begin{aligned} Q_s &= f_s\,\pi D\,L = 60 \times \pi \times 0.30 \times L = 56.55\,L \\ L &= \frac{500}{56.55} = 8.84\ \text{m} \end{aligned}
$$

**Answer: length of pile $\approx 8.9$ m, say 9 m.** (If no factor of safety is applied, i.e. the shaft carries 200 kN, $L = 3.54$ m.)

## Check yourself

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

1. The ultimate load is obtained from the allowable load by:
   A. Dividing by FS
   B. Multiplying by FS (2.5 assumed), giving 625 kN
   C. Adding 20%
   D. Using it directly
   Answer: B. Q_u = 2.5 x 250 = 625 kN. (Key point: Allowable to ultimate)

2. If the tip carries 20%, shaft load is:
   A. 125 kN
   B. 500 kN
   C. 250 kN
   D. 625 kN
   Answer: B. Q_s = 0.8 x 625 = 500 kN. (Key point: Shaft share)

3. With unit skin friction given, Q_s equals:
   A. f_s x pi D^2/4
   B. alpha c_u pi D L
   C. f_s x L
   D. f_s x pi D L
   Answer: D. Q_s = f_s pi D L = 56.55 L. (Key point: Shaft formula)

4. The pile length is about:
   A. 3.5 m
   B. 8.9 m
   C. 15 m
   D. 22 m
   Answer: B. L = 500 / 56.55 = 8.84 m, say 9 m. (Key point: Length)

## Related in Foundation Engineering

- [Friction pile length, 350 mm, 1200 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Afriction-pile-350mm-1200kn) — Foundation Engineering, chapter 8
- [Allowable load of 400 mm pile, layered soil](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Apile-400mm-layered-soil-chart) — Foundation Engineering, chapter 8
- [Negative skin friction on 16-pile group](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Aneg-skin-friction-group-16-piles) — Foundation Engineering, chapter 8
- [Allowable load of 4×4 pile group, layered sand](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Apile-group-4x4-figure-dcr) — Foundation Engineering, chapter 8
- [Negative skin friction on single pile and group](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Ansf-calculation-single-group) — Foundation Engineering, chapter 8, asked 2×
- [Negative skin friction: causes, effects, remedies](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Anegative-skin-friction-effects) — Foundation Engineering, chapter 8, asked 4×

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