---
title: "Allowable load of 4×4 pile group, layered sand"
subject: "Foundation Engineering"
chapter: 8
key: "semester_5/foundation_engineering:pile-group-4x4-figure-dcr"
source: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-4x4-figure-dcr
site: BCT Engineering
---

# Allowable load of 4×4 pile group, layered sand

Foundation Engineering · Chapter 8: Pile Foundations

- Asked in: 2081 Baisakh, 8 marks
- Video: [Ultimate and Allowable Capacity of Pile in Three Layered Soil With Water Table|Solved Problem](https://www.youtube.com/watch?v=hfbcwIJfAjw) — Master Civil Engineering
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/foundation_engineering/pile-group-4x4-figure-dcr
- Whole chapter: https://www.bctengineer.com/md/semester_5/foundation_engineering/chapter-8

## Question

Determine the allowable pile load capacity of 4×4 group of 40 cm diameter driven concrete pile shown in figure. Use FOS = 2.5, spacing = 2.5B and $D_{cr}/B = 12$. [Figure: water table at ground surface; loose sand 3 m thick, $\gamma = 16\ \text{kN/m}^3$, $\phi = 30^\circ$; medium dense sand 6 m thick, $\gamma_{sat} = 19\ \text{kN/m}^3$, $\phi = 35^\circ$; dense sand 5 m thick, $\gamma_{sat} = 20\ \text{kN/m}^3$, $\phi = 40^\circ$; pile passes through all layers.]

## Short answer

Skin friction and end bearing computed using effective stress and critical depth (Dcr=4.8m). Single pile capacity = 1024.8 kN, allowable per pile = 409.9 kN. For 4×4 group in sand at 2.5B spacing, efficiency η=1, group allowable load ≈ 6560 kN.

## Answer

**Given:** 16 driven concrete piles (4 × 4), $B = 0.40$ m, spacing $2.5B = 1.0$ m, FOS $= 2.5$, critical depth $D_{cr} = 12B = 4.8$ m, water table at ground level. Layers: loose sand 0–3 m ($\gamma = 16$, $\phi = 30^\circ$); medium dense sand 3–9 m ($\gamma_{sat} = 19$, $\phi = 35^\circ$); dense sand 9–14 m ($\gamma_{sat} = 20$, $\phi = 40^\circ$). The pile length is 14 m.

**Assumptions:** the given $\gamma$ values are saturated unit weights, $\gamma_w = 9.81$ kN/m³; concrete pile $\delta = 0.75\phi$; $K = 1$ (loose), $1.5$ (medium dense, interpolated) and $2$ (dense); $N_q$ for $\phi = 40^\circ$ is read from the Berezantsev curve as about 130 (an approximate chart reading). Effective stress is constant below $D_{cr}$.

#### Effective stresses
$\gamma' = 6.19,\ 9.19,\ 10.19$ kN/m³ for the three layers.

$$
\sigma'_v(3\text{ m}) = 6.19 \times 3 = 18.57,\qquad \sigma'_v(D_{cr}) = 18.57 + 9.19 \times 1.8 = 35.11\ \text{kN/m}^2
$$

#### Shaft friction ($Q_s = K\sigma'_{avg}\tan\delta\,P\,\Delta L$, $P = \pi \times 0.4 = 1.257$ m)
| Layer | $\sigma'_{avg}$ (kPa) | $K$ | $\delta$ | $f_s$ (kPa) | $Q_s$ (kN) |
|---|---|---|---|---|---|
| 0–3 m | 9.29 | 1 | 22.5° | 3.85 | 14.5 |
| 3–9 m | 32.63 | 1.5 | 26.25° | 24.14 | 182.0 |
| 9–14 m | 35.11 | 2 | 30° | 40.54 | 254.7 |

$Q_s = 451.2$ kN.

#### End bearing
$$
\begin{aligned} q' &= 35.11\ \text{kN/m}^2 \ (\text{at } D_{cr}) \\ q_p &= q'N_q = 35.11 \times 130 = 4564.6\ \text{kN/m}^2 \le 50N_q\tan\phi = 5454\ \text{kN/m}^2 \\ Q_p &= 4564.6 \times 0.1257 = 573.6\ \text{kN} \end{aligned}
$$

#### Capacity
$$
\begin{aligned} Q_u &= 451.2 + 573.6 = 1024.8\ \text{kN} \\ Q_{all} &= \frac{1024.8}{2.5} = 409.9\ \text{kN per pile} \end{aligned}
$$

For driven piles in sand at 2.5B spacing the group efficiency is taken as $\eta = 1$ (soil is compacted), so

$$
Q_{g,all} = 16 \times 409.9 = 6559\ \text{kN}
$$

**Answer: allowable load of the group $\approx 6560$ kN** (410 kN per pile). If the conservative Converse–Labarre efficiency ($\theta = 21.8^\circ$, $\eta = 0.637$) were used, the group value would be about 4176 kN.

## Check yourself

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

1. With the water table at ground level, effective stress is computed using:
   A. Total unit weight
   B. Submerged unit weight gamma' = gamma_sat - gamma_w
   C. Dry unit weight
   D. Zero
   Answer: B. Below water, gamma' = gamma_sat - 9.81 is used, e.g. 6.19 kN/m3 for loose sand. (Key point: Effective stress with water table)

2. Critical depth D_cr = 12B = 4.8 m means:
   A. Friction is zero below 4.8 m
   B. Effective stress is taken constant below 4.8 m
   C. End bearing is ignored
   D. The pile is shortened to 4.8 m
   Answer: B. Effective vertical stress stops increasing below D_cr and is held constant. (Key point: Critical depth)

3. The end bearing is q'N_q A_p, where q' is:
   A. Stress at the tip with no limit
   B. Effective stress at D_cr (35.11 kPa), checked against the limit 50 N_q tan(phi)
   C. Total stress at 14 m
   D. Average stress of all layers
   Answer: B. q' = 35.11 kPa at D_cr, and q_p is checked against the limiting value. (Key point: End bearing)

4. Why is eta = 1 used for the group?
   A. Sand has no skin friction
   B. Block failure governs
   C. Driven piles in sand at 2.5B compact the soil
   D. Clay is present
   Answer: C. Driven piles at this spacing compact sand, so no reduction is applied; group = 16 x 409.9 = 6559 kN. (Key point: Group efficiency)

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