---
title: "Compacted soil 18 kN/m³, saturation"
subject: "Soil Mechanics"
chapter: 5
key: "semester_4/soil_mechanics:max-dry-weight-18-saturation"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/max-dry-weight-18-saturation
site: BCT Engineering
---

# Compacted soil 18 kN/m³, saturation

Soil Mechanics · Chapter 5: Soil Compaction

- Asked in: 2075 Bhadra, 2 marks
- Video: [Chapter 6 Soil Compaction - Example 1 (Standard Proctor Test)](https://www.youtube.com/watch?v=v_1tQfxBowQ) — uSeeGeo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/max-dry-weight-18-saturation
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-5

## Question

The maximum dry density of a compacted soil mass is found to be 18 kN/m³ with optimum water content being 15%. Find the degree of saturation of this compacted soil if the specific gravity of soil solids is given as 2.65. What will be the value of the maximum dry density it can be further compacted to?

## Short answer

With γ_d = 18 kN/m³, w = 15%, G = 2.65: e = 0.444, S = 89.5%. The maximum possible dry unit weight at this water content (zero air voids) is 18.60 kN/m³.

## Answer

#### Given
$\gamma_{d} = 18$ kN/m³, $w = 15\%$, $G = 2.65$, $\gamma_w = 9.81$ kN/m³.

#### Degree of saturation
$$
e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{18} - 1 = 0.444
$$
$$
S = \frac{wG}{e} = \frac{0.15 \times 2.65}{0.444} = 0.895 = 89.5\%
$$

#### Further compaction
The maximum value the dry unit weight can reach at the same water content is when all air is expelled ($S = 100\%$), on the ZAV line:
$$
\gamma_{d,max} = \frac{G\gamma_w}{1 + wG} = \frac{2.65 \times 9.81}{1 + 0.15 \times 2.65} = \frac{25.99}{1.3975} = 18.60\ \text{kN/m}^3
$$

**Answer: $S = 89.5\%$; the soil can at most be compacted to 18.60 kN/m³ (in practice slightly less, since some air always remains).**

## Check yourself

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

1. gd = 18, G = 2.65 gives e:
   A. 0.25
   B. 0.65
   C. 1.0
   D. 0.444
   Answer: D. 2.65*9.81/18 - 1 = 0.444. (Key point: Void ratio)

2. S = wG/e with w = 0.15 gives about:
   A. 100%
   B. 95%
   C. 89.5%
   D. 72%
   Answer: C. 0.3975/0.444 = 0.895. (Key point: Saturation)

3. Maximum dry density reachable at same w is on the ZAV line:
   A. S = 100%, 18.60 kN/m3
   B. 25.99 kN/m3
   C. 18.0 kN/m3 (cannot increase)
   D. S = 50%, 16 kN/m3
   Answer: A. G gw/(1+wG) = 25.99/1.3975 = 18.60. (Key point: Further compaction)

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