---
title: "Compacted soil 17 kN/m³, OWC 16%"
subject: "Soil Mechanics"
chapter: 5
key: "semester_4/soil_mechanics:max-dry-weight-17-void-ratio"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/max-dry-weight-17-void-ratio
site: BCT Engineering
---

# Compacted soil 17 kN/m³, OWC 16%

Soil Mechanics · Chapter 5: Soil Compaction

- Asked in: 2079 Asoj, 3 marks
- 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-17-void-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-5

## Question

The maximum dry unit weight of a compacted soil mass is found to be 17 kN/m³ with optimum water content being 16%. Determine the void ratio and degree of saturation of this soil after compaction. Also, find the value of the maximum dry unit weight on the zero air void line at that optimum water content. Take specific gravity of soil solids as 2.65.

## Short answer

From γ_d = 17 kN/m³, w = 16%, G = 2.65: void ratio e = (2.65×9.81/17) - 1 = 0.529; degree of saturation S = 0.16×2.65/0.529 = 80.1%; zero air void dry unit weight = 2.65×9.81/(1+0.16×2.65) = 18.26 kN/m³.

## Answer

#### Given
$\gamma_{d} = 17$ kN/m³, $w = 16\%$ (OWC), $G = 2.65$, $\gamma_w = 9.81$ kN/m³.

#### Void ratio
$$
e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{17} - 1 = 0.529
$$

#### Degree of saturation
$$
S = \frac{wG}{e} = \frac{0.16 \times 2.65}{0.529} = 0.801 = 80.1\%
$$

#### Zero air void unit weight at $w = 16\%$
$$
\gamma_{d,zav} = \frac{G\gamma_w}{1 + wG} = \frac{2.65 \times 9.81}{1 + 0.16 \times 2.65} = \frac{25.99}{1.424} = 18.26\ \text{kN/m}^3
$$

**Answer: $e = 0.529$, $S = 80.1\%$, $\gamma_{d,zav} = 18.26$ kN/m³.**

## Check yourself

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

1. Void ratio from dry unit weight:
   A. e = G gw / gd - 1
   B. e = gd/(G gw)
   C. e = 1 - n
   D. e = wG/S
   Answer: A. gd = G gw/(1+e). (Key point: Void ratio formula)

2. gd = 17, G = 2.65, gw = 9.81 gives e closest to:
   A. 1.53
   B. 0.53
   C. 0.37
   D. 0.80
   Answer: B. 26.0/17 - 1 = 0.529. (Key point: Void ratio value)

3. S = wG/e with w = 0.16, G = 2.65, e = 0.529 gives:
   A. 100%
   B. 70%
   C. 60%
   D. 80%
   Answer: D. 0.424/0.529 = 0.80. (Key point: Saturation)

4. ZAV dry unit weight at 16% is greater than 17 because:
   A. Soil is fully saturated
   B. G is wrong
   C. Water content is zero
   D. Air remains in the compacted soil
   Answer: D. S = 80%, so ZAV 18.26 > 17. (Key point: ZAV)

## Close variants set in other exams

- [Compacted soil 18 kN/m³, porosity and ZAV](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Amax-dry-weight-18-porosity-zav) — Soil Mechanics, chapter 5, asked 2×

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