---
title: "Embankment 95% compaction, 950 cc mould"
subject: "Soil Mechanics"
chapter: 5
key: "semester_4/soil_mechanics:embankment-95-compaction-950cc-mould"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/embankment-95-compaction-950cc-mould
site: BCT Engineering
---

# Embankment 95% compaction, 950 cc mould

Soil Mechanics · Chapter 5: Soil Compaction

- Asked in: 2078 Baisakh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/embankment-95-compaction-950cc-mould
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-5

## Question

Evaluate the construction of the embankment if the required degree of compaction is 95% and the dry density of the embankment was found as 1.78 g/cc. The result of the compaction test performed in the laboratory for the same material using a 950 cc mould are as follows.

| w % | 7.7 | 11.5 | 14.6 | 17.5 | 19.7 | 21.2 |
|---|---|---|---|---|---|---|
| Mass of wet soil (kg) | 1.70 | 1.89 | 2.03 | 1.99 | 1.96 | 1.92 |

## Short answer

From the lab mould, dry densities peak at 1.865 g/cc at 14.6% water content. Required dry density at 95% compaction is 1.77 g/cc; field dry density 1.78 g/cc gives RC 95.4%, so embankment is satisfactory, though the margin is small.

## Answer

Degree of compaction (relative compaction) is $RC = \gamma_{d,field}/\gamma_{d,max}$. First find $\gamma_{d,max}$ from the lab test, then compare with the field value.

#### Lab dry density (mould volume = 950 cm³)

$$
\gamma_{bulk} = \frac{M}{950\ \text{cm}^3}, \qquad \gamma_d = \frac{\gamma_{bulk}}{1+w}
$$

| w (%) | Mass (kg) | $\gamma_{bulk}$ (g/cc) | $\gamma_d$ (g/cc) |
|---|---|---|---|
| 7.7 | 1.70 | 1.789 | 1.662 |
| 11.5 | 1.89 | 1.989 | 1.784 |
| 14.6 | 2.03 | 2.137 | 1.865 |
| 17.5 | 1.99 | 2.095 | 1.783 |
| 19.7 | 1.96 | 2.063 | 1.724 |
| 21.2 | 1.92 | 2.021 | 1.668 |

Plotting $\gamma_d$ against $w$ gives a peak at

- $\gamma_{d,max} \approx 1.865$ g/cc, OMC $\approx 14.6\%$

#### Check of the embankment

Required dry density $= 0.95 \times 1.865 = 1.77$ g/cc.

$$
RC = \frac{1.78}{1.865} \times 100 = 95.4\%
$$

**Answer: RC = 95.4% (> 95%), field dry density 1.78 g/cc > required 1.77 g/cc. The embankment construction is satisfactory.**

The margin is small, so the result is acceptable but only just; the moisture content of the field fill should also be checked to be near the OMC.

## Check yourself

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

1. Bulk density of test 1 (1.70 kg, 950 cc):
   A. 1.789 g/cc
   B. 1.66
   C. 0.558
   D. 1.70
   Answer: A. 1700/950 = 1.789. (Key point: Bulk density)

2. Lab MDD and OMC:
   A. 1.784 at 11.5%
   B. 2.137 at 14.6%
   C. 1.783 at 17.5%
   D. 1.865 g/cc at 14.6%
   Answer: D. Peak dry density at w = 14.6%. (Key point: MDD)

3. Required dry density = 0.95 x MDD:
   A. 1.95
   B. 1.77 g/cc
   C. 1.60
   D. 1.865
   Answer: B. 0.95*1.865 = 1.77. (Key point: Required)

4. Field 1.78 g/cc gives RC:
   A. 85%
   B. 90%, unsatisfactory
   C. 100%
   D. 95.4%, satisfactory
   Answer: D. 1.78/1.865 = 95.4% > 95%. (Key point: Evaluation)

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