---
title: "Cylindrical specimen, 95% compaction"
subject: "Soil Mechanics"
chapter: 5
key: "semester_4/soil_mechanics:cylindrical-specimen-relative-compaction"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/cylindrical-specimen-relative-compaction
site: BCT Engineering
---

# Cylindrical specimen, 95% compaction

Soil Mechanics · Chapter 5: Soil Compaction

- Asked in: 2078 Poush, 3 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/cylindrical-specimen-relative-compaction
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-5

## Question

A cylindrical specimen of a cohesive soil of 10 cm diameter and 20 cm height was prepared by compaction in a mold. Taking the specific gravity of soil solid as 2.65 and the wet weight of this specimen as 30 kN [as printed] and water content as 15%, find the following: (i) dry unit weight, void ratio and degree of saturation of this cylindrical specimen; (ii) if 95% of relative compaction is to be achieved in the field, what should be the dry unit weight of compacted soil of the same soil specimen in the field.

## Short answer

Assuming the printed wet weight is 30 N, volume is 1.571×10⁻³ m³. Dry unit weight γ_d = 19.10/1.15 = 16.61 kN/m³; void ratio e = 0.565; saturation S = 70.3%. For 95% relative compaction, field dry unit weight must be at least 15.78 kN/m³.

## Answer

#### Given
$D = 10$ cm, $H = 20$ cm, $G = 2.65$, $w = 15\%$, $\gamma_w = 9.81$ kN/m³.

**Assumption:** the weight "30 kN" is a misprint for **30 N**, since 30 kN in this small mould would give an impossible unit weight of about 19,000 kN/m³.

#### Volume and unit weights
$$
V = \frac{\pi}{4}(0.10)^2(0.20) = 1.571 \times 10^{-3}\ \text{m}^3
$$
$$
\gamma = \frac{30\ \text{N}}{1.571 \times 10^{-3}} = 19\,099\ \text{N/m}^3 = 19.10\ \text{kN/m}^3
$$
$$
\gamma_d = \frac{\gamma}{1+w} = \frac{19.10}{1.15} = 16.61\ \text{kN/m}^3
$$

#### Void ratio
$$
e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{16.61} - 1 = 0.565
$$

#### Degree of saturation
$$
S = \frac{wG}{e} = \frac{0.15 \times 2.65}{0.565} = 0.703 = 70.3\%
$$

#### (ii) Field requirement
Taking the specimen's dry unit weight as the maximum (laboratory) dry unit weight:
$$
\gamma_{d,field} = 0.95 \times \gamma_{d,max} = 0.95 \times 16.61 = 15.78\ \text{kN/m}^3
$$

**Answer: (i) $\gamma_d = 16.61$ kN/m³, $e = 0.565$, $S = 70.3\%$; (ii) field $\gamma_d \ge 15.78$ kN/m³.**

## Check yourself

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

1. The weight '30 kN' is taken as 30 N because:
   A. Kilonewtons are not used
   B. Volume is huge
   C. 30 kN gives an impossible unit weight (about 19,000 kN/m3)
   D. Soil is light
   Answer: C. Small mould cannot hold 30 kN. (Key point: Misprint)

2. Dry unit weight from bulk 19.10 kN/m3 and w = 15%:
   A. 16.61
   B. 15.78
   C. 19.10
   D. 21.97
   Answer: A. 19.10/1.15 = 16.61. (Key point: Dry unit weight)

3. S = wG/e with e = 0.565:
   A. 85%
   B. 100%
   C. 50%
   D. 70.3%
   Answer: D. 0.3975/0.565 = 0.703. (Key point: Degree of saturation)

4. Field dry unit weight for 95% RC:
   A. 16.61
   B. 16.61/0.95
   C. 0.95 x 16.61 = 15.78
   D. 0.95 x 19.10
   Answer: C. gd,field = RC x gd,max. (Key point: Field requirement)

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