---
title: "Core cutter field density numerical"
subject: "Soil Mechanics"
chapter: 2
key: "semester_4/soil_mechanics:core-cutter-density-numerical"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/core-cutter-density-numerical
site: BCT Engineering
---

# Core cutter field density numerical

Soil Mechanics · Chapter 2: Solids- Water –Air Relations and Index properties of soils

- Asked in: 2078 Baisakh, 6 marks
- Video: [Determination of Dry Density of Soil by Core Cutter Method](https://www.youtube.com/watch?v=5rDHjZ_RJq0) — NCTEL
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/core-cutter-density-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2

## Question

A field density test was conducted by core cutter method and the following data was obtained.
Weight of empty core-cutter = 22.80 N
Weight of soil and core-cutter = 50.05 N
Inside diameter of the core-cutter = 90.0 mm
Height of core-cutter = 180.0 mm
Weight of wet sample for moisture determination = 0.5405 N
Weight of oven-dry sample = 0.5112 N
Specific gravity of soil grains = 2.72
Determine (i) dry density, (ii) void ratio, and (iii) degree of saturation.

## Short answer

Dry density = 22.51 kN/m³ (2294 kg/m³), void ratio = 0.186, degree of saturation = 84.0%.

## Answer

#### Given
Empty cutter 22.80 N; soil + cutter 50.05 N; $D = 90$ mm; $H = 180$ mm; $G = 2.72$; moisture sample 0.5405 N wet and 0.5112 N dry. $\gamma_w = 9.81$ kN/m³.

#### Volume and bulk unit weight
$$
\begin{aligned}
W_{soil} &= 50.05 - 22.80 = 27.25\ \text{N} \\
V &= \frac{\pi}{4}(0.09)^2(0.18) = 1.1451 \times 10^{-3}\ \text{m}^3 \\
\gamma &= \frac{27.25}{1.1451 \times 10^{-3}} = 23\,797\ \text{N/m}^3 = 23.80\ \text{kN/m}^3
\end{aligned}
$$

#### Water content
$$
w = \frac{0.5405 - 0.5112}{0.5112} = 0.0573 = 5.73\%
$$

#### (i) Dry density
$$
\gamma_d = \frac{\gamma}{1+w} = \frac{23.80}{1.0573} = 22.51\ \text{kN/m}^3
$$
$$
\rho_d = \frac{22.51}{9.81} = 2.294\ \text{Mg/m}^3 = 2294\ \text{kg/m}^3
$$

#### (ii) Void ratio
$$
e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.72 \times 9.81}{22.51} - 1 = 0.186
$$

#### (iii) Degree of saturation
$$
S = \frac{wG}{e} = \frac{0.0573 \times 2.72}{0.1856} = 0.840 = 84.0\%
$$

**Answer: $\gamma_d = 22.51$ kN/m³ ($\rho_d = 2294$ kg/m³), $e = 0.186$, $S = 84.0\%$.**

## Check yourself

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

1. The volume of the core cutter is:
   A. 2.29 x 10^-3 m3
   B. 4.58 x 10^-3 m3
   C. 1.145 m3
   D. 1.1451 x 10^-3 m3
   Answer: D. V = (pi/4)(0.09)^2(0.18). (Key point: Cutter volume)

2. Water content from the moisture sample is:
   A. 5.4%
   B. 5.73%
   C. 94.6%
   D. 0.57%
   Answer: B. (0.5405 - 0.5112)/0.5112 = 5.73%, using dry weight as the base. (Key point: Water content)

3. Dry unit weight is:
   A. 12.3 kN/m3
   B. 23.80 kN/m3
   C. 25.2 kN/m3
   D. 22.51 kN/m3
   Answer: D. gamma = 27.25/1.1451e-3 = 23.80 kN/m3; divide by 1.0573. (Key point: Dry unit weight)

4. Void ratio is:
   A. 0.55
   B. 0.186
   C. 1.86
   D. 0.84
   Answer: B. 2.72 x 9.81/22.51 - 1 = 0.186. (Key point: Void ratio)

5. Degree of saturation is about:
   A. 16%
   B. 57%
   C. 100%
   D. 84%
   Answer: D. S = wG/e = 0.0573 x 2.72/0.1856 = 84%. (Key point: Saturation)

6. Weight of soil in the cutter is:
   A. 50.05 N
   B. 27.25 N
   C. 22.80 N
   D. 72.85 N
   Answer: B. Subtract the empty cutter weight: 50.05 - 22.80. (Key point: Soil weight)

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