---
title: "Relative density of sand numerical"
subject: "Soil Mechanics"
chapter: 2
key: "semester_4/soil_mechanics:relative-density-sand-numerical"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/relative-density-sand-numerical
site: BCT Engineering
---

# Relative density of sand numerical

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

- Asked in: 2074 Bhadra, 8 marks
- Video: [Relative Density of soil](https://www.youtube.com/watch?v=k3AIhoILWUk) — Elementary Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/relative-density-sand-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2

## Question

A relative density test conducted on a sandy soil obtained the following results: maximum void ratio = 1.25, minimum void ratio = 0.45, relative density = 40% and G = 2.65. Find the dry density of the soil in the present state. If a 3 m thickness of this stratum is densified to a relative density of 60%, how much will the soil reduce in thickness? What will be the new density in dry and saturated conditions?

## Short answer

Using Dr=40%, e0=0.93, present dry unit weight=13.47 kN/m³. After densifying to 60%, thickness reduces by 0.249 m, new dry unit weight=14.69 kN/m³ and saturated unit weight=18.95 kN/m³.

## Answer

#### Given
$e_{max} = 1.25$, $e_{min} = 0.45$, $D_r = 40\%$, $G = 2.65$, $H = 3$ m, $\gamma_w = 9.81$ kN/m³.

#### 1. Present state
$$
D_r = \frac{e_{max} - e}{e_{max} - e_{min}} \Rightarrow e_0 = 1.25 - 0.40(0.80) = 0.93
$$
$$
\gamma_d = \frac{G\gamma_w}{1+e_0} = \frac{2.65 \times 9.81}{1.93} = 13.47\ \text{kN/m}^3\ (\rho_d \approx 1373\ \text{kg/m}^3)
$$

#### 2. After densifying to $D_r = 60\%$
$$
e_1 = 1.25 - 0.60(0.80) = 0.77
$$
Reduction in thickness (volume of solids constant, area constant):
$$
\frac{\Delta H}{H} = \frac{e_0 - e_1}{1+e_0} \Rightarrow \Delta H = 3 \times \frac{0.93 - 0.77}{1.93} = 0.249\ \text{m}
$$
New thickness $= 3 - 0.249 = 2.751$ m.

#### 3. New unit weights
$$
\gamma_d = \frac{2.65 \times 9.81}{1.77} = 14.69\ \text{kN/m}^3
$$
$$
\gamma_{sat} = \frac{(G+e_1)\gamma_w}{1+e_1} = \frac{(2.65+0.77) \times 9.81}{1.77} = 18.95\ \text{kN/m}^3
$$

| Item | Value |
|---|---|
| Present $e$ | 0.93 |
| Present $\gamma_d$ | 13.47 kN/m³ |
| Reduction in thickness | 0.249 m (about 249 mm) |
| New $\gamma_d$ | 14.69 kN/m³ |
| New $\gamma_{sat}$ | 18.95 kN/m³ |

**Answer: present $\gamma_d = 13.47$ kN/m³; thickness reduces by 0.249 m; new $\gamma_d = 14.69$ kN/m³ and $\gamma_{sat} = 18.95$ kN/m³.**

## Check yourself

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

1. With emax=1.25, emin=0.45 and Dr=40%, what is the present void ratio?
   A. 0.57
   B. 0.93
   C. 0.80
   D. 1.05
   Answer: B. e = emax - Dr(emax-emin) = 1.25 - 0.4x0.8 = 0.93. (Key point: Present void ratio)

2. Which expression gives the present dry unit weight?
   A. G gamma_w/(1+e) = 2.65x9.81/1.93 = 13.47 kN/m3
   B. G gamma_w/e = 27.95 kN/m3
   C. (G+e) gamma_w/(1+e) = 18.2 kN/m3
   D. G gamma_w x (1+e) = 50.2 kN/m3
   Answer: A. Dry unit weight = G gamma_w/(1+e). (Key point: Present dry unit weight)

3. Why can the thickness reduction be found from void ratios as dH/H = (e0-e1)/(1+e0)?
   A. Voids volume stays constant while solids compress
   B. The water content stays constant
   C. Relative density is proportional to thickness
   D. Solids volume and area are constant, so only the void height changes
   Answer: D. With constant solids and area, the change in height of voids gives the settlement. (Key point: Thickness reduction)

4. How much does the 3 m layer reduce when densified to Dr=60% (e1=0.77)?
   A. 0.48 m
   B. 0.16 m
   C. 0.249 m
   D. 0.083 m
   Answer: C. dH = 3 x (0.93-0.77)/1.93 = 0.249 m. (Key point: Reduction value)

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