During the compaction test, the weight of compacted soil specimen along with mould is 38.2 N. The volume and weight of mould are 0.95×10-3 m³ and 20.5 N respectively and the water content is 12%. The dry unit weight of the compacted specimen will be nearly
In soil mechanics, the compaction test is crucial for determining the relationship between the water content and the dry unit weight of a soil. This relationship helps in finding the optimum water content at which maximum dry unit weight is achieved. The dry unit weight is a fundamental property that indicates the amount of solid particles per unit volume, excluding water.
To find the dry unit weight of the compacted soil specimen, we need to follow these steps:
The total weight of the compacted soil specimen along with the mould is given. To find the weight of the soil alone, we subtract the weight of the mould.
The volume of the compacted soil specimen is equal to the internal volume of the mould.
The bulk unit weight ($\gamma_b$) is the total weight of the soil (solids + water) per unit volume.
It is common practice to express unit weight in $\text{kN/m}^3$. We convert from $\text{N/m}^3$ to $\text{kN/m}^3$ by dividing by 1000.
The dry unit weight ($\gamma_d$) is the weight of soil solids per unit volume. It can be calculated from the bulk unit weight and the water content ($w$) using the formula:
$\gamma_d = \frac{\gamma_b}{1 + w}$
The water content is given as 12%, which should be used as a decimal in the formula.
Comparing this result to the given options, $16.63 \text{ kN/m}^3$ is nearest to $16.6 \text{ kN/m}^3$.
| Parameter | Value | Units |
|---|---|---|
| Weight of Soil + Mould | 38.2 | N |
| Weight of Mould | 20.5 | N |
| Weight of Soil Specimen | 17.7 | N |
| Volume of Soil Specimen | 0.95 $\times 10^{-3}$ | m³ |
| Bulk Unit Weight ($\gamma_b$) | $\approx 18.63$ | kN/m³ |
| Water Content ($w$) | 0.12 | (decimal) |
| Dry Unit Weight ($\gamma_d$) | $\approx 16.63$ | kN/m³ |
| Concept | Definition/Formula | Importance |
|---|---|---|
| Compaction Test | Laboratory test to determine the relationship between water content and dry unit weight for a soil. | Helps find optimum water content and maximum dry unit weight for field compaction specifications. |
| Bulk Unit Weight ($\gamma_b$) | Total weight (solids + water) per unit volume. $\gamma_b = W/V$. | Represents the in-situ weight-volume relationship of the soil. |
| Dry Unit Weight ($\gamma_d$) | Weight of solid particles per unit volume. $\gamma_d = W_s/V$. Also $\gamma_d = \gamma_b / (1+w)$. | Indicates the density of solid particles and is used as a measure of compaction effort effectiveness. |
| Water Content ($w$) | Ratio of weight of water to weight of solids, usually expressed as a percentage. $w = (W_w/W_s) \times 100\%$. | Affects the lubrication between soil particles, influencing the degree of compaction achievable. |
The dry unit weight is a critical parameter in geotechnical engineering, particularly concerning soil compaction. Achieving a target dry unit weight in the field is essential for constructing stable foundations, embankments, road bases, and other earth structures. A higher dry unit weight for a given soil type generally indicates a denser, stronger, and less permeable soil mass, which is desirable for engineering purposes.
The results from a compaction test (like the one described in the question, often a Standard Proctor or Modified Proctor test) are used to create a compaction curve. This curve plots dry unit weight versus water content. The peak of this curve gives the maximum dry unit weight and the corresponding optimum water content.
Understanding how to calculate dry unit weight from measured values of weight, volume, and water content is fundamental for interpreting compaction test results and ensuring that field compaction meets design specifications.
When once a pocket of smoke, containing air pollutants, is released into the atmosphere from a source like an automobile or a factory chimney, it gets dispersed into the atmosphere into various directions depending upon the
1. prevailing winds
2. temperature
3. pressure conditions
Select the correct answer.
In a flow net for a sheet pile wall, the number of flow paths is 5 and the number of equipotential drops is 10. If the coefficient of permeability is 6×10-3 mm/s and the head is 4.5 m, the seepage under the wall will be nearly