Soil Dry Density Calculation

The question asks us to calculate the theoretical maximum dry density for a soil sample given its specific gravity and optimum moisture content.

The theoretical maximum dry density of a soil is achieved when there are zero air voids in the soil mass. This means the soil is fully saturated with water at its optimum moisture content.

Given Information:

• Specific Gravity of soil solids (G) = 2.7
• Optimum Moisture Content (w) = 16% = 0.16
• Unit weight of water ($\gamma_w$) is typically taken as 1 g/cm³ for calculations involving g/cm³ units.

Formula for Theoretical Maximum Dry Density

The relationship between dry density ($\gamma_d$), specific gravity (G), moisture content (w), and unit weight of water ($\gamma_w$) for a soil with zero air voids (i.e., theoretical maximum dry density) is given by the formula:

$\gamma_{d,max} = \frac{G\gamma_w}{1+wG}$

This formula is derived from the basic phase relationships of soil, assuming the degree of saturation (S) is 1 (100%). The relationship $eS = wG$ becomes $e = wG$ when S=1, and then substituting this void ratio (e) into the dry density formula $\gamma_d = \frac{G\gamma_w}{1+e}$ gives the expression for maximum dry density.

Step-by-step Calculation:

Substitute the given values into the formula:

$G = 2.7$

$w = 0.16$

$\gamma_w = 1 \text{ g/cm}^3$

$\gamma_{d,max} = \frac{2.7 \times 1}{1 + 0.16 \times 2.7}$

First, calculate the term in the denominator:

$0.16 \times 2.7 = 0.432$

Now, add 1 to this value:

$1 + 0.432 = 1.432$

Finally, divide the numerator by the denominator:

$\gamma_{d,max} = \frac{2.7}{1.432}$

$\gamma_{d,max} \approx 1.88547 \text{ g/cm}^3$

Rounding to three decimal places, we get 1.885 g/cm³.

Result Summary:

The theoretical maximum dry density for the soil sample is approximately 1.885 g/cm³.

Comparing this calculated value with the given options, we find that it matches option 3.