4.2.2. Example 2

Example 2: The dry density of a sand with porosity of 0.387 is 1600 kg/m3. Find the void ratio of the soil and the specific gravity of the soil solids. [Take n = 0.387 = 1600 kg/m3]

This example introduces two important concepts in soil mechanics: dry density and porosity. Dry density is the mass of the dry soil per unit volume, typically expressed in kg/m³. In this case, we have a dry density of 1600 kg/m³. Porosity (n) is the ratio of the volume of voids to the total volume of the soil and is given as 0.387.

To find the void ratio (e), we will use the formula:

• e = n / (1 - n)

Substituting the given porosity (n = 0.387), we can calculate the void ratio.

After calculating void ratio, the next task is to find the specific gravity (G) of the soil solids. Specific gravity is a dimensionless quantity and is calculated using:

• G = (dry density of soil solids) / (density of water)

Assuming the density of water is approximately 1000 kg/m³, we can determine the specific gravity once we calculate the dry density of soil solids.

Think of a sponge. When dry, it has a certain 'dry density' based on how much material there is inside (the solid parts of the sponge). Now, when we consider porosity, it tells us how much empty space or 'voids' are present in the sponge where water could be held. The void ratio will tell us how much of the volume of the sponge is filled with these voids compared to the solid parts, and specific gravity will help us understand how heavy the sponge is when you compare it to an equal volume of water.

(a) e = = = 0.631

To calculate the void ratio (e), we use the formula: e = n / (1 - n). Substituting the value of n:

1. Start with n = 0.387.
2. Calculate (1 - n): (1 - 0.387) = 0.613.
3. Finally, divide n by (1 - n): e = 0.387 / 0.613.

This results in a void ratio of approximately 0.631. The void ratio indicates how much space is available for water in the soil compared to the soil's solid components. Higher void ratios mean more space for water, which is crucial for understanding soil behavior in construction and agriculture.

Imagine you have a jar filled with marbles (the solid parts of the soil) and there are spaces in between them (the voids). The void ratio helps us understand how many empty spaces there are compared to the number of marbles. If there are more spaces for air or water in the jar, it means the jar is less packed with marbles. This is similar to how soil can hold more water when it has a high void ratio.

(b) = G =

After finding the void ratio, the next step is to calculate the specific gravity (G) of the soil solids. Specific gravity is calculated using the dry density of soil solids. The formula used is: G = (Dry Density of Soil Solids) / (Density of Water).

Given the information, we assume density of water is approximately 1000 kg/m³, so: Once we have the dry density of soil solids, simply plug in the values to find G. Specific gravity helps understand how heavy the soil solids are compared to water and is fundamental in soil mechanics, especially for comparing different soils.

You can think of specific gravity like comparing the weight of different balls. If you have a basketball filled with air (less dense) compared to a basketball filled with sand (denser), specific gravity tells us how heavy each ball would be if both were the same size. This concept is similar to soil where different types of soil solids have varying weights contributing to their overall density and behavior.

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