Chapter 2, Problem 2.1P

(a)

To determine

Find the void ratio of soil specimen.

Answer to Problem 2.1P

The void ratio of soil is 0.76.

Explanation of Solution

Given information:

The mass (m) of the soil is $87.5\text{kg}$.

The volume (v) of the soil is $0.05{\text{m}}^{\text{3}}$.

The moisture content (w) is 15%.

The specific gravity $\left(G_s\right)$ of soil solids is 2.68.

Calculation:

Consider the unit weight of water $\left({\gamma }_w\right)$ is $9.81\text{kN}/{\text{m}}^{\text{3}}$.

Calculate the moist unit weight of the soil $\left(\gamma \right)$ using the relation.

$\begin{array}{c}\gamma =\frac{W}{V}\\ =\frac{mg}{V}\\ =\frac{87.5\text{kg}\times 9.81{\text{m/s}}^{\text{2}}}{0.05{\text{m}}^3}\times \frac{1\text{kN}}{1,000\text{N}}\\ =17.17\text{kN}/{\text{m}}^{\text{3}}\end{array}$

Calculate the dry unit weight of the soil $\left({\gamma }_d\right)$ using the relation.

$\begin{array}{c}{\gamma }_d=\frac{\gamma }{1+w}\\ =\frac{17.17}{1+\frac{15}{100}}\\ =14.93\text{kN}/{\text{m}}^{\text{3}}\end{array}$

Calculate the void ratio (e) using the relation as shown below:

$\begin{array}{l}{\gamma }_d=\frac{G_s{\gamma }_w}{1+e}\\ 14.93=\frac{\left(2.68\times 9.81\right)}{1+e}\\ 14.93+14.93e=26.29\\ e=0.76\end{array}$

Hence, the void ratio of soil is 0.76.

(b)

To determine

Calculate the porosity of soil.

Answer to Problem 2.1P

The porosity of soil is $\underset{\_}{0.43}$.

Explanation of Solution

Calculation:

Refer to part (a).

The void ratio of soil is $0.76$.

Calculate the void ratio $\left(e\right)$ using the relation.

$n=\frac{e}{1+e}$

Substitute $0.76$ for e.

$\begin{array}{c}n=\frac{0.76}{1+0.76}\\ =0.43\end{array}$

Hence, the porosity of soil is $\underset{\_}{0.43}$.

(c)

To determine

Calculate the dry unit weight of soil.

Answer to Problem 2.1P

The dry unit weight of soil is $\underset{\_}{14.93\text{kN}/{\text{m}}^{\text{3}}}$.

Explanation of Solution

Calculation:

Calculate the dry unit weight of the soil $\left({\gamma }_d\right)$ using the relation.

Refer to part (a).

Base on result, the dry unit weight of the soil is $14.93\text{kN}/{\text{m}}^{\text{3}}$.

Thus, the dry unit weight of soil is $\underset{\_}{14.93\text{kN}/{\text{m}}^{\text{3}}}$.

(d)

To determine

Calculate the moist unit weight of soil.

Answer to Problem 2.1P

The moist unit weight of soil is $\underset{\_}{17.17\text{kN}/{\text{m}}^{\text{3}}}$.

Explanation of Solution

Calculation:

Calculate the moist unit weight of the soil $\left(\gamma \right)$ using the relation.

Refer to part (a).

Base on result, the dry unit weight of the soil is $17.17\text{kN}/{\text{m}}^{\text{3}}$.

Thus, the moist unit weight of soil is $\underset{\_}{17.17\text{kN}/{\text{m}}^{\text{3}}}$.

(e)

To determine

Calculate the degree of saturation.

Answer to Problem 2.1P

The degree of saturation is $\underset{\_}{53\%}$.

Explanation of Solution

Calculation:

Refer to part (a).

The void ratio of soil is $0.76$.

Calculate the degree of saturation $\left(S\right)$ using the relation.

$\begin{array}{c}S=\frac{w{G}_s}{e}\\ =\frac{\frac{15}{100}\times 2.71}{0.76}\\ =0.534\times 100\%\\ =53\%\end{array}$

Hence, the degree of saturation is $\underset{\_}{53\%}$.