Chapter 3, Problem 3.7P

(a)

To determine

Calculate the moist unit weight of soil.

Answer to Problem 3.7P

The moist unit weight of soil is $\underset{\_}{125\text{lb}/{\text{ft}}^3}$.

Explanation of Solution

Given information:

The volume $\left(V\right)$ of the soil is $0.1{\text{ft}}^3$.

The moist mass $\left(W\right)$ of the soil is $12.5\text{lb}$.

The moisture content $\left(w\right)$ is $14\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.71$.

Calculation:

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

$\gamma =\frac{W}{V}$

Substitute $12.5\text{lb}$ for W and $0.1{\text{ft}}^3$ for V.

$\begin{array}{c}\gamma =\frac{12.5}{0.1}\\ =125\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Hence, the moist unit weight of soil is $\underset{\_}{125\text{lb}/{\text{ft}}^3}$.

(b)

To determine

Calculate the dry unit weight of soil.

Answer to Problem 3.7P

The dry unit weight of soil is $\underset{\_}{109.64\text{lb}/{\text{ft}}^3}$.

Explanation of Solution

Given information:

The volume $\left(V\right)$ of the soil is $0.1{\text{ft}}^3$.

The moist mass $\left(W\right)$ of the soil is $12.5\text{lb}$.

The moisture content $\left(w\right)$ is $14\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.71$.

Calculation:

Refer to part (a).

The moist unit weight of soil is $125\text{lb}/{\text{ft}}^3$.

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

${\gamma }_d=\frac{\gamma }{1+w}$

Substitute $125\text{lb}/{\text{ft}}^3$ for $\gamma$ and $14\%$ for w.

$\begin{array}{c}{\gamma }_d=\frac{125}{1+\frac{14}{100}}\\ =\frac{125}{1.14}\\ =109.64\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Hence, the dry unit weight of soil is $\underset{\_}{109.64\text{lb}/{\text{ft}}^3}$.

(c)

To determine

Calculate the void ratio of soil.

Answer to Problem 3.7P

The void ratio of soil is $\underset{\_}{0.54}$.

Explanation of Solution

Given information:

The volume $\left(V\right)$ of the soil is $0.1{\text{ft}}^3$.

The moist mass $\left(W\right)$ of the soil is $12.5\text{lb}$.

The moisture content $\left(w\right)$ is $14\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.71$.

Calculation:

Refer to part (b).

The dry unit weight of soil is $109.64\text{lb}/{\text{ft}}^3$.

Consider the unit weight of water $\left({\gamma }_w\right)$ is $62.4\text{lb}/{\text{ft}}^3$.

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

$e=\frac{G_s{\gamma }_w}{{\gamma }_d}-1$

Substitute $2.71$ for $G_s$, $109.64\text{lb}/{\text{ft}}^3$ for ${\gamma }_d$, and $62.4\text{lb}/{\text{ft}}^3$ for ${\gamma }_w$.

$\begin{array}{c}e=\frac{2.71\times 62.4}{109.64}-1\\ =\frac{169.104}{109.64}-1\\ =1.54-1\\ =0.54\end{array}$

Hence, the void ratio of soil is $\underset{\_}{0.54}$.

(d)

To determine

Calculate the porosity of soil.

Answer to Problem 3.7P

The porosity of soil is $\underset{\_}{0.35}$.

Explanation of Solution

Given information:

The volume $\left(V\right)$ of the soil is $0.1{\text{ft}}^3$.

The moist mass $\left(W\right)$ of the soil is $12.5\text{lb}$.

The moisture content $\left(w\right)$ is $14\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.71$.

Calculation:

Refer to part (c).

The void ratio of soil is $0.54$.

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

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

Substitute $0.54$ for e.

$\begin{array}{c}n=\frac{0.54}{1+0.54}\\ =\frac{0.54}{1.54}\\ =0.35\end{array}$

Hence, the porosity of soil is $\underset{\_}{0.35}$.

(e)

To determine

Calculate the degree of saturation.

Answer to Problem 3.7P

The degree of saturation is $\underset{\_}{70.2\%}$.

Explanation of Solution

Given information:

The volume $\left(V\right)$ of the soil is $0.1{\text{ft}}^3$.

The moist mass $\left(W\right)$ of the soil is $12.5\text{lb}$.

The moisture content $\left(w\right)$ is $14\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.71$.

Calculation:

Refer to part (c).

The void ratio of soil is $0.54$.

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

$S=\frac{w{G}_s}{e}$

Substitute $14\%$ for w, $2.71$ for $G_s$, and $0.54$ for e.

$\begin{array}{c}S=\frac{\frac{14}{100}\times 2.71}{0.54}\\ =\frac{0.3794}{0.54}\\ =0.702\times 100\%\\ =70.2\%\end{array}$

Hence, the degree of saturation is $\underset{\_}{70.2\%}$.

(f)

To determine

Calculate the volume of water.

Answer to Problem 3.7P

The volume of water is $\underset{\_}{0.024{\text{ft}}^3}$.

Explanation of Solution

Given information:

The volume $\left(V\right)$ of the soil is $0.1{\text{ft}}^3$.

The moist mass $\left(W\right)$ of the soil is $12.5\text{lb}$.

The moisture content $\left(w\right)$ is $14\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.71$.

Calculation:

Refer to part (a).

The moist unit weight of soil $\left(\gamma \right)$ is $125\text{lb}/{\text{ft}}^3$.

Refer to part (b).

The dry unit weight of soil $\left({\gamma }_d\right)$ is $109.64\text{lb}/{\text{ft}}^3$.

Consider the unit weight of water $\left({\gamma }_w\right)$ is $62.4\text{lb}/{\text{ft}}^3$.

Calculate the volume of water $\left(V_w\right)$ using the relation.

$V_w=\frac{\left(\gamma -{\gamma }_d\right)V}{{\gamma }_w}$

Substitute $125\text{lb}/{\text{ft}}^3$ for $\gamma$, $109.64\text{lb}/{\text{ft}}^3$ for ${\gamma }_d$, $62.4\text{lb}/{\text{ft}}^3$ for ${\gamma }_w$, and $0.1{\text{ft}}^3$ for V.

$\begin{array}{c}{V}_w=\frac{\left(125-109.64\right)\times 0.1}{62.4}\\ =\frac{1.536}{62.4}\\ =0.024{\text{ft}}^3\end{array}$

Therefore, the volume of water is $\underset{\_}{0.024{\text{ft}}^3}$.