Chapter 3, Problem 3.4P

(a)

To determine

Calculate the dry unit weight of the soil.

Answer to Problem 3.4P

The dry unit weight of the soil is $\underset{\_}{108.22\text{lb}/{\text{ft}}^{\text{3}}}$.

Explanation of Solution

Given information:

The diameter of the cylindrical mold $\left(d\right)$ is $4\text{in}\text{.}$.

The height of the cylindrical mold $\left(h\right)$ is $4.58\text{in}\text{.}$.

The weight of the compacted soil $\left(W\right)$ is $4\text{lb}$.

The moisture content $\left(w\right)$ is $12\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.72$.

Calculation:

Calculate the volume of the mold $\left(V\right)$ using the relation.

$V=\frac{1}{4}\pi d^{2}h$

Substitute $4\text{in}\text{.}$ for d and $4.58\text{in}\text{.}$ for h.

$\begin{array}{c}V=\frac{1}{4}\pi \times 4^2\times 4.58\\ =57.554\text{in}{\text{.}}^3\times {\left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)}^3\\ =0.033{\text{ft}}^3\end{array}$

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

${\gamma }_d=\frac{W}{V\left(1+w\right)}$

Substitute 4 lb for W, $0.033{\text{ft}}^3$ for V, and $12\%$ for w.

$\begin{array}{c}{\gamma }_d=\frac{4}{0.033\times \left(1+\frac{12}{100}\right)}\\ =\frac{4}{0.033\times 1.12}\\ =108.22\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Hence, the dry unit weight of the soil is $\underset{\_}{108.22\text{lb}/{\text{ft}}^{\text{3}}}$.

(b)

To determine

Calculate the void ratio of the soil.

Answer to Problem 3.4P

The void ratio of the soil is $\underset{\_}{0.568}$.

Explanation of Solution

Given information:

The diameter of the cylindrical mold $\left(d\right)$ is $4\text{in}\text{.}$.

The height of the cylindrical mold $\left(h\right)$ is $4.58\text{in}\text{.}$.

The weight of the compacted soil $\left(W\right)$ is $4\text{lb}$.

The moisture content $\left(w\right)$ is $12\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.72$.

Calculation:

Refer to part (a).

The dry unit weight of the soil $\left({\gamma }_d\right)$ is $108.22\text{lb}/{\text{ft}}^{\text{3}}$.

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

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

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

Substitute 2.72 for $G_s$, $108.22\text{lb}/{\text{ft}}^{\text{3}}$ for ${\gamma }_d$, and $62.4\text{lb}/{\text{ft}}^3$ for ${\gamma }_w$.

$\begin{array}{c}e=\frac{2.72\times 62.4}{108.22}-1\\ =\frac{169.728}{108.22}-1\\ =1.568-1\\ =0.568\end{array}$

Hence, the void ratio of the soil is $\underset{\_}{0.568}$.

(c)

To determine

Calculate the degree of saturation of the soil.

Answer to Problem 3.4P

The degree of saturation of the soil is $\underset{\_}{57.4\%}$.

Explanation of Solution

Given information:

The diameter of the cylindrical mold $\left(d\right)$ is $4\text{in}\text{.}$.

The height of the cylindrical mold $\left(h\right)$ is $4.58\text{in}\text{.}$.

The weight of the compacted soil $\left(W\right)$ is $4\text{lb}$.

The moisture content $\left(w\right)$ is $12\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.72$.

Calculation:

Refer to part (b).

The void ratio of the soil is 0.568.

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

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

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

Substitute 2.72 for $G_s$, $12\%$ for w, and 0.568 for e.

$\begin{array}{c}S=\frac{\frac{12}{100}\times 2.72}{0.568}\\ =\frac{0.3264}{0.568}\\ =0.574\times 100\%\\ =57.4\end{array}$

Hence, the degree of saturation of the soil is $\underset{\_}{57.4\%}$.

(d)

To determine

Calculate the additional water needed to achieve $100\%$ saturation in the soil sample.

Answer to Problem 3.4P

The required additional water to achieve $100\%$ saturation in the soil sample is $\underset{\_}{0.316\text{lb}}$.

Explanation of Solution

Given information:

The diameter of the cylindrical mold $\left(d\right)$ is $4\text{in}\text{.}$.

The height of the cylindrical mold $\left(h\right)$ is $4.58\text{in}\text{.}$.

The weight of the compacted soil $\left(W\right)$ is $4\text{lb}$.

The moisture content $\left(w\right)$ is $12\%$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.72$.

Calculation:

Refer to part (b).

The void ratio of the soil is 0.568.

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

Calculate the saturated unit weight of the soil $\left({\gamma }_{\text{sat}}\right)$ using the relation.

${\gamma }_{\text{sat}}=\frac{\left(G_s+e\right){\gamma }_w}{1+e}$

Substitute 2.72 for $G_s$, 0.568 for e, and $62.4\text{lb}/{\text{ft}}^3$ for ${\gamma }_w$.

$\begin{array}{c}{\gamma }_{\text{sat}}=\frac{\left(2.72+0.568\right)\times 62.4}{1+0.568}\\ =\frac{205.1712}{1.568}\\ =130.8\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

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

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

Substitute 4 lb for W and $0.033{\text{ft}}^3$ for V.

$\begin{array}{c}\gamma =\frac{4}{0.033}\\ =121.21\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Calculate the additional water needed to achieve $100\%$ saturation using the relation.

$\text{Additional water needed}=\left({\gamma }_{\text{sat}}-\gamma \right)V$

Substitute $130.8\text{lb}/{\text{ft}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, $121.21\text{lb}/{\text{ft}}^{\text{3}}$ for $\gamma$, and $0.033{\text{ft}}^3$ for V.

$\begin{array}{c}\text{Additional water needed}=\left(130.8-121.21\right)\times 0.033\\ =9.59\times 0.033\\ =0.316\text{lb}\end{array}$

Therefore, the additional water needed is $\underset{\_}{0.316\text{lb}}$.