Chapter 3, Problem 3.4P

(a)

To determine

Find the moist unit weight of the soil sample.

Answer to Problem 3.4P

The moist unit weight of the soil sample is $\underset{\_}{18.07{\text{kN/m}}^3}$.

Explanation of Solution

Given information:

The volume of moist soil sample is $V=5.66\times {10}^{-3}{\text{m}}^3$.

The moist weight of the soil is $W=102.3\times {10}^{-3}\text{kN}$.

The moisture content of the soil solid is $w=11\%$.

The specific gravity of soil solids is $G_s=2.7$.

Calculation:

Find the moist unit weight $\left(\gamma \right)$ using the equation:

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

Substitute $102.3\times {10}^{-3}\text{kN}$ for W and $5.66\times {10}^{-3}{\text{m}}^3$ for V.

$\begin{array}{c}\gamma =\frac{102.3\times {10}^{-3}}{5.66\times {10}^{-3}}\\ =18.07{\text{kN/m}}^3\end{array}$

Therefore, the moist unit weight of the soil sample is $\underset{\_}{18.07{\text{kN/m}}^3}$.

(b)

To determine

Find the dry unit weight of the soil sample.

Answer to Problem 3.4P

The dry unit weight of the soil sample is $\underset{\_}{16.28{\text{kN/m}}^3}$.

Explanation of Solution

Given information:

The volume of moist soil sample is $V=5.66\times {10}^{-3}{\text{m}}^3$.

The moist weight of the soil is $W=102.3\times {10}^{-3}\text{kN}$.

The moisture content of the soil solid is $w=11\%$.

The specific gravity of soil solids is $G_s=2.7$.

Calculation:

Refer part (a);

The moist unit weight of soil sample is $\gamma =18.07{\text{kN/m}}^3$.

Find the dry unit weight $\left({\gamma }_d\right)$ using the equation:

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

Substitute $18.07{\text{kN/m}}^3$ for $\gamma$ and 11% for w.

$\begin{array}{c}{\gamma }_d=\frac{18.07}{1+\frac{11}{100}}\\ =16.28{\text{kN/m}}^3\end{array}$

Therefore, the dry unit weight of the soil sample is $\underset{\_}{16.28{\text{kN/m}}^3}$.

(c)

To determine

Find the void ratio of the soil sample.

Answer to Problem 3.4P

The void ratio of the soil sample is $\underset{\_}{0.626}$.

Explanation of Solution

Given information:

The volume of moist soil sample is $V=5.66\times {10}^{-3}{\text{m}}^3$.

The moist weight of the soil is $W=102.3\times {10}^{-3}\text{kN}$.

The moisture content of the soil solid is $w=11\%$.

The specific gravity of soil solids is $G_s=2.7$.

Calculation:

Refer part (b);

The dry unit weight of soil sample is ${\gamma }_d=16.28{\text{kN/m}}^3$.

Find the void ratio (e) using the equation:

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

Here, the specific gravity of soil solids is $G_s$.

Substitute $16.28{\text{kN/m}}^3$ for ${\gamma }_d$, 2.7 for $G_s$, and $9.81{\text{kN/m}}^3$ for ${\gamma }_w$.

$\begin{array}{l}16.28=\frac{2.7\times 9.81}{1+e}\\ 1+e=\frac{26.487}{16.28}\\ e=0.626\end{array}$

Therefore, the void ratio of the soil sample is $\underset{\_}{0.626}$.

(d)

To determine

Find the porosity of the soil sample.

Answer to Problem 3.4P

The porosity of the soil sample is $\underset{\_}{0.385}$.

Explanation of Solution

Given information:

The volume of moist soil sample is $V=5.66\times {10}^{-3}{\text{m}}^3$.

The moist weight of the soil is $W=102.3\times {10}^{-3}\text{kN}$.

The moisture content of the soil solid is $w=11\%$.

The specific gravity of soil solids is $G_s=2.7$.

Calculation:

Refer part (c);

The void ratio of soil sample is $e=0.626$.

Find the porosity (n) using the equation:

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

Substitute 0.626 for e.

$\begin{array}{c}n=\frac{0.626}{1+0.626}\\ =0.385\end{array}$

Therefore, the porosity of the soil sample is $\underset{\_}{0.385}$.

(e)

To determine

Find the degree of saturation of the soil sample.

Answer to Problem 3.4P

The degree of saturation of the soil sample is $\underset{\_}{47.4\%}$.

Explanation of Solution

Given information:

The volume of moist soil sample is $V=5.66\times {10}^{-3}{\text{m}}^3$.

The moist weight of the soil is $W=102.3\times {10}^{-3}\text{kN}$.

The moisture content of the soil solid is $w=11\%$.

The specific gravity of soil solids is $G_s=2.7$.

Calculation:

Refer part (c);

The void ratio of soil sample is $e=0.626$.

Find the degree of saturation (S) using the relation:

$S=\frac{w{G}_s}{e}\times 100\%$

Substitute 0.11 for w, 2.7 for $G_s$, and 0.626 for e.

$\begin{array}{c}S=\frac{0.11\times 2.7}{0.626}\times 100\%\\ =47.4\%\end{array}$

Therefore, the degree of saturation of the soil sample is $\underset{\_}{47.4\%}$.

(f)

To determine

Find the volume occupied by water.

Answer to Problem 3.4P

The volume occupied by water is $\underset{\_}{0.001{\text{m}}^3}$.

Explanation of Solution

Given information:

The volume of moist soil sample is $V=5.66\times {10}^{-3}{\text{m}}^3$.

The moist weight of the soil is $W=102.3\times {10}^{-3}\text{kN}$.

The moisture content of the soil solid is $w=11\%$.

The specific gravity of soil solids is $G_s=2.7$.

Calculation:

Find the weight of soil $\left(W_s\right)$ solids using the equation:

$W_s=\frac{W}{1+w}$

Substitute $102.3\times {10}^{-3}\text{kN}$ for W and 0.11 for w.

$\begin{array}{c}{W}_s=\frac{102.3\times {10}^{-3}}{1+0.11}\\ =92.16\times {10}^{-3}\text{kN}\end{array}$

Find the weight of water $\left(W_w\right)$ using the relation:

$W_w=W-W_s$

Substitute $102.3\times {10}^{-3}\text{kN}$ for W and $92.16\times {10}^{-3}\text{kN}$ for $W_s$.

$\begin{array}{c}{W}_w=102.3\times {10}^{-3}-92.16\times {10}^{-3}\\ =10.14\times {10}^{-3}\text{kN}\end{array}$

Find the volume occupied by water $\left(V_w\right)$ using the relation:

${\gamma }_w=\frac{W_w}{V_w}$

Substitute $9.81{\text{kN/m}}^3$ for ${\gamma }_w$ and $10.14\times {10}^{-3}\text{kN}$ for $W_s$.

$\begin{array}{l}9.81=\frac{10.14\times {10}^{-3}}{V_w}\\ V_w=0.001{\text{m}}^3\end{array}$

Therefore, the volume occupied by water is $\underset{\_}{0.001{\text{m}}^3}$.