Chapter 6, Problem 6.1P

To determine

Find the degree of saturation at optimum moisture content.

Answer to Problem 6.1P

The degree of saturation at optimum moisture content is $\underset{\_}{78.1\%}$.

Explanation of Solution

Given information:

The maximum dry unit weight of the soil ${\gamma }_d$ is $16.8{\text{kN/m}}^{\text{3}}$.

The optimum moisture content of the soil w is 17 %.

The specific gravity of soil $G_s$ is 2.73.

Calculation:

Determine the degree of saturation (S) using the relation.

${\gamma }_d=\frac{G_s{\gamma }_w}{1+\frac{G_{s}w}{S}}$

Here, ${\gamma }_w$ is the unit weight of water.

Substitute $16.8{\text{kN/m}}^{\text{3}}$ for ${\gamma }_d$, 2.73 for $G_s$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and 17 % for w.

$\begin{array}{l}16.8=\frac{2.73\times 9.81}{1+\frac{2.73\times \left(\frac{17}{100}\right)}{S}}\\ 1+\frac{0.4641}{S}=\frac{26.78}{16.8}\\ S-1.594S=-0.4641\end{array}$

$\begin{array}{l}-0.594S=-0.4641\\ S=\frac{0.4641}{0.594}\\ S=0.781\times 100\\ S=78.1\%\end{array}$

Therefore, the degree of saturation at optimum moisture content is $\underset{\_}{78.1\%}$.