Chapter 17, Problem 1P

To determine

Void ratio, dry density and degree of saturation of the soil.

Answer to Problem 1P

  $e=0.70$

  ${\gamma }_d=100.81\frac{\text{lb}}{{\text{ft}}^3}$

  $S=0.94$

Explanation of Solution

Given information:

  $\text{Bulk density}=125{\text{lb/ft}}^3$

  $\text{Moisture content}=24\%$

  $\text{Specific gravity}=2.75$

Concept used:

  ${\gamma }_d=\frac{\gamma }{1+w}$

  $\begin{array}{c}{\gamma }_d=\frac{G_s{\gamma }_w}{1+e}\\ e=\frac{G_s{\gamma }_w}{{\gamma }_d}-1\end{array}$

  $S=\frac{w{G}_s}{e}$

  $\begin{array}{c}{\gamma }_d\to \text{Dry density}\\ \gamma \to \text{Bulk density}\\ w\to \text{Moisture content}\\ G_s\to \text{Specific gravity}\\ {\gamma }_w\to \text{Density of water}\\ e\to \text{Void ratio}\\ S\to \text{Degree of saturation}\end{array}$

Calculation:

The dry density is determined using the formula

  $\begin{array}{c}{\gamma }_d=\frac{\gamma }{1+w}\\ =\frac{125}{1+0.24}\\ =100.81\frac{\text{lb}}{{\text{ft}}^{\text{3}}}\end{array}$

The void ratio is determined using the relation

Take, density of water as $62.4\frac{\text{lb}}{{\text{ft}}^3}$.

  $\begin{array}{c}e=\frac{G_s{\gamma }_w}{{\gamma }_d}-1\\ =\left(\frac{2.75\times 62.4}{100.81}\right)-1\\ =0.70\end{array}$

The degree of saturation is determined using the relation

  $\begin{array}{c}S=\frac{w{G}_s}{e}\\ =\frac{0.24\times 2.75}{0.70}\\ =0.94\end{array}$

Conclusion:

The void ratio, dry density and degree of saturation of the soil is $0.70$, $100.81\frac{\text{lb}}{{\text{ft}}^3}$ and $0.94$.