Chapter 3, Problem 3.1P

(a)

To determine

Show that ${\gamma }_{\text{sat}}={\gamma }_d+\left(\frac{e}{1+e}\right){\gamma }_w$.

Explanation of Solution

Express the formula for saturated unit weight $\left({\gamma }_{\text{sat}}\right)$ as follows.

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

Here, $G_s$ is the specific gravity of the soil, e is the void ratio, and ${\gamma }_w$ is the unit weight of the water.

Rearrange equation (1) as shown below.

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

Here, $\left(\frac{G_s{\gamma }_w}{1+e}\right)$ is the equation of dry unit weight $\left({\gamma }_d\right)$ of the soil.

Substitute ${\gamma }_d$ for $\frac{G_s{\gamma }_w}{1+e}$.

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

Hence, the saturated unit weight of the soil $\underset{\_}{{\gamma }_{\text{sat}}={\gamma }_d+\left(\frac{e}{1+e}\right){\gamma }_w}$ is proved.

(b)

To determine

Show that ${\gamma }_d=\frac{eS{\gamma }_w}{\left(1+e\right)w}$.

Explanation of Solution

Express the formula for dry unit weight of the soil $\left({\gamma }_d\right)$ as follows.

${\gamma }_d=\frac{G_s{\gamma }_w}{1+e}$ (2)

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

Express the relation between specific gravity of soil solids $\left(G_s\right)$, degree of saturation (S), void ratio (e) and moisture content (w) as shown below.

$G_s=\frac{Se}{w}$

Substitute $\frac{Se}{w}$ for $G_s$ in Equation (2).

$\begin{array}{c}{\gamma }_d=\frac{\left(\frac{Se}{w}\right){\gamma }_w}{1+e}\\ =\frac{eS{\gamma }_w}{\left(1+e\right)w}\end{array}$

Hence, the dry unit weight of the soil $\underset{\_}{{\gamma }_d=\frac{eS{\gamma }_w}{\left(1+e\right)w}}$ is proved.

(c)

To determine

Show that $e=\frac{{\gamma }_{\text{sat}}-{\gamma }_d}{{\gamma }_d-{\gamma }_{\text{sat}}-{\gamma }_w}$.

Explanation of Solution

Refer to part (a).

The saturated unit weight of the soil ${\gamma }_{\text{sat}}={\gamma }_d+\left(\frac{e}{1+e}\right){\gamma }_w$.

Rearrange Equation as shown below.

$\begin{array}{l}{\gamma }_{\text{sat}}\left(1+e\right)={\gamma }_d\left(1+e\right)+e{\gamma }_w\\ {\gamma }_{\text{sat}}+{\gamma }_{\text{sat}}e={\gamma }_d+{\gamma }_{d}e+e{\gamma }_w\\ {\gamma }_{\text{sat}}-{\gamma }_d={\gamma }_{d}e+e{\gamma }_w-{\gamma }_{\text{sat}}e\\ {\gamma }_{\text{sat}}-{\gamma }_d=e\left({\gamma }_d-{\gamma }_{\text{sat}}+{\gamma }_w\right)\end{array}$

$e=\frac{{\gamma }_{\text{sat}}-{\gamma }_d}{{\gamma }_d-{\gamma }_{\text{sat}}+{\gamma }_w}$

Hence, the void ratio $\underset{\_}{e=\frac{{\gamma }_{\text{sat}}-{\gamma }_d}{{\gamma }_d-{\gamma }_{\text{sat}}+{\gamma }_w}}$ is proved.

(d)

To determine

Show that $w_{\text{sat}}=\frac{n{\gamma }_w}{{\gamma }_{\text{sat}}-n{\gamma }_w}$.

Explanation of Solution

Express the formula for saturated unit weight $\left({\gamma }_{\text{sat}}\right)$ as follows.

${\gamma }_{\text{sat}}=\frac{\left(1+w_{\text{sat}}\right)G_s{\gamma }_w}{1+e}$ (3)

Express the formula for specific gravity of soil solids $\left(G_s\right)$ as shown below.

$G_s=\frac{Se}{w}$

The degree of saturation $S=1$ for saturated sample.

Substitute 1 for S.

$G_s=\frac{e}{w_{\text{sat}}}$

Substitute $\frac{e}{w}$ for $G_s$ in Equation (3).

$\begin{array}{c}{\gamma }_{\text{sat}}=\frac{\left(1+w_{\text{sat}}\right)\left(\frac{e}{w_{\text{sat}}}\right){\gamma }_w}{1+e}\\ =\frac{\left(1+w_{\text{sat}}\right)}{w_{\text{sat}}}\frac{e}{\left(1+e\right)}{\gamma }_w\end{array}$

Here, $\frac{e}{\left(1+e\right)}$ is the equation of porosity $\left(n\right)$ of the soil.

Substitute n for $\frac{e}{\left(1+e\right)}$.

$\begin{array}{l}{\gamma }_{\text{sat}}=\frac{\left(1+w_{\text{sat}}\right)}{w_{\text{sat}}}n{\gamma }_w\\ {\gamma }_{\text{sat}}{w}_{\text{sat}}=n{\gamma }_w+n{\gamma }_w{w}_{\text{sat}}\\ \left({\gamma }_{\text{sat}}-n{\gamma }_w\right)w_{\text{sat}}=n{\gamma }_w\\ w_{\text{sat}}=\frac{n{\gamma }_w}{{\gamma }_{\text{sat}}-n{\gamma }_w}\end{array}$

Hence, the moisture content $\underset{\_}{w_{\text{sat}}=\frac{n{\gamma }_w}{{\gamma }_{\text{sat}}-n{\gamma }_w}}$ is proved.