Chapter 3, Problem 3.12P

(a)

To determine

Calculate the saturated unit weight of soil.

Answer to Problem 3.12P

The saturated unit weight of soil is $\underset{\_}{125.89\text{lb}/{\text{ft}}^3}$.

Explanation of Solution

Given information:

The diameter of the sample $\left(d\right)$ is $1.375\text{in}\text{.}$.

The length of the sampler $\left(L\right)$ is $18\text{in}\text{.}$.

The length of the sampler for performing shear strength $\left(L_{\text{shear}}\right)$ is $3\text{in}\text{.}$.

The moist weight $\left(W\right)$ of the sample is $1.85\text{lb}$.

The oven-dried weight $\left(W_S\right)$ of the sample is $1.5\text{lb}$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.74$.

Calculation:

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

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

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

Substitute $1.375\text{in}\text{.}$ for d and $18\text{in}\text{.}$ for L.

$\begin{array}{c}V=\frac{1}{4}\pi \times {1.375}^2\times 18\\ =26.728\text{in}{\text{.}}^3\times {\left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)}^3\\ =0.015{\text{ft}}^3\end{array}$

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

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

Substitute $1.85\text{lb}$ for W and $0.015{\text{ft}}^3$ for V.

$\begin{array}{c}\gamma =\frac{1.85}{0.015}\\ =123.33\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Calculate the moisture content $\left(w\right)$ using the relation.

$w=\frac{W-W_s}{W_s}$

Substitute $1.85\text{lb}$ for W and $1.5\text{lb}$ for $W_s$ .

$\begin{array}{c}w=\frac{1.85-1.5}{1.5}\\ =\frac{0.35}{1.5}\\ =0.2333\times 100\%\\ =23.33\%\end{array}$

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

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

Substitute $123.33\text{lb}/{\text{ft}}^3$ for $\gamma$ and $23.33\%$ for w.

$\begin{array}{c}{\gamma }_d=\frac{123.33}{1+\frac{23.33}{100}}\\ =\frac{123.33}{1.2333}\\ =100\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

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

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

Substitute $2.74$ for $G_s$, $100\text{lb}/{\text{ft}}^3$ for ${\gamma }_d$, and $62.4\text{lb}/{\text{ft}}^3$ for ${\gamma }_w$.

$\begin{array}{c}e=\frac{2.74\times 62.4}{100}-1\\ =\frac{170.976}{100}-1\\ =1.7098-1\\ =0.71\end{array}$

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 $2.74$ for $G_s$, $0.71$ for e, and $62.4\text{lb}/{\text{ft}}^3$ for ${\gamma }_w$.

$\begin{array}{c}{\gamma }_{\text{sat}}=\frac{\left(2.74+0.71\right)\times 62.4}{1+0.71}\\ =\frac{3.45}{1.71}\times 62.4\\ =125.89\text{lb}/{\text{ft}}^3\end{array}$

Hence, the saturated unit weight of soil is $\underset{\_}{125.89\text{lb}/{\text{ft}}^3}$.

(b)

To determine

Calculate the moisture content at $100\%$ saturation.

Answer to Problem 3.12P

The moisture content at $100\%$ saturation is $\underset{\_}{25.9\%}$.

Explanation of Solution

Given information:

The diameter of the sample $\left(d\right)$ is $1.375\text{in}\text{.}$.

The length of the sampler $\left(L\right)$ is $18\text{in}\text{.}$.

The length of the sampler for performing shear strength $\left(L_{\text{shear}}\right)$ is $3\text{in}\text{.}$.

The moist weight $\left(W\right)$ of the sample is $1.85\text{lb}$.

The oven-dried weight $\left(W_S\right)$ of the sample is $1.5\text{lb}$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.74$.

Calculation:

Refer to part (a).

The void ratio of the soil is $0.71$.

Calculate the moisture content at $100\%$ saturation $\left(w_{\text{sat}}\right)$ using the relation.

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

Substitute $0.71$ for e and $2.74$ for $G_s$.

$\begin{array}{c}{w}_{\text{sat}}=\frac{0.71}{2.74}\\ =0.259\times 100\%\\ =25.9\%\end{array}$

Hence, the moisture content at $100\%$ saturation is $\underset{\_}{25.9\%}$.

(c)

To determine

Calculate the amount of water needed to achieve full saturation.

Answer to Problem 3.12P

The amount of water needed to achieve full saturation is $\underset{\_}{0.0066\text{lb}}$.

Explanation of Solution

Given information:

The diameter of the sampler $\left(d\right)$ is $1.375\text{in}\text{.}$.

The length of the sampler $\left(L\right)$ is $18\text{in}\text{.}$.

The length of the sampler for performing shear strength $\left(L_{\text{shear}}\right)$ is $3\text{in}\text{.}$.

The moist weight $\left(W\right)$ of the sample is $1.85\text{lb}$.

The oven-dried weight $\left(W_S\right)$ of the sample is $1.5\text{lb}$.

The specific gravity of the soil solids $\left(G_s\right)$ is $2.74$.

Calculation:

Refer to part (a).

The saturated unit weight of soil $\left({\gamma }_{\text{sat}}\right)$ is $125.89\text{lb}/{\text{ft}}^3$.

The moist unit weight of soil $\left(\gamma \right)$ is $123.33\text{lb}/{\text{ft}}^3$.

Calculate the volume of the sampler for performing shear strength $\left(V\right)$ using the relation.

$V=\frac{1}{4}\pi d^2{L}_{\text{shear}}$

Substitute $1.375\text{in}\text{.}$ for d and $3\text{in}\text{.}$ for $L_{\text{shear}}$.

$\begin{array}{c}V=\frac{1}{4}\pi \times {1.375}^2\times 3\\ =4.45468\text{in}{\text{.}}^3\times {\left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)}^3\\ =0.00258{\text{ft}}^3\end{array}$

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

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

Substitute $125.89\text{lb}/{\text{ft}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, $123.33\text{lb}/{\text{ft}}^{\text{3}}$ for $\gamma$, and $0.00258{\text{ft}}^3$ for V.

$\begin{array}{c}\text{Amount of water needed}=\left(125.89-123.33\right)0.00258\\ =2.56\times 0.00258\\ =0.0066\text{lb}\end{array}$

Hence, the amount of water needed to achieve full saturation is $\underset{\_}{0.0066\text{lb}}$.