Chapter 5, Problem 5.17CTP

(a)

To determine

Plot the compaction curve along with the air void curve.

Explanation of Solution

Given information:

The specific gravity of solids is $\left(G_s\right)$ 2.71.

The volume of mold V is $943.3{\text{cm}}^{\text{3}}$.

Calculation:

Determine the moisture content for specimen 1 $\left(w\right)$ using the relation:

${\left(w\right)}_1=\frac{m_s-m_d}{m_d}$

Here, $m_s$ is the mass of the wet specimen and $m_d$ is the mass of dried specimen.

Substitute $1,652\text{g}$ for $m_s$ and $1,430\text{g}$ for $m_d$.

$\begin{array}{c}w=\frac{1,652-1,430}{1,430}\\ =0.155\\ =15.5\%\end{array}$

Similarly calculate the moisture content values for remaining samples and tabulate the values in Table 1.

Determine the moist unit weight of specimen 1$\left(\gamma \right)$ using the relation:

$\begin{array}{c}\gamma =\frac{W}{V}\\ =\frac{m_s\times g}{V}\end{array}$

Here, V is the volume of the mold and g is the acceleration due to gravity.

Substitute 1,652 g for $m_s$, $9.81{\text{m/s}}^{\text{2}}$ for g, and $943.3{\text{cm}}^{\text{3}}$ for V.

$\begin{array}{c}{\gamma }_d=\frac{\left(1,652\text{g}\times \frac{1\text{kg}}{1,000\text{g}}\right)\times 9.81{\text{m/s}}^{\text{2}}}{943.3{\text{cm}}^{\text{3}}\times {\left(\frac{1\text{m}}{100\text{cm}}\right)}^3}\\ =\frac{\left(16.206\text{N}\times \frac{1\text{kN}}{1,000\text{N}}\right)}{943.3\times {10}^{-6}{\text{m}}^{\text{3}}}\\ =17.18{\text{kN/m}}^{\text{3}}\end{array}$

Similarly calculate the moist unit weight values for remaining samples and tabulate the values in Table 1.

Find the dry unit weight ${\gamma }_d$ for specimen 1 using the formula:

${\gamma }_d=\frac{\gamma }{1+\frac{w\left(\%\right)}{100}}$

Substitute $17.18{\text{kN/m}}^{\text{3}}$ for $\gamma$ and 15.5% for w.

$\begin{array}{c}{\gamma }_d=\frac{17.18}{1+\frac{15.5}{100}}\\ =14.87{\text{kN/m}}^{\text{3}}\end{array}$

Similarly calculate the dry unit weight values for remaining samples and tabulate the values in Table 1.

Determine the zero air-void unit weight using the relation:

${\gamma }_{zav}=\frac{{\gamma }_w}{w+\frac{1}{G_s}}$

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

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, 15.5% for w, and 2.71 for $G_s$.

$\begin{array}{c}{\gamma }_{zav}=\frac{9.81}{\left(\frac{15.5}{100}+\frac{1}{2.71}\right)}\\ =18.72{\text{kN/m}}^{\text{3}}\end{array}$

Similarly calculate the zero air void unit weight values for remaining samples and tabulate the values in Table 1.

Find the dry unit weight at 90% saturation for sample 1.

Find the void ratio (e) at 90% saturation using the relation:

$se=w{G}_s$

Substitute 0.9 for s, 15.5% for w, and 2.71 for $G_s$.

$\begin{array}{l}se=w{G}_s\\ 0.9e=\left(\frac{15.5}{100}\right)\times 2.71\\ e=0.466\end{array}$

Find the dry unit weight at 90% saturation using the formula:

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

Substitute 2.71 for $G_s$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and 0.466 for e.

$\begin{array}{c}{\gamma }_d=\frac{2.71\times 9.81}{1+0.466}\\ =18.13{\text{kN/m}}^{\text{3}}\end{array}$

Similarly calculate the dry unit weight $\left(s=90\%\right)$ values for remaining samples and tabulate the values in Table 1.

Find the dry unit weight at 80% saturation for sample 1.

Find the void ratio (e) at 80% saturation using the relation:

$se=w{G}_s$

Substitute 0.8 for s, 15.5% for w, and 2.71 for $G_s$.

$\begin{array}{l}se=w{G}_s\\ 0.8e=\left(\frac{15.5}{100}\right)\times 2.71\\ e=0.525\end{array}$

Find the dry unit weight at 80% saturation using the formula:

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

Substitute 2.71 for $G_s$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and 0.525 for e.

$\begin{array}{c}{\gamma }_d=\frac{2.71\times 9.81}{1+0.525}\\ =17.43{\text{kN/m}}^{\text{3}}\end{array}$

Similarly calculate the dry unit weight $\left(s=80\%\right)$ values for remaining samples and tabulate the values in Table 1.

Show the dry unit weight, zero-air void unit weight, dry unit weight at 90% saturation, dry unit weight at 80% saturation as in Table 1.

$m_s$	$m_d$	w	$\gamma$ $\left({\text{kN/m}}^{\text{3}}\right)$	${\gamma }_d$ $\left({\text{kN/m}}^{\text{3}}\right)$	${\gamma }_{zav}$	$e$ $\left(s=90\%\right)$	${\gamma }_d$ $\left(s=90\%\right)$	$e$ $\left(s=80\%\right)$	${\gamma }_d$ $\left(s=80\%\right)$
1652	1430	15.5	17.18	14.87	18.72	0.467	18.13	0.525	17.43
1799	1541	16.7	18.71	16.03	18.30	0.503	17.69	0.566	16.98
1938	1637	18.4	20.15	17.02	17.74	0.554	17.11	0.623	16.38
1936	1604	20.7	20.13	16.68	17.03	0.623	16.38	0.701	15.63
1895	1546	22.6	19.71	16.07	16.49	0.681	15.82	0.766	15.06
1864	1507	23.7	19.38	15.67	16.19	0.714	15.51	0.803	14.75

Plot the compaction curve along with the zero air void curve as shown in Figure 1.

(b)

To determine

Find the optimum moisture content and maximum dry unit weight.

Answer to Problem 5.17CTP

The optimum moisture content is 18.5%.

The maximum dry unit weight is $\underset{\_}{17.1{\text{kN/m}}^{\text{3}}}$.

Explanation of Solution

Given information:

The specific gravity of solids is $\left(G_s\right)$ 2.71.

The volume of mold V is $943.3{\text{cm}}^{\text{3}}$.

Calculation:

Refer to Figure 1:

The maximum dry unit weight occurs at optimum moisture content.

The optimum moisture content is 18.5%.

The maximum dry unit weight is $\underset{\_}{17.1{\text{kN/m}}^{\text{3}}}$.

(c)

To determine

Find the degree of saturation at optimum moisture content.

Answer to Problem 5.17CTP

The degree of saturation at optimum moisture content is 90.3%.

Explanation of Solution

Given information:

The specific gravity of solids is $\left(G_s\right)$ 2.71.

The volume of mold V is $943.3{\text{cm}}^{\text{3}}$.

Calculation:

Consider the maximum dry unit weight as $17.1{\text{kN/m}}^{\text{3}}$ and the optimum moisture content is 18.5%.

Find the void ratio using the formula.

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

Substitute 2.71 for $G_s$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and $17.1{\text{kN/m}}^{\text{3}}$ for ${\gamma }_d$.

$\begin{array}{c}e=\frac{2.71\times 9.81}{17.1}-1\\ =0.555\end{array}$

Find the degree of saturation using the formula.

$se=wG$

Substitute 0.555 for e, 18.5% for w, and 2.71 for G.

$\begin{array}{l}s\left(0.555\right)=\left(\frac{18.5}{100}\right)\times 2.71\\ s=0.903\times 100\\ s=90.3\%\end{array}$

Therefore, the degree of saturation at optimum moisture content is 90.3%.

(d)

To determine

Plot the theoretical curve for $s=90\%$.

Explanation of Solution

Given information:

The specific gravity of solids is $\left(G_s\right)$ 2.71.

The volume of mold V is $943.3{\text{cm}}^{\text{3}}$.

Calculation:

Refer to Figure 1.

The theoretical curve for $S=90\%$ is plotted in Figure 1.

(e)

To determine

Find the moisture content when compact at 80% degree of saturation.

Answer to Problem 5.17CTP

The moisture content when compact at 80% degree of saturation is 17.6%.

Explanation of Solution

Given information:

The specific gravity of solids is $\left(G_s\right)$ 2.71.

The volume of mold V is $943.3{\text{cm}}^{\text{3}}$.

Refer to Figure 1:

Find the moisture content from Figure 1 as follows:

• The theoretical curve for $s=80\%$ is plotted in Figure 1.
• Mark the intersection point of curve for $s=80\%$ and compaction curve.
• Obtain the moisture content value.

Therefore, the moisture content when compact at 80% degree of saturation is 17.6%.