Chapter 3, Problem 3.21CTP

(a)

To determine

Find the number of truckloads required to construct the fill.

Answer to Problem 3.21CTP

The number of truckloads required to construct the fill is $\underset{\_}{10,539\text{trips}}$.

Explanation of Solution

Given information:

Borrow pit:

The water content of the undisturbed soil is $w_{\text{pit}}=15\%$.

The unit weight of the undisturbed soil in the borrow pit is ${\left(\gamma \right)}_{\text{pit}}=19.1{\text{kN/m}}^3$.

The specific gravity of the undisturbed soil is $G_s=2.70$.

Loading truck:

The capacity of the truck is $V_{\text{truck}}=4.80{\text{m}}^3$.

The net weight of the soil and water when the truck is loaded with is $W=72.7\text{kN}$.

Compacted fill:

The dry unit weight of the compacted fill is ${\left({\gamma }_d\right)}_{\text{fill}}=17.3{\text{kN/m}}^3$.

The finished volume of the compacted fill is $V_{\text{fill}}=38,500{\text{m}}^3$.

The moisture content of the fill is $w_{\text{fill}}=15\%$.

Calculation:

Find the weight of the solids ${\left(W_s\right)}_{\text{truck}}$ in each truck using the relation:

${\left(W_s\right)}_{\text{truck}}=\frac{W}{1+w_{\text{pit}}}$

Substitute 72.7 kN for W and 15% for $w_{\text{pit}}$.

$\begin{array}{c}{\left(W_s\right)}_{\text{truck}}=\frac{72.7}{1+\frac{15}{100}}\\ =\frac{72.7}{1.15}\\ =63.2\text{kN}\end{array}$

Find the weight of the solids in the compacted fill ${\left(W_s\right)}_{\text{fill}}$ using the relation:

${\left(W_s\right)}_{\text{fill}}={\left({\gamma }_d\right)}_{\text{fill}}{V}_{\text{fill}}$

Substitute $17.3{\text{kN/m}}^3$ for ${\left({\gamma }_d\right)}_{\text{fill}}$ and $38,500{\text{m}}^3$ for $V_{\text{fill}}$.

$\begin{array}{c}{\left(W_s\right)}_{\text{fill}}=17.3\times \text{38,500}\\ =\text{666,050}\text{kN}\end{array}$

Find the number of truckloads required using the relation:

$\text{Number}\text{of}\text{truckloads}=\frac{{\left(W_s\right)}_{\text{fill}}}{{\left(W_s\right)}_{\text{truck}}}$

Substitute 666,050 kN for ${\left(W_s\right)}_{\text{fill}}$ and 63.2 kN for ${\left(W_s\right)}_{\text{truck}}$.

$\begin{array}{c}\text{Number}\text{of}\text{truckloads}=\frac{\text{666,050}}{63.2}\\ =10,539\text{trips}\end{array}$

Therefore, the number of truckloads required to construct the fill is $\underset{\_}{10,539\text{trips}}$.

(b)

To determine

Find the volume of the pit that remains in the borrow area.

Answer to Problem 3.21CTP

The volume of the pit that remains in the borrow area is $\underset{\_}{40,102.5{\text{m}}^3}$.

Explanation of Solution

Given information:

Borrow pit:

The water content of the undisturbed soil is $w_{\text{pit}}=15\%$.

The unit weight of the undisturbed soil in the borrow pit is ${\left(\gamma \right)}_{\text{pit}}=19.1{\text{kN/m}}^3$.

The specific gravity of the undisturbed soil is $G_s=2.70$.

Loading truck:

The capacity of the truck is $V_{\text{truck}}=4.80{\text{m}}^3$.

The net weight of the soil and water when the truck is loaded with is $W=72.7\text{kN}$.

Compacted fill:

The dry unit weight of the compacted fill is ${\left({\gamma }_d\right)}_{\text{fill}}=17.3{\text{kN/m}}^3$.

The finished volume of the compacted fill is $V_{\text{fill}}=38,500{\text{m}}^3$.

The moisture content of the fill is $w_{\text{fill}}=15\%$.

Calculation:

Find the weight of the soil $\left(W_{\text{soil}}\right)$ using the relation:

${\left(W_s\right)}_{\text{fill}}=\frac{W_{\text{soil}}}{1+w_{\text{pit}}}$

Substitute 666,050 kN for ${\left(W_s\right)}_{\text{fill}}$ and 15% for $w_{\text{pit}}$.

$\begin{array}{c}666,050=\frac{W_{\text{soil}}}{1+\frac{15}{100}}\\ W_{\text{soil}}=666,050\left(1+0.15\right)\\ =765,957.5\text{kN}\end{array}$

Find the volume of the pit $\left(V_{\text{pit}}\right)$ using the relation:

$V_{\text{pit}}=\frac{W_{\text{soil}}}{{\left(\gamma \right)}_{\text{pit}}}$

Substitute 765,957.5 kN for $W_{\text{soil}}$ and $19.1{\text{kN/m}}^3$ for ${\left(\gamma \right)}_{\text{pit}}$.

$\begin{array}{c}{V}_{\text{pit}}=\frac{765,957.5}{19.1}\\ =40,102.5{\text{m}}^3\end{array}$

Therefore, the volume of the pit that remains in the borrow area is $\underset{\_}{40,102.5{\text{m}}^3}$.

(c)

To determine

Find the liters of water will have to be added per truck load.

Answer to Problem 3.21CTP

The liters of water will have to be added per truck load is $\underset{\_}{193.3\text{liters}}$.

Explanation of Solution

Given information:

Borrow pit:

The water content of the undisturbed soil is $w_{\text{pit}}=15\%$.

The unit weight of the undisturbed soil in the borrow pit is ${\left(\gamma \right)}_{\text{pit}}=19.1{\text{kN/m}}^3$.

The specific gravity of the undisturbed soil is $G_s=2.70$.

Loading truck:

The capacity of the truck is $V_{\text{truck}}=4.80{\text{m}}^3$.

The net weight of the soil and water when the truck is loaded with is $W=72.7\text{kN}$.

Compacted fill:

The dry unit weight of the compacted fill is ${\left({\gamma }_d\right)}_{\text{fill}}=17.3{\text{kN/m}}^3$.

The finished volume of the compacted fill is $V_{\text{fill}}=38,500{\text{m}}^3$.

The moisture content of the fill is $w_{\text{fill}}=15\%$.

Calculation:

Neglect the moisture lost by evaporation during excavation, haulage, and handling.

The moisture content increased from 15% to 18%.

Find the weight of the water $\left(W_w\right)$  to be added using the relation:

$W_w={\left(W_s\right)}_{\text{truck}}\left(w_{\text{fill}}-w_{\text{pit}}\right)$

Substitute 63.2 kN for ${\left(W_s\right)}_{\text{truck}}$, 18% for $w_{\text{fill}}$, and 15% for $w_{\text{pit}}$.

$\begin{array}{c}{W}_w=63.2\times \left(\frac{18}{100}-\frac{15}{100}\right)\\ =1.896\text{kN}\end{array}$

Consider the unit weight of water as ${\gamma }_w=9.81{\text{kN/m}}^3$.

Find the volume of water $\left(V_w\right)$ to be added using the relation:

$V_w=\frac{W_w}{{\gamma }_w}$

Substitute 1.896 kN for $W_w$ and $9.81{\text{kN/m}}^3$ for ${\gamma }_w$.

$\begin{array}{c}{V}_w=\frac{1.896}{9.81}\\ =0.1933{\text{m}}^3\times \frac{1,000\text{liter}}{1{\text{m}}^3}\\ =193.3\text{liters}\end{array}$

Therefore, the liters of water will have to be added per truck load is $\underset{\_}{193.3\text{liters}}$.

(d)

To determine

Find the water content at saturation.

Answer to Problem 3.21CTP

The water content at saturation is $\underset{\_}{19.7\%}$.

Explanation of Solution

Given information:

Borrow pit:

The water content of the undisturbed soil is $w_{\text{pit}}=15\%$.

The unit weight of the undisturbed soil in the borrow pit is ${\left(\gamma \right)}_{\text{pit}}=19.1{\text{kN/m}}^3$.

The specific gravity of the undisturbed soil is $G_s=2.70$.

Loading truck:

The capacity of the truck is $V_{\text{truck}}=4.80{\text{m}}^3$.

The net weight of the soil and water when the truck is loaded with is $W=72.7\text{kN}$.

Compacted fill:

The dry unit weight of the compacted fill is ${\left({\gamma }_d\right)}_{\text{fill}}=17.3{\text{kN/m}}^3$.

The finished volume of the compacted fill is $V_{\text{fill}}=38,500{\text{m}}^3$.

The moisture content of the fill is $w_{\text{fill}}=15\%$.

Calculation:

Find the void ratio $\left(e_{\text{fill}}\right)$ in the compacted fill using the relation:

${\left({\gamma }_d\right)}_{\text{fill}}=\frac{G_s{\gamma }_w}{1+e_{\text{fill}}}$

Substitute $17.3{\text{kN/m}}^3$ for ${\left({\gamma }_d\right)}_{\text{fill}}$, 2.70 for $G_s$, and $9.81{\text{kN/m}}^3$ for ${\gamma }_w$.

$\begin{array}{l}17.3=\frac{2.70\times 9.81}{1+e_{\text{fill}}}\\ 1+e_{\text{fill}}=\frac{26.487}{17.3}\\ e_{\text{fill}}=0.531\end{array}$

Find the water content (w) using the relation:

$S{e}_{\text{fill}}=w{G}_s$

Substitute 1 for S, 0.531 for $e_{\text{fill}}$, and 2.70 for $G_s$.

$\begin{array}{c}1\times \text{0}\text{.531}=w\times 2.70\\ w=0.197\\ =19.7\%\end{array}$

Therefore, the water content at saturation is $\underset{\_}{19.7\%}$.