Chapter 5, Problem 5.1P

To determine

Show that the compactive energy in method C is $2,700\text{kN}\cdot {\text{m/m}}^{\text{3}}$.

Explanation of Solution

Given information:

The mold volume (V) is $2,124{\text{cm}}^{\text{3}}$.

Mass of the hammer (m) 4.54 kg.

The hammer drop height (h) is 457.2 mm.

The number of layers of compaction (N) is 5.

The number of blows per layer is 56.

Calculation:

Determine the energy per blow.

$\begin{array}{c}E=mg\times h\\ =\left(4.54\times 9.81\right)\frac{\text{kg-m}}{{\text{s}}^{\text{2}}}\times \frac{1\text{N}}{1\frac{\text{kg-m}}{{\text{s}}^{\text{2}}}}\times 457.2\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\\ =20.36\text{N}\cdot \text{m}\end{array}$

Determine the total energy imparted to the soil in the mold.

$\begin{array}{c}{E}_{\text{Tot}}=\text{Blows}\text{per}\text{layer}\times N\times E\\ =5\times 56\times 20.36\\ =5,700.8\text{N}\cdot \text{m}\end{array}$

Determine the energy per unit volume.

$\begin{array}{c}\text{Energy}\text{per}\text{unit}\text{volume}=\frac{E_{\text{Tot}}}{V}\\ =\frac{5,700.8}{2,124{\text{cm}}^3\times {\left(\frac{1\text{m}}{100\text{cm}}\right)}^3}\\ =2,683,992\text{N}\cdot {\text{m/m}}^3\times \frac{1\text{kN}}{1,000\text{N}}\\ \approx 2,700\text{kN}\cdot {\text{m/m}}^3\end{array}$

Therefore, the compactive energy in method C is $2,700\text{kN}\cdot {\text{m/m}}^{\text{3}}$ is proved.