Compaction of Soil

\begin{aligned} (V_S)_{\text{heap}}&=(V_S)_{\text{compacted soil}}\\ \left ( \frac{V_T}{1+e} \right )_{\text{heap}}&=\left ( \frac{V_T}{1+e} \right )_{\text{compacted soil}}\\ \frac{10000}{1+1}&=\frac{7500}{1+e}\\ e&=0.5 \end{aligned}

\begin{array}{r} \mathrm{OMC}-\mathrm{SP} \gt \mathrm{OMC}-\mathrm{MP} \\ \mathrm{MDD}-\mathrm{SP} \lt \mathrm{MDD}-\mathrm{MP} \end{array}

Optimum moisture content of the soil decreases with increase compactive efforts.

The bulk unit weight of soil is given by
\gamma=\frac{(G+S e) \gamma_{w}}{1+e}
For dry unit weight, degree of saturation,
S=0 \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+e}
But we know that
G w=\text { Se } \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+\frac{G w}{S}}
For zero air voids, degree of saturation,
S=100 \% \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+G w}

Flocculated structure is formed when clay particles settle on sea bed, since sea bed is polar in nature.

Bulk density of the soil
=\frac{1.8 \times 1000}{944}=1.907 \mathrm{gm} / \mathrm{cc}
Water content of soil
=\left(\frac{23-20}{20}\right) \times 100=15 \%
For, theoretical maximum dry unit weight, degree of saturation should be 100 %.
\begin{aligned} S e &=G w \\ \Rightarrow \quad e &=\frac{2.7 \times 0.15}{1} \Rightarrow e=0.405 \end{aligned}
The theoretical maximum dry unit weight may be given as
\gamma_{d, \max }=\frac{G \gamma_{w}}{1+e}=\frac{2.7 \times 9.81}{1+0.405}=18.85 \mathrm{kN} / \mathrm{m}^{3}

As per Indian Standard light compaction test, a hammer of 2.6 kg is allowed to fall from a height of 310 mm and 3 layers are tamped 25 times in a mould of volume 1000 cc .
\therefore Energy imparted
\begin{array}{l} =2.6 \times 25 \times 3 \times 9.81 \\ \times \frac{310}{1000}=593.01 \mathrm{N}-\mathrm{m} \end{array}
For embankment compaction, the volume of soil covered by rammer
\begin{aligned} &=0.05 \times 10^{4} \times \frac{500}{10} \\ &=25000 \mathrm{cc} \end{aligned}
\therefore Energy imparted in every drop of hammer
\begin{aligned} &=\left(1+\frac{50}{100}\right) \times 400\\ &=600 \mathrm{N}-\mathrm{m} \end{aligned}
If 'n' number of passes are required to develop equivalent energy to Indian Standard light compaction test, then
n \times \frac{600}{25000}=\frac{593.01}{1000}
\Rightarrow \quad n=24.71 \simeq 25\text{( approx.)}
However, If the data of Standard Proctor Test are taken i.e. weight of hammer =2.495 kg; Volume of mould =944cc, then the value of 'n' will be close to 26.