written 19 months ago by
Chandan15
• 300

•
modified 19 months ago

Volume of soil at natural condition.
$V_{1}$ = $V_{S}$ + $V_{w1}$
w=32 3%
$W_{W1}$ =0.323 $W_{S}$
$=$
$V_{w}=\frac{0.323 W_{s}}{\gamma_{w}}=\frac{0.323 W_{s}}{1}$
$\left[\gamma_{w}=1 \mathrm{~g} / \mathrm{cc}\right]$
Thus
$$
V_{1}=V_{s}+0.323 W_{s}
$$
Similarly, volume of soil at shrinkage limit
$$
V_{d}=V_{s}+0.23 W_{s}
$$
If water content of soil is reduced to $16.4 \%$, it is below the shrinkage limit. Hence the volume of soil at this water content will remain same as shrinkage limit.
$$
\begin{aligned}
\therefore V_{2} &=V_{d} \\
\text { % Volume reduction } &=\frac{V_{1}V_{2}}{V_{1}} \times 100 \\
&=\frac{\left(V_{s}+0.323 W_{s}\right)\left(V_{s}+0.23 W_{s}\right)}{V_{s}+0.323 W_{s}} \times 100 \\
&=\frac{9.3 W_{s}}{V_{s}+0.323 W_{s}}=\frac{9.3 W_{s}}{W_{s}\left(\frac{V_{s}}{W_{s}}+0.323\right)} \\
&=\frac{9.3}{\frac{V_{s}}{W_{s}}+0.323}=\frac{9.3}{\left(\frac{V_{s}}{V_{s} \gamma_{s}}\right)+0.323} \\
&=\frac{9.3}{\left(\frac{1}{G \gamma_{w}}\right)+0.323}=\frac{9.3}{\left(\frac{1}{2.65 \times 1}+0.323\right)}=13.28 \% \quad\left[\because G=\frac{\gamma_{s}}{\gamma_{w}}\right]
\end{aligned}
$$