written 2.4 years ago by
Chandan15
• 300

•
modified 2.4 years ago

Given:
(i) Void ratio at $150 \mathrm{kN} / \mathrm{m}^{2}$ stress
$$
e_{0}=1.95
$$
Let $\Delta e$ be the change in void ratio when stress is increased from $150 \mathrm{kN} / \mathrm{m}^{2}$ to $210 \mathrm{kN} / \mathrm{m}^{2}$.
We know,
$$
C_{c}=\frac{\Delta e}{\log \frac{\sigma_{2}}{\sigma_{1}}}
$$
Impervious
$$
0.28=\frac{\Delta e}{\log _{10} \frac{210}{150}}
$$
$$
\therefore \quad \Delta e=0.28 \times \log _{10} \frac{210}{150}=0.041
$$
Negative sign shown void ratio decreases with increased pressure.
(ii) Settlement,
$$
\begin{aligned}
\Delta H &=\frac{C_{c} H_{0}}{1+e_{0}} \log _{10}\left[\frac{\bar{\sigma}_{1}+\Delta \sigma}{\bar{\sigma}_{1}}\right]=\frac{0.28 \times 6}{1+1.95} \log _{10} \frac{210}{150} \\
&=0.0832 \mathrm{~m} \text { or } 83.20 \mathrm{~mm}
\end{aligned}
$$
(iii) Given,
$$
T_{v}=0.2(\text { for } 50 \%)
$$
Let ' $t$ ' be the time required for $50 \%$ consolidation.
Using,
where,
$T_{v}=\frac{C_{v} \cdot t}{d^{2}}$
$\Rightarrow \quad d=$ length of drainage path
where, $C_{v}$ can be found $d=H_{0}=6 \mathrm{~m}$ (for single drainage)
where,
Now, substituting value of $C_{v}$ and $T_{v}$ in equation (i), we get
$$
\begin{aligned}
0.2 &=\frac{1.54 \times 10^{3} \times t}{6^{2}} \
t &=\frac{0.2 \times 6^{2}}{1.54 \times 10^{3}}=4675.32 \text { sec or } 1.30 \text { hours } .
\end{aligned}
$$
(iv) Let $U$ be the degree of consolidation that will take place in 1 hour. Since $50 \%$ consolidation completed in $1.30$ hours. Hence degree of consolidation occurred in one hour must be less than $50 \%$.
Time factor,
$$
T_{v}=\frac{C_{v} \times t}{d^{2}}=\frac{\left(1.54 \times 10^{3}\right) \times(1 \times 60 \times 60)}{6^{2}}=0.154
$$
Also,
$$
\begin{aligned}
\therefore \quad 0.154 &=\frac{\pi}{4} U^{2} \\
U &=\sqrt{\frac{0.154 \times 4}{\pi}}=0.44 \text { or } 44 \%
\end{aligned}
$$
$[U \leq 60 \%]$
Let $\Delta$h be the amount of settlement after the end of one hour using,
$$
U=\frac{\Delta h}{\Delta H}
$$
where $\Delta H=$ Ultimate settlement of soil stratum
$$
=72 \mathrm{~cm}
$$
$$
\therefore \quad 0.44=\frac{\Delta h}{72}
$$
or
$$
\Delta h=0.44 \times 72=31.68 \mathrm{~cm}
$$