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Problem based on Consolidation:A staturated soil stratum 6m thick lies above an impervious
• A saturated soil stratum 6 metres thick lies above an impervious stratum and below a pervious stratum. It has a compression index of $0.28$ and a coefficient of permeability of $3.5 \times 10^{-4} \mathrm{~cm} / \mathrm{sec}$. Its void ratio at a stress of $150 \mathrm{kN} / \mathrm{m}^{2}$ is $1.95$.

Determine:

(i) The change in void ratio due to an increase in stress to $210 \mathrm{kN} / \mathrm{m}^{2}$ (ii) Settlement of the soil stratum due to the above increase in stress (iii) Time required for 50 per cent consolidation. Assume time factor $T$ for 50 per cent consolidation as $0.20$ (iv) The amount of settlement which will take place in one hour. The ultimate settlement of stratum was found $72 \mathrm{~cm}$. Assume time factor $T$ for $50 \%$ consolidation as $0.20$.

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 $$\Delta h=0.44 \times 72=31.68 \mathrm{~cm}$$