A strata of normally consolidated clay of thickness 10 ft is drained on one side only. It has a hydraulic conductivity of k = 1.863 x 10^-8 in/sec and a coefficient of volume compressibility mv = 8.6 x 10^-4 in^2/lb. Determine the ultimate value of the compression of the stratum by assuminga

Question

A strata of normally consolidated clay of thickness 10 ft is drained on one side only. It has a hydraulic conductivity of [latex]k=1.863 imes 10^{-8}[/latex] in/sec and a coefficient of volume compressibility [latex]m_{v}=8.6 imes 10^{-4} \mathrm{in}^{2} / \mathrm{lb}[/latex]. Determine the ultimate value of the compression of the stratum by assuming a uniformly distributed load of [latex]5250 \mathrm{lb} / \mathrm{ft}^{2}[/latex] and also determine the time required for 20 percent and 80 percent consolidation

Accepted Answer

Total compression,

[latex]S_{t}=m_{v} H \Delta p=8.6 imes 10^{-4} imes 10 imes 12 imes 5250 imes \frac{1}{144}=3.763 \mathrm{in}[/latex]

For determining the relationship between U% and T for 20% consolidation use the equation

[latex]T=\frac{\pi}{4} \frac{U \%}{100}^{2} ext { or } T=\frac{3.14}{4} imes \frac{20}{100}^{2}=0.0314[/latex]

For 80% consolidation use the equation

[latex]T=1.781-0.933 \log (100-U \%)[/latex]

Therefore [latex]T=1.781-0.933 \log _{10}(100-80)=0.567[/latex]

The coefficient of consolidation is

[latex]c_{v}=\frac{k}{\gamma_{w} m_{v}}=\frac{1.863 imes 10^{-8}}{3.61 imes 10^{-2} imes 8.6 imes 10^{-4}}=6 imes 10^{-4} \mathrm{in}^{2} / \mathrm{sec}[/latex]

The times required for 20% and 80% consolidation are

[latex]t_{20}=\frac{H_{d r}^{2} T}{c_{v}}=\frac{(10 imes 12)^{2} imes 0.0314}{6 imes 10^{-4} imes 60 imes 60 imes 24}=8.72 ext { days }[/latex]

[latex]t_{80}=\frac{H_{d r}^{2} T}{c_{v}}=\frac{(10 imes 12)^{2} imes 0.567}{6 imes 10^{-4} imes 60 imes 60 imes 24}=157.5 ext { days }[/latex]

Question 7.10: A strata of normally consolidated clay of thickness 10 ft is...

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