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

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 \times 10^{-8} in/sec and a coefficient of volume compressibility m_{v}=8.6 \times 10^{-4} \mathrm{in}^{2} / \mathrm{lb}. Determine the ultimate value of the compression of the stratum by assuming a uniformly distributed load of 5250 \mathrm{lb} / \mathrm{ft}^{2} and also determine the time required for 20 percent and 80 percent consolidation

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Total compression,

 

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

 

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

 

T=\frac{\pi}{4} \frac{U \%}{100}^{2} \text { or } T=\frac{3.14}{4} \times \frac{20}{100}^{2}=0.0314

 

For 80% consolidation use the equation

 

T=1.781-0.933 \log (100-U \%)

 

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

 

The coefficient of consolidation is

 

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

 

The times required for 20% and 80% consolidation are

 

t_{20}=\frac{H_{d r}^{2} T}{c_{v}}=\frac{(10 \times 12)^{2} \times 0.0314}{6 \times 10^{-4} \times 60 \times 60 \times 24}=8.72 \text { days }

 

t_{80}=\frac{H_{d r}^{2} T}{c_{v}}=\frac{(10 \times 12)^{2} \times 0.567}{6 \times 10^{-4} \times 60 \times 60 \times 24}=157.5 \text { days }

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