Question

A 20 m thick clay layer is sandwiched between a silty sand layer and a gravelly sand layer. The layer experiences 30 mm settlement in 2 years.

Given :

$T_V=\{\begin{array}{cc}\frac{\pi }{4}{\left(\frac{U}{100}\right)}^2& forU\le {60}^{\circ {/}_{\circ }}\\ 1.784-0.933lo{g}_{10}(100-U)& U>{60}^{\circ {/}_{\circ }}\end{array}$



Where $T_V$ is the time factor and U is the degree of consolidation in %

If the coefficient of consolidation of the layer is $0.003c{m}^2/s$, the deposit will experience a total of 50 mm settlement in the next _____ years.

• A. 4.43

Answer 1

The correct option is A 4.43

$\frac{C_{v}t}{H^2}=T_v$

$\Rightarrow T_v=\frac{0.003\frac{c{m}^2}{sec}\times 2\times 86400\times 365sec}{{(\frac{20}{2}\times 100)}^{2}c{m}^2}$

$T_v=0.1892=\frac{\pi }{4}\left(U^2\right)$

$T_v<0.2826$

$\Rightarrow U<60\%$

$\Rightarrow U=0.4908$

$\Rightarrow Totalconsolidation=\frac{30}{0.4908}$

= 61.12 mm

Degree of consoildation for 50mm settlement

$=\frac{50}{61.12}=U=0.818=81.8\%$

$\Rightarrow T_v=0.60835=C_{v}t/H^2$

$\Rightarrow t=\frac{0.60835\times (10{)}^2}{0.003\times {10}^{-4}}sec$

= 6.43 yr

$\Rightarrow Additionalnumberofyears=4.43yrs$