Question

A 10 m thick clay layer is resting over a 3 m thick sand layer and is submerged. A fill of 2 m thick sand with unit weight of 20$kN/m^3$ is placed above the clay layer to accelerate the rate of consolidation of the clay layer. Coefficient of consolidation of clay is $9\times {10}^{-2}{m}^2/year$ and coefficient of volume compressibility of clay is $2.2\times {10}^{-4}{m}^2/kN$. Assume Taylor's relations between time factor and average degree of consolidation.


The settlement (in mm, round off to two decimal places) of the clay layer, 10 years after the construction of the fill, is

• A. 18.83

Answer 1

The correct option is A 18.83
Given

Coefficient consolidation $\left(C_v\right)$

$=9\times {10}^{-2}{m}^2/year$

$Volumecompressibility\left(m_v\right)$

$=2.2\times {10}^{-4}{m}^2/kN$

Thickness of clay layer (H) = 10 m

Change in effective stress due to surcharge

$(i.e.,sandlayer)\Delta \overline{\sigma }$ = 2 $\times$ 20

$=40kN/m^2$

$\left(Totalsettlement\Delta H\right)=m_{v}H\Delta \overline{\sigma }$

$=2.2\times {10}^{-4}\times 10\times 40$

= 0.088 m

Now, According to Taylor relation

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

$d=\frac{10}{2}=5m$

(double drainage both side sand layer)

t = 10 year

$T_v=\frac{9\times {10}^{-2}\times 10}{5^2}=0.036$

We know that

$T_v=0.197for50\%Consolidation$

Here $T_v<0.197$

So, we use

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

$\frac{\pi }{4}\left(U^2\right)=0.036$

U = 0.214

$NowU=\frac{\Delta h}{\Delta H}$

$\Delta h=settlementafter10years$

$\Delta h=\Delta H\times 0.214$

$=0.088\times 0.214$

= 0.01883 m

= 18.83 mm