Question

An undisturbed soil sample was taken from the middle of a clay layer (i.e., 1.5 m below GL), as shown in figure. The water table was at the top of clay layer. Laboratory test result are as follows:

Natural water content of clay : 25%

Preconsolidation pressure of clay : 60 kPa

Compression index of clay : 0.50

Recompression index of clay : 0.05

Specific gravity of clay : 2.70

Bulk unit weight of sand : $17kN/m^3$

A compacted fill of 2.5 m height with unit weight of $20kN/m^3$ is placed at the ground level.


Assuming unit weight of water as $10kN/m^3$, the ultimate consolidation settlement (expressed in mm) of the clay layer is .

• A. 37.04

Answer 1

The correct option is A 37.04
$\overline{\sigma }=17\times 1+0.5\times {\gamma }_{subclay}$


For clay

w = 0.25

S = 1

g = 2.7

$\Rightarrow Se=Gw$

$\Rightarrow e=\frac{2.7\times 0.25}{1}$

e = 0.675

$\Rightarrow {\gamma }_{sub}=\frac{G-1}{1+e}{\gamma }_w=\frac{2.7-1}{1+0.675}\times 10$

$\Rightarrow {\gamma }_{sub}=10.15kN/m^3$

$\overline{\sigma }=17+0.5\times 10.15$

$=22.07kN/m^2$

$\Delta \overline{\sigma }=2.5\times 20=50kN/m^2$


Alternate Method

$\Delta H_1=\frac{C_{r}H}{1+e_0}log\frac{60}{22.07}$

$=\frac{0.05\times 1000}{1.675}log\frac{60}{22.07}$

$\Delta H_1=12.966mm$

$\Delta H_2=\frac{C_{c}H}{1+e^{\prime }}log\left(\frac{72.07}{60}\right)$

$C_r=\frac{\Delta e}{log\left(\frac{60}{22.07}\right)}=0.05$

$\Delta e=0.0217$

$\Rightarrow e^{\prime }=0.653$

$\Delta H_2=\frac{0.5\times 1000}{1.653}log\left(\frac{72.07}{60}\right)$

$\Delta H_2=24.075mm$

$\Rightarrow \Delta H=\Delta H_1+\Delta H_2=37.04mm$