Question

In a 6 m thick stratum of fine sand having submerged density of $11kN/m^3$ , quicksand condition occurred at a depth of 4.2 m of excavation. What is the depth of lowering of groundwater table required for making an excavation 5 m deep? Take density of water as $10kN/m^3$.

• A. 3.85 m
• B. 1.68m
• C. 1.1 m
• D. 0.897 m

Answer 1

The correct option is B 1.68m

Let x be the depth of ground water table initially. Total upward water force on the sand stratum at the bottom of excavation


Total downward force at the bottom of excavation
= Weight of soil in saturated condition

$={\gamma }_{sat}\times (6-4.2)$

$=({\gamma }_{sub}+{\gamma }_w)\times 1.8$

When quicksand condition occurs, then total upward water force becomes equal to total downward force i.e.

$(6-x)\times {\gamma }_w=({\gamma }_{sub}+{\gamma }_w)\times 1.8$

$\Rightarrow$ $(6-x)\times 10=(11+10)\times 1.8$

$\Rightarrow$ $x=6-\frac{21\times 1.8}{10}$

$\Rightarrow$ $x=2.22m$


Similarly, if the depth of displaced ground water table is y from the surface, then

$(6-y)\times {\gamma }_w=(6-5)\times ({\gamma }_{sub}+{\gamma }_w)$

$\Rightarrow$ $(6-y)\times 10=1\times (11+10)$

$\Rightarrow$ $y=6-2.1$

$\Rightarrow$ $y=3.9m$

$\therefore$ Lowering of water table = y - x

$=3.9-2.2$

$=1.68m$