Question

A multistory building with a basement is to be constructed. Top 4 m consists of loose silt, below which dense sand layer is present upto a great depth. ground water table is at the surface. The foundation consist of the basement slab of 6 m width which will rest on the top of dense sand as shown in the figure. For dense sand, saturated unit weight = 20 kN/$m^3$, and bearing capacity factors $N_q$ = 40 and $N_{\gamma }$ = 45. For loose slit, saturated unit weight = 18 kN/$m^3$, $N_q$= 15, $N_{\gamma }$ = 20. Effect cohesion C' is zero for both soils. Unit weight of water is 10 kN/ $m^3$. Neglect shape factor and depth factor. Average elastic modulus E and Poisson's ratio (1/m) of dense sand is 60 $\times$ ${10}^3$ kN/$m^3$ and 0.3 respectively.


The foundation slab is subjected to net safe bearing capacity derived in the above in the above question. Using influence factor $I_f$ = 2.0, and neglecting embedment depth and rigidity corrections, the immediate settlement of the dense sand layer will be:

• A. 159.55 mm
• B. 111 mm
• C. 126 mm
• D. 179 mm

Answer 1

The correct option is A 159.55 mm

$q_u=C{N}_c+q{N}_q+\frac{1}{2}\gamma B{N}_{\gamma }$

Here, $\gamma$ = ${\gamma }_{su{b}_1}$ and q = ${\gamma }_{su{b}_2}$ $\times$4
${\gamma }_{su{b}_1}$ = ${\gamma }_{sat\left(sand\right)}$- ${\gamma }_w$
= 20-10 = 10 kN/ $m^3$

${\gamma }_{su{b}_2}={\gamma }_{sat\left(silt\right)}-{\gamma }_w=18-10=8kN/m^3$

$\therefore q_u=0+8\times 4\times 40+0.5\times 10\times 6\times 45=2630kN/m^3$

Since in this case, soil is not filled
$q_{nu}=q_u$

$q_{ns}=\frac{q_{nu}}{F}=\frac{2630}{3}$

$=876.66kN/m^2$

$S_i=\frac{q_{ns}B(1-{\mu }^2)I_f}{E_s}$

$=\frac{876.66\times 6[1-(0.3{)}^2]\times 2}{(60\times {10}^3)}$

= 159.55 mm