Question

A square footing of 4 m. side is placed at 1 m depth in a sand deposit. The dry unit weight ($\gamma$) of sand is 15 kN/$m^3$. This footing has an ultimate bearing capacity of 600 kPa. Consider the depth factor; $d_q=d_{\gamma }=1.0$ and the bearing capacity factor : $N_{\gamma }=18.75$ . This footing is placed at a depth of 2 m in the same soil deposit. For a factor of safety of 3.0 for Terzaghi's theory, the safe bearing capacity (in kPa) of this footing would be

• A. 270

Answer 1

The correct option is A 270
(270 kPa)
Skide of square footing = 4 m
depth of footing = 1m
Unit weight of soil = 15 KN/$m^3$
Ultimate bearing capacity = 600 KPa
Depth factors, $d_q$ = $d_{\gamma }$ = 1
$N_{\gamma }$ = 18.75
According to terzaghi, the ultimate bearing capacity of square footing is given as at depth of footing = 1 m
$q_u$ = 1.3 C$N_c$ + q$N_q$$d_q$ + 0.4$B_{\gamma }$$N_{\gamma }$$d_{\gamma }$
For sand, C = 0, q = $\gamma$$D_f$ = 15 $\times$ 1 = 15 KN/$m^2$
600 = 0 + 15 $\times$ $N_q$ $\times$ 1 + 0.4 $\times$ 4 $\times$ 15 $\times$ 18.75 $\times$ 1
$\Rightarrow$ $N_q$ = 10
Now at depth of footing at 2 m
$q_u=1.3C{N}_c$ +q$N_q$ + 0.4B$\gamma$$N_{\gamma }$ $d_{\gamma }$
$q_u$ = 0 + (2 $\times$ 15)10 $\times$ 1 + 0.4 $\times$ 4 $\times$15 $\times$ 18.75 $\times$ 1
$q_u$ = 750 KPa
$\because$ We know that
$q_{nu}$ = $q_u$ - $\gamma$$D_f$
$q_{nu}$ = 750 -15 $\times$ 2
$q_{nu}$ = 720 KPa
and safe bearing capacity $q_{safe}$

$q_{safe}$ = $\frac{q_{nu}}{FOS}$ + $\gamma$$D_f$

= $\frac{720}{3}$ + 15 $\times$ 2

= 270 KPa