Question

The bearing capacity of a rectangular footing of plane dimension 1.5 m $\times$ 3 m resting on the surface on a sand deposit was estimated as 600 kN/$m^2$ when the water table is far below the base of the footing. The bearing capacities in kN/$m^2$ when the water level rises to depths of 3 m, 1.5 m and 0.5 m below the base of the footing are

• A. 600, 600, 400
• B. 600, 450, 350
• C. 600, 500, 250
• D. 600, 400, 250

Answer 1

The correct option is A 600, 600, 400

Size of footing = 1.5 m $\times$ 3 m.


width B = 1.5 m
Conhesion,
C = 0
Footing rests on the surface so,
q = $\gamma$$D_f$ = $\gamma$ $\times$ 0 = 0
When water table is far below base of footing
$q_u$ = 0.4$\gamma$ B$N_{\gamma }$
600 = 0.4$\gamma$ B$N_{\gamma }$
Case I : when water table rises to depth of 3m, $D_w$ = 3 m. when water table below is below the detph equal to width of footing, it will not affect the bearing capacity so bearing capacity = 600 kN/$m^2$

Case II : $D_w$ = 1.5 m
($D_w$ $\nless$ B)
$q_u$ = 600 kN/$m^2$

Case III : $D_w$ = 0.5 m
($D_w$ < B)

So, $q_u$ = 0.4 [${\gamma }_t$$D_w$ + ${\gamma }_{sub}$ (1.5 - $D_w$)]$N_{\gamma }$
value of $\gamma$${}_t$ and ${\gamma }_{sub}$ is not given in question but we can definetley say that the value of $q_u$ in this case will be less than 600 kN/$m^2$
Hence we can choose the correct answer.