Question

A square footing (2 m $\times$ 2 m ) is subject to an inclined point load, P as shown in the figure below. The water table is located well below the base of the footing. Considering one way eccentricity, the net safe load carrying capacity of the footing for a factor of safety of 3.0 is __ kN.

The following factor may be used:
Bearing capacity factors : $N_q$ = 33.3, $N_{\gamma }$ = 37.16,
shape factors : $F_{qs}$ = $F_{\gamma }$${}_s$ = 1.314
Depth factors : $F_{qd}$ = $F_{\gamma }$${}_d$ = 1.113
Inclination factors : $F_{qi}=0.444,F_{\gamma }$${}_i$ = 0.02

• A. 441.71

Answer 1

The correct option is A 441.71
$q_{ns}=\frac{1}{3}\left(q_{nu}\right)$

$q_{nu}=C{N}_c+(N_q-1)+\frac{1}{2}B\gamma N_{\gamma }\text{(for normal case)}$
For inclined loading, net safe load carrying capacity

$q_{ns}=\frac{1}{3}\left(q\right(N_q-1)\times F_{qs}\times F_{qd}\times F_{qp}+\frac{1}{2}B\gamma N_{\gamma }\times F_{\gamma s}$ $\times F_{\gamma }$${}_d\times$ $F_{\gamma }$${}_p$)

$q_{ns}=\frac{1}{3}(18\times 1(33.3-1)\times 1.314\times 1.113\times 0.444+\frac{1}{2}\times 2\times 18\times 37.16\times 1.314\times 1.113\times 0.02)$

$=\frac{397.09}{3}=132.364kN/m^2$

For one way eccentrically, we will have to take reduced area of footing


$\Rightarrow \text{Load carrying capacity}$

$=\frac{132.364\times 2.89}{cos{30}^0}=441.71kN$