Question

A 4 m wide strip footing is founded at a depth of 1.5 m below the ground surface in a $c-\varphi$ soil as shown in the figure. The water table is at a depth of 5.5 m below ground surface. the soil properties are c' = 25 kN/$m^2$, ${\varphi }^{\prime }$ = ${28.63}^o$, ${\gamma }_{sat}$ = 19 kN/$m^3$, ${\gamma }_{bulk}$ = 17 kN/$m^3$ and ${\gamma }_w$ = 9.81 kN/$m^3$. the values of bearing capacity factors for different ${\varphi }^{\prime }$ are given below.

$\begin{array}{cccc}{\varphi }^{{}^{\prime }}& N_c& N_q& N_{\gamma }\\ {15}^{\circ }& 12.9& 4.4& 2.5\\ {20}^{\circ }& 17.7& 7.4& 5.0\\ {25}^{\circ }& 25.1& 12.7& 9.7\\ {30}^{\circ }& 37.2& 22.5& 19.7\end{array}$

Using Terzaghi's bearing capacity equation and a factor of safety $F_s=2.5,$ the net safe bearing capacity (expressed in kN/$m^2$) for local shear failure of the soil is

• A. 298.48

Answer 1

The correct option is A 298.48
Local shear failure
$tan{\varphi }^{\prime }=\left(\frac{2}{3}tan\varphi \right)$

$=\frac{2}{3}\times 0.54589=0.36393$

${\varphi }^{\prime }={20}^o$

$\text{So, corresponding}{N}_c=17.7,N_q=7.4,N_{\gamma }=5$

$q_{nu}=\frac{2C^{\prime }}{3}{N}_c+q(N_q-1)+\frac{1}{2}B\gamma N_{\gamma }$

$=\frac{2\times 35\times 17.7}{3}+1.5\times 17\times 6.4+\frac{1}{2}\times 4\times 17\times 5$

$=746.2kN/m^2$

$\text{Safe bearing capacity}=\frac{746.2}{2.5}=298.48kN/m^2$