Question

A square footing of 2 m sides rests on the surface of a homogeneous soil bed having the properties: cohesion c = 24 kPa, angle of internal friction $\varphi$ = ${25}^o$, and unit weight $\gamma$ = $18kN/m^3$. Terzaghi's bearing capacity factor $\varphi$= ${25}^o$ are $N_c$ = 25.1, $N_q$ = 12.7, $N_{\gamma }$ = 9.7, $N_c^{\prime }$ = 14.8, $N_q^{\prime }$ = 5.6, and $N_{\gamma }^{\prime }$ = 3.2. The ultimate bearing capacity of the foundation (in kPa, round off to 2 decimal places) is

• A. 353.92

Answer 1

The correct option is A 353.92

$Q_{ultimate}=1.3C^{\prime }{N}_c^{\prime }+q{N}_q^{\prime }+0.4B\gamma N_{\gamma }^{\prime }$

$=1.3\times (\frac{2}{3}\times 24)\times 14.8+0+0.4\times 2\times 18\times 3.2$

= 353.92 kPa

$Here,\varphi <{29}^o$
Footing rests on the ground
Hence, q = 0
$so,c^{\prime }=\frac{2}{3}c$
$use,N_q^{\prime }{N}_c^{\prime }{N}_{\gamma }^{\prime }accordingly$