Question

The permissible settlement of an isolated foundation on a granular soil is 40 mm. A plate load test is performed with 300 mm square plate undergoing a settlement of 20 mm under the same load intensity. The ultimate bearing capacity of soil with plate is $25t/m^2$. The ultimate bearing capacity of soil for foundation will be $t/m^2$

• A. 60.42

Answer 1

The correct option is A 60.42
$\frac{S_f}{S_F}={(\frac{B_f}{B_p}\times \frac{B_p+0.3}{B_f+0.3})}^2$
$\Rightarrow \frac{40}{20}={(\frac{B_f}{0.3}\times \frac{0.3+0.3}{B_f+0.3})}^2$
$\Rightarrow B_f=0.7243m\simeq 752\text{mm}$
Now, $q_{uf}=q_{up}\left(\frac{B_f}{B_p}\right)=25\times \frac{725}{300}=60.42t/m^2$