The water table is very much low. The $\gamma=18kN/m^{3}, $N_{q}=137$. Calculate the safe load capacity of pile K=2.0. Assume the critical depth as 15 times the diameter of pile.

Here,

• Diameter of pile, $D=0.3 \mathrm{~m}$
• Vertical stress at Critical Depth $=15 D$ $=4.5 \mathrm{~m}$

Ultimate Bearing Capacity, $$ \begin{aligned} Q_{u}=& Q_{pu}+Q_{f} \\ \Rightarrow Q_{u}=&\left(q N_{q} A_{b}\right)+\left(f_{s}A_{1}\right)+\left(f_{2} A_{2}\right) \\ \Rightarrow Q_{u}=&\left(81 \times 14 7 \times \frac{\pi}{4} \times 0.3^{2}\right) \\ &+\left(2 \times \frac{81+0}{2} \times 4.5 \times \pi \times 0.3\right) \tan 22.5^{\circ} \\ &+\left(2 \times \frac{81+81}{2} \times 7.5 \times \pi \times 0.3\right) \times \tan 22 .5^{\circ} \end{aligned} $$ Hence,

• Ultimate Bearing Capacity, $Q_{u}=1401.017 \mathrm{kN}$
• Safe load Capacity, $Q_{S}=\frac{Q_{u}}{FO S}=\frac{1401.017}{2.5}$

$$ =560.407 \mathrm{KN} $$