written 19 months ago by
Chandan15
• 300

•
modified 19 months ago

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, $Q_{S}=\frac{Q_{u}}{FO
S}=\frac{1401.017}{2.5}$
$$
=560.407 \mathrm{KN}
$$
Being watched by a moderator
I'll actively watch this post and tag someone who might know the answer.