1
5.9kviews
A 12m long and 300mm diameter concrete pile is driven in a uniform deposit of sand.

The water table is very much low. The $\gamma=18kN/m^{3},$N_{q}=137$. Calculate the safe load capacity of pile K=2.0. Assume the critical depth as 15 times the diameter of pile. Being watched by a moderator I'll actively watch this post and tag someone who might know the answer. 1 Answer 0 615views Here, • Diameter of pile,$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}\$

$$=560.407 \mathrm{KN}$$