Question

A single vertical friction pile of diameter 500 mm and length 20 m is subjected to a vertical compressive load. The pile is embedded in a homogenous sandy strata where angle of internal friction $\left(\varphi \right)={30}^{\circ }$, dry unit weight $\left({\gamma }_d\right)=20{\text{kN/m}}^3$ and angle of wall friction $\left(\delta \right)=\frac{2\varphi }{3}$. Considering the coefficient of lateral earth pressure $\left(K\right)=2.7$ and the bearing capacity factor $\left(N_q\right)=25$, the ultimate bearing capacity of the pile (in kN).

• A. 4499

Answer 1

The correct option is A 4499
$\varphi ={30}^{\circ }$
${\gamma }_d=20{\text{kN/m}}^3$
$\delta =\frac{2\varphi }{3}$
$K=2.7$
$N_q=25$
$Ultimatebearingcapacity=q_{pu}.A_{\text{base}}+f_s{A}_s$
$=\overline{\sigma }{N}_q.A_{\text{base}}+K{\overline{\sigma }}_{av}\tan \delta .A_s$

Assuming loose sand (as $\varphi$ is less)
The eff. stress will vary upto $15D$ and thereafter it is constant.

Point bearing strength
$\overline{\sigma }{N}_q\times A_{\text{base}}$
$=150\times 25\times \frac{\pi }{4}(0.5{)}^2\text{kN}$
$=736.31\text{kN}$
Frictional Strength
$=f_{s_1}{A}_{s_1}+f_{s_2}{A}_{s_2}$
$f_{s_1}=K_1{\overline{\sigma }}_{a{v}_1}\tan \delta$
$=2.7\times \frac{150}{2}\times \tan (\frac{2}{3}\times {30}^{\circ })$
$=73.704{\text{kN/m}}^2$
$f_{s_2}=k{\overline{\sigma }}_{av}\tan \delta$
$=2.7\times 150\times \tan 20$
$=147.408{\text{kN/m}}^2$
$\Rightarrow$ Frictional Strength
$=(73.704\times \pi D\times 7.5+147.408\times \pi D\times 12.5)\text{kN}$
$=3762.654\text{kN}$
$\Rightarrow$ Ultimate Bearing Capacity
$=736.31+3762.654$
$=4498.96$
$=4499\text{kN}$