If undrained cohesion is 45 kN/m$^{2}$ and the pile spacing is 1 m centre to centre. What is the capacity of group? Take factor of safety of 2 and adhesion factor as 0.7.

Given Data: $$ \begin{array}{l} D=0.3 \mathrm{~m} \\ L=9 \mathrm{~m} \\ C_{U}=45 \mathrm{kN} / \mathrm{m}^{2} \\ \text { FOS }=2 \\ \alpha=0.7 \end{array} $$

• Case I: Considering as individual pile, we get,

$$ Q_{u}=Q_{PU}+Q_{f} $$ $$ \begin{array}{l} Q_{U}=Q_{P U}+Q_{f} \\ Q_{U}=C_{U} N_{C} A_{b}+alpha*C_{U}A_{S} \\ Q_{U}=\left(45 \times 9 \times \frac{\pi}{4} \times 0.3^{2}\right)+(0.7 \times 45 \times \pi \times 0.3\times 9) \\ =295.82 \mathrm{KN} \\ \end{array} $$

• For Pile group,
• $Q\mathrm{ug}_{1}=n \times Q_{u}$

$=9 \times 295.82$ $=2662.38 \mathrm{kN}$ $=2662.38 \mathrm{kN}$

• Case II: Considering as Block Failure,

$$ \begin{array}{l} Q_{ug2}=C_{U} N_{C}A_{b}+alpha*C_{U} A_{s} \\ =45 \times 9 \times 2.3^{2}+(1 \times 45 \times 4 \times 2.3 \times 9) \end{array} $$ $=5868.45 \mathrm{KN}$

• Allowable load Capacity

$$ \begin{array}{l} =\frac{min \left\{Q u g_{1}, Q_{ug_{2}}\right\}}{FOS} \\ =\frac{2662.38}{2}=1331.19 \mathrm{kN} \end{array} $$