The unit cohesion of material is 18kN/m$^{2}$ and unit weight of soil is 15kN/m$^{3}$. Compute negative skin friction given adhesion coefficient is 0.4.

• For individual piles: $$ \therefore \quad F_{n g}=16 \cdot F_{n}=16 \cdot \alpha \cdot \bar{c}_{u} \cdot\left(P_{s}\right) L_{c} $$ where $P_{s}=$ perimeter surface of the pile of dia $0.25 \mathrm{~m}$ $$ \begin{aligned} \pi d &=3.14 \times 0.25=0.785 \mathrm{~m} \\ L_{c} &=3 \mathrm{~m} \\ \bar{c}_{u} &=c_{u}=18 \mathrm{kN} / \mathrm{m}^{2} \\ \alpha &=0.4 \\ F_{n g} &=16 \times 0.4 \times 18 \mathrm{kN} / \mathrm{m}^{2} \times 0.785 \mathrm{~m} \times 3 \mathrm{~m} \\ &=271.3 \mathrm{kN} \end{aligned} $$
  • For pile group :

where $$ \begin{aligned} F_{n g} &=\bar{C}_{u g} \cdot P_{s g} L_{c}+\gamma \cdot L_{c} \cdot A_{b g} \\ \bar{C}_{u g} &=c_{u g}=c_{u}=18 \mathrm{kN} / \mathrm{m}^{2} \\ P_{s g} &=4 \times 2.5 \mathrm{~m}=10 \mathrm{~m} \\ L_{c} &=3 \mathrm{~m} \end{aligned} $$ $$ \begin{aligned} \gamma &=15 \mathrm{kN} / \mathrm{m}^{2} \\ A_{bg} &=(2.5 \mathrm{~m} \times 2.5 \mathrm{~m})=6.25 \mathrm{~m}^{2} \\ F_{n g} &=18 \mathrm{kN} / \mathrm{m}^{2} \times 10 \mathrm{~m} \times 3 \mathrm{~m}+15 \mathrm{kN} / \mathrm{m}^{3} \times 3 \mathrm{~m} \times 6.25 \mathrm{~m}^{2} \\ &=540+281.25=821.25 \mathrm{kN} \end{aligned} $$ The greater of individual and group gives negative skin friction, $$ F_{n g}=821.25 \mathrm{kN} $$