2076 Chaitra · 6 marks

Raft 20 m × 10 m, factor of safety

Short answer

For raft 20×10 m, Df=2.5 m on clay cu=80 kPa, Skempton Nc=5.80, net ultimate 463.7 kPa. Net applied pressure 152.5 kPa gives FS=3.04.

A raft foundation is 20 m × 10 m exerts a gross pressure of $200\ \text{kN/m}^2$ at the foundation level. The depth of foundation is 2.5 m. If the soil is clay $C_u = 80\ \text{kN/m}^2$ and $\gamma = 19\ \text{kN/m}^3$ . Determine the factor of safety. Use Skempton's equations.

Given: raft $20\ \text{m} \times 10\ \text{m}$ , gross pressure $q = 200$ kN/m², $D_f = 2.5$ m, clay $c_u = 80$ kN/m², $\gamma = 19$ kN/m³.

Skempton's equation: $q_{nu} = c_uN_c$ , with $B = 10$ m, $L = 20$ m.

\begin{aligned} \frac{D_f}{B} &= \frac{2.5}{10} = 0.25 < 2.5 \\ N_{c(sq)} &= 6\left(1 + 0.2 \times 0.25\right) = 6.30 \\ N_{c(rect)} &= \left(0.84 + 0.16 \times \frac{10}{20}\right)(6.30) = 0.92 \times 6.30 = 5.80 \\ q_{nu} &= 80 \times 5.796 = 463.7\ \text{kN/m}^2 \end{aligned}

Net applied pressure:

q_{net} = q - \gamma D_f = 200 - 19 \times 2.5 = 152.5\ \text{kN/m}^2

FS = \frac{q_{nu}}{q_{net}} = \frac{463.7}{152.5} = 3.04

Answer: factor of safety $\approx 3.04$ .

✓ Answer checked by Claude Sonnet 5.5

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