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Mat 10 m × 15 m with basement, FOS

Short answer

For mat 10×15 m, Df=2.5 m, cu=30 kPa, total load 4000 kN. Net ultimate capacity qnu=cu Nc=30×5.96=178.9 kPa. Gross pressure q=26.67 kPa gives FS=6.71; mat is overcompensated if excavation relief considered.

The 10 m × 15 m size mat is constructed at 2.5 m depth having basement for underground parking. The site consists of highly compressible saturated clay having cohesion of 30 kN/m230\ \text{kN/m}^2. If the mat carries the total load of 4000 kN. Calculate the factor of safety.

Given: mat B=10B = 10 m, L=15L = 15 m, Df=2.5D_f = 2.5 m, saturated clay cu=30c_u = 30 kN/m², total load Q=4000Q = 4000 kN. The unit weight of soil is not given.

Assumption: the relief due to excavation is ignored (conservative) because γ\gamma is not given, so the net applied pressure is taken equal to Q/AQ/A.

Bearing capacity factor (Skempton)

Nc(sq)=6(1+0.2×2.510)=6.30Nc(rect)=(0.84+0.16×1015)(6.30)=5.96qnu=cuNc=30×5.964=178.9 kN/m2\begin{aligned} N_{c(sq)} &= 6\left(1 + 0.2 \times \frac{2.5}{10}\right) = 6.30 \\ N_{c(rect)} &= \left(0.84 + 0.16\times\frac{10}{15}\right)(6.30) = 5.96 \\ q_{nu} &= c_uN_c = 30 \times 5.964 = 178.9\ \text{kN/m}^2 \end{aligned}

Applied pressure and factor of safety

q=400010×15=26.67 kN/m2FS=178.926.67=6.71\begin{aligned} q &= \frac{4000}{10 \times 15} = 26.67\ \text{kN/m}^2 \\ FS &= \frac{178.9}{26.67} = 6.71 \end{aligned}

Answer: factor of safety ≈6.7\approx 6.7. If the weight of the soil excavated for the basement (γDf≈18×2.5=45\gamma D_f \approx 18 \times 2.5 = 45 kN/m²) is taken into account, it exceeds the applied pressure of 26.7 kN/m², so the mat is over-compensated and the true factor of safety is even higher.

✓ Answer checked by Claude Sonnet 5.5

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