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Raft 20 m × 30 m, depth of foundation

Short answer

Fully compensated depth Df = total load/(γA) = 100000/(600×19) = 8.77 m. For FS=3, solve Skempton's equation: 10×Nc = 3(166.67−19Df) with Nc=5.68(1+0.01Df) gives Df=7.70 m.

A raft foundation is 20 m × 30 m. The raft is constructed over a soft clay stratum having Cu=10 kN/m2C_u = 10\ \text{kN/m}^2 and γ=19 kN/m3\gamma = 19\ \text{kN/m}^3. If the live load and dead load on the raft are 100 MN, find the depth of foundation if (a) the foundation is fully compensated, (b) determine the depth of foundation for a factor of safety of 3.

Given: raft B=20B = 20 m, L=30L = 30 m, A=600A = 600 m², cu=10c_u = 10 kN/m², γ=19\gamma = 19 kN/m³, total load Q=100Q = 100 MN =100 000= 100\,000 kN.

Gross pressure: q=100 000600=166.67q = \dfrac{100\,000}{600} = 166.67 kN/m².

(a) Fully compensated

Net pressure is zero: q=γDfq = \gamma D_f

Df=166.6719=8.77 mD_f = \frac{166.67}{19} = 8.77\ \text{m}

(b) Depth for FS = 3

Use Skempton's equation: qnu=cuNcq_{nu} = c_uN_c and FS=qnuq−γDfFS = \dfrac{q_{nu}}{q - \gamma D_f}. For Df/B≤2.5D_f/B \le 2.5:

Nc=(0.84+0.162030)×6(1+0.2Df20)=5.68 (1+0.01Df)N_c = \left(0.84 + 0.16\frac{20}{30}\right)\times 6\left(1 + 0.2\frac{D_f}{20}\right) = 5.68\,(1 + 0.01D_f) 10×5.68(1+0.01Df)=3 (166.67−19Df)56.8+0.568Df=500−57DfDf=443.257.568=7.70 m\begin{aligned} 10 \times 5.68(1 + 0.01D_f) &= 3\,(166.67 - 19D_f) \\ 56.8 + 0.568D_f &= 500 - 57D_f \\ D_f &= \frac{443.2}{57.568} = 7.70\ \text{m} \end{aligned}

Check: Nc=6.12N_c = 6.12, qnu=61.2q_{nu} = 61.2 kN/m², qnet=166.67−19(7.70)=20.4q_{net} = 166.67 - 19(7.70) = 20.4 kN/m², FS=3.0FS = 3.0 (O.K.; Df/B=0.385D_f/B = 0.385).

Answer: (a) Df=8.77D_f = 8.77 m for a fully compensated raft; (b) Df=7.70D_f = 7.70 m for FS = 3.

✓ Answer checked by Claude Sonnet 5.5

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