Why must an equation be solved for Df?

Both Nc and net pressure depend on Df

2074 Chaitra · 6 marks

Raft 14 m × 21 m, depth of foundation

Short answer

Using Skempton: FS = cu Nc/(q−γ Df) with cu=7 kPa, q=135 kPa, γ=19 kN/m³. Nc=5.68(1+0.0143Df). Solving 7Nc = 3(135−19Df) yields required depth Df=6.34 m.

A building is to be supported on a reinforced concrete raft covering an area of 14 m × 21 m. The subsoil is clay with an unconfined compressive strength of $14\ \text{kN/m}^2$ . The pressure on the soil due to weight of the building and loads it will carry will be $135\ \text{kN/m}^2$ , at the base of the raft. If the unit weight of excavated soil is $19\ \text{kN/m}^3$ , at what depth should the bottom of the raft be placed to provide a factor of safety of 3? Use Skempton's bearing capacity formula.

Given: raft $B = 14$ m, $L = 21$ m, $q_u$ (unconfined) $= 14$ kN/m² so $c_u = 7$ kN/m², gross pressure at base $q = 135$ kN/m², $\gamma = 19$ kN/m³, FS = 3.

Skempton: $FS = \dfrac{c_uN_c}{q - \gamma D_f}$ (net ultimate capacity over net pressure). For $D_f/B \le 2.5$ :

N_c = \left(0.84 + 0.16\times\frac{14}{21}\right)\times 6\left(1 + 0.2\frac{D_f}{14}\right) = 5.68\,(1 + 0.01429D_f)

\begin{aligned} 7 \times 5.68(1 + 0.01429D_f) &= 3\,(135 - 19D_f) \\ 39.76 + 0.568D_f &= 405 - 57D_f \\ D_f &= \frac{365.24}{57.568} = 6.34\ \text{m} \end{aligned}

Check: $D_f/B = 0.45 < 2.5$ ; $N_c = 6.19$ ; $q_{nu} = 43.4$ kN/m²; $q_{net} = 135 - 19(6.34) = 14.5$ kN/m²; $FS = 3.0$ .

Answer: the base of the raft should be placed at a depth of about 6.3 m.

✓ Answer checked by Claude Sonnet 5.5

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