The figure is missing. What does the answer assume so the stresses can be calculated?

Water table at the ground surface and the given gammas are saturated unit weights

No water table, all soil is dry

Water table at the top of the clay

Water table at the bottom of the clay

Which expression gives sigma0' at mid-height of the clay?

(20 - 9.81)(4) + (18 - 9.81)(2.5) = 61.2 kPa

What is substituted in the settlement equation?

S = [0.22 x 2.5/(1 + 1.30)] log10(81/51), about 48 mm

S = [0.22 x 2.5/1.30] log10(81/51)

S = [0.22 x 2.5/2.30] log10(30/51)

S = [0.22 x 1.25/2.30] log10(81/51)

2078 Chaitra · 3 marks

Final settlement of clay, +30 kN/m²

Short answer

Effective stress at mid-height of the clay layer is (20−9.81)×4 + (18−9.81)×1.25 = 51.0 kPa. Final settlement Sc = (0.22×2.5/2.30) log10(81/51) ≈ 0.048 m, i.e., about 48 mm.

Calculate the final settlement of the clay layer as shown in the figure below due to an increase of pressure of 30 kN/m² at mid height of the layer. [Figure: sand layer 4 m thick, γ = 20 kN/m³, above a clay layer 2.5 m thick, γ = 18 kN/m³, e = 1.30, $C_c$ = 0.22.]

Assumptions: the figure is not available, so the water table is taken at the ground surface and the given $\gamma$ values are saturated unit weights. $\gamma_w = 9.81$ kN/m³. The clay is normally consolidated and settlement is computed at mid-height of the clay (1.25 m below its top).

Ground  ------------------------
  Sand  4 m   gamma = 20 kN/m3
        ------------------------
  Clay  2.5 m gamma = 18, e = 1.30
        Cc = 0.22        <- mid-height

Effective stress at mid-clay

\begin{aligned} \sigma_0' &= (20-9.81)(4) + (18-9.81)(1.25) \\ &= 40.76 + 10.24 = 51.0\ \text{kPa} \end{aligned}

Settlement, with $\Delta\sigma = 30$ kPa:

\begin{aligned} S_c &= \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_0'} \\ &= \frac{0.22 \times 2.5}{2.30}\log_{10}\frac{81.0}{51.0} \\ &= 0.2391 \times 0.2009 = 0.048\ \text{m} \end{aligned}

Answer: final settlement $\approx 48\ \text{mm}$ .

If there were no water table (both layers above it, $\sigma_0' = 80 + 22.5 = 102.5$ kPa), the same formula gives about 26.7 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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