Exercise 1 Solution

Let’s denote with γ_w = 1000 kg/m³ the water specific weight and with γ_grain = 2700 kg/m³ the specific weight of the solid. The specific weight of the saturated clay γ_s can be computed as follows:

γ_s = ϕ γ_w + (1 − ϕ)γ_grain

γ_s = 0.4 (1000) + (1 − 0.4)2700 = 2020 kg/m³

With a 6 m deep lake, pressure p at 15 m below the lake bottom amounts to

p = (6 + 15) m 1000 kg/m³ = 21,000 kg/m²

Pressure is often reported in bars. There are 9.8067 × 10^–5 bars per 1 kg/m², so

p = 21,000 kg/m² = 2.06 bars.

The geostatic stress is:

σ_c = 6 m 1000 kg/m³ + 15 m 2020 kg/m³ = 6000 kg/m² + 30,300 kg/m² = 36,300 kg/m²

σ_c = 3.56 bar

Rearranging Terzaghi’s principle (Equation 3), the effective vertical stress σ_z is equal to:

σ_z = σ_c − p = 3.56 bar − 2.06 bar = 1.50 bar

If the water level drops to 4 m, both the pressure and the total vertical stress reduce the same amount ( = 0.2 bar). Hence, σ_z does not vary.

Return to Exercise 1