Calculate the effective stress at Point X in the soil profile shown below when (g = 10 m/s2):

Thickness of the Sand Layer = 2 m (i.e. A = 2 m),
Thickness of the Gravel Layer = 7 m (i.e. B = 7 m),
Thickness of the Silt Layer to Point X = 3 m (i.e. C =3 m), and
Depth of Water Table from the Surface = 6 m (i.e. D = 6 m)

Question

Calculate the effective stress at Point X in the soil profile shown below when (g = 10 m/s2):

Thickness of the Sand Layer = 2 m (i.e. A = 2 m),
Thickness of the Gravel Layer = 7 m (i.e. B = 7 m),
Thickness of the Silt Layer to Point X = 3 m (i.e. C =3 m), and
Depth of Water Table from the Surface = 6 m (i.e. D = 6 m)

BlackTom AI · Accepted Answer

We need to compute the effective vertical stress at Point X in a layered soil profile with a water table at 6 m, using given unit weights for dry and saturated states and g = 10 m/s^2.

First, restate the data: Sand layer thickness A = 2 m with γ_dry = 14.6 kN/m^3 and γ_sat = 17.8 kN/m^3; Gravel layer thickness B = 7 m with γ_dry = 16.9 kN/m^3 and γ_sat = 19.8 kN/m^3; Silt layer thickness C = 3 m with γ_dry = 15.9 kN/m^3 and γ_sat = 18.8 kN/m^3. Water table depth D = 6 m. Point X is at the bottom of the Silt layer, i.e., at z = A + B + C = 12 m.

Next, determine how much of the depth is above and below the water table to use the appropriate unit weights:
- From 0 to 2 m (Sand, above water table): use γ_dry of Sand = 14.6 kN/m^3.
- From 2 to 6 m (Gravel, still above water table): use γ_dry of Gravel = 16.9 kN/m^3.
- From 6 to 9 m (Gravel, below water table): use γ_sat of Gravel = 19.8 kN/m^3.
- From 9 to 12 m (Silt, below water table): use γ_sat of Silt = 18.8 kN/m^3.

Now compute the total vertical stress σ at 12 m by summing the contributions of each segment with the appropriate unit weight:
- Sand: 2 m × 14.6 kN/m^3 = 29.2 kN/m^2
- Gravel above WT: 4 m × 16.9 kN/m^3 = 67.6 kN/m^2
- Gravel saturated: 3 m × 19.8 kN/m^3 = 59.4 kN/m^2
- Silt saturated: 3 m × 18.8 kN/m^3 = 56.4 kN/m^2
Total σ = 29.2 + 67.6 + 59.4 + 56.4 = 212.6 kN/m^2 (kPa).

Compute the pore water pressure u at Point X. Since the water table is at 6 m and Point X is at 12 m, the depth of water above the point is z − D = 12 − 6 = 6 m. Using g = 10 m/s^2 gives γ_w ≈ 10 kN/m^3. Therefore u = γ_w × (z − D) = 10 × 6 = 60 kPa.

Finally, obtain the effective stress σ′ by subtracting pore pressure from total vertical stress: σ′ = σ − u = 212.6 − 60 = 152.6 kPa.

Thus, the calculated effective stress at Point X is 152.6 kPa, matching the provided answer when γ_w is taken as 10 kN/m^3 (consistent with g = 10 m/s^2).

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