Effective vertical stress beneath a river — sand bed saturated unit weight given A river has 5 m of water overlying a sand bed. The sand has saturated unit weight γ_sat = 20 kN/m^3 and the unit weight of water γ_w = 9.81 kN/m^3. Determine the effective vertical stress at a point located 5 m below the top surface of the sand bed (i.e., 5 m into the sand, beneath 5 m of water).

Introduction / Context:
In geotechnical engineering, effective stress governs soil strength and compressibility. Beneath water bodies, a soil element is subjected to total overburden stress from both the overlying water and the soil above, while the pore-water pressure reduces the stress carried by the soil skeleton. Correct separation of total stress and pore pressure is essential to compute the effective stress σ′ = σ − u.

Given Data / Assumptions:

• Water depth above the sand bed: 5 m.
• Point of interest: 5 m below the top of the sand bed.
• Sand saturated unit weight γ_sat = 20 kN/m^3.
• Water unit weight γ_w = 9.81 kN/m^3.
• Hydrostatic conditions; same elevation datum for total stress and pore pressure; no seepage.

Concept / Approach:
Total vertical stress at the point equals weight of the overlying water column plus weight of the overlying saturated sand column. Pore-water pressure equals γ_w multiplied by the vertical distance from the free water surface to the point. Effective stress is the difference: σ′ = σ − u.

Step-by-Step Solution:
Overburden due to water: σ_w = γ_w * 5 = 9.81 * 5 = 49.05 kN/m^2.Overburden due to sand above the point: σ_s = γ_sat * 5 = 20 * 5 = 100 kN/m^2.Total stress at the point: σ = σ_w + σ_s = 49.05 + 100 = 149.05 kN/m^2.Pore pressure at the point: u = γ_w * (water depth + sand depth to point) = 9.81 * (5 + 5) = 98.1 kN/m^2.Effective stress: σ′ = 149.05 − 98.1 ≈ 50.95 kN/m^2 ≈ 51 kN/m^2.

Verification / Alternative check:
Using submerged unit weight concept for the sand portion gives the same result: γ′ = γ_sat − γ_w = 20 − 9.81 = 10.19 kN/m^3. Then σ′ = γ′ * 5 = 10.19 * 5 ≈ 50.95 kN/m^2, confirming the computation.

Why Other Options Are Wrong:
41, 49, 53, 55 kN/m^2 do not match σ′ computed from consistent σ and u; they reflect arithmetic errors or misuse of γ_sat and γ_w.

Common Pitfalls:

• Subtracting γ_w5 instead of γ_w10 for pore pressure (forgetting the water head plus saturated sand head above the point).
• Using dry unit weight instead of saturated unit weight below water table.

Final Answer:
51 kN/m^2