If the groundwater table at a site rises from 5 meters below the ground surface to the ground surface, how does the effective stress at a point 10 meters deep change?

The effective stress at a point 10 meters deep changes significantly when the groundwater table rises from 5 meters below the ground surface to the ground surface. This change is governed by the relationship between total stress, pore water pressure, and effective stress. We define total stress (σ) as the total pressure exerted by the weight of both soil solids and water above a given point. Pore water pressure (u) is the pressure exerted by the water within the soil's pore spaces. Effective stress (σ') is the stress carried by the soil solid particles, which is responsible for the soil's strength and deformation, and it is calculated as the total stress minus the pore water pressure: σ' = σ - u.

To determine the change, we will consider two scenarios at a depth of 10 meters, assuming a typical dry/moist unit weight of soil (γ_dry) as 18 kN/m³, a typical saturated unit weight of soil (γ_sat) as 20 kN/m³, and the unit weight of water (γ_w) as 9.81 kN/m³.

Scenario 1: Groundwater table at 5 meters below the ground surface.
At a depth of 10 meters, the soil from 0 to 5 meters is above the groundwater table, and the soil from 5 to 10 meters is saturated below the groundwater table.
1. Total Stress (σ) at 10m: The total weight of the soil and water column above the point. This is the sum of the stress from the 5 meters of dry/moist soil and the 5 meters of saturated soil. So, σ = (5 meters γ_dry) + (5 meters γ_sat) = (5 m 18 kN/m³) + (5 m 20 kN/m³) = 90 kPa + 100 kPa = 190 kPa.
2. Pore Water Pressure (u) at 10m: This is the pressure from the column of water from the groundwater table down to the point. The water table is at 5 meters, so the height of the water column above 10 meters is 10 m - 5 m = 5 meters. So, u = 5 meters γ_w = 5 m 9.81 kN/m³ = 49.05 kPa.
3. Effective Stress (σ') at 10m: Using the formula σ' = σ - u, we get σ' = 190 kPa - 49.05 kPa = 140.95 kPa.

Scenario 2: Groundwater table rises to the ground surface (0 meters below the ground surface).
Now, all 10 meters of soil from the ground surface down to the point are saturated.
1. Total Stress (σ) at 10m: Since the entire 10-meter column is saturated, the total stress is calculated using the saturated unit weight for the full depth. So, σ = 10 meters γ_sat = 10 m 20 kN/m³ = 200 kPa.
2. Pore Water Pressure (u) at 10m: With the groundwater table at the surface, the water column above 10 meters is the full 10 meters. So, u = 10 meters γ_w = 10 m 9.81 kN/m³ = 98.1 kPa.
3. Effective Stress (σ') at 10m: Using the formula σ' = σ - u, we get σ' = 200 kPa - 98.1 kPa = 101.9 kPa.

Change in Effective Stress:
The effective stress changes from 140.95 kPa (initial) to 101.9 kPa (final). The change is 101.9 kPa - 140.95 kPa = -39.05 kPa.

Therefore, the effective stress at the point 10 meters deep decreases by 39.05 kPa. This reduction in effective stress occurs because while the total stress increases by 10 kPa (from 190 kPa to 200 kPa) due to the soil above 5 meters becoming saturated, the pore water pressure increases by a much larger amount, 49.05 kPa (from 49.05 kPa to 98.1 kPa). The larger increase in pore water pressure relative to the increase in total stress leads to a net decrease in effective stress. This decrease in effective stress can reduce the shear strength of the soil and impact its stability.