Effective Stress in Soils

In geotechnical engineering, soil is treated as a multiphase material consisting of solid particles and void spaces filled with water and/or air. Because the soil skeleton and the water respond differently to applied loads, we must separate the stress carried by the soil particles from the stress carried by the water.

Total Stress and Pore Water Pressure

Consider a horizontal plane at some depth beneath the ground surface. The Total Stress ($\sigma$) is the total pressure acting on this plane from the weight of everything above it (soil solids, water, and structural loads). The Pore Water Pressure ($u$) is the pressure of the water filling the void spaces between the soil particles.

Terzaghi's Effective Stress Principle

In 1925, Karl Terzaghi formulated the principle of Effective Stress ($\sigma'$) for fully saturated soils. Effective stress is the portion of the total stress that is transmitted through the physical contact points between the soil particles.

The principle is defined by the fundamental equation:

$\sigma' = \sigma - u$

where:

• $\sigma'$ = effective stress [kPa or kN/m²]
• $\sigma$ = total stress [kPa]
• $u$ = pore water pressure [kPa]

The effective stress is not a physical force you can measure directly; it is a calculated quantity. However, it is the most important concept in soil mechanics because all measurable effects of a change in stress, such as compression (settlement) and changes in shear strength, are exclusively due to changes in effective stress.

Worked Example

Problem: A soil profile consists of a 4-meter thick layer of fully saturated clay beneath the ground surface. The water table is exactly at the ground surface. The saturated unit weight of the clay ($\gamma_{sat}$) is $19.0 \text{ kN/m}^3$. Calculate the total stress, pore water pressure, and effective stress at a depth of 4 meters. (Assume the unit weight of water $\gamma_w = 9.81 \text{ kN/m}^3$).

Solution:

1. Calculate the total stress ($\sigma$) at depth $z = 4 \text{ m}$: $\sigma = \gamma_{sat} \times z = 19.0 \times 4 = 76.0 \text{ kPa}$
2. Calculate the pore water pressure ($u$) at depth $z = 4 \text{ m}$: $u = \gamma_w \times z = 9.81 \times 4 = 39.24 \text{ kPa}$
3. Calculate the effective stress ($\sigma'$): $\sigma' = \sigma - u = 76.0 - 39.24 = 36.76 \text{ kPa}$

Alternatively, effective stress can be calculated directly using the buoyant (submerged) unit weight $\gamma' = \gamma_{sat} - \gamma_w$: $\gamma' = 19.0 - 9.81 = 9.19 \text{ kN/m}^3$ $\sigma' = \gamma' \times z = 9.19 \times 4 = 36.76 \text{ kPa}$

Engineering Check

If the water table in the above example were to permanently drop by 2 meters, the pore water pressure would decrease, causing the effective stress to increase. This increase in effective stress is exactly what causes ground subsidence when groundwater is over-pumped from aquifers.\n