The importance of the forces transmitted through the soil skeleton from particle to particle was recognised by Terzaghi (1943), who presented his Principle of Effective Stress, an intuitive rela- tionship based on experimental data. The principle applies only to fully saturated soils, and relates the following three stresses:
1 the total normal stress (σ) on a plane within the soil mass, being the force per unit area transmitted in a normal direction across the plane, imagining the soil to be a solid (single- phase) material;
2 the pore water pressure (u), being the pressure of the water filling the void space between the solid particles;
3 the effective normal stress (σ′) on the plane, representing the stress transmitted through the soil skeleton only (i.e. due to interparticle forces).
The relationship is:
s s= ¢ +u (3.1)
The principle can be represented by the following physical model. Consider a ‘plane’ XX in a fully saturated soil, passing through points of interparticle contact only, as shown in Figure 3.1. The
Figure 3.1 Interpretation of effective stress.
Effective stress wavy plane XX is really indistinguishable from a true plane on the mass scale due to the relatively small size of individual soil particles. A normal force P applied over an area A may be resisted partly by interparticle forces and partly by the pressure in the pore water. The interparticle forces are very random in both magnitude and direction throughout the soil mass, but at every point of contact on the wavy plane may be split into components normal and tangential to the direction of the true plane to which XX approximates; the normal and tangential components are N′ and T, respectively. Then, the effective normal stress is approximated as the sum of all the components N′ within the area A, divided by the area A, i.e.
¢ = ¢ s SN
A (3.2)
The total normal stress is given by s= P
A (3.3)
If point contact is assumed between the particles, the pore water pressure will act on the plane over the entire area A. Then, for equilibrium in the direction normal to XX
P=SN¢ +uA or
P A
N
A u
= S ¢+ i.e.
s s= ¢ +u
The pore water pressure which acts equally in every direction will act on the entire surface of any particle, but is assumed not to change the volume of the particle (i.e. the soil particles themselves are incompressible); also, the pore water pressure does not cause particles to be pressed together.
The error involved in assuming point contact between particles is negligible in soils, the actual contact area a normally being between 1% and 3% of the cross-sectional area A. It should be understood that σ′ does not represent the true contact stress between two particles, which would be the random but very much higher stress ΣN′/a.
Effective vertical stress due to self-weight of soil
Consider a soil mass having a horizontal surface and with the water table at surface level. The total vertical stress (i.e. the total normal stress on a horizontal plane) σv at depth z is equal to the weight of all material (solids + water) per unit area above that depth, i.e.
sv =gsatz
The pore water pressure at any depth will be hydrostatic since the void space between the solid particles is continuous, so at depth z
u=gwz
Hence, from Equation 3.1, the effective vertical stress at depth z in this case will be
¢ = -
=
(
-)
s s
g g
v v
sat w
u z
The parameter in brackets is often referred to as the buoyant unit weight, γ′ (γ′ = γsat - γw). A common critical design condition for geotechnical systems is when the soil is fully saturated and under positive pore water pressure everywhere (i.e. after heavy rain) as this will represent the condition of minimum effective stress (and hence minimum strength). Under such condi- tions, the buoyant unit weight may be used to directly calculate effective stress. For other groundwater conditions, however (e.g. when steady-state seepage is occurring – Chapter 2), the pore water pressures and total stresses should be calculated independently and used in Equation 3.1.
Example 3.1
A layer of saturated clay 4 m thick is overlain by sand 5 m deep, the water table being 3 m below the surface, as shown in Figure 3.2. The saturated unit weights of the clay and sand are 19 and 20 kN/m3, respectively; above the water table the (dry) unit weight of the sand is 17 kN/m3. Assuming initially that all of sand above the water table is dry, plot the values of total vertical stress and effective vertical stress against depth. If sand to a height of 1 m above the water table is saturated with capillary water, how are the above stresses affected?
Figure 3.2 Example 3.1.
Effective stress
Solution
The total vertical stress is the weight of all material (solids+water) per unit area above the depth in question. Pore water pressure is the hydrostatic pressure corresponding to the depth below the water table. The effective vertical stress is the difference between the total vertical stress and the pore water pressure at the same depth. The stresses need only be calculated at depths where there is a change in unit weight (Table 3.1).
In all cases the stresses would normally be rounded off to the nearest whole number. The stresses are plotted against depth in Figure 3.2.
From Section 2.1, the water table is the level at which pore water pressure is atmospheric (i.e. u = 0). Above the water table, water is held under negative pressure and, even if the soil is saturated above the water table, does not contribute to hydrostatic pressure below the water table. The only effect of the 1-m capillary rise, therefore, is to increase the total unit weight of the sand between 2 and 3 m depth from 17 to 20 kN/m3, an increase of 3 kN/m3. Both total and effective vertical stresses below 3 m depth are therefore increased by the constant amount 3 × 1 = 3.0 kPa, with pore water pressures below this depth being unchanged.