EG&G (AMEC2121) 1 PROF. MARTIN FAHEY
1. Stresses in the Ground: the Principle of Effective Stress
Saturated soil is a two-phase material – a network of soil particles in contact, with water-filled voids in between the particles. At any depth in the ground, there will be a total vertical stress, resulting from the total weight of material (soil + water) above that point.
3Assume we have a sand with a void ratio e = 0.6, and ！ = 26 kN/m. This gives a dry unit weight of ！ = sd33316.25 kN/m, and a saturated unit weight ！ = 20 kN/m (if ！ = 10 kN/m). (Refer to notes on Phase satw
Suppose we have a layer of this sand, 10 m thick, placed on a smooth level concrete floor. If the sand is dry, the vertical stress at a depth of z = 10 m would be:
2( = ！.z = 16.26 × 10 = 162.5 kPa (i.e. 162.5 kN on each 1 m of floor area) vd
This is somewhat artificial, because it assumes that the total weight of a 10 m high column of sand is 2distributed equally over an area of 1 m. However, if we looked in detail at the plane where the sand 2touches the floor, we would see that we have actual contact only on a very small part of each 1 m of area,
so that the actual contact stress between individual sand grains and the floor would be much higher than 162.5 kPa.
Column of sand, 1 m x Column of sand, 1 m x
1 m in plan, 10 m high1 m in plan, 10 m highActual contact stress on a plane in the soil is ? Actual contact stress on a plane in the soil is ? 33！= 16.25 kN/m= 16.25 kN/m. . ！ddTotal weight = 162.5 Total weight = 162.5 22kN, resting on 1 mkN, resting on 1 m
Vertical stress = Vertical stress = 2222..Actual contact stress on floor is >>> 162.5 kN/mActual contact stress on floor is >>> 162.5 kN/m162.5 kN/m162.5 kN/m(kPa).(kPa).
When we talk about stress in soil, we are referring to this somewhat artificial notion of the force being averaged over the whole area, and it does not refer to the actual contact stresses or forces between soil 1particles.
For the same layer of soil, but this time fully saturated (the water table is at the ground surface), the total
vertical stress at a depth of 10 m would be
2( = ！.z = 20 × 10 = 200 kPa (i.e. 200 kN on each 1 m of floor area) vsat
If we installed a standpipe (an open-ended tube) into the soil to a depth of 10 m, the water level in the standpipe would be at the ground surface (corresponding to the position of the water table). At the bottom of the tube, the water pressure (u) would then be:
u = ！.z = 10 × 10 = 100 kPa w
12 For stress in a steel bar, the same applies. If the bar is 100 mm in cross-sectional area, with an axial force of 1000 N, we 2imagine the axial stress to uniform across the whole cross section = 10 N/mm. However, if we looked down at the atomic
level, we would see a much more complicated interaction between atoms. At this scale, the concept of stress is meaningless.
School of Civil & Resource Engineering, The University of Western Australia
EG&G (AMEC2121) 2 PROF. MARTIN FAHEY
So, in this case, the water is carrying 100 kPa of the total of 200 kPa, so we can imagine that the soil is
carrying the remaining 100 kPa. Total stress Total stress ((
The principle of effective stress
states that at any stage, the effective