The floor under water is usually pretty light despite having hundreds of pounds of water above it, how is this possible?
In: Physics
Sediment is individual grains that touch each other but also with space between them, which in this case is 100% filled with water. Water pressure when there is no flow (hydrostatic) pushes equally in all directions (is isotropic), meaning the upward stress (force per unit area) from the water is the same as the downward stress at every location. The effective vertical stress that soil is under at the bottom of the lake is equal to the stress caused by the weight of the overburden minus the stress caused by pore pressure. In hydrostatic, drained (no excess pore water pressure) conditions, the pore pressure in any situation is always equal to the depth of the water multiplied by the density of water (1Mg/m³ – and 1Mg = 1000kg) multiplied by acceleration due to gravity (we geotechnical engineers tend to take 10m/s² as the value for this, even though it is not strictly accurate). The overburden pressure in general is calculated by taking the thickness of the overburden, multiplied by the density, multiplied by acceleration due to gravity. In this case the overburden is just water, so the density is 1Mg/m³. Therefore in a 20m deep lake, for example, we have an overburden pressure of 20x1x10=200kPa (kPa = kilopascal, units of stress. 10Pa = 10 Newtons per square meter. 10 Newtons = 1kg x acceleration due to gravity 10m/s²) and a pore pressure of 20x1x10=200kPa. The effective stress on the bed of the lake is 0, so the sediment is not compacted.
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