Foundation, Concrete and Earthquake Engineering

### Land Subsidence Due to Groundwater Extraction

Pumping large volumes of water from a confined aquifer at rates substantially larger than the natural recharge causes a contraction of the aquifer which can result in a downward movement of the land surface. In unconfined aquifers, groundwater pumping causes a downward movement of the water table which likewise can lead to a downward movement of the land surface. This downward movement is called subsidence or consolidation. This movement can be a few centimeters to several meters. If the subsidence is not uniform, the differential settlement can produce severe damage to structures. Important subsidence has occurred in the San Joaquin Valley in California, in Mexico City, in Venice, around Shanghai, and in southern Taiwan. These large subsidences tend to occur in thick deposits containing fine sands, silt, and clays and often are the result of excessive pumping. Decrease of the groundwater pumpage can reduce and sometimes reverse the subsidence as is currently the case in the Harris-Galveston area in south-central Texas (Hibbs, 1997). UNESCO (United Nations Educational, Scientific, Cultural Organization) has published a guidebook to studies of land subsidence due to groundwater withdrawals (Poland, 1984).
Calculation of Subsidence
Consider a unit area of a horizontal plane at a depth Z below the ground surface. The total downward pressure Pt due to the weight of the overburden on the plane is resisted partly by the upward hydrostatic pressure Ph and partly by the intergranular pressure Pi exerted between the grains of the material:
Pt = Ph + Pi or Pi = Pt – Ph.
A lowering of the water table results in a decrease of the hydrostatic pressure and a corresponding increase of the intergranular pressure. If Pi1 and Pi2 denote the intergranular pressures before and after a drop in the water table or piezometric surface, the vertical subsidence can be calculated as
where Z is the thickness of the soil layer and E is the modulus of elasticity of the soil. Typical ranges of values of E are given in Table 1. In general, the modulus of elasticity increases nonlinearly with the intergranular pressure. If there are layers of different soil types, the subsidences are calculated separately for each layer and added to obtain the total subsidence. As the modulus of elasticity of clayey materials is much less than that of sand or gravel, most of the settlement occurs in the clayey layers.
TABLE 1. Modulus of Elasticity of Soils and Rocks
The previous equation can also be used to calculate the rebound when the intergranular pressure decreases. Caution must be exercised because the modulus of elasticity is not the same for decompression as for compression. This is particularly the case for clays. For Boston blue clay the rebound modulus of elasticity is only about 50% of that for compression. If subsidence has occurred for a long time, complete rebound is unlikely

Example 1: Consider a 50-m-thick sand layer. The water table is located at a depth of 11 m below the ground surface. Calculate the total and the intergranular pressures at 11 m depth and at the bottom of the sand layer, given that the porosity of the sand is n = 0.35, its volumetric water content above the water table is θ = 0.08, the specific weight of the solids is γs = 25.5 kN/m3, and the specific weight of the water is γ = 9.81 kN/m3.

Solution: At the water table the intergranular pressure, which is also the total pressure, is Pt = 11[(1 – 0.35)25.5 + 0.08*9.81] = 190.96 kPa. The total pressure at the bottom of the sand layer is Pt = 173.6 + 39[(1 – 0.35)25.5 + 0.35*9.81] = 953.912 kPa. The hydrostatic pressure at the bottom of the sand layer is Ph = 9.81*39 =382.59 kPa. The intergranular pressure is thus 953.912 – 382.59 = 571.32 kPa.

Example 2: If in the previous problem the water table drops 30 m, what is the change in intergranular pressure at the bottom of the sand layer? See Figure 1.

Solution: The depth to the water table would then be 41 m. The total pressure at the bottom of the sand layer is Pt = 41[(1 – 0.35)25.5 + 0.08*9.81] + 9[(1 – 0.35)25.5 + 0.35*9.81] = 891.82 kPa. The hydrostatic pressure is Ph = 9.81*9 = 88.29 kPa. The intergranular pressure is 891.82 – 88.29 = 803.53 kPa. The increase in intergranular pressure due to the 40 m drop in the water table is 803.53 – 571.32 = 232.21 kPa.

Example 3: Calculate the subsidence for the situation depicted in the previous problem if the modulus of elasticity of the sand is 10,000 N/cm2 or 105 kN/m2.

Solution: The drop in the water table produces a linear increase in the intergranular pressure varying from 0.0 at the 11-m depth to 232.21 kPa at the 41-m depth. The average increment in intergranular pressure is (0 + 232.21)/2 = 116.17 kPa, and the settlement in the layer from 11 to 41 m is Su1 = 30*116.17/105 = 0.0348 m. The subsidence in the layer from 41 to 50 m is Su2 = 9*232.21/105 = 0.0208 m. The total subsidence is thus 0.0348 + 0.0208 = 0.0556 m.