Effect of Ground Water Table
Courses > Foundation Analysis and Design > Bearing Capacity of Shallow Foundations > Effect of Ground Water TableIntroduction
If the water table is located at an elevation such that part of the slip mechanism for the footing would be below the water table, higher pore pressures will exist in that zoen, implying lower shear strength and consequently lower bearing capacity. The effect of the water table is best accounted for by suitably choosing the value of the weight for use in the bearing capacity equation as the buoyant unit weight, the wet (or material) unit weight, or some number in between.
Concepts and Formulas
The buoyant unit weight must be used if the groundwater table is at or above the base of the footing (that is z_{w} < D), for in this case the whole of the potential slip mechanism below the footing base is under water. Since the depth of the slip mechanism is of the order of B below the base of the foundation (that is, a total depth of D + B), if the water table is below this depth, the material unit weight must be used. A simple interpolation can be used for water table depths between D and D + B.
definitions:
z_{w} = depth to the water table from ground
B = footing width
D = depth of embedment
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Solved sample problems
Example 1: Determination of equivalent unit weight of soil to calculate soil bearing capacity with the effect of ground water table (English units)
Given:

Moist unit weight of soil above ground water table: 120 lb/ft^{3}.

Moist content = 20%

Friction angle, f = 25 degree

Cohesion of soil above ground water table: 1000 lb/ft^{2}.

Cohesion of soil below ground water table: 500 lb/ft^{2}.

Footing: 8 feet wide square footing, bottom of footing at 2 ft below ground surface.

Location of ground water table: 6 ft below ground water surface.
Requirement: Determine equivalent unit weight of soil to be used for calculating soil bearing capacity.
Solution:
Determine equivalent unit weight:
Dry unit weight of soil, g_{dry} = g_{m} /(1+ w) = 120/(1+0.2) = 100 lb/ft^{3}.
Volume of solid for 1 ft^{3} of soil, V_{s }= g_{dry} / (G_{s}g_{w}) = 100 / (2.65*62.4) = 0.6 ft^{3}.
Volume of void for 1 ft^{3} of soil, V_{v} = 1V_{s}=10.6=0.4 ft^{3}.
Saturate unit weight of soil, g_{sat} = g_{dry} + g_{w} V_{v} = 100+62.4*0.4=125 ft^{3}.
Buoyant unit weight of soil = g_{sat}  g_{w} = 12562.4=62.6 ft^{3}.
z_{w}= 6
D = 2
B = 8
D < z_{w} < D + B
Thus, equivalent unit weight of soil:
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