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Effect of water table on soil bearing capacity

When the water table is above the wedge zone, the soil parameters used in the bearing capacity equation should be adjusted.  Bowles proposed an equation to adjust unit weight of soil as follows:

ge=(2H-Dw)(Dw/H2)gm+(g’/H2)(H-Dw)2                              [1.16]

Where

ge = Equivalent unit weight to be used in bearing capacity equation,

H = 0.5Btan(45+f/2), is the depth of influence zone,

Dw= Depth from bottom of footing to ground water table,

gm = Moist unit weight of soil above ground water table,

g’ = Effective unit weight of soil below ground water table.

Conservatively, one may use the effective unit water under ground water table for calculation.  Equation 1.16 can also used to adjust cohesion and friction angle if they are substantially differences.

Example 9: Determine equivalent unit weight of soil to calculate soil bearing capacity with the effect of ground water table

Given:

• Moist unit weight of soil above ground water table: 120 lb/ft3.

• Moist content = 20%

• Friction angle, f = 25 degree

• Cohesion of soil above ground water table: 1000 lb/ft2.

• Cohesion of soil below ground water table: 500 lb/ft2.

• 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.

RequirementDetermine equivalent unit weight of soil to be used for calculating soil bearing capacity.

Solution:

Determine equivalent unit weight:

Dry unit weight of soil, gdry = gm /(1+ w) = 120/(1+0.2) = 100 lb/ft3.

Volume of solid for 1 ft3 of soil, Vs = gdry / (Gsgw) = 100 / (2.65*62.4) = 0.6 ft3.

Volume of void for 1 ft3 of soil, Vv = 1-Vs=1-0.6=0.4 ft3.

Saturate unit weight of soil, gsat = gdry + gw Vv = 100+62.4*0.4=125 ft3.

Effective unit weight of soil = gsat - gw = 125-62.4=62.6 ft3.

Effective depth, H = 0.5B tan(45+f/2) = 0.5*8*tan (45+30/2) = 6.9 ft

Depth of ground water below bottom of footing, Dw= 6-2 = 4 ft

Equivalent unit weight of soil,

ge = (2H-Dw)(Dw/H2)gm+(g’/H2)(H-Dw)2

=(2*6.9-4)(4/6.92)*100+(62.6/6.92)(6.9-4)2

= 93.4 lb/ft3.