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## Monday, May 28, 2018

Soil investigation is one of the most important activities carried out before the commencement of any construction project. In the soil test report, the geotechnical engineer is expected to state the strength of the soil at different layers, and ultimately recommend a suitable foundation. One of the parameters used in describing the strength of a soil formation for purposes of foundation design is the soil bearing capacity, which is based on the shear strength of the soil.

In this post, we are going to present an example on how to determine the bearing capacity of a soil using the general bearing capacity equation.

Background
Terzaghi in 1943 extended the plastic failure theory Bof Prandtl to evaluate the bearing capacity for shallow strip footings. After the development of Terzaghi's bearing capacity equation, severtal scholars such as Meyerhof (1951 and 1963), Vesic (1973), Hansen (1970) etc worked on this area and refined the solution to what is known as the general bearing capacity equation. This modification allowed for depth factor, shape factor and inclination factors.

The modified general ultimate bearing capacity equation can be written as;

qu = c'FcsFcdFciNc + qFqsFqdFqiNq + 0.5FÎ³sFÎ³dFÎ³iÎ³BNÎ³

Where;
Fcs, Fqs, FÎ³s are shape factors which account for the shearing resistance developed along the surface in soil above the base of the footing
Fcd, Fqd, FÎ³d are depth factors to determine the bearing capacity of rectangular and circular footings
Fci, Fqi, FÎ³i are inclination factors to determine the bearing capacity of a footing on which the direction of load application is inclined at a certain angle to the vertical

Solved Example
Let us determine the bearing capacity of a simple pad foundation with the following data;

Depth of foundation Df = 0.9m
Width of foundation B = 1.0m
Effective cohesion of soil c' = 12 kN/m2
Angle of internal friction Ï†' = 27°
Unit weight of soil = 18.5 kN/m2
Water table is about 9m below the surface

From table, we can determine the bearing capacity factors;

(culled from Das and Sobhan, 2012)

Angle of internal friction Ï†' = 27°
Nc = 23.94; Nq = 13.20; NÎ³ = 14.47

Fcs = 1 + (B/L)(Nq /Nc) = 1 + (1.0/1.0)(13.2/23.94) = 1.551
Fqs = 1 + (B/L)tanÏ†'  = 1 + (1.0/1.0)tan 27 = 1.509
FÎ³s = 1 + 0.4(B/L)  = 1 + 0.4(1.0/1.0) = 1.4

Fcd = 1 + 0.4(Df/B)  = 1 + 0.4(0.9/1.0) = 1.36
Fqd = 1 + 2tanÏ†'(1 - sin Ï†')2(Df/B)  = 1 + 2tan27(1 - sin 27)2 (0.9/1.0) = 1.273
FÎ³d = 1.0

Since we are assuming vertical loads, take Fci = Fqi = FÎ³i = 1.0

q = (18 kN/m3 × 0.9m) = 16.2 kN/m2

qu = (12 × 23.94 × 1.551 × 1.36 × 1.0) + (16.2 × 1.509 × 1.273 × 1.0 × 13.20) + (0.5 × 1.4 × 1.0 × 1.0 × 1.0 × 14.47) = 1026.883 kN/m2

Using a factor of safety (FOS) of 3.0
qallowable = qu /FOS = 1026.883/3.0 = 342.294 kN/m2

So with this, the bearing capacity of the soil at that layer can be established.