A rectangular or circular shallow footing has an ultimate bearing capacity that can be calculated as a function of the footings physical dimensions and soil properties.  These physical dimensions include the footings width, length or diameter and height from top of soil surface to bottom of foundation.  Also soil properties that affect bearing capacity are the soil cohesion, unit weight and soil friction angle.

This formulation was originally based on work performed by K. Terzaghi for the bearing capacity of strip footings, also with shallow footings.  Further development of this formula was performed by G. Meyerhof for the appropriate use on non-continuous footings.  With the introduction of shape, depth and load inclination factors, this formula was refined for the application of rectangular and circular footings.  Further contributions of these factors where made by E. Debeer, J. Hansen and A. Hanna.

The Meyerhof bearing capacity formulation for rectangular or circular footings is as follows:

Qu = (C)(Nc)(Fcs)(Fcd)(Fci) + (qo)(Nq)(Fqs)(Fqd)(Fqi)  +(1/2)(γ)(B)(Nγ)(Fγs)(Fγd)(Fγi)

C = soil cohesion

qo = (γ)(d); d = bottom of footing surface to top of soil surface

γ = soil unit weight

B = foundation width (rectangular footing) or diameter (circular footing)

Bearing capacity factors:

Nc = (Nq - 1)cot(φ) = (Nq - 1)[cos(φ)/sin(φ)]

Nq = tan2(45 + φ/2)eπtan(φ)

Nγ = 2(Nq + 1)tan(φ)

Shape factors: L = footing length, where L > B

Fcs = 1 + (B/L)(Nq / Nc)

Fqs = 1 + (B/L)tan(φ)

Fγs = 1 – 0.4(B/L)

Depth factors:

Where d/B </= 1

Fcd = 1 + 0.4(d/B)

Fqd = 1 + 2tan(φ)[1 – sin(φ)]2(d/B)

Fγd = 1

Where d/B > 1

Fcd = 1 + 0.4tan-1(d/B)

Fqd = 1 + 2tan(φ)[1 – sin(φ)]2tan-1(d/B)

Fγd = 1

Load Inclination factors: β is the angle of inclination between vertical and the direction of applied load to the footing

Fci = Fqi = (1 – β° /90°)2

Fγi = (1 – β / φ) 2

We’ll now see this formulation used in an example application.

Let’s say we have a circular footing that’s 8 ft in diameter and its depth from the top of soil surface to the base of the foundation is 8 ft.  The soil is a dense sand that has a unit weight of 120 lb/ft3, soil cohesion value of zero and a soil friction angle of 37°.  The load is applied to the footing vertically with no inclination angle.  What is the ultimate soil bearing capacity?

We will begin by calculating each of the different factors.

Bearing capacity factors:

Nq = tan2(45 + φ/2)eπtan(φ)

Nq = tan2(45 + 37°/2)eπtan(37°)

Nq = 42.92

Nc = (Nq - 1)[cos(φ)/sin(φ)]

Nc = (42.92 - 1)[cos(37°)/sin(37°)]

Nc = 55.63

Nγ = 2(Nq + 1)tan(φ)

Nγ = 2(42.92 + 1)tan(37°)

Nγ = 66.19

Shape factors:

Fcs = 1 + (B/L)(Nq / Nc)

Fcs = 1 + (1)( 42.92/ 55.63)

Fcs = 1.772

Fqs = 1 + (B/L)tan(φ)

Fqs = 1 + (1)tan(37°)

Fqs = 1.754

Fγs = 1 – 0.4(B/L)

Fγs = 1 – 0.4(1)

Fγs = 0.6

Depth factors: d = 8 ft = B, so d/B >/= 1

Fcd = 1 + 0.4(d/B)

Fcd = 1 + 0.4(1)

Fcd = 1.4

Fqd = 1 + 2tan(φ)[1 – sin(φ)]2(d/B)

Fqd = 1 + 2tan(37°)[1 – sin(37°)]2(1)

Fqd = 1.239

Fγd = 1

Fci = Fqi = Fγi = 1, since β = 0°

So the ultimate bearing capacity can be calculated as follows:

C = 0

γ = 120 lb/ft3

qo = (γ)(d) = (120 lb/ft3)(8 ft) = 960 lb/ft2

B = 8 ft

Qu = (C)(Nc)(Fcs)(Fcd)(Fci) + (qo)(Nq)(Fqs)(Fqd)(Fqi)  +(1/2)(γ)(B)(Nγ)(Fγs)(Fγd)(Fγi)

Qu = (960 lb/ft2)( 42.92)(1.754)(1.239)(1) +(1/2)( 120 lb/ft3)( 8 ft)(66.19)(0.6)(1)(1)

Qu = 89,543 psf + 19,062.7 psf

Qu = 108,605.7 psf

While somewhat lengthy, this calculation is very straight forward and an easy way to determine the ultimate bearing capacity of a rectangular or circular shallow foundation.

REFERENCE:

Das, Braja M. (2000).  Fundamentals of Geotechnical Engineering.  Pacific Grove, CA: Brooks/Cole – Thomson Learning.