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How To Determine The Ultimate Bearing Capacity Of A Strip Footing With A Shallow Foundation Using Terzaghi Capacity Factors

By Edited Nov 13, 2013 0 0

A strip footing can be defined as having a shallow foundation when the height of the soil surface to the bottom of the foundation is equal to as or no more than 4 times the foundation width.  Upon this classification, along with the fact that the footing is continuous such as with a wall footing, the application of Terzaghi’s soil bearing capacity calculation can be used. 


The Terzaghi bearing capacity formulation for ultimate bearing capacity is as follows:


Qu = (C)(Nc) +(qo)(Nq) +(1/2)(γ)(B)(Nγ)


C = soil cohesion

Nc = (Nq - 1)cot(φ) = (Nq - 1)[cos(φ)/sin(φ)]; bearing capacity factor

qo = (γ)(d); d = height from top of soil surface to bottom of foundation

Nq = tan2(45 + φ/2)eπtan(φ)

γ = soil unit weight

B = width of foundation base

Nγ = 2(Nq + 1)tan(φ)


This formula is based on the equilibrium of a general strip footing with different geometric variables and the soil type present.  A soil surcharge of qo is assumed to be acting from the top of the soil surface downward towards the soil level below the footing foundation for simplicity.  The bearing capacity factors of Nc, Nq and Nγ were developed by Prandtl, Reissner, Caquot, Kerisel and Vesic.


Let’s see an example of how to calculate the bearing capacity for a strip footing:


We have a footing with a 5 ft wide foundation and the bottom of this foundation to the top soil surface is 5 ft.  The soil unit weight is 115 lb/ft3, the friction angle is 22° and the cohesion is 150 psf.


We can begin to calculate the soil bearing capacity factors.

Nq = tan2(45 + φ/2)eπtan(φ)

Nq = tan2(45 + 22/2)eπtan(22)

Nq = tan2(56)eπtan(22)

Nq = 7.82


Nc = (Nq - 1)[cos(φ)/sin(φ)]

Nc = (7.82 - 1)[cos(22)/sin(22)]

Nc = 16.88


Nγ = 2(Nq + 1)tan(φ)

Nγ = 2(7.82 + 1)tan(22)

Nγ = 7.13


Now the final bearing capacity calculation can be performed with the following variables:


C = 150 psf

γ = 115 lb/ft3

qo = (γ)(d) = (115 lb/ft3)(5 ft) = 575 lb/ft2

B = 5 ft


Qu = (C)(Nc) +(qo)(Nq) +(1/2)(γ)(B)(Nγ)

Qu = (150 psf)(16.88) +(575 lb/ft2)( 7.82) +(1/2)( 115 lb/ft3)( 5 ft)( 7.13)

Qu = 2,532 psf + 4,496.5 psf + 2,049.9 psf

Qu = 9,078.4 psf


Using this formulation is a handy and useful method for determining ultimate soil bearing capacity.  It can be applied to different soil types as it is based on the cohesion and friction angle of the soil.



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




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