Shallow Foundations - Eccentricity - Solution 1 (c)

(c) e = M/P = 100 ft-kips/100 kips = 1.0

Therefore, e > B/6  i.e., 3 ft./6 = 0.5
qmin = P/A(1 - 6e/B) = P/A(1 - 6x1/3) < 0
qmax = P/A(1 + 6e/B) = 3P/A = (3 x 100 kips)/(3.0 ft. x 3.0 ft.)
or, qmax = 33.33 kips/ft^2

Ultimate Bearing Capacity, Tult


Here, we use the concept of effective width, Beff


Beff is the effective width where the pressure on the foundation is acting (see figure)


In this case, Beff = B - 2e = 3.0 - 2(1) = 1 ft
Leff = L - 2e = 3 - 2(0) = 3.0 ft. ...because its just one-way eccentricity in this problem

Tult = yDfNq + (0.85)(0.5)(Beff)yNy         ......Square Footing/Sand

Tult = (120 pcf)(3.0 ft.)(22.5) + (0.85)(0.5)(1.0 ft.)(120 pcf)(19.7)


Please note that although the effective footing is now a rectangle with Beff/L = 0.33, I've still not changed the shape factor multiplier. I've not done so because a B/L ratio of 0.33 will yield a higher shape factor multiplier and hence, this is more conservative.


Tult = 9104.7 psf

Let's go a step further and evaluate the factor of safety against bearing capacity failure,
Total ultimate load the foundation can sustain is Qult = Tult x Effective Area = Tult x Beff x B = 27314.1 lbs or 27.3 kips.

The factor of safety against bearing capacity failure is, FS = Qult/P = 27.3 kips/100 kips = 0.273


And this foundation, my friends, is doomed until something is done about it. 


A few things that we can do to improve the factor of safety are discussed below:


1. Increase the width of the foundation: Let's see how this affects the FS - 


Earlier our foundation was 3' x 3' , now make it 5' x 5'
so, B/6 = 5/6 = 0.834' which is still less than eccentricity, e.   


Tult = (120 pcf)(3.0 ft.)(22.5) + (0.85)(0.5)(3.33 ft.)(120 pcf)(19.7)


Tult = 11445.65 psf or 11.4 ksf


Qult = Tult x Effective Area = 11445.65 psf x 5 ft. x 3.33 ft. = 190570 psf or 190.6 ksf


FS = 190.6/100 = 1.9 which is a great improvement from 0.27, however, still less than what I would like to have for shallow foundations.


2. Increase the depth of the foundation to 5 feet (keep the dimensions of the foundation same): Let's see how this affects the FS - 

Tult = (120 pcf)(5.0 ft.)(22.5) + (0.85)(0.5)(1 ft.)(120 pcf)(19.7)

Tult = 14504.7 psf or 14.5 ksf

Qult = Tult x Effective Area = 14504.7 psf x 3 ft. x 1 ft. = 43514.1 psf or 43.5 ksf

FS = 43.5/100 = 0.435 which is an increase from 0.27, however, still too less.


Notice that the factor of safety is more sensitive to the increase in the width as opposed to the depth of the foundation.

3. A combination of the above, i.e. increase the footing width and lower the foundation. I'll leave this for you to try yourself.

I hope the above exercises were helpful in understanding how load eccentricity affects shallow foundations and ways to improve factors of safety.