Shallow Foundations - Eccentricity - Solution 1 (b)

(b) e = M/P = 20 kips-ft/100 kips = 0.2 ft.
Therefore, e < B/6  i.e., 3 ft./6 = 0.5
qmin = P/A(1 - 6e/B) = P/A(1 - 6x0.2/3) = 6.67 kips/ft^2
qmax = P/A(1 + 6e/B) =  (100 kips)(1.4)/(3.0 ft. x 3.0 ft.) or, qmax = 15.55 kips/ft^2

Ultimate Bearing Capacity, Tult
Tult = yDfNq + (0.85)(0.5)(B)yNy         ......Square Footing/Sand

Tult = (120 pcf)(3.0 ft.)(22.5) + (0.85)(0.5)(3.0 ft.)(120 pcf)(19.7)
Tult = 11114.1 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 Area = Tult x B x B = 100026.9 lbs or 100 kips.

The factor of safety against bearing capacity failure is, FS = Qult/P = 100 kips/100 kips = 1.0


Thus, we can see that the factor of safety is same as that of scenario a earlier. However, since we have a different pressure distribution between a and b, we know where we would want more reinforcement.