Shallow Foundations - Eccentricity - Solution 1 (a)

a. eccentricity, e = M/P = 50 kips-ft/100 kips = 0.5 ft.
    Therefore, e = B/6  i.e., 3 ft./6 = 0.5
     qmin = P/A(1 - 6e/B) = P/A(1 - 6x1/6) = 0
     qmax = P/A(1 + 6e/B) = 2P/A = (2 x 100 kips)/(3.0 ft. x 3.0 ft.)
     or, qmax = 22.2 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

Do you think this Factor of Safety is good enough in such a case?

It is my opinion that this factor of safety is very inadequate for such a foundation.

Your comments and insights are welcome.
I'll post solution for rest of the problem tomorrow.