Shallow Foundations - Solution 1 (Part a, b, c, d)

Solution:
a. Ultimate bearing capacity with deep GWL
Tult = cNc + yDfNq + 0.5ByNy
c = 0 (sand)
For phi = 30 deg; Nq = 22.5, Ny = 19.7 ...(Table 36.2 CERM 11th ed.)
Tult = 0 + (120 pcf)(3 feet)(22.5) + 0.5(3 feet)(120 pcf)(19.7)
Tult = 8100 psf + 3546 psf
Tult = 11646 psf or 11.6 ksf

b. Net bearing capacity with deep GWL
Tnet = Tult - yDf = 11646 psf - (120 pcf)(3 feet) = 11286 psf or 11.3 ksf

c. Allowable bearing capacity with deep GWL
Tall = Tnet/(factor of safety) = 11286 psf/3.0 = 3762 psf or 3.8 ksf

d. Net bearing capacity with GWL = 2 feet deep below the ground surface
Here, we use the concept of effective unit weight to the submerged part.
y' = effective unit weight = y - yw where yw = unit weight of water = 62.4 pcf
Therefore, the second part of the equation becomes, y(2 feet)Nq + y'(1 feet)Nq
And the last part of the equation becomes 0.5(B)(y')Ny

Tult = (120 pcf)(2 feet)(22.5) + (120 pcf - 62.4 pcf)(1 feet)(22.5) + (0.5)(3
feet)(120 pcf - 62.4 pcf)(19.7)
Tult = 5400 psf + 1296 psf + 1702.08 psf
Tult = 8398.08 psf or 8.4 ksf
Tnet = Tult - overburden = 8398.08 psf - [(120 pcf)(3 feet) + (62.4 pcf)(1 feet)]
Tnet = 7975.68 psf or 8 ksf

Note how the presence of water has reduced the bearing capacity.

I'll post the solutions to rest of the problem tomorrow. I'll also add another problem on shallow foundation tomorrow with increased complexity, i.e. soil layers, eccentric loads etc. Also, please let me know if you find any errors on my solution, I'll try to rectify it.