Soil Stress

In general,

q_{s} = \frac{P_{t}}{A_{f}} \pm \frac{M_{t}}{S_{f}}

For Soil Load Combination,

q_{s, \max} = \frac{459.4}{225} + \frac{857.1}{562.5} = 3.57 ksf

q_{s, \min} = \frac{459.4}{225} - \frac{857.1}{562.5} = 0.52 ksf

Since q_s,min is greater than zero, the entire footing is in contact with the soil and no uplift occurs.

For Concrete Load Combination,

P_u = 1.2 ∙ 250 k + 0.5 ∙ 125 k = 362.5 k

M_u = 1200 k-ft

q_{s, \max} = \frac{362.5}{225} + \frac{1, 200}{562.5} = 3.74 ksf

q_{s, \min} = \frac{362.5}{225} - \frac{1, 200}{562.5} = -0.52 ksf

Since q_s,min is less than zero, part of the footing is in uplift and we must use equilibrium rather than flexure theory to determine soil stresses.

Sum moments at right edge of footing to find location of resultant footing reaction,

\frac{P_{u} \cdot b_{f}}{2} - R \cdot r - M_{u} = 0

From vertical equilibrium, R = P_u

$r = \frac{\frac{362.5 (15)}{2} - 1, 200}{362.5} = 4.19 ft$ from right edge of footing

The resulting soil pressure is assumed to vary linearly and take the form of a prism with centroid at r, thus,

$q_{\max} = \frac{2 P_{u}}{3 \cdot r {\cdot b}_{f}} = \frac{2 (362.5)}{3 (4.19) (15)} = 3.85 ksf$ 3.85 ksf (at right footing edge)

Zero stress line is at 3r = 3 · 4.19' = 12.57' from right footing edge