Soil pressure under rigid mat with column loads

The spacings and the points $A, B, C, D$ are not legible, so these assumptions are made: the mat is $13\ \text{m}$ (x) $\times$ $15\ \text{m}$ (y); columns are at $x = -5, 0, +5$ m and $y = +6, 0, -6$ m (top row at $y = +6$); and $A, B, C, D$ are the corners $A(-6.5, +7.5)$, $B(+6.5, +7.5)$, $C(+6.5, -7.5)$, $D(-6.5, -7.5)$. The column loads include the mat weight.

Total load and centroid

$$\begin{aligned} Q &= 600+700+600+1100+1300+1400+700+800+600 = 7800\ \text{kN} \\ \sum Qx &= (-5)(600+1100+700) + (5)(600+1400+600) = +1000\ \text{kN·m} \\ \sum Qy &= 6(1900) - 6(2100) = -1200\ \text{kN·m} \\ e_x &= \frac{1000}{7800} = 0.128\ \text{m},\quad e_y = \frac{-1200}{7800} = -0.154\ \text{m} \end{aligned}$$

So $M_y = Qe_x = 1000$ kN·m and $M_x = Qe_y = -1200$ kN·m.

Section properties

$$\begin{aligned} A &= 13 \times 15 = 195\ \text{m}^2 \\ I_x &= \frac{13 \times 15^3}{12} = 3656.25\ \text{m}^4,\quad I_y = \frac{15 \times 13^3}{12} = 2746.25\ \text{m}^4 \\ \frac{Q}{A} &= \frac{7800}{195} = 40.0\ \text{kN/m}^2 \end{aligned}$$

Soil pressure

$$q = \frac{Q}{A} + \frac{M_x\,y}{I_x} + \frac{M_y\,x}{I_y} = 40 - 0.3282\,y + 0.3641\,x$$

Answer: $q_A = 35.2$, $q_B = 39.9$, $q_C = 44.8$, $q_D = 40.1$ kN/m² (all compressive, so no tension under the mat; maximum at C).