Consider 1-m length perpendicular to the drawing

$W_1 = 2400 \times \frac{1}{2}(6)(25)(1)$
$W_1 = 180,000 ~ \text{kg}$

$W_2 = 2400 \times 3(25)(1)$

$W_2 = 180,000 ~ \text{kg}$

$F_w = 1000(7.5) \times 15(1)$

$F_w = 112,500 ~ \text{kg}$

Moment About *m*

Righting Moment, *RM*

$RM = 4W_1 + 7.5W_2 = 4(180,000) + 7.5(180,000)$
$RM = 2,070,000 ~ \text{kg}\cdot\text{m}$

Overturning Moment, *OM*

$OM = 5F_w = 5(112,500)$

$OM = 562,500 ~ \text{kg}\cdot\text{m}$

Reactions at the Base

$R_y = W_1 + W_2 = 360,000 ~ \text{kg}$
$R_x = F_w = 112,500 ~ \text{kg}$

Location of *R*_{y}

$\bar{x} R_y = RM - OM$
$\bar{x}(360,000) = 2,070,000 - 562,500$

$\bar{x} = 4.1875 ~ \text{m}$

Eccentricity

$e = 4.5 - \bar{x} = 4.5 - 4.1875$
$e = 0.3125 ~ \text{m}$

$M = R_y e = 360,000(0.3125)$

$M = 112,500 ~ \text{kg}\cdot\text{m}$

$\sigma_a = \dfrac{P}{A} = \dfrac{360,000}{1(9)}$

$\sigma_a = 40,000 ~ \text{kg/m}^2$

$\sigma_f = \dfrac{6M}{bd^2} = \dfrac{6(112,500)}{1(9^2)}$

$\sigma_f = 8,333.33 ~ \text{kg/m}^2$

$q_{max} = \sigma_a + \sigma_f = 40,000 + 8,333.33$

$q_{max} = 48,333.33 ~ \text{kg/m}^2$ *answer*