Solution to Problem 525 | Flexure Formula

$\Sigma F_V = 0$

$2.4w = 240(0.2) + 240(0.2)$

$w = 40 \, \text{kN/m}$
 

$(\,f_b\,)_{max} = \dfrac{Mc}{I}$
 

Where:
$f_b = 8 \, \text{MPa}$

$M = 6 \, \text{kN}\cdot\text{m}$

$c = \frac{1}{2}x$

$I = \dfrac{bh^3}{12} = \dfrac{x(x^3)}{12} = \frac{1}{12}x^4$
 

Thus,
$8 = \dfrac{6(\frac{1}{2}x)(1000^2)}{\frac{1}{12}x^4}$

$x^3 = 4\,500\,000$

$x = 165.1 \, \text{ mm square}$           answer