Solution to Problem 548 | Unsymmetrical Beams

$M_{max} = \frac{1}{8} w_o L^2$

$M_{max} = \frac{1}{8} w_o (4^2)$

$M_{max} = 2w_o$
 

$M_r = \dfrac{f_b I}{y}$
 

$M_t = \dfrac{40(30 \times 10^6 )}{80}$

$M_t = 15\,000\,000 \, \text{N}\cdot\text{mm}$

$M_t = 15 \, \text{kN}\cdot\text{m}$
 

$M_c = \dfrac{80(30 \times 10^6 )}{200}$

$M_c = 12\,000\,000 \, \text{N}\cdot\text{mm}$

$M_c = 12 \, \text{kN}\cdot\text{m}$
 

The section is stronger in tension and weaker in compression, so compression governs in selecting the maximum moment.
 

$M_{max} = M_r$

$2w_o = 12$

$w_o = 6 \, \text{kN/m}$           answer