If the maximum permissible bending stress is 125 MPa. Find the moment of resistance of beam. Also find the maximum intensity of UDL over an entire span.

$\sigma = 125\hspace{0.05cm}N/mm^2\\ y = \frac{220}{2} = 110\hspace{0.05cm}mm\\ I = \frac{bd^3}{12} - 2(\frac{bd^3}{12})\\ \hspace{0.05cm} = \frac{120\hspace{0.05cm}\times\hspace{0.05cm}22063}{12} - 2\hspace{0.05cm}(\frac{55\hspace{0.05cm}\times\hspace{0.05cm}200^3}{12})\\ \hspace{0.05cm} = 33.15\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm^4$

$\frac{M}{I} = \frac{\sigma}{y}\\ M = \frac{125}{110}\hspace{0.05cm}\times\hspace{0.05cm}33.15\hspace{0.05cm}\times\hspace{0.05cm}10^6 = 3.89\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}Nmm$

$M_{max} = \frac{WL^2}{8}\\ 3.8\hspace{0.05cm}\times\hspace{0.05cm}10^6 = \frac{W\hspace{0.05cm}\times\hspace{0.05cm}(5\hspace{0.05cm}\times\hspace{0.05cm}10^3)^2}{8}\\ W = 1.2\hspace{0.05cm}N/mm$