The flange is at top and the web is vertical. Determine the maximum tensile and the compressive stress developed and locate their position.

$a_1 = 80\hspace{0.05cm}\times\hspace{0.05cm}10 = 80\hspace{0.05cm}mm^2\\ a_2 = 10\hspace{0.05cm}\times\hspace{0.05cm}120 = 1200\hspace{0.05cm}mm^2\\ y_1 = 125\\ y_2 = 60\\ y = \frac{a_1y_1 + a_2y_2}{a_1 + a_2} = 86\hspace{0.05cm}mm$

$M_{max} = \frac{WL^2}{2}\\ \hspace{0.5cm} = \frac{2\hspace{0.05cm}\times\hspace{0.05cm}2^2}{2}\\ \hspace{0.5cm} = 4\hspace{0.05cm}KNm\\ \hspace{0.5cm} = 4\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}Nmm$

$\frac{M}{I} = \frac{\sigma}{y}$

Now, $I_{xx} = I_{NA}\\ I_{NA} = I_{Flange} + I_{Web}\\ I_{Flange} …