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, $I_{xx} = I_{NA}\\ I_{NA} = I_{Flange} + I_{Web}\\ I_{Flange} = \frac{bd^3}{12} + Ah^2\\ \hspace{0.5cm} = \frac{80\hspace{0.05cm}\times\hspace{0.05cm}10^3}{12} + (80\hspace{0.05cm}\times\hspace{0.05cm}10)\hspace{0.05cm}\times\hspace{0.05cm}39^2\\ \hspace{0.5cm} = 1.2\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm^4$

$I_{Web} = \frac{bd^3}{12} + Ah^2\\ \hspace{0.5cm} = \frac{10\hspace{0.05cm}\times\hspace{0.05cm}120^3}{12} + (10\hspace{0.05cm}\times\hspace{0.05cm}120)\hspace{0.05cm}\times\hspace{0.05cm}26^2\\ \hspace{0.5cm} = 2.25\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm^4$

$I_{NA} = I_{Flange} + I_{Web}\\ \hspace{0.25cm} = 3.47\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm^4$

$\frac{M}{I} = \frac{\sigma_t}{y_t}\\ \frac{4\hspace{0.05cm}\times\hspace{0.05cm}10^6}{3.47\hspace{0.05cm}\times\hspace{0.05cm}10^6} = \frac{\sigma_t}{44}\\ \sigma_t = 50.66\hspace{0.05cm}N/mm^2$

$\frac{M}{I} = \frac{\sigma_c}{y_c}\\ \frac{4\hspace{0.05cm}\times\hspace{0.05cm}10^6}{3.47\hspace{0.05cm}\times\hspace{0.05cm}10^6} = \frac{\sigma_c}{86}\\ \sigma_c = 99.13\hspace{0.05cm}N/mm^2$