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

Data: L = 2 m, w = 2 kN/m

To find: $\sigma_{c}(max)$ and $\sigma_{t}(max)$

$\mathrm{M}_{\max }=\frac{w L^{2}}{2}=\frac{\left(2 \times 10^{3}\right) \times 2^{2}}{2}=4 \times 10^{3} N-m=4 \times 10^{6} N-m m$

$\overline{Y}_{b a s e}=\frac{a_{1} y_{1}+a_{2} y_{2}}{a_{1}+a_{2}}=\frac{(80 \times 10) \times 125+(10 \times 120) \times 60}{(80 \times 10)+(10 \times 120)}=86 \mathrm{mm}$

$\mathrm{I}_{\mathrm{NA}}=\mathrm{I}_{\mathrm{XX}}=\left[\left(\frac{b d^{3}}{12}\right)+A h^{2}\right]_{1}+\left[\left(\frac{b d^{3}}{12}\right)+A h^{2}\right]_{2}$

$\mathrm{I}_{\mathrm{XX}}=\left[\left(\frac{80 \times 10^{3}}{12}\right)+(80 \times 10) \times(44-5)^{2}\right]_{1}+\left[\left(\frac{10 \times 120^{3}}{12}\right)+(10 \times 120) \times(86-60)^{2}\right]_{2}$

$\mathrm{I}_{\mathrm{NA}}=\mathrm{I}_{\mathrm{XX}}=347.466 \mathrm{x} 10^{4} \mathrm{mm}^{4}$

$\overline{Y}_{b a s e}=\mathrm{Y}_{\mathrm{C}}=86 \mathrm{mm}$

$\mathrm{Y}_{\mathrm{t}}=130-86=44 \mathrm{mm}$

$Maximum Compressive and Tensile Stress developed:$

$\frac{M_{\max }}{I}=\frac{\sigma_{C}}{Y_{C}}=\frac{\sigma_{t}}{Y_{t}}$

$\frac{4 \times 10^{6}}{347.466 \times 10^{4}}=\frac{\sigma_{C}}{86}=\frac{\sigma_{t}}{44}$

$\sigma_{C}=\frac{4 \times 10^{6} \times 86}{347.466 \times 10^{4}}=99.002 N / m m^{2}$ (At Bottom fiber)

$\sigma_{t}=\frac{4 \times 10^{6} \times 44}{347.466 \times 10^{4}}=50.652 N / m m^{2}$ (At Top fiber)