Find: $I_{xx}=?$

$$ \begin{array}{l}{\overline{X}=\frac{\text { Flange Width }}{2}=\frac{100}{2}=50 \mathrm{mm}} \\ {\mathrm{A}_{1}=100 \times 30=3000 \mathrm{mm}^{2} \mathrm{A}_{2}=120 \times 30=3600 \mathrm{mm}^{2}} \\ {Y_{1}=\frac{30}{2}=15 \mathrm{mm} \quad Y_{2}=30+\frac{120}{2}=90 \mathrm{mm}}\end{array} $$

$$ \begin{array}{l}{\overline{Y}=\frac{A_{1} Y_{1}+A_{2} Y_{2}}{A_{1}+A_{2}}=\frac{(3000 \times 15)+(3600 \times 90)}{3000+3600}=55.91 \mathrm{mm}} \\ {\text { To find } \mathrm{I}_{\mathrm{xx}}} \\ {\mathrm{I}_{\mathrm{xx}=} \mathrm{I}_{\mathrm{xx} 1+} \mathrm{I}_{\mathrm{xx} 2}} \\ {\mathrm{I}_{\mathrm{xx}}=\left(\mathrm{IG}_{+} \mathrm{Ah}^{2}\right) 1+\left(\mathrm{IG}_{+} \mathrm{Ah}^{2}\right)_{2}}\end{array} $$

$$ \begin{array}{l}{\text { Here, } \mathrm{h}_{1}=\overline{Y}-Y_{1}=55.91-15=40.91 \mathrm{mm}} \\ {\qquad \mathrm{h}_{2}=Y_{2}-\overline{Y}=90-55.91=34.09 \mathrm{mm}}\end{array} $$

$$ \begin{array}{l}{\mathrm{I}_{\mathrm{xx}}=\left(\frac{100 \times 30^{3}}{12}+3000 \times 40.91^{2}\right)_{1}+\left(\frac{30 \times 120^{3}}{12}+3600 \times 34.09^{2}\right)_{2}} \\ {\mathrm{I}_{\mathrm{xx}}=(5245884.3)_{1}+(8503661.16)_{2}} \\ {\mathrm{I}_{\mathrm{xx}}=13.75 \times 10^{6} \mathrm{mm}^{4}}\end{array} $$