Calculate the moment of inertia for an inverted T-section about its horizontal centroidal axis. Take the size of flange 100mm x 30 mm and vertical web 120mm x 30mm , overall depth-150mm

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written 4.7 years ago by

Find: $I_{xx}=?$

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

$$ \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) …

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