Determin the moment of inertia of an angle section 100 mm x 80 mm about vertical centroidal axis. Longer leg is vertical.

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written 5.5 years ago by

teamques10 ★ 69k

Find: MI about vertical centroidal axis = $I_{yy}=?$

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$\begin{array}{ll}{\mathrm{A}_{1}=10 \times 90=900 \mathrm{mm}^{2}} & {\mathrm{A}_{2}=80 \times 10=800 \mathrm{mm}^{2}} \\ {\mathrm{X}_{1}=10 / 2=5 \mathrm{mm}} & {\mathrm{X}_{2}=80 / 2=40 \mathrm{mm}} \\ {\overline{X}=\frac{\mathrm{A}_{1} X_{1}+\mathrm{A}_{2} X_{2}}{A_{1}+A_{2}}=} & {\frac{(900 \times 5)+(800 \times 40)}{900+800}=21.47 \mathrm{mm}}\end{array}$

$\begin{array}{l}{\text { To find } \mathrm{I}_{\mathrm{YY}}=\left(\mathrm{I}_{\mathrm{YY}}\right)_{1}+\left(\mathrm{I}_{\mathrm{YY}}\right)_{2}} \\ {\mathrm{I}_{\mathrm{YY}}=\left(\mathrm{IG}_{+} \mathrm{Ah}^{2}\right)_{1}+\left(\mathrm{IG}_{+} \mathrm{Ah}^{2}\right)_{2}}\end{array}$

$\begin{array}{l}{\text { Here, } \mathrm{h}_{2}=X_{1}-\overline{X}=40-21.47=18.53 \mathrm{mm}} \\ {\qquad \begin{array}{l}{\mathrm{h}_{1}=\overline{X}-X_{2}=21.47-5=16.47 \mathrm{mm}} \\ {\mathrm{I}_{\mathrm{yy}}=\left(\frac{90 \times 10^{3}}{12}+900 \times 16.47^{2}\right)_{1}+\left(\frac{10 \times 80^{3}}{12}+800 \times 18.53^{2}\right)_{2}}\end{array}}\end{array}$

$\begin{array}{l}{\mathrm{I}_{y y}=(251634.81)_{1}+(701355.39)_{2}} \\ {\mathrm{I}_{y y}=9.53 \times 10^{3} \mathrm{mm}^{4}}\end{array}$

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