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Moment of Inertia Problems # 2 COMPOSET AREA

Ex. Determine the moment of inertia for the beam's  cross-sectional area about the  $x^{\prime}$  axis.

$① 40 * 160$

$② 40 * 120$

$③ 40 * 120$

$I_{\overline{\mathbf{x}}} = I _ {axis} + A \cdot \overline{y}^{2}$

$I_{\overline{x}_,{1}}=\frac{1}{12} (40)(160)^{3}+ (160 *40)*0^{2}=13.65 * 10^{6} \mathrm{mm}^{4}$

$I_{\overline{x}_,{2}}=\frac{1}{12} (120)(40)^{3}+ (120 *40)*(60)^{2}=17.92 * 10^{6} \mathrm{mm}^{4}$

$I_{\overline{x}_,{3}}=\frac{1}{12} (120)(40)^{3}+ (120 *40)*(60)^{2}=17.92 * 10^{6} \mathrm{mm}^{4}$

$I_{\overline{x},{total}} = I _ {\overline{x},{1}} +I_{\overline{x},{2}}+I_{\overline{x},{3}}$

$=13.65 * 10^{6}+17.92 * 10^{6}+17.92 * 10^{6}=49.5 * 16^{6} \mathrm{mm}^{4}$

Determine the distance  $\overline{x}$   to the centroid  $C$   of the  beam's cross-sectional area and then compute the moment $\overline{I}_{y^{\prime}}$  about the $y^{\prime}$
axis.

$\overline{x}=\frac{813000}{13200}=61.6 mm$

$I_{\overline{y}.{1}}=\frac{1}{12} (30)(200)^{3}+6000 (100-61.6)^{2}=28.85 * 10^{6} \mathrm{mm}^{4}$

$I_{\overline{y}.{2}}=\frac{1}{12} (140)(30)^{3}+ 4200 (15-61.6)^{2}=9.44 * 10^{6} \mathrm{mm}^{4}$

$I_{\overline{y}.{3}}=\frac{1}{12} (30)(100)^{3}+ 3000 (50-61.6)^{2}=2.904 * 10^{6} \mathrm{mm}^{4}$

$I_{\overline{y},{total}} = I _ {1} +I_{2}+I_{3}=41.2*10^{6} mm^{4}$

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