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Actual question is Find the moment of Inertia of the cross-sectional area of an I section about its centroidal axis:

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My solution: Here $A_{1}, A_{2}$ and $A_{3}$ are the areas:

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Centriod at X axis; $ X_{C} = \large \frac {A_{1}X_{1} + A_{2}X_{2}+A_{3}X_{3}}{A_{1} + A{2}+ A_{3}}= 15$ cm

Centriod at Y axis; $ Y_{C} = \large \frac {A_{1}Y_{1} + A_{2}Y_{2}+A_{3}Y_{3}}{A_{1} + A{2}+ A_{3}}= 10.96$ cm

Moment of Inertia w.r.t Centroid X-X: $$ I_{XX} = I_{XX1} + I_{XX2} + I_{XX3}$$ $$ I_{XX1} = I_{G1.X} + A_{1}Y^{2}$$ $$ = I_{G1.X} + A_{1}.(Y_{1} - \overline {Y})^{2}$$ $$= \large\frac {30*5^{3}}{12}+ 150(25 - 10.96^{2}) (\;Here I_{G1.X} = \large \frac {b.d^3}{12}) $$

$$= $ 11048.24 cm^{4}$$ $$ I_{XX2} = I_{G2.X} + A_{2}Y^{2}$$ $$ = I_{G1.X} + A_{2}.(Y_{2} - \overline {Y})^{2}$$ $$ = \large \frac {5*15^{3}}{12} + 75(12.5 - 10.96^{2})$$ $$ =1584.12 cm^{4}$$ $$ I_{XX3} = I_{G3.X} + A_{3}.Y^{2}$$ $$ = I_{G3.X} + A_{3}.(Y_{3} - \overline {Y})^{2}$$ $$= \large \frac {20*5^{3}}{12} + 100(22.5 - 10.96^{2})$$ $$ = 13525.25 cm^{4}$$ $$ I_{xx} = 11048.28 + 1584.12 + 13525.5 = 26137.86 cm^{4}$$ Moment of Inertia w.r.t Centroid Y-Y: $$ I_{YY} = I_{YY1} + I_{YY2} + I_{YY3}$$ $$ I_{YY1} = I_{G1 }+ A_{1}.X^{2}$$ $$ =I_{G1.Y} + A_{1}. (X_{1} - \overline X)^{2} (Here I_{G1.X} = \large \frac{d.b^{3}}{12}) $$ $$ = \large \frac {5*30^{3}}{12} + (150)(15 - 15)^{2}$$ $$ = 11250 cm^{4}$$ $$ I_{YY2} = I_{G2} + A_{2}.X^{2}$$ $$ = I_{G2.Y} + A_{2}. (x^{2} - \overline X)^{2}$$ $$ = \large \frac {15*5^{3}}{12} + (75)(15-15)^{2}$$ $$ = 156.25 cm^{4}$$ $$ I_{YY3} = I_{G3} + A_{3}.X^{2}$$ $$ = I_{G3.Y} + A_{3}. (x^{2} - \overline X)^{2}$$ $$ = \large \frac {5*20^{3}}{12} + (100)(15-15)^{2}$$ $$= 3333.33 cm^{4}$$ $$ I_{yy} = 11250 + 156.25 + 3333.33 = 14739.58 cm^{4}$$ I following this lengthy method to proceed, is there any easy method to solve it?

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