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June 15th, 2019, 11:10 PM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 403 Thanks: 1  Is it possible to solve this moment of Inertia by integration?
Find the moment of Inertia of the crosssectional area of an I section about its centroidal axis: Is it possible to solve this moment of Inertia by integration? 
June 16th, 2019, 12:59 PM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff. 
Yes. The formula is $\displaystyle I = \int m^2 dr$. The usual way to do this is to write m as a function of r (using any density information you have. Usually at this level the density is constant.) However you have three basic objects. All of them are essentially cylinders (The disks at top and bottom are simply solid cylinders which aren't very tall) so you can just look up the formula. The length of the cylinders don't matter. Here's a list of standard moments of inertia. It's good practice to calculate them on your own but then you can freely use the tables. Moments of inertia are additive along the same axis. Dan 
June 17th, 2019, 12:40 AM  #3 
Senior Member Joined: Aug 2014 From: India Posts: 403 Thanks: 1 
Sol: Here A1, A2 and A3 are the areas: Centriod ; $\displaystyle X_{C} = \large \frac {A_{1}X_{1} + A_{2}X_{2}+A_{3}X_{3}}{A_{1} + A{2}+ A_{3}}= 15 cm$ Centriod ; $\displaystyle 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 XX: $\displaystyle I_{XX} = I_{XX1} + I_{XX2} + I_{XX3}$ $\displaystyle I_{XX1} = I_{G1.X} + A_{1}Y^{2}$ $\displaystyle = I_{G1.X} + A_{1}.(Y_{1}  \overline {Y})^{2}$ = $\displaystyle \large \frac {30*5^{3}}{12}+ 150(25  10.96^{2}$) (Here $\displaystyle I_{G1.X} = \large \frac {b.d^3}{12}$) = $\displaystyle 11048.24 cm^{4}$ $\displaystyle I_{XX2} = I_{G2.X} + A_{2}Y^{2}$ $\displaystyle = I_{G1.X} + A_{2}.(Y_{2}  \overline {Y})^{2}$ $\displaystyle = \large \frac {5*15^{3}}{12} + 75(12.5  10.96^{2})$ $\displaystyle =1584.12 cm^{4}$ $\displaystyle I_{XX3} = I_{G3.X} + A_{3}.Y^{2}$ $\displaystyle = I_{G3.X} + A_{3}.(Y_{3}  \overline {Y})^{2}$ =$\displaystyle \large \frac {20*5^{3}}{12} + 100(22.5  10.96^{2})$ $\displaystyle = 13525.25 cm^{4}$ $\displaystyle I_{xx} = 11048.28 + 1584.12 + 13525.5 = 26137.86 cm^{4}$ Moment of Inertia w.r.t Centroid YY: $\displaystyle I_{YY} = I_{YY1} + I_{YY2} + I_{YY3}$ $\displaystyle I_{YY1} = I_{G1 }+ A_{1}.X^{2}$ = $\displaystyle I_{G1.Y} + A_{1}. (X_{1}  \overline X)^{2}$ (Here$\displaystyle I_{G1.X} = \large \frac{d.b^{3}}{12}$) =$\displaystyle \large \frac {5*30^{3}}{12} + (150)(15  15)^{2}$ = $\displaystyle 11250 cm^{4}$ $\displaystyle I_{YY2} = I_{G2} + A_{2}.X^{2}$ = $\displaystyle I_{G2.Y} + A_{2}. (x^{2}  \overline X)^{2}$ = $\displaystyle \large \frac {15*5^{3}}{12} + (75)(1515)^{2}$ = $\displaystyle 156.25 cm^{4}$ $\displaystyle I_{YY3} = I_{G3} + A_{3}.X^{2}$ =$\displaystyle I_{G3.Y} + A_{3}. (x^{2}  \overline X)^{2}$ = $\displaystyle \large \frac {5*20^{3}}{12} + (100)(1515)^{2}$ = $\displaystyle 3333.33 cm^{4}$ $\displaystyle I_{yy} = 11250 + 156.25 + 3333.33 = 14739.58 cm^{4}$ I following this lengthy method to proceed, is there any easy method to solve it? Last edited by Ganesh Ujwal; June 17th, 2019 at 12:52 AM. 
June 17th, 2019, 12:08 PM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Sorry about that. Dan Last edited by topsquark; June 17th, 2019 at 12:16 PM.  
June 17th, 2019, 12:15 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
The new problem is that I didn't notice before that you aren't given any masses. The best you can do is assume that the density of each piece is the same. So you would have $\displaystyle I_{tot} = \dfrac{1}{2} \rho (0.150)^1 + \dfrac{1}{2} \rho (0.025)^2 + \dfrac{1}{2} \rho (0.100)^2$ where $\displaystyle \rho$ is the density of the material. Factoring out the density to simplify a bit: $\displaystyle I_{tot} = \rho \left ( \dfrac{1}{2} (0.150)^2 + \dfrac{1}{2}(0.025)^2 + \dfrac{1}{2} (0.100)^2 \right )$ Dan  

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