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 June 15th, 2019, 11:10 PM #1 Senior Member   Joined: Aug 2014 From: India Posts: 476 Thanks: 1 Is it possible to solve this moment of Inertia by integration? Find the moment of Inertia of the cross-sectional 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: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timey-wimey 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: 476 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 X-X: $\displaystyle I_{XX} = I_{XX1} + I_{XX2} + I_{XX3}$ $\displaystyle I_{XX1} = I_{G1.X} + A_{1}Y^{2}$ $\displaystyle = I_{G1.X} + A_{1}.(Y_{1} - \overline {Y})^{2}$ = $\displaystyle \large \frac {30*5^{3}}{12}+ 150(25 - 10.96^{2}$) (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 Y-Y: $\displaystyle I_{YY} = I_{YY1} + I_{YY2} + I_{YY3}$ $\displaystyle I_{YY1} = I_{G1 }+ A_{1}.X^{2}$ = $\displaystyle I_{G1.Y} + A_{1}. (X_{1} - \overline X)^{2}$ (Here$\displaystyle I_{G1.X} = \large \frac{d.b^{3}}{12}$) =$\displaystyle \large \frac {5*30^{3}}{12} + (150)(15 - 15)^{2}$ = $\displaystyle 11250 cm^{4}$ $\displaystyle I_{YY2} = I_{G2} + A_{2}.X^{2}$ = $\displaystyle I_{G2.Y} + A_{2}. (x^{2} - \overline X)^{2}$ = $\displaystyle \large \frac {15*5^{3}}{12} + (75)(15-15)^{2}$ = $\displaystyle 156.25 cm^{4}$ $\displaystyle I_{YY3} = I_{G3} + A_{3}.X^{2}$ =$\displaystyle I_{G3.Y} + A_{3}. (x^{2} - \overline X)^{2}$ = $\displaystyle \large \frac {5*20^{3}}{12} + (100)(15-15)^{2}$ = $\displaystyle 3333.33 cm^{4}$ $\displaystyle I_{yy} = 11250 + 156.25 + 3333.33 = 14739.58 cm^{4}$ 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
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Quote:
 Originally Posted by topsquark 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.)
This is what I get for not double checking the equation. It should be $\displaystyle I = \int r^2 dm$.

-Dan

Last edited by topsquark; June 17th, 2019 at 12:16 PM. June 17th, 2019, 12:15 PM   #5
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Quote:
 Originally Posted by Ganesh Ujwal I following this lengthy method to proceed, is there any easy method to solve it?
Yes. The moment of inertia of a solid cylinder is $\displaystyle I = \dfrac{1}{2} mr^2$ where r is the radius of the cylinder.

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 Tags inertia, integration, moment, solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post markosheehan Physics 5 September 4th, 2016 11:59 AM avi Physics 3 December 30th, 2015 05:11 AM Busybee Physics 0 May 22nd, 2014 04:32 AM s.nataniel1 Calculus 0 December 18th, 2011 04:52 AM MasterOfDisaster Physics 9 June 30th, 2011 05:29 PM

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