- #1
ichiro_w
- 13
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Area moment of inertia--circular cross section
From the bending beam calculation, the moment of inertia of the cross section with regard to a coplanor axis of rotation is used. If we have a circular "beam", the area moment of inertia of a circular disk of radius a about a diameter is [tex] I_d = \frac{\pi a^4}{4}[/tex] according to two separate references. I believe the integral involved can be generally stated as [tex]I_d = \int y^2 dA [/tex] if y is the distance to the diameter d perpendicular to y which diameter (as all diameters of a uniformly mass distributed disk) passes through the centroid of the disk.
Now my stab at actually evaluating this is to do a double integral in polar coordinates and long story short the only way I can come up with the agreed upon answer is [tex] \int_0^\pi\int_0^a r^2\; r\; \mathrm{dr}\;\mathrm{d\theta}[/tex] where [tex] r\;\mathrm{dr}\;\mathrm{d\theta}=dA[/tex] and [tex]r^2[/tex] is the distance to the origin (centroid)
If I were in a creative writing glass I might get a passing grade for this fudge but I really would like to understand what I am doing better than backing into an answer like this. Any help would truly be appreciated.
From the bending beam calculation, the moment of inertia of the cross section with regard to a coplanor axis of rotation is used. If we have a circular "beam", the area moment of inertia of a circular disk of radius a about a diameter is [tex] I_d = \frac{\pi a^4}{4}[/tex] according to two separate references. I believe the integral involved can be generally stated as [tex]I_d = \int y^2 dA [/tex] if y is the distance to the diameter d perpendicular to y which diameter (as all diameters of a uniformly mass distributed disk) passes through the centroid of the disk.
Now my stab at actually evaluating this is to do a double integral in polar coordinates and long story short the only way I can come up with the agreed upon answer is [tex] \int_0^\pi\int_0^a r^2\; r\; \mathrm{dr}\;\mathrm{d\theta}[/tex] where [tex] r\;\mathrm{dr}\;\mathrm{d\theta}=dA[/tex] and [tex]r^2[/tex] is the distance to the origin (centroid)
If I were in a creative writing glass I might get a passing grade for this fudge but I really would like to understand what I am doing better than backing into an answer like this. Any help would truly be appreciated.