Moment of inertia of a branched cantilever beam

[dp_033]

Hey there!
I am working on a project concerning the mathematical modeling of nano-swimmers in a viscous medium. Assuming the nano-swimmers to be cantilever beams, the project involves calculation of moment of inertia of the said nano-swimmers. While calculating the moment of inertia of a simple or a tapered cantilever beam is quite easy, I was facing problems dealing with a branched cantilever beam. The branches are placed in pairs and each pair has the same dimensions, protrude at the same distance from the support and face the opposite direction from each other. Any sort of help would be appreciated!

 

[berkeman]

Welcome to the PF.

Can you post a sketch or diagram of what you are asking about? Use the Upload button in the lower right of the Edit window to attach a PDF or JPEG image. Thanks

 

[A.T.]

Hey there!
I am working on a project concerning the mathematical modeling of nano-swimmers in a viscous medium. Assuming the nano-swimmers to be cantilever beams, the project involves calculation of moment of inertia of the said nano-swimmers. While calculating the moment of inertia of a simple or a tapered cantilever beam is quite easy, I was facing problems dealing with a branched cantilever beam. The branches are placed in pairs and each pair has the same dimensions, protrude at the same distance from the support and face the opposite direction from each other. Any sort of help would be appreciated!

This might help:
https://en.wikipedia.org/wiki/Parallel_axis_theorem

 

[coquelicot]

With respect to which axis do you need the moment of inertia?
If it is with respect to the main longitudinal axis, the problem is simple, because the moment of inertia of a rigid composite system is the sum of the moments of inertia of its component subsystems with respect to the same axis. So, all you need to know is the moment of inertia of the central cylinder (Wikipedia), and to compute the integral of $\rho \pi r^2 y dy$, where $\rho$ is the mass/volume density, $r$ is the radius of the branch, and y is the distance from the center of the branch to the integration point on the axis of the branch (I hope you know how to compute this integral).
Then you simply sum the moments of inertia.

Added: While it is not exactly true that the branches consists in cylinders "glued" to the main central cylinder, this is an excellent approximation that much simplify the computations. I think this approximation is quite reasonable for your project.
If you need a very exact theoretical computation, this is somewhat more complex.

 

[dp_033]

I'm sorry, I didn't mention the axis. The torsional moment of Inertia can be calculated as you mentioned, but I have been facing problem calculating the area moment of inertia. I've attached the file which shows the axis in question

 

[coquelicot]

Here is how I would handle the problem.
1) make the following approximation : the branches are pure cylinders glued to the main cylinder. As mentioned above, this is a very good approximation.
2) for each cylinder, compute the 3 principal moments of inertia, with respect to the axes passing through their center of gravity, and parallel to the 3 desired axes. Since these axes are also the principal axes of each cylinder, this may even be found in Wikipedia
3) for each cylinder, use the parallel axis theorem to compute the 3 moments of inertia with respect to the same system of axes, but whose origin translated to the center of mass of the whole body (or any other chosen point). Be sure not to confuse the axes.
4) for each axis, add all the corresponding moments of inertia of all the cylinders, and let M1, M2, M3 be the 3 obtained overall moments of inertia.
5) if necessary, use the parallel axis theorem to compute the 3 moments of inertia with respect to the same axes but whose origin is translated to any point you want.

 

[dp_033]

Here is how I would handle the problem.
1) make the following approximation : the branches are pure cylinders glued to the main cylinder. As mentioned above, this is a very good approximation.
2) for each cylinder, compute the 3 principal moments of inertia, with respect to the axes passing through their center of gravity, and parallel to the 3 desired axes. Since these axes are also the principal axes of each cylinder, this may even be found in Wikipedia
3) for each cylinder, use the parallel axis theorem to compute the 3 moments of inertia with respect to the same system of axes, but whose origin translated to the center of mass of the whole body (or any other chosen point). Be sure not to confuse the axes.
4) for each axis, add all the corresponding moments of inertia of all the cylinders, and let M1, M2, M3 be the 3 obtained overall moments of inertia.
5) if necessary, use the parallel axis theorem to compute the 3 moments of inertia with respect to the same axes but whose origin is translated to any point you want.

This is more or less what I've been trying to do.
1. The area moment of inertia of the beam w.r.t to the given axis would be simple(circular cross-section).
2. Use the parallel axis theorem to find the moment of inertia of branches(circular cross-section as well).
3. Shift them to CG of the whole body.
(for area moment of inertia of different cross-sections)
https://en.wikipedia.org/wiki/List_of_second_moments_of_area

Thanks!

 

[coquelicot]

This is more or less what I've been trying to do.

So, what is exactly your problem ? please, be precise.