# Moment of inertia of a branched cantilever beam

• dp_033
In summary, the conversation discusses a project involving the mathematical modeling of nano-swimmers in a viscous medium. The project requires the calculation of the moment of inertia of cantilever beams, including a branched cantilever beam with paired branches facing opposite directions. The conversation provides helpful resources and outlines a method for calculating the moment of inertia with respect to different axes.

#### 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!

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dp_033 said:
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

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.

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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.

coquelicot said:
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!

dp_033 said:
This is more or less what I've been trying to do.

## 1. What is the moment of inertia of a branched cantilever beam?

The moment of inertia of a branched cantilever beam refers to its resistance to bending when subjected to an external force. It is a measure of the beam's ability to resist deformation and depends on the beam's shape, size, and distribution of mass.

## 2. How is the moment of inertia calculated for a branched cantilever beam?

The moment of inertia is calculated by integrating the products of each infinitesimal mass element and its distance from the axis of rotation squared. For a branched cantilever beam, this calculation can be more complex as it involves multiple arms or branches.

## 3. What factors affect the moment of inertia of a branched cantilever beam?

The moment of inertia of a branched cantilever beam is affected by its shape, size, and distribution of mass. The longer and thicker the beam, the higher its moment of inertia will be. Additionally, the distribution of mass along the beam's length and branches will also impact the moment of inertia.

## 4. How does the moment of inertia impact the structural stability of a branched cantilever beam?

The higher the moment of inertia, the more resistant the beam will be to bending and deformation, leading to increased structural stability. A larger moment of inertia also means that the beam can withstand larger external forces without breaking or buckling.

## 5. How can the moment of inertia be manipulated in a branched cantilever beam?

The moment of inertia can be manipulated in a branched cantilever beam by changing its shape, size, and distribution of mass. For example, increasing the beam's thickness or adding additional support at key points can increase its moment of inertia and improve its structural stability.