Moment of inertia of a composite object in Solidworks

[Stephen Bulking]

TL;DR Summary

I'm having trouble finding the moment of inertia of a rotary drum and 1177 peanut centers within.

Hey guys do you know how Solidworks allow us to estimate the moment of inertia of the 3D model given the material properties? Well I'm trying to find the moment of inertia of my rotary drum with 1177 peanuts centers inside the drum and obviously I can't draw 1177 peanut centers inside the drum... so do you guys have anyway to solve this problem

Currently my solution is to create 1 single wedge in place of all the peanuts knowing approximately how much space (volume and mass) the peanuts currently occupy and then slide that wedge into the drum and re evaluate the whole thing

 

[Stephen Bulking]

I forgot to mention, the drum is tilted at a 30 degree angle so I chose a wedge at the bottom

 

[Filip Larsen]

Perhaps you have some specific questions? Are you unsure about if your mechanical assumptions (which are?) can be applied to the calculation problem (which is?) you are trying to solve? Or perhaps you are more asking about how to solve this (or similar) problems in Solidworks?

 

[Stephen Bulking]

Perhaps you have some specific questions? Are you unsure about if your mechanical assumptions (which are?) can be applied to the calculation problem (which is?) you are trying to solve? Or perhaps you are more asking about how to solve this (or similar) problems in Solidworks?

Hmm... I thought I made myself clear enough... welp no problem...

I'm trying to design a rotary drum for a household panning application which includes a motor directly turning the drum (of course when I say directly I mean with bearings and intermediate shaft, please see included kinematic diagram)

In order to calculate the required torque for the motor, I need to know the moment of inertia of the drum and 1177 peanut centers inside it about the motor shaft axis and I plan to use Solidworks to estimate this. However, I can't draw 1177 peanut centers inside the drum because I don't know how they are distributed.

Therefore my solution is to replace all 1177 peanut centers with a wedge-shape object with the same mass and volume, this is the assumption I'm unsure about. Do you guys have any other idea?

Do you understand it now, Filip?

 

[Lnewqban]

Rotating your drawing 90 degrees around a vertical axis, you will see the shaft torque having to overcome the initial moment of inertia of the rotary drum and peanut centers within plus the resistive torque of the tilted center of mass of the beans adopting a dynamic angle of repose.

For slow rotating speed, the angle of repose of the beans will give you how much the weight will be off-center.
Please, see:
https://en.wikipedia.org/wiki/Angle_of_repose

 

[Stephen Bulking]

Rotating your drawing 90 degrees around a vertical axis, you will see the shaft torque having to overcome the initial moment of inertia of the rotary drum and peanut centers within plus the resistive torque of the tilted center of mass of the beans adopting a dynamic angle of repose.

For slow rotating speed, the angle of repose of the beans will give you how much the weight will be off-center.
Please, see:
https://en.wikipedia.org/wiki/Angle_of_repose

View attachment 339323

Wow, cool I didn't even know of the angle of repose... Can you please show me the book from which you read this?

Also in my model, I'm assuming that the torque required to turn the drum of the motor consists of 2 components: torque to counteract gravitational torque and torque to overcome the drum's moment of inertia. I'm not sure if this model is right, how do you suppose I should change my model?

Thanks

 

[Lnewqban]

Wow, cool I didn't even know of the angle of repose... Can you please show me the book from which you read this?

Also in my model, I'm assuming that the torque required to turn the drum of the motor consists of 2 components: torque to counteract gravitational torque and torque to overcome the drum's moment of inertia. I'm not sure if this model is right, how do you suppose I should change my model?

Thanks

Sorry, I can't remember the book from which I learned about it.
It happened many years ago.
It is an everyday thing for people who have to work with piles of dirt, sand, gravel, etc.

Your approach is correct, I believe.
Your original question is difficult to respond.
If possible, I would do a practical test, or consider the worst conditions regarding angle, weight, amount of beans, etc.
A safety factor is a wonderful thing.

 

[Filip Larsen]

I'm assuming that the torque required to turn the drum of the motor consists of 2 components: torque to counteract gravitational torque and torque to overcome the drum's moment of inertia.

Assuming the drum (with no peanuts) has its center of mass (CoM) on the rotation axis, the motor only needs to provide torque to lift the CoM for the peanuts, which I assume will be maximum when the "surface" of the peanuts has reach the angle of repose Lnewqban mentioned (i.e. the drum rotates slow enough that the penauts surface has time to maintain angle of repose).

The motor also needs to provide torque to overcome friction torque from the bearings, if that is what you mean by "overcome the drum's moment of inertia". Any available torque above these two "needed" torques will go into spin the drum faster, with higher drum MoI only meaning lower angular acceleration.

To the extend you already have a drum on an axis available I will also recommend that you make practical measurements. By attaching weights of varying mass to the outside of the drum you should be able to make a reasonable estimate for the minimum torque needed to turn the loaded drum up past the angle of repose and overcome static friction in the bearings. If the engine is electric with the usual RPM/torque relationship once its picks up speed its available torque will decrease which should mean the drum will settle at some rotation rate after some time that depends on the drums moment of inertia.

 

[jack action]

I would simply model the peanuts as an internal friction resistance.

Imagine that, instead of peanuts, you have a block of ice following the shape of your drum. With a very low friction coefficient, the block of ice will not move at all. Its contribution to the moment of inertia is null.

Replace the block of ice with a steel block and the block will move upward and stop at a certain level, thus not contributing to the moment of inertia either at that point. But the friction resistance will be constant and much higher than with the ice.

Now use a block of rubber with a coefficient of friction high enough to stick to the drum wall. Then the block of rubber will follow the drum rotation and then it will contribute to the moment of inertia. Of course, at that point, the sliding friction may or may not be present.

If things tumble back (like your peanuts), then the angle of repose (post #5) gives you an equivalent of the internal friction going on, assuming no moment of inertia, i.e. the steel block case.

 

[Stephen Bulking]

Thanks for your comments guys, I think I'll just do a simpler model not accounting the angle of repose for the extent of this project

@jack action yeah, I get your idea but instead of the block of ice, I modeled a wedge of peanuts with the same volume to find the moment of inertia of the composite object and also to find the shifted COM when the peanuts are at the desired 10 o'clock position (as described in a paper I found), please see figure

@Filip Larsen, oh you don't need to take the bearings into account, I already did in another equation. By rotational inertia of the drum I mean there's going to be some resistance of the drum to it start spinning, not the bearings, see the figure and equations I have embedded. I don't think torque above the required will spin the drum faster... The drum needs to spin at 30-45 RPM and I'm building a PID controller to keep it spinning at the desired speed.

 

[Stephen Bulking]

Model I built in Solidworks to assess the moment of inertia of the composite object that is the drum and peanuts inside

Equation I used to evaluate the torque