Motion System Design, Stepper Motor Sizing

Sizing: Adding a rubber sleeve as a primary roller

Author: Bob Parente | May 11, 2018

Recap of the last posting

In our last posting, we were comparing the inertia of a 12” long, 0.25” diameter steel rod (0.0013 lb-in^2) to a 1” long, 1” diameter aluminum coupling (0.00957 lb-in^2).

We asked why the lighter weight aluminum coupling would have a larger inertia than a heavier steel rod and the person in the back said “it’s because the coupling’s radius is larger and inertia is a function of the radius squared” and they were correct.

Let’s continue with our calculations by adding another inertial element.

Adding a roller (rubber sleeve)

Figure 1: Calculating inertia for a primary roller (rubber sleeve)
Figure 1: Calculating inertia for a primary roller (rubber sleeve)

I’m going to add a rubber sleeve over the steel shaft. I’m going to make it 10” long. That’s 2” shorter than the steel shaft; because the shaft needs to be supported by bearings on each end and we have to have some shaft length for the coupling to clamp on.

I Googled “density of rubber” and found that hard rubber is 74 pounds per cu-ft. and soft rubber is 69 pounds per cu-ft. This gives us a range of 0.0399 lbs/in^3 to 0.0428 lbs/in^3.

So someone picks a number for us to use.

Do you like 0.04 lbs/in^3? Great, let’s go with that.

Calculate the volume of the sleeve

I’m going to pick a circumference that is 2” with an outer diameter of 2/\pi or  0.63662”. Why that value you ask?

Well, one revolution would produce two inches of linear motion if a piece of material was sitting on the surface of the roller. We’ll get into that inertial calculation latter.

Using our trusty volume equation again \pi(r^2) \times length we get:

    \[ Volume = \pi\left(\frac{0.63662}{2}\right)^2 \times 10 = 3.1831 in^3\]

Multiply the volume by the rubber’s density and we get a weight (w_1) of 0.1273 lbs.

Using our inertia equation

    \[J = \frac {(w_1r^2)}{2} \]

we get:

    \[J_1 = \frac{(0.1273\times0.31831^2)}{2} = 0.00645 lb-in^2\]

But wait. Our rubber sleeve has a hole in it that allows it slip over the steel shaft. So let’s calculate that inertia and subtract it from our total.

    \[Volume = \pi\left(\frac{0.250}{2}\right)^2 \times 10 = 0.491 in^3\]

    \[Weight (w_2) = 0.491 \times 0.04 = 0.0196 lbs\]

    \[Inertia (J_2) = \frac{(0.0196 \times 0.125^2)}{2} = 0.000153 lb-in^2\]

Subtract  J_2 from  J_1 and we get:

    \[J_T = 0.00645 - 0.000153 = 0.006297 lb-in^2\]

There’s an another way

Now having done all that we could have looked up the inertial equation for a thick-walled hollow tube which is:

    \[J_T = \left(\frac{(w_1 - w_2)}{2}\right) \times (r_2^2 + r_1^2)\]

Where r_1 is the inner radius and r_2 is the outer radius.

Plug and chug our values:

    \[J_T = \left(\frac{(0.1273 - 0.0196)}{2}\right) \times (0.125^2 + 0.31831^2) = 0.006297 lb-in^2\]

The same as we calculated before.

Note that the weight is its actual weight, by that I mean the weight of the missing center is subtracted from the total (as if it was a solid) weight

A more efficient way

Now, I don’t know about you, but the first time I looked at that equation I would have said that it should “r_2^2 - r_1^2” not “r_2^2 + r_1^2”, but you do add the two squared radii together. I’ll leave the derivation of that up to you.

The equation that makes more sense to me is the following:

    \[J =  \left(\frac{\pi\rho h}{2}\right) \times (r_2^4 - r_1^4)\]

Where \rho is the density and h is the height or length of the hollow tube.

This is exactly what we did by using two separate equations. One for the whole rubber sleeve and one for the 0.25” section we removed.

Now that we have calculated the inertia of the rubber sleeve, let’s calculate the total inertia that the motor shaft sees.

Our 1”x 1” coupling has an inertia of 0.00957 lb-in^2.

Our 0.25”x 12” steel rod has an inertia of 0.0013 lb-in^2

And our 0.6366” x 10” rubber sleeve has an inertia of 0.006297 lb-in^2

Since everything is directly attached to the motor’s shaft, all we need to do is add the three inertias together to get our total inertia of 0.017167 lb-in^2

More next time.

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