In our last posting, we calculated the inertia of a helical shaft coupling. The coupling is used to connect the motor shaft to the load shaft.
We calculated the inertia of:
- a coupling that was 0.5” in diameter and 0.5” long at 0.0003 lb-in2
- a coupling that was 1” in diameter and 1” long at 0.00957 lb-in2
The notable difference between the two couplings is that the large one, even though it’s only twice as large in both its length and diameter, has an inertia that’s 32 times greater.
So what, you might ask? Well if we are striving to match the motor’s rotor inertia to the load’s inertia, then a larger coupling might force us to choose a larger motor and larger motors cost more. A large coupling may very well be needed to meet the torque requirements of the application, so don’t just go by the inertia of the coupling when choosing one.
Steel load shaft inertia
Okay, so what’s next, now that we have a coupling attached to the motor’s shaft? How about we connect a steel load shaft to the coupling and see what impact that has on the total inertia.
Figure 1: Calculating the inertia of a steel rod used as a load shaftLet’s pick some numbers out of the air and make our steel shaft 0.25” in diameter and 12” long. If you Google “density of steel lbs/in3” you get at the very top of the page:
“Approximately 7.85 g/cm3”. Not quite the units that I wanted, but if you look down the page a bit you’ll see steel has a density of 0.283 lbs/in3. Yea, yea there are different carbon steels and other alloys to consider, but ours is the one that’s 0.283 pounds per cu-in.
The inertia (J) equation for this shaft is the same as the one we used for the coupling in the previous posting and that is:
![\[ J = (mr^2)/2 \]](https://www.novantaims.com/wp-content/ql-cache/quicklatex.com-b42d9496fc7a78a41765acf45feba972_l3.png "Rendered by QuickLaTeX.com")
where 
is the material density and 
is the shaft radius.
First we need to calculate the weight of the shaft first with:
![\[ Mass_l_b_s = \pi \times r^2 \times length \times density \]](https://www.novantaims.com/wp-content/ql-cache/quicklatex.com-2dae60a15c0dc43342f16f59cb2731e7_l3.png "Rendered by QuickLaTeX.com")
![\[ 3.14159 \times .125^2 \times 12 \times 0.283 =\boxed {0.167} \]](https://www.novantaims.com/wp-content/ql-cache/quicklatex.com-555cfa03af81c9ef512b180e0620c9ef_l3.png "Rendered by QuickLaTeX.com")
Then we will calculate the inertia of the shaft:
![\[ (0.167 \times 0.125^2)/2 = \boxed{0.0013} \]](https://www.novantaims.com/wp-content/ql-cache/quicklatex.com-64efff72ba164f0ee6f19334874561d3_l3.png "Rendered by QuickLaTeX.com")
Load shaft inertia equals 0.0013 lb-in2
Why does the steel shaft have a lower inertia than the aluminum coupling?
An interesting thing to note, ok I think it’s interesting, is that the one-inch diameter aluminum coupling weighing only 0.0766 lbs (2.18 times lighter than the steel shaft) and yet it has an inertia (.00957 lb-in2) that is 7.36 times larger than the steel shaft.
How can a lightweight aluminum coupling have seven times the inertia of a steel rod that’s 12 inches long and weighs more?
Yes, you got it! Inertia is a function of the radius squared.
- The coupling has a one-inch diameter, a 0.5” radius. Square it and you get 0.25
- The steel shaft has a 0.25” in diameter, a 0.125” radius. Square it and you get 0.015625.
This produces a 16:1 inertial factor just by having a larger diameter. So even though the aluminum coupling is lighter than the steel shaft, its diameter is what makes its inertia greater.
I think we’re getting up to speed with this. Or, We’re overcoming the inertia of these calculations.
More next time.
