Simple max shaft torque calculation

AI Thread Summary
To calculate the maximum torque for a 2" diameter shaft rotating at 8 RPM with a 60" diameter, 6500 lb cylinder, first determine the rotational moment of inertia of the cylinder. The torque required to start or stop the shaft depends on this inertia and can be calculated using angular acceleration formulas. If the cylinder is offset by 2", an additional torque from the weight times the moment arm must be included. The peak torque on the brakes or gearbox is what needs to be evaluated, ignoring variations in torque amplitude. Understanding these calculations is crucial for selecting an appropriate brake and gear reducer.
AQUAPOP
Messages
4
Reaction score
0
Hello.

I'm trying to add a mechanical (pneumatic or electrical) brake to a shaft.

I have a 2" diameter shaft rotating at 8 rpm's max.

The shaft has and 60" diameter, 6500lb. cylinder attached to it axially.

How do I calculate the maximum torque of the shaft?

Thanks in advance,
Frank
 
Engineering news on Phys.org
Rotating shafts generally fail from fatigue. Do you know how to calculate stresses and predict fatigue?
 
At this point I'm not really concerned with the shaft failing.

I had a grossly undersized gear reducer fail a few times on this application,
so I'm trying to add a brake, and then find a more appropriate gear reducer.

thanks for the quick reply. :)
 
Sorry, I think I get you now. You're trying to figure out how much torque this 60" diameter, 6500 pound cylinder requires to stop or start it moving at 8 RPM? If that's correct, do you have dimensions on it and do you know how to calculate the rotational moment of inertia? I presume also that there are rolling element bearings supporting this shaft so that friction can be neglected?
 
You are correct.
I have steel cylinders (heat exchangers) on a horizontal rotisserie.
Both ends are supported on shafts in bearings.
The cylinders vary in diameter and length, and internal structure.
Another wrinkle is because of varying internal structure some rotate 2" offset from their center of gravity.
 
The torque required to start or stop the shaft is a function of it's rotational inertia. Imagine a car and you have to push it. The lighter the car is, the easier it is to accelerate it from a dead stop by pushing it. The torsional analogue of that is what you're interested in. Just as the car has some mass which resists accelerating, the shaft with the attached cylinder has rotational resistance due to its rotational inertia. You can calculate the rotational moment of inertia using standard formulas for simple shapes as shown here: https://webspace.utexas.edu/cokerwr/www/index.html/RI.htm
or if you can do the math, try a more advanced calculation following the equations provided by Wikipedia here:
http://en.wikipedia.org/wiki/Moment_of_inertia#Example_calculation_of_moment_of_inertia

Once you know the rotational inertia, and you want to find the torque required to change the rotational speed, use the formula for angular acceleration,
9be08b9254aaacbc0386b26bf137f2ae.png

http://en.wikipedia.org/wiki/Angular_acceleration
http://hyperphysics.phy-astr.gsu.edu/hbase/n2r.html
http://theory.uwinnipeg.ca/physics/rot/node5.html
I've provided a few different sites but they all say the same thing. Note that angular acceleration is simply the change in angular rotational rate divided by the time taken to change, ie: [PLAIN]http://upload.wikimedia.org/math/f/d/9/fd97cb711276815954e9824fabee8baf.png. So for example, if you want to accelerate the cylinder from 0 to 8 RPM in 1 second, that's an acceleration rate of 8/60*2*pi*radians/s2. If it takes longer or shorter than 1 second, divide by the number of seconds.

For the torque exerted by the cylinder being off center by 2", just add an additional torque of the cylinder weight times this 2" moment arm (ie: 6500 lb x 2" = 13,000 lb in). The amplitude of that torque is obviously going to vary sinusoidally but for your purposes (trying to determine the peak torque on the brakes or gearbox) you really don't care about that.
 
Last edited by a moderator:
  • Like
Likes 2 people
Thread 'Physics of Stretch: What pressure does a band apply on a cylinder?'
Scenario 1 (figure 1) A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material. The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm). At this level of stretch, the load cell reads 45N tensile force. Key numbers...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Back
Top