How Much Torque is Required to Rotate a 1500lb Steel Drum?

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SUMMARY

The required torque to rotate a 1500lb steel drum with an 8-foot diameter at 0.02 RPM is calculated using the moment of inertia and angular acceleration. The moment of inertia for a hollow cylinder was initially miscalculated, leading to a torque estimate of 8.81 lb-in. However, factoring in bearing friction and using the correct polar moment of inertia equation reveals that the torque required is significantly higher, approximately 25.9 N·mm, based on a friction coefficient of 0.0010 for self-aligning ball bearings. The wrapping mechanism's load on the drum must also be considered for accurate torque calculations.

PREREQUISITES
  • Understanding of torque calculations and formulas
  • Knowledge of moment of inertia for hollow cylinders and polar moment of inertia
  • Familiarity with bearing friction coefficients and their impact on torque
  • Basic principles of angular acceleration and RPM conversions
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  • Research the polar moment of inertia for various cylinder geometries
  • Learn about bearing friction coefficients and their application in torque calculations
  • Investigate the effects of load from wrapping mechanisms on rotating drums
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Mechanical engineers, industrial equipment designers, and anyone involved in the design and operation of rotating machinery, particularly those working with large drums and bearing systems.

SevenToFive
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We have a 1500lb steel drum that has a 8 foot diameter that we need to rotate to wrap up air filter material. The drum is supported by bearings on both sides and needs to turn at 0.02rpm.
I calculated the torque, using T=Moment of Inertia * Angular Acceleration. For the moment of inertia I used the equation for a hollow cylinder.
Gravity =386.088in/sec^2
For the weight I figured 2500lbs to factor in losses through friction(0.5). Outside diameter with the roll of air filter material would be 109 inches, inside diameter =100 inches. Final rpm = 0.02, Acceleration time is 5 seconds.

The angular acceleration =0.02*2*3.14/60/5 = 0.00042Rad/sec^2
Moment of inertia = 21041.835lb-in-sec^2
T=I*AA = 21041.835lb-in-sec^2 * 0.00042rad/sec^2
T=8.81lb-in

However that value seems very low to me. Any help is greatly appreciated.
 
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SevenToFive said:
T=8.81lb-in
... and, in "ft-lbs?" Can you turn it by hand? With a "cheater?"
 
Bystander said:
... and, in "ft-lbs?"
Went the wrong way, sorry. The "by hand" inquiry stands.
 
Calculate the torque required to turn the drum based strictly upon the bearing friction as a comparison for judging if your calculated starting torque is greater than the basic required rotating torque.
 
JBA said:
Calculate the torque required to turn the drum based strictly upon the bearing friction as a comparison for judging if your calculated starting torque is greater than the basic required rotating torque.
I would have 1500lbs, 0.50 for bearing efficiency, and 48" from the center of the drum to the outside, which would give me 36000lb-in, which seems a bit on the high side.
 
I am not sure to what the term "bearing efficiency" refers, since the rolling resistance of an anti-friction bearing is generally based upon the "bearing friction factor" and 0.50 is an extremely high "bearing friction factor" for any type of anti-friction (rolling type) bearing.

See the below website for a typical anti-friction (rolling bearing) resistance torque calculation.

http://www.skf.com/us/products/bear...ction/estimating-frictional-moment/index.html

Also, I believe you used the wrong Moment of Inertia equation for your calculation; and, should have used the "Polar Moment of Inertia" equation for a cylinder rotating about its axis.
 
.02 RPM is VERY VERY slow (once an hour?).. I think in this case the amount of energy needed to accelerate it can pretty much be ignored, you're BIG one is friction which will be orders of magnitude higher.

Looking at this http://www.skf.com/ca/en/products/b...ction/estimating-frictional-moment/index.html and getting some guidelines, the formula is
M = 0,5 μ P d
It seems to me you've mistaken 0.5 as the friction coefficient μ, it is NOT.

From their table, I'm going to choose "Self aligning ball bearings" since most pillow block bearings in industrial applications would fall into that category.. μ for them is 0.0010. You haven't mentioned the shaft size, for that kind of load I'm going to take a stab at guessing a 3" bearing (the d component in the formula)
So that brings me to the following
M = 0.5 * .0010 * (1500 lb * .454kg/lb) * (3in * 25.4 mm/in)
M = 0.5 * .0010 * 681 * 76.2
M = 25.9N*mm (This is far less than I had estimated in my mind)

However, is the wrapping mechanism going to put any load on the drum? that may be the largest force working against you.
 
Rx7man said:
However, is the wrapping mechanism going to put any load on the drum? that may be the largest force working against you.

+1.

What is the tension in the filter material?
 

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