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Torque Required to Rotate a Drum

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  1. Mar 25, 2015 #1

    BSZ

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    I need to figure out the torque required to rotate a drum. The drum is being rotated on its perpendicular axis. I am designing a drum dumper so the drum starts upright and is rotated 135 degrees to dump its contents into a hopper in the floor. I want it to take about 5 seconds to do the rotation at about 30rpm. The pivot point is about 7.5" up vertically off center. I thought I was using the right equations but my answer is not logical. Thanks!

    Mass 600lbs
    Outer diameter 17.5"
    Length 26"
     
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  3. Mar 25, 2015 #2

    SteamKing

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    Well, are we supposed to guess what you did? Why your calculations are "not logical"?

    If you want help, give us something to review. :smile:
     
  4. Mar 25, 2015 #3

    BSZ

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    Sorry my thoughts were to see what formulas you guys came up with before sharing mine. I attached my calculation sheet. The top half is my motor and gear box calculations.
     

    Attached Files:

  5. Mar 25, 2015 #4

    jack action

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    First of all, you cannot rotate 135° at 30 rpm in 5 s. 135° is 0.375 revolution (= 135/360) so it will either take 5 s at 4.5 rpm (5/60*4.5 = 0.375) or take 0.75 s at 30 rpm (0.75/60*30 = 0.375).

    To find the torque needed, you have 2 possible sources for resistance: Inertia when accelerating and bearing friction.

    For inertia, T = Iα.

    For friction, you need to define the friction torque (from the bushings or roller bearings), http://www.skf.com/group/products/bearings-units-housings/spherical-plain-bearings-bushings-rod-ends/general/friction/index.html [Broken].

    Usually, in cases like yours, the friction torque is a lot larger compared to the acceleration torque, so the later can be neglected.
     
    Last edited by a moderator: May 7, 2017
  6. Mar 26, 2015 #5

    BSZ

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    I wrote down the wrong RPM. The 30rpm is the output of the gear box I am looking at, the out put of the driven sprocket is about 9rpm which will be turning the drum. I want to make sure the drive system that was picked has enough torque to turn the drum. But it looks like I will have to pick out a less rpm motor. Did you have a chance to look at the calculation page? Thanks!
     
  7. Mar 26, 2015 #6

    jack action

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    It is not that your calculations are wrong, it is the fact that you go backward.

    You define the torque and rpm you need for your system to work. This will give you the power required by your motor. If you have to add other components (gears, belts, chains, ect.) in between to adapt the torque-rpm relation between your system and your motor, you will also need to consider the power losses due to these components (just like you did), so a bigger motor will be needed.

    So you begin at your system and end up with your motor, not the other way around.
     
  8. Mar 26, 2015 #7

    BSZ

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    I was just typing out calculations for motors, gear boxes and sprockets while I tried to figure out why my calculation come out with 3 ft*lb to rotate this drum. I am going to motors, gear boxes and sprockets as needed. Alpha is what is going to change with different mechanical components. Just don't know if I am using the wright formulas. Looking for some guidance with what formulas I need to be using. Thanks
     
  9. Mar 26, 2015 #8

    jack action

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    I have to say that nothing appears below «Torque Required to Turn Drum» when I open your file (If there was suppose to be something), but every calculation above that is OK.

    The formulas that should be used after that are the ones I presented in my previous post. They will tell you the combinations of P and α you can handle with that motor.
     
  10. Apr 2, 2015 #9

    BSZ

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    The thing is I know it is going to take way more than 3 ft-lb to rotate the drum. I am rotating it on the perpendicular axis not the parallel axis. I haven't bothered with figuring out the extra torque need due to friction because the amount of friction from the bearings is minor. That is why I feel that my equations or calculation are incorrect. Thanks!
     
  11. Apr 2, 2015 #10

    jack action

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    Then use the perpendicular axis:

    320px-Moment_of_inertia_solid_cylinder.svg.png
    e7d1b33e1e2a39ed115b6d7a3adb4a35.png
    (source)​

    The real challenge in you case is that the moment of inertia is constantly changing because the liquid moves around as the drum is turned; it even gets out at one point.

    As the drum turns, you have to recalculate the moment of inertia:

     
  12. Apr 2, 2015 #11

    Baluncore

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    So BSZ, you are going to empty a drum. Simple, but;

    1. What material does it contain, liquid or solid?
    2. What is the viscosity, or the rest angle of the material surface?
    3. What mass, volume or level of material is in the drum at the start?
    4. What is the mass of the empty drum?
    5. Is the drum completely open at the top end, or is there only one hole and a vent?
    6. Might you rotate it beyond 90° before opening/release, to reducing the torque needed by better content balance?
     
  13. Apr 6, 2015 #12

    BSZ

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    1. What material does it contain, liquid or solid? Solid
    2. What is the viscosity, or the rest angle of the material surface? 30 degrees
    3. What mass, volume or level of material is in the drum at the start? drum is full, 375lbs
    4. What is the mass of the empty drum? its a fiber drum so only a couple of pounds
    5. Is the drum completely open at the top end, or is there only one hole and a vent? completely open
    6. Might you rotate it beyond 90° before opening/release, to reducing the torque needed by better content balance? The drum is in a carriage and the carriage has a spout on it help control the flow of material.
     
  14. Apr 6, 2015 #13

    Baluncore

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    At each point during rotation, the position of the Centre of Gravity must be calculated for the appropriate content geometry. The mass and the horizontal offset of the C of G from the pivot generates a torque that must be opposed. During the rotation, the C of G of the mass is also being lifted relative to the pivot and therefore energy must be expended. The supply of that energy will require an additional torque.

    You can break the rotation down into four distinct geometric phases. The first thing to note is that the material will not begin to move until the drum has rotated through the rest angle of 30°. Secondly, note that being 26” high and 17.5” diameter, the drum diagonal angle is Atan( 26” / 17.5” ) = 56.06°, lets call that 56°.

    Situation A. 0° to 30°.
    The drum begins in a balanced state, but as the C of G starts to move off centre it generates a torque due to the unbalanced cylindrical mass of material in the drum. At the start of that initial period, the acceleration of rotation of the eccentric mass of the drum and contents must be considered. The total torque is the sum of those two torques. Choice of initial acceleration could keep that component less than the torque expected later from the unbalanced mass.

    Situation B. 30° to 86°.
    At 30° the material starts to flow. The simple geometry then changes from a cylindrical mass to have a diagonally cut upper end surface. The C of G continues to move and the mass is progressively reduced as rotation angle increases until the geometry changes again when the intersection of the rest surface begins to cut the base of the drum. That is at 56° + 30° = 86°.

    Situation C. 86° to 120°
    During this period the diagonal flowing surface intersects the flat base of the drum. This geometry will hold until the drum reaches 90° + 30° = 120° at which point it should, given time, be empty.

    Situation D. 120° to 135°.
    You continue rotation of the emptying drum to 135°. I have ignored the less important dynamic forces and the time lag or delay due to material flow.

    Now you can plot the sum of all those torques against angle and look for the maximum. It is really just a problem of diagonally truncated cylindrical sections.
     
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