Max. torque on a shaft with an unbalanced load

  • #1
I'm new to the forum and hope someone can help me out. I've attached a picture so hope it works.

I have a ladle for pouring metal, the mass of the ladle itself is 60kg and the max. mass of material that can fit in the ladle is 90kg. Total mass 150kg. The ladle has a tapered profile, so the top diameter is 340mm, bottom diameter is 270mm. The overall height is 460mm. The ladle sits in a ring and this ring is attached to the end of a shaft. As the shaft rotates it pours the ladle in a teapot fashion in the direction of rotation of the shaft. The point of pivot on the ladle is centred in the x-axis and 150mm from the top of the ladle in the y-axis.

To find the torque applied by the ladle on the shaft can I treat the ladle like a beam, ie. treat the total mass as a distributed load, and therefore being able to calculate a point load and position. Then use this position as my distance from the shaft and apply the point load.

I was going to treat the change in diameter as negligable as it is very small compared to the height of the ladle.

The purpose of this is to select an appropriate size shaft and motor/gearbox to move it. I hope i have been clear and thanks in advance for any help.


  • ladle and splined shaft 2.bmp
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Answers and Replies

  • #2
All you need to do is find the center of mass of the ladle + fluid for different angles, If you are pouring slowly the surface of the fluid will be horizontal, but finding the volume and CM of the fluid will be hard unless you have some CAD software.

The torque is the just the weight times the horizontal offset of the CM from the rod.
  • #3
WallyLux: Yes, you can compute an equivalent point load and position (CG). You can use an equivalent nontapered cylinder, which would have an outside diameter (OD) of 305 mm. You could say the molten metal became solidified in the ladle. The maximum torque will occur when the fluid level is even with the ladle holding ring (shaft centreline); i.e., you fill the ladle only up to the shaft centreline. Then the molten metal inadvertently partially solidifies, and you rotate the ladle 90 deg. This gives you a shaft maximum torque of T = -cg*(m1 + m2)*g = -(-114.4 mm)(60 kg + 67.35 kg)(9.81 m/s^2) = 142 900 N*mm, where cg = centre of gravity y-coordinate location relative to the ring (negative means below the ring), m1 = ladle mass, and m2 = molten metal mass. (Notice the ladle is filled to only about 75 % of its capacity; hence, m2 = 67.35 kg.) The simultaneous, vertically downward shear force on the shaft is V = (m1 + m2)*g = 1249 N.

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 60 kg, not 60kg. See the international standard for writing units (ISO 31-0).
  • #4
AlephZero: The pours are relatively fast so it does make it difficult. I use Solidworks but haven't really used Fluidmotion much. Do you know if this software capable of what you are suggesting.
  • #5
nvn: That is along the lines of what I was doing but where I'm getting stuck is deciding what load to use. Do I assume that the entire mass of the ladle and material (whether it is full or just filled to the ring) acts at the CG or can I calculate a point load from the 'distributed load'. ie. (assuming the ladle is full) Rotate the ladle 90 deg so that its longest section (its height) is now horizontal.

x (length of ‘beam’) = 0.46 m
L (distance from left hand side that R, the reaction force (the shaft) applies) = 0.15 m
w (distributed load) = m1 + m2 = (60 kg + 90 kg) * 9.81 = 1471.5 N
P (point load) = w.x = 1471.5 * 0.46 = 676.89N acting at, 0.5x (CG)

Distance from R to CG = (0.46/2) – 0.15 = 0.08 m

So now I have 676.89N @ 0.08 m
T = 676.89N * 0.08 m = 54.15 N.m
  • #6
WallyLux: Yes, you can rotate the ladle 90 deg, or less. The ladle mass is not really uniformly distributed along the ladle height. Instead of assuming it is uniform, you could compute the correct ladle CG, if you wish. For the fluid, continue to do it like you did, for now, and show us how it turns out. Last, you made a mistake when you computed P. The point load P is w; i.e., the entire weight, w, acts at the system CG.
  • #7
Thanks for clarifying that, it's been bugging me that my load force suddenly halved.
Thanks for everyones help and I'll get back to you soon.
  • #8
Just an update, and a result that I’m so far much happier with.
I haven’t done a CAD model to analyse the flow yet but I did build a much more accurate model of the ladle in solidworks, and assumed the material in the ladle was solid. I cheated a bit and used the CAD model data to show me the CM, this allowed me to do a few iterations with different material volume much quicker than calculating it by hand. From this I determined that the maximum torque occurs when the ladle is about 50-60% full, where the surface of the material in the ladle, when the ladle is upright, is about 50 mm below the line of the shaft.

Another mistake I found in my calculations was the mass of the empty ladle, which I have now corrected.

If you’re interested my results at max. torque are now:
Mass of ladle = 101 kg (big difference, I know)
Mass of material = 85.6 kg
CM occurs at 232.9 mm from the bottom of the ladle, or 156.1 mm below the shaft axis.
Giving me a load of 839.7 N, with working radius 156.1 mm

T(max)= 839.7 N * 0.1561 m = 285.6 N.m
  • #9
WallyLux: I want to make sure I understand the updated problem. Are the dimensions of the ladle still the same as listed in post 1? And are those outside dimensions of the ladle? Or inside dimensions? What is the wall thickness of the ladle? What is the density of the ladle material? What is the density of the fluid? Is the ring and shaft centerline still 150 mm from the top of the ladle? Is the shape of the ladle still as shown in your attached image? Is the ladle wall thickness uniform?
  • #10
Sorry i got ahead of myself and forgot to state the changes I've made. Since the first post the criteria has changed. I'm now designing for a 180 kg ladle instead of 90 kg, just a small increase. They are the new outside dimensions. The pivot is now 195 mm below the rim of the ladle. The wall thickness of the shell of the ladle is 5 mm and the thickness of the packing inside it is 42mm. Total wall thickness of the ladle = 47 mm. Shape is the same, its portionally larger and wall thickness is uniform.
  • #11
WallyLux: What are the outside dimensions of the new ladle (outside top diameter, outside bottom diameter, and outside height)? What is the density of the ladle material, the density of the packing material, and the density of the fluid? And what are these three materials?

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