Solving Conveyor Motor Tripping Problem for Foundry

In summary: I'm guessing they're dumping the castings onto the belt, which would raise the normal force quite a bit.The output of the gearbox is doing 29rpm. So the belt speed is 0.42m/s (1.4ft/s).
  • #1
SevenToFive
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A company called the other day with a question about their conveyor. My company did not build the conveyor but did supply the gearbox. The conveyor is 8 feet long, and 4 feet wide, it is level and transports castings to the grinding department of a foundry. The motor they are using is a 4 pole 5HP motor, assuming at 60Hz. I was told the castings get "dumped" onto this conveyor which is always turning, so there is a bit of a shock load. The foundry wants to be able to move 3000lbs of castings on this conveyor, however when they put the load on the conveyor the motor trips out. The gearbox that is on their conveyor is rated for 9000in-lbs with a 5HP motor and 29 rpm. The conveyor shaft is 11 inches in diameter and coupled directly to the output shaft of the gearbox.

I used an equation that I found but had to convert the values to metric and then back to standard.
The force is MU * Mass* Gravity, where MU is 0.5, the 3000lbs is converted to Newtons:13344.66 Newtons, gravity is 9.81
So F=0.5*13344.66*9.81
Force = 65455.6Newtons

Torque = 65455.6N * 0.2794m : the 0.2794 is the 11 inch shaft converted to meters
Torque = 18288.3Nm; 161865.2in-lbs

Am I on the right path with this? If so the current gearbox is greatly undersized.

Thanks for the help it is greatly appreciated.
 
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  • #2
What does MU stand for?

You converted 3000 pounds to 13344.66 Newtons. That is correct. But then in your F = MU*Mass*Gravity equation, you are not using the mass; you are using the weight of 13344.66 Newtons. So you end up multiplying the mass by gravity 2 times. To find the mass of a 13344.66 Newton weight, you have to divide by 9.81 to yield 1360.3 kg. If I use that as the mass in your equation, I get:
F = 0.5 * 1360.3 * 9.81 = 6672.3 Newtons.

So it appears that your "Force" result is almost 10 times higher than it should be - unless I am misunderstanding something.
 
  • #3
TomHart said:
What does MU stand for?

You converted 3000 pounds to 13344.66 Newtons. That is correct. But then in your F = MU*Mass*Gravity equation, you are not using the mass; you are using the weight of 13344.66 Newtons. So you end up multiplying the mass by gravity 2 times. To find the mass of a 13344.66 Newton weight, you have to divide by 9.81 to yield 1360.3 kg. If I use that as the mass in your equation, I get:
F = 0.5 * 1360.3 * 9.81 = 6672.3 Newtons.

So it appears that your "Force" result is almost 10 times higher than it should be - unless I am misunderstanding something.
MU was a coefficient of friction. Thanks for the help.
 
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  • #4
Well the customer called, it seems that many of the original things have changed. Now the castings will be sliding onto this conveyor that will be 4 feet wide and 10 feet long and level with a 14 inch roller. Can my same calculation as above used to determine the torque requirement?

Thanks to all of those who reply.
 
  • #5
You're working with a distributed load. It's been several years, but I recall that this is a standard model of most gearmotor manufacturer's Engineering Guides. I know SEW Eurodrive has their online sizing tool PTPilot, but probably the others have something similar (Dodge, etc.). Worst case, this is a standard calculation to be found in PDF Engineering Guides.

Offhand, 5HP seems a little light for this load situation. These calculations all deal with Peak Torque, which accounts for starting from zero velocity and accelerating to max speed. Which is a lot different from steady state torque. In any case, you must take into account all of the inertias, both translational & rotational.
 
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  • #6
If they dump/drop the castings onto the belt the normal force could be quite a bit higher than it is "at rest". Could they visually estimate how far the belt moves before the castings are up to speed/not sliding? That would allow you to calculate the acceleration of the casting and hence the force needed top accelerate them.
 
  • #7
SevenToFive said:
The gearbox that is on their conveyor is rated for 9000in-lbs with a 5HP motor and 29 rpm. The conveyor shaft is 11 inches in diameter

I take it the output of the gearbox is doing 29rpm? In which case I make the belt speed 0.42m/s (1.4ft/s). Perhaps I'm wrong but that sounds a bit fast for heavy castings. Do they really want a 3000lb casting coming off the end at that speed or do they mean there are 3000lb in total of smaller castings on the belt at once?

What exactly are they doing when it trips? Are they dropping one 3000lb casting? Are they putting 3000lb of smaller castings on the belt and then turning it on? Are they putting 9 * 300lb castings on the belt and dropping another one on?
 
  • #8
SevenToFive said:
Torque = 65455.6N * 0.2794m : the 0.2794 is the 11 inch shaft converted to meters
Torque = 18288.3Nm; 161865.2in-lbs
In addition to @TomHart 's comment, the torque is obtained by multiplying the force with the radius of the shaft, not the diameter. Thus your number is not 10 times, but 20 times larger than expected.
 

FAQ: Solving Conveyor Motor Tripping Problem for Foundry

1. What is the cause of conveyor motor tripping in foundries?

The most common cause of conveyor motor tripping in foundries is overloading. This can happen due to excessive material on the conveyor, improper belt tension, or issues with the motor itself.

2. How can I prevent my conveyor motor from tripping?

To prevent your conveyor motor from tripping, it is important to regularly maintain and inspect the motor and conveyor system. This includes checking belt tension, ensuring proper lubrication, and addressing any potential issues before they become larger problems.

3. Can environmental factors contribute to conveyor motor tripping?

Yes, environmental factors such as extreme temperatures, humidity, and dust can all impact the performance of a conveyor motor. It is important to choose a motor that is suitable for the specific conditions of a foundry environment.

4. How can I troubleshoot a tripping conveyor motor?

If your conveyor motor is tripping, you can troubleshoot the issue by first checking for any obvious signs of damage or wear on the motor and conveyor components. You can also use a multimeter to test the electrical connections and voltage, and make sure the motor is receiving enough power. Additionally, you can consult the manufacturer's manual or contact a professional for assistance.

5. What are some long-term solutions for preventing conveyor motor tripping?

Some long-term solutions for preventing conveyor motor tripping include investing in a more powerful motor, regularly maintaining and inspecting the motor and conveyor system, and implementing safety measures to prevent overloading. It is also important to address any issues or warning signs as soon as they arise to prevent them from becoming larger problems in the future.

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