Conveyor Torque 1

[seventofour]

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 that is connected to the gearbox is 11 inches in diameter.

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.

 

[3DDave]

The instantaneous load is much higher - dropping the casting onto the conveyor will be multi-G impact, so the friction will be much higher than 1. You didn't specify the support for the belt. If it isn't all rollers there is friction between the belt and whatever is underneath that will increase the load even higher.

What you may need to look at is the acceleration and power requirements, not torque/force. If they put a ramp to cause the castings to meet the belt at the belt speed there would be no acceleration, but if they just drop them and expect the belt to accelerate the castings, then you need to decide how fast they accelerate, et al.

Periodic impact systems often have large flywheels to provide the instantaneous power rather than building a gearbox/motor to do so.

 

[seventofour]

3DDave, thanks for the reply.

Well the customer got back to me, now it seems everything has changed. The roller diameter is now 14 inches, width is 4 feet, and has a length of 10 feet, zero angle, and the conveyor has to handle 10000lbs. It is a level conveyor and the castings will not be any acceleration.

Does anyone have equations to help solve for this? I have been searching the internet for them but so far no luck.

Thanks in advance to those who reply.

 

[3DDave]

Not without a diagram of how the conveyor is constructed; possibly a video.

 

[RolMec]

If this is a conveyor as in "belt conveyor", basics may be found in here (link points only to one possible source):

http://rtodds-eng.com/Published Articles/6th Edition of CEMA Belt Book Belt Con 13.pdf

But, if changing the spec., can they ask you (as the gearbox OEM) to supply their conveyor design??
However, for the sizing of a conveyor drive there's at least to be considered:
roller and / or pulley dia & bearings
belt type & strength
load characteristics (throughput, type of loading)
a sketch with (sufficient) dimensions
then
drive characteristics (i, J, n) from the applicable range of options, with required P and P-rating of drivetrain to be the design output.
So, why not come back on the original conveyor designer & supplier? Due to the higherloading, at least for the belt (if it is a belt) there should be re-consideration of its loadbering capac.

Regards

Roland Heilmann
Lpz FRG