OHL OverHung Load explanation

[SonicFlow]

This is just something I'm curious about. What is the Overhung Load calculation really telling us?

I can't justify the OHL calculation in my head. Why would increasing the rpm reduce the OHL (be more favorable) for a gearmotor? If one doubles the output rpm then the OHL is reduced by a factor of 2.

From Dodge gearmotor catalog and Rexnord:
OHL (lbf) = (126000*HP*Fc*Lf)/(PD*RPM).
HP = motor horse power, Fc = Load Connection Factor (1.0 for a chain on sprocket), Lf = Load location factor(1.0 for Dodge gearboxes), PD = pitch diameter.

Obviously this formula is a simplification of some factors at work on motor durability. Each gearmotor is rated for a certain Overhung load, and one uses the formula to help select the correct package.

Certainly the fatigue cycles would worse at higher rpm, so this formula must not be addressing that issue. What useful info does it tell us?

 

[MikeHalloran]

The formula gives you the overhung load limit/rating/maximum, based on the motor's HP rating.

It does not tell you what the overhung load actually is for a given situation.



Mike Halloran
Pembroke Pines, FL, USA

 

[SonicFlow]

MikeH,

That kinda seems like how it should work. In other words you are saying we are aiming for a larger number, or higher rating as output from the equation, if I understand your point.

However, that doesn't match what the 2009 Dodge Quantis catalog is saying in the How To Order section (p.15).
"The calculated OHL must be less than the allowable OHL."

http://www.baldor.com/support/CA1603catalog.asp

They are suggesting this equation as a way to check an application, and compare it to their gearmotors. For example a 1 hp right angle with an 88 body size has 4512 lbf listed as the allowable OHL.

So we prefer a smaller number in the calculated OHL.

Thanks.

 

[MikeHalloran]

The equation given produces an 'allowable OHL'.

They must be expecting you to come up with a 'calculated OHL' using a different equation, e.g. one based on the kinematics and kinetics of your particular application.





Mike Halloran
Pembroke Pines, FL, USA

 

[dvd]

What you need to recognize is that the reducer size is not staying the same. For example, a 10,000:1 ratio reducer may be powered by a fractional horsepower motor but have an eight inch diameter output shaft. While a 10:1 ratio reducer with the same fractional horsepower motor might have a 5/8" diameter shaft. The torque being transmitted by the output shaft is the reducer ratio times the motor torque. The ohl for each respective reducer is the the output shaft torque divided by the pitch diameter/2.

 

[PeterCharles]

Typically you get an overhung load on a gearbox output shaft if the gearbox drives a machine through a transmission chain drive.

You work back from the absorbed power of your machine to get the tension in the chain. Then you check the gearbox overhung load rating from the manufacturers catalogue to see that it exceeds the chain tension. Oh, and make sure that you make allowance for the position of the chain drive if it's centreline is not where the gearbox manufacturer has specified his overhung load will act.

If your overhung load exceeds the gearbox rating (not uncommon even if the absorbed power is within the gearbox power rating) then expect failed output shaft bearings, even failed output shafts!

 

[SonicFlow]

Thanks for the replies. dvd and PeterCharles you've helped me to picture the problem better.

This is the first time I've sized a gearbox (~220:1 reduction to a chain driven sprocket. Probably a 2 hp motor. Only 8 rpm output.) The 2nd equation [OHL = torq/(pitch radius)] definitely makes more intuitive sense, but when you start out not knowing the torq you can use the 1st OHL equation I guess.

And I think I get the presence of the rpm denominator in the 1st OHL equation now. If you have a high output rpm (most of the input values will be 1750 rpm) then you don't have much reduction and as much torq on the output shaft.

 

[Acajun]

Actually the formula for calculating OHL is OHL= 2* Torque/Pitch diameter of sprocket * Transmission element factor.

Transmission element factors are the following:
Gears < 17 teeth = 1.15
Chain sprokets < 13 teeth = 1.40
Chain sprockets < 20 teeth = 1.25
Narrow V-Belt pulleys = 1.75
Flat Belt Pulleys = 2.50

Also torque is calculated by multiplying the motors output torque by the reducers ratio.

The units OHL rating is calcultated at the output shafts mid point. If you are beyond the mid-point of the output shaft then another set of calculations are required.

Hope this helped.

Jp

 

[electricpete]

If you have a high output rpm (most of the input values will be 1750 rpm) then you don't have much reduction and as much torq on the output shaft.

Yup that's it. Assuming hp is treated as a constant or input to the equation, which it is.

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[PeterCharles]

Acajun,
Can you please explain the derivation of the "Transmission element factors" you have quoted. And surely the output torque is determined by the driven machine not the motor unless the required machine power is equal to the motor power??


Also, my experience is that the position of the overhung load is where the manufacturers catalogue says it is, not always at the mid point of the output shaft.

Regarding the gearbox output torque, in general terms if the output speed is LESS than the input speed then the output torque is GREATER than the input torque, and vice versa. Of course as I said previously the controlling factor is the torque required to drive the machine.

 

[Acajun]

"Transmission element factor", also called fz, is a constant that is used to describe the friction between two elements. ie. chain and sprocket, v-belt and pulley etc.

If you have a motor that develops 100 in-lbs of torque @1750rpm coupled to a speed reducer with a ratio of 100:1 you will have 10,000 in-lbs of torque available on the output shaft of the reducer @17.5rpm. Even if the required driven torque is less that what is available I've always calculated using worst case, you have to remember that the start-up torque of the motor is always higher than the operating torque especially with a dynamic load.

The reason that OHL is calculated at mid point is that this is usually where the sprocket or pulley is installed. As you go further out on the shaft then you have a lever effect that increases the radial force on the shaft, output bearing etc. Remember that the unit for OHL is pounds.

Actually the reason I calculate OHL when I'm designing an application is to know what the minimun size of sprocket or pulley to use is.

What I do is take the catalog OHL rating of the reducer I select and use the following formula to get the size of the sprocket. D=2T/OHL*fz
D=Diameter
T= Torque
OHL = Overhung Load
fz = Transmission Element Factor.


I hope that I was able to answer your questions.

Jp

 

[SonicFlow]

The "Transmission element factor" is also referred to as the "Load Connection Factor".

The Dodge Quantis Gear Engineering Catalog gives some slightly different values than what Acajun showed.
Chain Drive: 1.00
Spur or Helical Gear: 1.25
Synchronous Belt Drive: 1.30
V-Belt Drive: 1.50
Flat Belt: 2.50
Same general pattern. I figure we must keep a flat belt tight or it will slip, adds to forces. I imagine the derivation of the factor is empirical, and pretty rough.

Also I agree with Peter Charles ""the position of the overhung load is where the manufacturers catalogue says it is, not always at the mid point of the output shaft."" and most are going to say put it as close to the base of the shaft as you can. But standard gearbox output shafts are pretty short, so midpoint is not far off the closest position (given sprocket thickness).

"locating the overhung load close to the seal retainer will minimize the overhung load and could turn a potential failure into a successful operation. "

http://pt.rexnord.com/customer_support/technical_information/tech6.asp

If you have a really long gearbox shaft you could support it near the sprocket with another bearing to reduce the OHL, no?

Thanks for the posts everyone.

 

[PeterCharles]

Acajun
The reason I asked for the derivation of the Transmission Element Factor was that I could not understand how for a chain drive it could be anything other than 1. I suppose you took it from a publication somewhere as no derivation is quoted.
The value of 1 that Sonicflow quotes for a chain drive is something I can accept. As for the flat belt factor this is easily derived since the tension ratio across the pulley (T1/T2) = e^mu*theta at the point of slip, mu being the coefficient of friction between the belt and pulley, and theta being the angle of wrap. The spur gear factor must allow for the pressure angle which effectively reduces the PCR.