Torque on an Operating Shaft

[Busman]

I am a structural engineer assisting a client in determining the cause of a shaft failure. We normally strain gage components, but this is an inaccessible rotating shaft. I can analyze the shaft and key, but need to determine the shaft torque. I have limited motor info so far (25 HP, 1200RPM), though the in service speed is 7 RPM. I have the formula T=HPx5252/RPM. It can't be that simple to determine the in service torque, is it? Do I need a chart of torque vs speed for the motor? I'll spend more time researching, thought I could get a little insight here as well. Not looking for specifics, just to be pointed in the right direction.

 

[MintJulep]

I have the formula T=HPx5252/RPM. It can't be that simple to determine the in service torque, is it?

Well, yes and no.

Just because the motor's nameplate says "25 HP" doesn't mean that it's producing 25 HP at your operating condition.

You'll be able to get motor curves from the manufacturer (although it's a good bet that they didn't bother to develop them a 7 rpm).

At any rate, you'll need to measure input voltage and current.

 

[JJPellin]

The appearance of the fracture surface of the shaft should give you a very strong indication of the cause. A shaft that fails from high torque would tend to fracture at a 45 degree angle from the shaft center-line. A shaft that failed from fatigue in bending would tend to fracture squarely with characteristic beach marks on the fracture surface. Describe the appearance of the break and attached pictures if available and you should get some good responses to help you understand the fracture mechanism.

Johnny Pellin

 

[desertfox]

Hi Busman

Yeah the formula you have will give you the maximum torque attainable excluding losses ie friction etc for 25HP.
There are clues as to the type of failure of a shaft by looking at the fracture surfaces as mentioned by others and some clues are given at this site:-

http://www.plant-maintenance.com/articles/rcfa.shtml

Now from your post I calculate the torque to be about 25000Nm which is a fair old torque.
To expand on JJPellins comments on failures, a failure at 45 degrees to the shaft centerline, can be caused by fatigue cracks growing perpendicular to the max principle stresses, generated by a shaft under torsion or by a brittle failure in a material that is weaker in tension than in shear.
A shaft fracture at perpendicular to the shaft centerline is usually indicative of a shear failure.
If you can post some pictures that would be good.

Regards

desertfox

 

[IceStationZebra]

If they are using a variable speed drive you may want to consider torque ripple. As far as actual HP goes you will need to measure the amperage and voltage like MintJulep was alluding to. If if you can't get a curve from the manufacturer you should be able to get in the ballpark.

ISZ

 

[EngineerTex]

Hang on a minute here. Are you really running a motor that is rated for 1200 rpm at 7 rpm? VFD is capable of doing a lot of things, but this sounds a little bit off.

The problem that I see with your formula is not that I think the formula is incorrect -- it is not. Rather, I think it is the wrong formula. A motor which produces 25 hp at 1200 rpm will not put out 25 hp at 7 rpm. What type of motor is this? If it's an induction motor (almost guaranteed if it's 25 hp), then you will NOT get 25 hp when you're running it at 7 rpm. When you run induction motors (and most all electric motors) below the nameplate speed, they produce a torque at or usually well below the torque produced at the motor's nameplate speed. So if you determine the torque produced at 25 hp and 1200 rpm, you're going to be dealing with that torque or less (probably a WHOLE LOT less) at 7 rpm.

But when I step back from the problem and look at what you're asking, I think that we're missing something here. My first question to you would be: is there a gearbox between the motor and the 7 rpm shaft?

Engineering is not the science behind building. It is the science behind not building.

 

[321GO]

Quote:
"Well, yes and no.

Just because the motor's nameplate says "25 HP" doesn't mean that it's producing 25 HP at your operating condition."

Ok, but the motor IS capable of 25HP, so should this not be used a upper design limit?



"If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack."
Winston Churchill

 

[desertfox]

I think the OP needs to clarify whether or not there is a VFD.
Also the motor might well not be running at full load but in the light of no other information we at least know its maximum torque available.

desertfox

 

[Ron]

EngineerTex...the motor is coupled to a gearbox....that's how it gets from 1200 rpm to 5 rpm. The motor is still running at 1200 rpm.

 

[MintJulep]

The torque vs speed curve for induction motors is not linear.

Ok, but the motor IS capable of 25HP, so should this not be used a upper design limit?

The shaft is apparently failing at much lower power, so it's probably a fatigue failure, not an overload.

 

[Busman]

Ron is correct, the motor is coupled to a gearbox (ratio 70.32) with a 15 tooth sprocket at the box connected by a chain to a 52 tooth sprocket at the shaft in question. I was told 7 RPM, but field measurement and the above info suggests 5 RPM. Beach marks on the shaft (from a stress riser at a sunk key) do indicate fatigue failure. I can't attached pics, they are in a failure analysis (material) report which I cannot provide. From the posts and other discussion, we will plan to, among other things, measure and monitor the input voltage and current for fluctuations that may be leading to the variation in torque and fatigue failure. Thanks for all of the input.

 

[PeteDB]

If you are transmitting 25HP at 1200RPM at the motor the torque at the motor is
25*198000/pi/1200 = 1313 in-lbs.
The gearbox increases that to 92,332 in-lbs.
The sprockets increases the torque from the gearbox to 320,084 in-lbs on the shaft in question.

In addtion to the torque on the shaft there is an overhung load caused by the chain tension whose value depends upon the pitch radius of the 52 tooth sprocket. The overhung load will add a cyclical bending stress to the shaft in addition to the torsional stress it sees.

 

[chicopee]

Have you tried to calculate torque within the elastic range using the torsion formula that includes torsion stress? Also where is the fracture? Is this a stepped shaft with different diameters? Any misalignment or vibration? There are many scenatios that can cause shaft failure. Send us a picture of the shaft break and a sketch showing shaft, motor, gearbox, bearing supports and drives.