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Estimating stresses in a aluminium housing for an electric motor with cooling fins

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Sondrev

Mechanical
Nov 29, 2017
4
Hi

I am trying to estimate the stresses developed in an aluminium cylinder exposed to an internal pressure from shrink fitting it on to a steel cylinder. For reference think of a housing for an electric motor.
The outer tube was heated to 120degC under assembly and the steel core was lowered in without any issues. The yield strength in the housing is stated to be 200Mpa but we experienced a rupture when the parts cooled.

I have done a hoop stress calulation and checked it with using stress-strain relation (Sigma = E*epsilon) and both give approx. 120Mpa stress. Here i have assumed that the outer circumerence of the tube is smooth and 640mm.

Inner diameter = 580mm
Outer diamter (circular part) = 640mm
Outer diameter fins = 780mm
the thickness of the fins is equal to the section without fins
The size of the interference fit is known. Approx 0.7mm over the diameter
¨

The question is: Does the fins affect the stresses developed in the "non-fin" part ? In other words: would the stress on the inner surface be smaller if the tube was constructed with a uniform cross section of ((640-580)/2 = 30mm) rather than having a variable cross section of 30mm and 100mm ?


Cheers.





 
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The unfortunate lack of pictures is going to create the need for dozens of questioning responses, and possibly drive off some helpful responders before they even begin.
Specifically good clear close up pictures of the fracture and the fracture surface would move the discussion along nicely.

.7 mm / .028" diametral interference. Do you need that much for your function?
Did the fracture occurred in a jagged line between 2 fins ?
Is the finned housing a sand casting?
The thick/thin geometry I wistfully presume your part has could cause geometrical stress concentrations.
That same wistfully presumed geometry could cause immense casting quality issues.

3 down, 17 to go.

 
A shrink fit is usually 0.006 to 0.010 Often, as little as 0.002 to 0.004 Your design is metric, but 0.002 tolerances are possible in metric-machine shops.

That 0.028 is interference fit is likely the probable.

The motor housing primary stresses - WITHOUT the shrink fit! - are to be taken by the steel motor housing right? The startup torque, running torques, vibration, weights, bearing loads from the load and shaft - everything, right? I'd expect that 0.028 shrink to have fouled up the bearing alignment or bearing cups as well.
 
Thanks for the replies.
My first post so I will take that into concideration the next time.

Regarding the lack of pictures/information about the fracture:

I didnt want to move the discussion into finding out what caused the fracture. My main objective was to find out if my approach to the strength calculation of the housing was correct or not. We are hopefully going to do this again some time and it would be nice to have confidence in the calculations that has been done.

This is the formula that was used to estimate the surface pressure from the shrink fit. But I am, as initialy stated, a but uncertain to what effect the fins have in this case.

We Choose an interference of 0.7mm to make sure that we obtain a shrink fit when the temperature of motor rose to operating condition of around 70 degrees. Since the aluminium expands 3 times that of the steel core (assuming uniform temperature)

Regarding the fracture:
I have attached a photo of the fraacture and SEM taken of the fracture face. The conclusion is that there is a casting defect. My knowledge of fracture mechanics is limited, but that SEM photo does not look like anything i have seen before....

Fracture:
SEM:






 
It looks like a powder metal part. The fins will affect the elasticity of the part and cause the strain to be concentrated in the gaps between the fins. By eye, I'd estimate that this caused the stresses to be doubled to tripled over what would be expected in a smooth cylinder that would result if the fins were machined off. Then there are cutouts which also serve to increase the stress concentration by allowing the related groove areas to see even more strain than those areas without cutouts.
 
You can't overcome the difference in CTE with a shrink fit.
When you shrink fit you have to yield the Al, it will then expand from its new size.
After enough cycles the Al will be loose on the steel.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube
 
I had not seen that problem: With two different materials with such a great thermal expansion difference, the Al (outside) will always be pulling away from the steel inside. It will not be a long-lived solid fit. (If the Al were inside, the two metals might work.) But 0.028 interference is still too much.
 
I suppose it matters how much temperature difference from nominal there is, plus the contribution of the temperature gradient over the structure when heated. It might be a low temperature soak that causes the largest problem. With powdered metals, the properties are so process dependent that understanding them allows only gross generalizations.
 

I don't think that it is a powder metal. The Company we hired in for the manufacturing uses pure alumunium where they add the alloying elements themself to obtain the correct material composition. We checked the composition and found it to be similar to AL-6063. They did a stress test on a sample from the same cast and it had a yield strenght of 245MPa.

We normally buy machined components and they don't have cooling fins and this approach with shrink fit has worked just fine for us. When we get the equuipment in for refurbishment after 10 years of operation we still have to use heat/force to extract the core from the housings. Although, i agree that it is too large in this case.

I dont see why the Aluminium has to yield, as long as we are in the elastic range of the material? To some extent there will be some yielding occuring, but we wont loose the whole shrink fit.

Does anyone have of you know of some sourch of knowledge on calculating stress in a variable cross section pipe under stress?
 
That fracture surface is not from a cast material or that material has huge amounts of porosity. Is that the wrong photo?

Did they also do a stress-strain curve on the material? Without knowing the modulus you can't make any useful calculation.
 
The photo is taken from a report created by the 3rd party inspecting the fracture for us, so yes it is. We were able to peel chips of the material on the fracture surface so it's safe to assume that it was very porous...

They did a sample parallel to the casting in which they have stated the strength of the material.

 
Fix the porosity problem before worrying about the strain.

A stress-strain curve is still needed to calculate the expected stress. Tensile yield strength by itself is useless.

Is there an alloy it was supposed to be made of?

3 down, 17 to go.
 
If you keep the shrink light enough to not yield the Al, then my guess is at hot conditions the parts will be loose. Of course when cooled they will again be tight. Even if the Al yields at hot conditions it will be tight at RT.

Run the numbers and see. One of the problems with trying to hold this fine balance is that it forces you to work to tighter tolerances which drive up costs. If you want the parts still tight at 70C then you need to figure out the sizes at that temp and see if you can get an elastic fit that will work.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube
 
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