Need pointers for calculating 6061 shrink fit 4

[JimLou]

Hi! I'm new here, and a shade-tree mechanic.

My current project is to shrink fit a sleeve over a drive pulley hub. Both parts are 6061 T6. The hub has an I.D of 1.6" and an O.D. of 2.1". The sleeve will have an O.D. of 2.88" with 0.10" deep serpentine belt grooves in it.

In my copy of Machinery's Handbook are formulas for calculating the shrink fit allowance for steel and cast iron components, but not for aluminum.

Can someone please point me in the right direction? Thanks!

 

[GregLocock]

http://en.wikipedia.org/wiki/Coefficient_of_thermal_expansion

aluminium expands about twice as much as steel, and the pressure after the shrinking for a given interference will be about 1/3 as much.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[JimLou]

Thanks, Greg! So should I figure on about 3 times the interference that I would use for steel parts? That seems too simple.

I'll go check the FAQ.

 

[JimLou]

Thanks, Greg. That Wikipedia link was what I needed.

The FAQ suggests that I don't know enough to be here, so I'll just lurk until I get smarter.

Thanks again.

Jim

 

[MikeHalloran]

I assume there is some overbearing reason why you can't just make it a one- piece part.

Rather than relying on the press fit to transmit torque, I'd make it a light press or transitional joint, and Dutch pin it.


It takes a long time to get smart by lurking. Asking your own questions is much faster. Go ahead; nobody here will intentionally insult you.



Mike Halloran
Pembroke Pines, FL, USA

 

[JimLou]

Thanks, Mike!

The reason I don't want to make it as one piece is that it's a single complicated part that I have to turn on a manual-control lathe. There are two multi-groove serpentine belt sheaves and a very precise press fit onto a 5/8" driven shaft. That bore is deep in the part and would be very difficult to do with my machines and tools.

My plan is to turn off one set of grooves, shrink fit the sleeve over it and cut a new set of grooves in the sleeve. That's going to be significantly less work than cutting two sets of grooves, boring the shaft hole, and making all the other, non-critical features of the part.

By 'Dutch pin' do you mean pins inserted axially into the joint? Good idea! But why not shrink-fit and then pin it? Either a shrink fit or a light press fit is of similar difficulty with my machines.

I could also TIG weld it, but don't want to mess up the metallurgy. The last couple of grooves would be in the HAZ, and I don't want them annealed. The added cost of having the part heat treated would price it into the range of making a complete part from scratch.

This is a drive for a supercharger. It's a tight press fit onto the alternator shaft with an axial retaining bolt. The driven sheave is next to the alternator. The sheave that drives the supercharger is outboard of the driven sheave, and is currently the same diameter. It can't transmit quite enough power, so I'm increasing its diameter by 5/8".

The only redeeming feature of the setup is that the belt tensions are applied in opposite directions, so that the load on the alternator shaft and bearing is minimized. The arrangement is far less than ideal, but it's what my customer has to work with.

Thanks again!

 

[MikeHalloran]

Go ahead, shrink fit it if you prefer.

Yes, I mean axial pins through the joint, so you're not carrying torque with the shrink alone.

Instead of smooth pins, I'd use setscrews, so the joint can also carry some axial force incidental to pressing the pulley on the alternator.

Next time you have a lot of machining to do, consider aluminum alloy 2024.



Mike Halloran
Pembroke Pines, FL, USA

 

[JimLou]

Thanks again, Mike! Another good idea with the screws.

I almost always use 6061 for machining because it's usually available in the configuration I need. My supplier in St. Louis carries lots of 6061, but only a few 2024 square bars and a modest inventory of sheet and plate. Round stock I would have to get mail order. For a single part, which is almost all of my business, the difference is usually not worth the bother and expense. 6061 cuts slowly but very well with the fluid and tooling I use.

When you're designing a part for production, you have a lot of options available that I, as a one-man job shop, just don't have. But then I have the luxury of seldom getting close to the limits of the material. If in doubt, I can usually add a little more.

 

[IFRs]

I'd be concerned that a shrink fit will really hold up in your environment. Can you make splines or a key or a square hole / square bar?

 

[JimLou]

Splines would involve more work than making a complete part. And there is very little wall thickness on the sleeve - 1/8" to 3/16" between the ID and the minor diameter of the sheave - I'm still working out the details. I fear that splines or keyways would weaken it too much.

I would think that the shrink fit would be pretty secure unless the belt slips. That would heat the sleeve more than the hub and loosen the joint.

 

[desertfox]

Hi JimLou

Shouldn't the shrink fit be capable of taking the required
torque without slipping? you make no mention of the power and hence the torque your trying to transmit your basis for an interference fit should based on this.
Secondly will the joint be subject at any time to oil ingress which would reduce the friction in the joint and make it slip easier under load.

Regards

desertfox

 

[GregLocock]

"So should I figure on about 3 times the interference that I would use for steel parts? That seems too simple."

It is. That would give you the same stress as in the steel part... so you may crack the aluminium one. Getting a good press fit in aluminium is not as straightforward as it might seem.

Here's a link to a calculator faq404-1230



Incidentally that comment in my sig isn't aimed at the you, it's there for everyone. Including me!

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[JimLou]

Unfortunately I have no way to get the power requirement. Trying to calculate it from the RPM and displacement of the supercharger isn't practical because there are lots of unknowns.

I could build a belt tensioner with a strain gauge on it and calculate torque from that, but again, that's more work than making a complete pulley.

As for oil entry, that won't be a problem; it will probably never be exposed to oil, but it will get water spray, and might get detergent or other cleaning agents.

That calculator is interesting. I'll run the dimensions through it later and see what it says. BTW, so far my experience tells me that I want .002" to .003" interference. Now's my big chance to splatter egg all over my face.

 

[rb1957]

i don't quite know why we started discussing thermal expansion, [maybe it somes form the consideration of freeze fitting the parts together] the pressure resulting from an interference fit is due to E (and that's where i see the 1/3rd ratio coming in, Esteel = 30Mpsi, Eal = 10Mpsi).

i think greg's later post is quite correct, in that if you install an Al fitting with the same interference as a Steel fitting, i think you'll likely see plastic stresses in the Al, which are probably undesirable. note, the 1/3 ratio applies only if both parts are elastic.

0.002" to 0.003" interference on a 2" diameter doesn't sound like much.

i'd worry about the groves, the wall of the sleeve is 0.39" thick, with a 0.1" grove in it, with a bunch (that's not a mathematical term) of tension stress from the interference.

 

[JimLou]

If I'm using it properly, the calculator linked above says that at .003" interference it will generate 295 Nm of torque. With the ratio between the engine and the driven sheave, that is almost 150% of what the engine puts out.

It also says that I need about 44 degrees K differential to get it together. I think. It shows the inner part at -44 degrees K and the outer part at +44 degrees. If it's 'or' that's about what I got earlier by completely unrelated calculation. If it's 'and' I'm totally screwed up.

As for the mechanical implications, I don't believe that crushing the inner part is a significant concern; it will see only compressive forces and is well shaped to resist them.

Deforming the outer part is a concern, though. The calculator shows stresses on it from 6 to 9 MPa, or a max value of 11.7K psi. My reference shows yield strength of 40K psi for 6061T6.

This sure has been a fun learning experience, and the numbers make sense to my feeble mind. But if anyone would care to check my work I'd appreciate it.

 

[rb1957]

not quite sure i like your usage of "generate", in "an interference of 0.003" will generate 295 inlbs of torque". i think you mean "will absorb" or "react" or "withstand".

does 150% mean your motor generates 200 in.lbs *1.5 = 300 in.lbs or 100 in.lbs (200 in.lbs is 100% more than 100 in.bs; 150% more is 300 in.lbs) ?

i don't know that the calculator stresses take the groove into account.

my data (mil hdbk 5) says ftu = 40 ksi, fty = 36 ksi for 6061T6.

is 9 Mpa = 12 ksi ? i always go back to 14.7psi = 0.101325 MPa so that 1ksi = 7 MPa ... maybe there's a decimal out of place ?

 

[JimLou]

Agreed that generate is a bad word here, but you understand my meaning correctly. "Withstand" is a much better choice.

And I meant foot pounds of torque. Dope slap self-administered. I've been translating metric and American values back and forth until my head spins. I deal with metric terms too little to have a feel for what a million Pascals or a hundred Newton-meters is.

The engine has put out 430 foot pounds on a chassis dyno, and there's about a 3:1 ratio between the crank pulley and the driven alternator pulley. Allowing for the loss between the engine and the chassis dyno, 150% is close enough. In any case there is no way that the sleeve will see enough torque to make it slip; the drive belts can only transmit a fraction of what's available anyway.

For the stress calculation I used the minor dimension of the grooves, so the those numbers should be conservative.

As for being a decimal off, absolutely a possibility; as I said, I'm not used to dealing with these units, and could have very easily dropped a 10 in the translation.

 

[unclesyd]

If you are talking any RPM I definately would run a few calculations of the affects of Spin on the outer cylinder especially being Aluminum. The small radius helps but from my understanding you will have a relatively thin section in the outer cylinder.

 

[JimLou]

OW! I'm getting brain-cramps and you throw that in!

Good point! Its max speed could be 18,000 RPM or a bit more. I'll need to sheck the engine data and confirm the drive pulley size; a 3:1 ratio is just a reasonable guess. The section should be about .26" thick, not counting the larger diameter of the ribs.

I think I know where to look for a calculator.

Thanks!

 

[rb1957]

thx for your clarifications.

a word of caution, calculating the stress based on the net area (the cross-section through the base of the groove) is not as conservative as it appears. the stress concentration peak at the base of the groove will be higher than the thickness ratio, if you can see what i mean. but it's probably not a problem, given the very loaw stresses you're anticipating.

if i read the posts correctly, i think you've considered adding shear pins ... this would be a much better way of transmitting torque, but harder to accomodate with your groove.

thinking about your stress, maybe you don't have enough interference ?

as i understand things, you've got a 450 ft.lb motor working at 6000 rpm. there is a step-down so that the hub is expected to see 150 ft.lbs of torque at 18,000 rpm. 150 ft.lbs is equivalent to a couple of 1200 lbs (acting on an arm of 2"). the shear area supporting this force is pi*d*w, about 6w, a shear stress of 200/w psi, w maybe 1" (or more), shear stress = 200 psi, not much.