Shrink Fit

[Kpar]

Hi all, I'm about to have a go at heat shrinking a brass ring onto a cast iron wheel. I need some advice before I start.
Brass ring is 5.5325 ID x 6.4300 OD x 1.3265 Wide.
Two questions,
1. what diameter should I machine the cast, 004" - 005" larger than then the ID of the brass ring?
2. what temp and heat source to expand the brass ring, will oven temp be enough ?
Thanks for any input
Kpar

 

[jack620]

Hi,
while researching many websites on shrink fits (in steel) I came across the following rule of thumb:
allow 0.025mm (1 thou) per 25mm of diameter for a shrink fit
I heated the outer piece to 200C in my kitchen oven and cooled the inner piece in the freezer. I still needed to give the inner piece a tap with a hammer to make it fit.

Be aware brass has roughly double the coeff of expansion than steel. The table below gives linear coefficient of expansion for various metals. I assume you would need to divide your calculated linear expansion by pi to get the increase in diameter of the ring.

 

[Kpar]

Thanks Jack, maybe I should do a test and measure the ID after heating. In the Webber Q if the wife lets me.
Kpar

 

[jack620]

No worries. I made some edits while you were posting. I reckon a test is a very good idea. The Q sounds like the perfect place to do the heating.

 

[Richard Carlstedt]

Forgive me for using Inch and Fahrenheit as a guideline, but I did this extensively when working .
The basic formula for Steel is called 5 ball 63 so it is easy to remember ( 5 ball means 5 zeros) .
That means that steel expands at the rate of .0000063 per inch per degree F.
Now that is tough to handle by most machinists so they start by multiplying by 100 degrees , which really is moving the decimal over two places , so 100 F change results in .00063, a more manageable number for some folks .
Your steel is 5.5325, so that times .00063 will show the material growth for each 100 degrees --which is .00348 " ( Per 100 F)

Now Brass has a much higher rate of expansion and is 5 Ball 99 which if you run that number , you will get .005468" for the ID of the brass ring. Now please know that this number or relationship is From Ambient Temps.

So you can see it only takes 100 degrees F more for the Brass to be .005" larger or drop on the steel disk.
First, it is common to heat the outer ring and drop it on the center disk. You could freeze the steel for the 100 degrees for a .0034" reduction, or even - 150 degrees ( .0034 x 1.5=) for .0051"

But Wait, there is another 2 factors to consider.
First is will the assembly stay at the same temperature ? If the ring can get hot, while the disk stays cool, the IF (Interference Fit) will decrease and maybe it will fail
Second is the thinner the ring (thickness) the greater the IF should be . Increasing the IF means the ring is under greater tension and less chance of loosening.. It is normal for such thin rings to have double the IF compared to a heavier structure.

The Normal IF used is .0005 to .001" per inch of Diameter at the IF point.
So your initial thought of .004-.005" is good , but a bit heavier will help assure you will not have failure.
In review , for a heavier fit you want .008 to .010" IMHO.
Heating the ring 200 degrees F will get you size to size , but you want it bigger so it will drop in and that means another 200 degrees for a total on the brass ring of 400 F
Say your room temp is 54 deg F , and you add 400 to 454...... The temperature that paper burns !
So if the ring is on a hot plate with a piece of paper on top, when the paper burns , you are there.

Good luck, Good question and I hope I have answered it to your desire
Rich

 

[Kpar]

Rich, well explained in terms that even I can understand. I might do a test as I would be disappointed if the ring stuck half way due to not enough heat.
Kpar

 

[Zeb]

Great answers. I know from doing wing attach fittings or other oddball stuff that size (nose gears), the setup needs to be square and fast. Even if the temps are liquid nitrogen to torch the gradient will change fast within the second. Kinda obvious I guess, but it's tears trying to X-Y etch and sketch jammed parts out with an endmill.

 

[BaronJ]

Hi Guys,

Many years ago, as a seven or eight year old, I remember watching a group of men fitting steel tires to cast railway wagon wheels for the coal mining industry. It took a few seconds for the red hot tire to be fitted. It was placed on the wheel and hammered into place by several men. Fascinating to watch how it was taken from a furnace and placed on the wheel. The foundry, not there any more, was in the next street to where I lived at the time.

I used to stand looking out of the bedroom window on a night watching the red glow in the sky when they were running the furnace.

 

[clockworkcheval]

Maybe I get it all wrong, but I reckon in calculating expansion the expansion of a disc is the same as the expansion of a ring. So the lineair expansion is the diameter expansion.

 

[jack620]

I reckon in calculating expansion the expansion of a disc is the same as the expansion of a ring.

Yes, you're right. This video explains: https://www.khanacademy.org/science...n-of-ring-class-11-india-physics-khan-academy

 

[goldstar31]

My father started or finished his blacsmith/farrier's thing at 17 and then became a sapper-- and set fire to rel
uctany Army horses and smashed up carts--- but he shrunk steel wheels on cart wheels pulling the spokes into place. Later, he tyred the colliery locomotive and coal railway waggons.
But as a toddler, I recalled the huge iron cone which shaped the steel bands. Thet were quite common-- THWN.

I suppose that coopers still use Ir-on Bands and set fire to port wine or is it whisky oak casks.

 

[Richard Carlstedt]

clockworkcheval- "So the linear expansion is the diameter expansion. "

Absolutely correct ! -but there is one thing to keep in mind, and that is differential expansion.
If the disk and the ring are the same material, like steel/steel then a .005 shrink fit for instance is the same regardless of temperature
When the ring has a higher expansion rate, such as in the case above with a brass ring on steel disk, and the whole assembly is heated, the "Interference Fit" becomes less, and depending on the load could fail ( come off). Let me do a simple example here
We have a 1" diameter steel rod and a brass ring with a ID of .999, so that is nominally a .001" interference fit ..OK ?
so if the ring was heated to 100 (F) degrees over the disk temp, they would be the same size ( .0000099 x 100= .000999= .001") - Lets say you put it together at that point .
Now we heat the whole assembly to 500 degrees
The shaft grows at .0000063 x 500= .00315"
The ring grows at .0000099 x 500= .00495 ---a difference of .0018"---which means the ring comes loose since it only had a .001" Interference Fit .
If it had a .002" IF, they would still be together
Now if the ring was Aluminum (.000011) , it would be even worse as Aluminum has almost twice the expansion rate of steel
Rich

 

[SmithDoor]

The first shrink fit I toll 0.005" on 10" ring.
Heat up the ring and fell in place cool down and the ring was on for ever.

I have 0.005 shrink on most parts and has worked out.
Do not try that with roller/ball bearing.

Dave

clockworkcheval- "So the linear expansion is the diameter expansion. "

Absolutely correct ! -but there is one thing to keep in mind, and that is differential expansion.
If the disk and the ring are the same material, like steel/steel then a .005 shrink fit for instance is the same regardless of temperature
When the ring has a higher expansion rate, such as in the case above with a brass ring on steel disk, and the whole assembly is heated, the "Interference Fit" becomes less, and depending on the load could fail ( come off). Let me do a simple example here
We have a 1" diameter steel rod and a brass ring with a ID of .999, so that is nominally a .001" interference fit ..OK ?
so if the ring was heated to 100 (F) degrees over the disk temp, they would be the same size ( .0000099 x 100= .000999= .001") - Lets say you put it together at that point .
Now we heat the whole assembly to 500 degrees
The shaft grows at .0000063 x 500= .00315"
The ring grows at .0000099 x 500= .00495 ---a difference of .0018"---which means the ring comes loose since it only had a .001" Interference Fit .
If it had a .002" IF, they would still be together
Now if the ring was Aluminum (.000011) , it would be even worse as Aluminum has almost twice the expansion rate of steel
Rich

 

[kwoodhands]

My father started or finished his blacsmith/farrier's thing at 17 and then became a sapper-- and set fire to rel
uctany Army horses and smashed up carts--- but he shrunk steel wheels on cart wheels pulling the spokes into place. Later, he tyred the colliery locomotive and coal railway waggons.
But as a toddler, I recalled the huge iron cone which shaped the steel bands. Thet were quite common-- THWN.

I suppose that coopers still use Ir-on Bands and set fire to port wine or is it whisky oak casks.

My grandfather told me how the family put an iron band on wagon wheels. He was Polish but lived in Ukraine til he was 16. They made wagons, carts buggys etc.
I recall him telling me that the band was wrapped around a wheel and held in place with clamps akin to a clothespin. Then the over lapping band was marked and the band released and cut short. The ends were hammered to forge weld. The completed band was placed in a bed of coals.
The wooden wheel was nearby. Two or three men lifted the band with rods that were forged welded to the band edge. The band was placed over the wooden wheel and tapped into place if needed. Water was poured over the wheel. The lift rods were cut off with a chisel and filed flush.
mike

 

[Kpar]

Very interesting comments from somewhat more experienced members than me.
Well I was going to do a test today, but now out of action for a few days.
It seems my 4 jaw chuck has put on weight or, I'm getting weaker in my old age.
will let all know when I get back to it
Kpar

 

[TSutrina]

Hi all, I'm about to have a go at heat shrinking a brass ring onto a cast iron wheel. I need some advice before I start.
Brass ring is 5.5325 ID x 6.4300 OD x 1.3265 Wide.
Two questions,
1. what diameter should I machine the cast, 004" - 005" larger than then the ID of the brass ring?
2. what temp and heat source to expand the brass ring, will oven temp be enough ?
Thanks for any input
Kpar
View attachment 126300

That chart on coefficient of expansion is the starting point to calculate the temperature difference needed. ( steel OD x coef x delta temp = new OD, brass ID x coef x delta temp = new ID ) Then add some additional clearance. You can heat heat brass to a high temperature. In aerospace we usually had a housing of aluminum that has a limit upper temperature and a steel motor stator or a ring liner for a ball bearing.
I can tell you that it is extremely easy to **** the parts relative to each other causing them to stick and loose them by a temperature difference before getting them part. Long stators have this problem often. And with the end turns you can not us an arbor press to force the fit. So have to use a puller to get it off with the help of heat since the aluminum is a much better heat conductor then steel.
I did learn a trick that you can use to get the ring off it cocks. Had an aluminum housing with a shrunk in stator and needed three copper slip rings to connect the winding to three phases. The diameter was about 18 inches ID. and Kapton tap was between the two as insulation. We knew that you couldn't heat the ring up in a oven and hand heating it with a torch trying to by hand evenly spread out the heat will not work.

So we put the unit on a turn table, record play rotating speed. Applied heat with a torch on the ring. Done this dozens of time and each time the ring dropped off by themselves. This may even work to heat the ring to install it. Set the cast iron on a level turn table and make it level then set the chamfer of steel and brass in contact and level the brass ring. Now with the parts turning heat with a torch only the brass ring until it falls onto the steel.

 

[Zeb]

Well, speak of the Devil...my wife recommended we watch this video together tonight on steam train history, and they shrink-fitted a wheel just as described above! Here's a link fast-forwarded to the timestamp.

 

[KatherineKing]

Thank you for your video you shared, I just started watching it but it has already gripped me!!

 

[Howder1951]

As a millwright we did plenty shrink fits over the years. Most often the calculations were on a blue print and interpreted by the machinist on the job. Seat of the pants checks usually allowed .001" per inch of diameter was very doable, allowing .0005" per inch was really easy (at about 250-300 degree F). A thin layer of oil or light grease to lubricate and keep the heat from transferring to the internal parts quickly. Lots of patience to heat apart is necessary as hot spots will deform some parts and thin wall parts cool way faster. If possible it is wise to calculateytarget measurements, and if possible measure the heated part prior to installation. Do everything you can to assure it will go on clean and you won't have as many failures. Always have a backup plan if things go south on an expensive part. Good luck!

 

[Iampappabear]

When large power presses are assembled, the tie rods (they hold everything together in layers) are shrunk to give the appropriate tension. These tie rods typically have 1" holes bored 4 feet deep in the upper end. The press is then assembled and the lower nuts are hand tightened against the lower surface of the machine. Rod heaters are dropped into the holes and turned on. As the tie rod heats up the lower nuts drops away the gap is measured and when the gap reaches the correct dimension(usually around 1/8" on a 100 ton press ) the nut is spun up to the lower surface and the heaters turned off. Hope this all makes sense.

Colin