What's new

shrink connection

Kunstmaan

Alibre Super User
Hi People,
Its maybe a strange question in this forum, but I have to create a presentation about howto place a round piece of metal in a square block of metal making a firm connection whit no welding.

I have learned this some 40 years ago but cant find my books to consult.
Tryed allready searcing the web wiht no result as I am not familiar how its called in english.

Can someone point me to a source on the web that I can consult.

Many thanks
Berdien Mareike
 

Ken226

Alibre Super User
Are you referring to a press (interference) fit situation?

I use the data from engineers edge, which I believe they got from McDonnel Douglas.

Standard interference fits chart (imperial).
 

idslk

Alibre Super User
For those connection you heat up the part with the hole (and for strong connections you can cool down the shaft in addition, depending on the calculation you can use liquid nitrogen to get the shaft "small enough" (done this a long time ago during my education)) put the shaft into the wheel and than you let cool them down to room temprature and the parts are "shrinked" together. Often used to connect the driving ring to the wheel for trains. Called shrink-fitting...

Regards
Stefan
 

Ken226

Alibre Super User
For those connection you heat up the part with the hole (and for strong connections you can cool down the shaft in addition, depending on the calculation you can use liquid nitrogen to get the shaft "small enough" (done this a long time ago during my education)) put the shaft into the wheel and than you let cool them down to room temprature and the parts are "shrinked" together. Often used to connect the driving ring to the wheel for trains. Called shrink-fitting...

Regards
Stefan

I use this method for installing bearing races in crankcases and wrist pins into connecting rods, with my kitchen stove and freezer. :)

It works quite well on larger diameter parts, but as diameters get smaller, the dimension change per degree temp Change gets quite small, and you start needing specialize stuff like the liquid nitrogen you mentioned.



If I remember correctly, the coefficient of cubic expansion for steel is about 0.000007 inches per °fahrenheit. So, a .25" diameter steel part would need to be heated 200° above ambient temp to increase its hole diameter to the appropriate .251" press fit diameter. Still doable without special equipment.

I only remembered that because it's the size and method I used to install the driving pins in my floating reamer holder design, pictured in the media gallery.
 
Last edited:

Ken226

Alibre Super User
Your wife would be correct!

:-0
 

Attachments

  • Screenshot_20220614-141706.png
    Screenshot_20220614-141706.png
    1,014.4 KB · Views: 25

Ken226

Alibre Super User

DavidJ

Alibre Super User
Staff member
Surely the linear coefficient of thermal expansion is simpler for calculating shrink fit temperatures? (and easier to find) - relates to hole/shaft diameter
 

Ken226

Alibre Super User
Surely the linear coefficient of thermal expansion is simpler for calculating shrink fit temperatures? (and easier to find) - relates to hole/shaft diameter



Yes, it is easier. But also less accurate.

Not so much that it's a big deal for larger parts, but can be an issue for small parts. Linear expansion with regard to holes generally gives the circumferential expansion, which is an even smaller change in diameter.

It goes the same place, as long as one is aware that calculating the linear expansion will give about 1/3 the diameter change value for holes.
 
Last edited:

ctcboater

Member
Yes, it is easier. But also less accurate.
At least in the situation of a diametrical shrink fit, volumetric expansion isn't the best way to calculate shrink strength. The length is only marginally changed and the change has little relevance to the total stresses.
 

DavidJ

Alibre Super User
Staff member
Ken - can you point me to any reference that supports your claim ?

Volume CTE must = (Linear CTE) ^ 3 (for a material behaving isotropically) how can using Linear CTE be less accurate ...

Linear CTE will give diametrical and circumferential delta, as they are related by a constant (Pi).
 

Ken226

Alibre Super User
Ken - can you point me to any reference that supports your claim ?

Volume CTE must = (Linear CTE) ^ 3 (for a material behaving isotropically) how can using Linear CTE be less accurate ...

Linear CTE will give diametrical and circumferential delta, as they are related by a constant (Pi).


You are probably right. It's been over a decade since school and I was going from possibly flawed memory.

But, this is what I was basing it on. I admit, I could have misread, as I was multitasking when I replied.

 

DavidJ

Alibre Super User
Staff member
There could be some debate perhaps if localised chilling or heating was involved - that potentially gets much more complex as there could be restriction of movement by surrounding/adjacent material.
 
Top