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Pulling off a axle


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#1 MatsNorway

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Posted 01 November 2011 - 14:01



I trying to calculate the force needed to pull of a axle with 73.245mm dia and a bore hole with the dia of 73.025mm.

The shaft length is 108mm.

Both are steel.

outer dim on the bore piece is never less than 126.8mm

Ra:0.8 (surface roughness)

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#2 cheapracer

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Posted 01 November 2011 - 17:03

Mate I've been putting shafts in and out of bores all my life and no 2 are the same ever.

You get old worn bores that slide in and out easy and new tight ones that may need getting some lubrication onto although I find if you get them hot first its a major help.

Ice or cold water can help to shrink the shaft too although at only 4" it can't really get much smaller.

Edited by cheapracer, 01 November 2011 - 17:06.


#3 sharo

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Posted 01 November 2011 - 20:56

:lol:
In my humble experience sometimes a bit too much heating happens to make the bore never shrink back. A friend of mine uses liquid Nitrogen when for example pressing new valve leads (sorry not sure it's the correct English term) into a cylinder head.

Edited by sharo, 01 November 2011 - 20:56.


#4 Greg Locock

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Posted 01 November 2011 - 21:00

Big problem is the coefficient of friction, anywere between 0.18 and 0.3 is seen in practice, and if you were disassembling it can go higher.

Her's several answers::

Then you need an equation called Lame's equation for press fits.

http://www.tribology...lators/e3_8.htm
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It's a tricky piece of maths to derive that. If the geometry of the outer is complex you'll need FEA or tests.

//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
If that calculator is no good (I haven't checked it lately) then any good machinist's handbook should have a table.
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Here's a formula I grabbed off the web, but I have been told it is incomplete.

Alternatively desertfox looked up the equations, so you can roll your own:

Here are the formulas you need:-



Pa= f*3.142*d*L*Pc

where f=friction coefficient
Pc=contact pressure between the two members
d= nominal shaft dia
L=length of external member.
Pa= axial force required to interference fit

to calculate Pc for a given interference use the formula:-

Pc=x/[Dc*[((Dc^2+Di^2)/(Ei(Dc^2-Di^2))+((Do^2+Dc^2)/(Eo*(Do^2-Dc^2))-((Ui/Ei)+Ui/Eo))]

where x = total interference
Dc=dia of shaft
Di = inner dia of shaft(this is zero for solid shaft)
Do= outside dia of collar
Uo=poissons ratio for outer member
Ui=poissons ratio for inner member
Eo=modulus of elasticity for outer member
Ei=modulus of elasticity for inner member

This formula for Pc will simplify if the materials are the same.

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http://www.roymech.c.../Cylinders.html
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Here's the sort of thing that might be more use

http://www.eassistan...HTML_ench7.html

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take your pick!

Edited by Greg Locock, 02 November 2011 - 03:27.


#5 MatsNorway

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Posted 02 November 2011 - 07:52

http://imageshack.us...esspasning.jpg/

Looks right then?

Im Assuming the δ1 is the pressfit (0.22mm)

All the talk about interference is about that i should add it to the pressfit or something?

I find it odd that i don`t need to add the surface roughness.

Regarding the machinist handbook.
I tried to calculate it but i have not been teached well enought to do this.

Pa= f*3.142*d*L*Pc

where f=friction coefficient
Pc=contact pressure between the two members
d= nominal shaft dia
L=length of external member.
Pa= axial force required to interference fit


Pa= f*3.142*d*L*Pc
friction factor x pi x dia(mm) x Length (mm) x N/mm^2 = Newton

Pa = 0.12 x 3.142 x 73 x 108 x 221.63 = 658 728.2N

65 872.8 kg or
66tonnes ....

Edit:
I see know that i should have added the surface roughness to the pressfit. total -> 0.2206mm

Edited by MatsNorway, 13 February 2012 - 07:47.


#6 Kelpiecross

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Posted 02 November 2011 - 12:43

[quote name='MatsNorway' date='Nov 2 2011, 18:52' post='5376208']

Why do you want to calculate the force needed?
I would have thought the usual procedure would be to use a puller, if that didn't work, then a press, then heat, liquid N2, bad language etc. Doesn't seem to be much point in calculating the force needed.



#7 MatsNorway

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Posted 02 November 2011 - 15:40

Why do you want to calculate the force needed?
I would have thought the usual procedure would be to use a puller, if that didn't work, then a press, then heat, liquid N2, bad language etc. Doesn't seem to be much point in calculating the force needed.


When you are they guy that are supposed to design the puller or the press it is.

And designing tools without knowing the forces involved is a lottery.

Worth mentioning is that this part is 2.65meters long.

Edited by MatsNorway, 02 November 2011 - 15:55.


#8 Kelpiecross

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Posted 03 November 2011 - 02:04

Worth mentioning is that this part is 2.65meters long.


I thought it was 108mm?

#9 Greg Locock

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Posted 03 November 2011 - 03:54

I thought it was 108mm?

the intereference length, ie the bore in the wheel, is 108 long, the axle itself is 2m long.

#10 MatsNorway

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Posted 03 November 2011 - 08:23

correct.

Edit: the Civ ing next to me got 68 tonnes..
Edit2: he used wrong values.. i should start a blog... damn near a fulltime job to explain things for him.

I think i got im going in the right direction. hes bad at Norwegian.

btw: That was funny cheapy i did not get it the first time.

Edited by MatsNorway, 03 November 2011 - 09:12.


#11 MatsNorway

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Posted 08 November 2011 - 12:12

After finding some minor errors in the civ ing`s math.
He arrived with a pressmonn (Pm) at no more than 163,5/mm^2

Thats on Gmin.

Even with this low number he got a Force of 609KN

He used a friction factor of 0.15

I redid the numbers for Gmax: 224.36um
and got

Pm= 204.2N/mm^2
Fmax: 750.6KN or 76tonnes..

i don`t know if we have that strong a press designed for parts.

Edit: whe have but in a different division.

Edited by MatsNorway, 08 November 2011 - 13:06.


#12 Greg Locock

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Posted 08 November 2011 - 22:23

i don`t know if we have that strong a press designed for parts.

Edit: whe have but in a different division.


So use heat and coolth and probably large hammers as well as the press. Hmm, have you thought about a slide hammer?

#13 fredeuce

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Posted 08 November 2011 - 23:29

Matts,
If it theoretically is going to take 76 tonnes to dismantle these components then I think a press is going to be the only way.

If you have seen a 100 tonne workshop press then you will know how robustly they are built to withstand the maximum operating force.

That said, any puller you might want to fabricate is going to resemble the press in order to withstand the forces involved.

I would be trotting over to your other division and ask them nicely about using their press. (Here in Australia the custom is to offer beer as an inducement. The usual currency is a "slab" or carton)

As mentioned some heat would be useful . In addition when parts like this are loaded in a press a shock load introduced via a hammer at a suitable spot can often be enough to get things to start moving.

Caution ! When pressing components with loads of this magnitude use all safety protection available. Loads like this can be very destructive. Things like bearing races can shatter causing shrapnel to fly about your ears. Take care.

#14 Greg Locock

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Posted 08 November 2011 - 23:59

Can either component be sacrificed or modified?

#15 MatsNorway

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Posted 09 November 2011 - 11:37

Can either component be sacrificed or modified?


Nope

I was hoping the numbers where wrong. this is to assemble/disassemble a stabiliser to a train.

This train.Posted Image

dry facts:
Top speed 160km`t
maximum passengers: 310
86meters long total.
2550kW

Source: http://www.tu.no/nyh...rticle20749.ece

Sounds like divison is the wrong word. It will not be a big issue to use the press not in the area where this component actually belongs.

Fredeuce:
We have a press that does 150tonnes probably more. Its a old one.

Good advices. this will be one of my most advanced and heavy tools so far.

Icing and heating is fairly likely to take place as well.

I tought about situations where i might be needing to give it a knock.


But safety protection?

how would you secure this apart from doing a good design, healthy safety margin and lots of double checking?

The ones i will do is to make end stops to secure the tool to the pressplate.(thingy where the hammer hits) so that it does not slide away.

It will slide to the side without a anchoring to the press.


#16 Tony Matthews

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Posted 09 November 2011 - 12:18

I tought about situations where i might be needing to give it a knock.


But safety protection?

Get someone else to hit it.

#17 mariner

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Posted 13 November 2011 - 13:05

I don't want to critisize anybody's engineering as I am not an engineer but if it for a train where such high forces are involved is an interference fit the best overall solution?

I see Greg has quoted co-efficients of friction in a 2:1 ratio and there is ( maybe) a presumption that the interference fit will be consistent over 108mm on a what is , I assume, a large bore item.

Again , I am not qualifed to give opinions here but how safety critical is the joint? I know train tyres ( the actual steel band touching the track) are mounted with heat and cold in an interference fit but that is strict radial load and IIRC safety locks are ofen fitted as there is a history of rail tyres working loose with nasty results

Forgive me if I have misunderstood the issue.

#18 MatsNorway

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Posted 13 November 2011 - 15:12

I don't want to critisize anybody's engineering as I am not an engineer but if it for a train where such high forces are involved is an interference fit the best overall solution?


Spline is used on a different location on the same train..

We don`t use rings on the wheels in norway.

train tyres is called wheelset when they are with shaft and wheel assembled.

Edited by MatsNorway, 13 November 2011 - 15:13.


#19 Greg Locock

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Posted 13 November 2011 - 21:21

Marier's point is good, there is no particular reason why you should use parallel shafts and bores in an interference fit, but I think experience suggests they work well, and things might wobble if you had a barrel shaped shaft, or burst if you had a barrel shaped bore. I suppose you might think about tapers, but that gives me the creeps (literally, the wheel would tend to wiggle off the shaft).



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#20 MatsNorway

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Posted 15 November 2011 - 20:07

Marier's point is good, there is no particular reason why you should use parallel shafts and bores in an interference fit, but I think experience suggests they work well, and things might wobble if you had a barrel shaped shaft, or burst if you had a barrel shaped bore. I suppose you might think about tapers, but that gives me the creeps (literally, the wheel would tend to wiggle off the shaft).




Dunno what a barrel shaped bore is? you mean the part got a oval outer shape? it does..

Do not know anything about tapered pressfits. But i think they get used on either boat propellers or electric water turbines. guessing with the cone towards the jet then and it would be fine.

Edit: 13.02.2012

Its a parallel shaft and bore.

"But i think they get used on either boat propellers or electric water turbines."
Thats correct. I saw some a few years back. Rather big taper too. And they had a oil channel for the dismantling.

Edited by MatsNorway, 13 February 2012 - 07:58.


#21 MatsNorway

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Posted 13 February 2012 - 07:28

To shrinkfit it on it seems i need to have a temperature difference of 262K or Celsius.

pressfit max = 0.2206
Delta Temperature = ?
expantion coificient for steel = 11,5 x 10^-6
inner dia bore = 73mm

.......................pressfit (0.2206)
Delta T = ---------------------------------------------------------------- = 262.6 Celsius
.................. coeficient (11.5 x 10^-6) x inner dia bore (73)

Sounds doable to me. we just freeze the shaft and heat the bore. Add a press as well and we can perhaps reduce the temperature needed somewhat.

Edited by MatsNorway, 13 February 2012 - 12:42.


#22 Kelpiecross

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Posted 14 February 2012 - 13:02

To shrinkfit it on it seems i need to have a temperature difference of 262K or Celsius.

pressfit max = 0.2206
Delta Temperature = ?
expantion coificient for steel = 11,5 x 10^-6
inner dia bore = 73mm

.......................pressfit (0.2206)
Delta T = ---------------------------------------------------------------- = 262.6 Celsius
.................. coeficient (11.5 x 10^-6) x inner dia bore (73)

Sounds doable to me. we just freeze the shaft and heat the bore. Add a press as well and we can perhaps reduce the temperature needed somewhat.


Liquid nitrogen and a press works well - I have used this method a couple of times. I have seen railway-type wheels (from a travelling crane) fitted to a axle by heating the wheel with oxy-acetylene torches. I don't know what temp was used - but it was very, very hot - maybe something over 300 degrees C. From memory, I think the wheel just dropped on to the axle under its own weight.

#23 cheapracer

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Posted 14 February 2012 - 13:29

... fitted to a axle by heating the wheel with oxy-acetylene torches. I don't know what temp was used - but it was very, very hot - maybe something over 300 degrees C.


We used to standard fit axle bearing retainers by getting them cherry and dropping them on then at some stage they went to cold drift fitment about 30 years ago or so.


#24 MatsNorway

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Posted 14 February 2012 - 15:31

Most engineers and so on go crazy if you suggest heating something beond 120degrees celsius. while there is no structural chance going on below 200Celsius. And personally i don`t think its a Mega issue going beond that either... anehealing does need a long long time to take out the hardness. so a in comparison sudden heating and air cooling should not be a big issue. But im not metallurist. Few of those here btw.

#25 MatsNorway

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Posted 12 March 2012 - 07:32

Turns out it can`t be done.

one part is that the manufacturer of these parts does not sell them individually.

Second is that there will be tearing if you dismantle the parts. So neither part would be reusable. Supposedly.

And working in norway means its highly likely to be cheaper to buy the parts premounted from Germany.

its supposedly 6 on each train set. 36 sets in norway, price is around 5000-7000$ so far pr part if the ratio is 6.5:1 Kroner/Dollar
Thats a lot of money so its not unlikely the train company decides to alter the design to spline if possible.

Edited by MatsNorway, 12 March 2012 - 08:20.