Request for advice regarding dowel press fit (jpg drawing included )

[John2004]

Hi everyone,

I have a small cantilevered mounted bearing housing that rotates about it's longitudinal axis. The bearing housing is loaded perpendicular to it's rotation axis, as is usually the case.

I am using a standard hardened steel dowel pin as the bearing shaft. The dowel pin is just pressed into a steel support, and then the bearing housing is installed onto the dowel pin / shaft. The steel support can slide up or down on two dowel pins and can be adjusted and locked as needed. This allows for adjustment of the vertical location of the bearing housing.

The bearing housing shaft is pressed in-between two other dowel pins that the steel support slides on, which have a very close tolerance slide fit. I think I should probably use a light press on the bearing shaft, so that the press of the bearing shaft does not affect the hole size of the slide fit dowel holes located on each side of the bearing shaft. I have no choice but to located all the dowels fairly close to each other.

The bearing housing is oscillated manually by hand and the device is always used at room temperature. There are no vibrations present since the device moves so slow, but there are reversing loads on the bearing housing, and cyclic loading with variable loads.

I was hoping to use a light press fit for the dowel of 0.0001" minimum to 0.0012" maximum.

I have a picture of the housing arrangement at the following link. It is "housing # 2" i.e., the one shown on the left, that I am using...

http://www.ice9.zoomshare.com/

1. Can anyone please give me a recommendation for a minimum amount of press fit that I can probably get away with for the bearing shaft ? Is 0.0001" minimum to 0.0012" maximum probably OK ?

2. Due to the way the housing is loaded, I think for the most part the load would not tend to pull the dowel / shaft out of the support and would actually make it bind in the support ? However, there still may be some tendency for the housing load to pull the dowel out of the support. Is there a way to calculate or get a close estimate how much axial force will be applied to the 3/8" OD shaft due to the housing load ?

3. Can anyone please tell me the method or formula for calculating how much axial force it takes to pull the dowel out of the hole, based on the amount of interference of the press fit ? The dowel and support are both steel.

4. Do I need to be concerned with changes in the minimum interference of the press fit, just due to normal changes in ambient air temperature ? The device is always used a room temperature.

Thanks for your help.

Sincerely,
John

 

[MikeHalloran]

If both parts are steel, temperature shouldn't be a problem.

Machinery's Handbook can help you estimate how much force it will take to assemble the press fit. It will take roughly as much force to disassemble it, axially. There may be other ways to disassemble it with much less force.

You didn't specify the thickness of the steel support. If it's thin in any direction, the applied load may distort it. The moment product of force 'LA' and its distance from the support, is resisted by radial loads at the left and right ends of the support housing bore, i.e., the force is magnified, by the ratio of the lever arm lengths, press fit or not.

On the other hand, it's expensive to maintain a total tolerance range of .0002" in production. You might want to make the support axially thick and radially thin, and use a heavier press fit, so the support yields during assembly and you can use more tolerance.

On the other, other hand, you might consider using a clearance fit and Loctite, or mechanical retention of the pin in the support.





Mike Halloran
Pembroke Pines, FL, USA

 

[John2004]

Hi Mike,

Thanks for your reply.

Mike Wrote:
>On the other hand, it's expensive to maintain a total >tolerance range of .0002" in production.

John:
The minimum interference I was shooting for is .0001" and the maximum is .0012", so I have a total tolerance range of about .001" on the dowel hole.

Loc-tite is an option, but I would rather just use a press fit instead, to keep costs down.

I may be able to press the bearing shaft into the steel support first with a heavier press, then drill the holes on each side of the bearing shaft that the support slides on.

The support is 5/8" thick, and the dowel is pressed in the full 5/8", it actually sticks out the other side about 1/4".

Space is really limited for mechanical retention, even a set screw would be hard to implement due to various design constraints. The only mechanical retention I can think of is to use a palnut or "push nut" on the dowel where it comes through the 5/8" support. However, being hardened, I doubt the push nut will have much bite or holding power on the dowel ?

A shrink fit is an option, but at added expense.

Perhaps it will work with a light press. The load force LA shown in the drawing won't create much axial "pull out" force on the dowel pin will it ? It seems difficult to estimate what this force might be, but it does not appear it would be much.

Thanks again,
John

 

[GregLocock]

If you look in the FAQ

faq404-1230

there is a link to a press fit calculator

Cheers

Greg Locock

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

 

[MikeHalloran]

How many of these assemblies will you be making, say in a year?



Mike Halloran
Pembroke Pines, FL, USA

 

[John2004]

Hi everyone,

Thanks for the link Greg, that site is really good.

Mike, per year it would probably be lots of 100 to 1,000 to start, but if it works out OK, then perhaps a few thousand, I am not 100 % sure.

In my previous message, I said the support was 5/8" thick, but it is actually 7/8" thick X 1.570" long X .828" high.

I also found a press fit calculator here...

http://www.engineerstoolbox.com/index.html

The calculator lets you work both ways, i.e., you can calculate the force produced by a certain amount of interference, or put in a force and calculate the required interference.

Using this calculator and selecting AISI 4130 steel for both parts with an interference of 0.0001", I got a radial pressure of 7474.445 PSI. This pressure seems to be linear, i.e., it goes up in 7474.445 increments per each added 0.0001" of interference, so an interference of .001" produces 74,744.45 PSI.

How do I convert the radial pressure to axial force needed to remove the dowel pin?

Do I multiply the area of contact of the press-fit by radial pressure? Do I take the circumference of the .375" dowel and multiply it by the 7/8" length of contact of the press fit, to get the effective area ? This did not seem to produce meaningful results, i.e., 1.038 * 7474.445 = 7758.47

I also tried to take the length of contact * the diameter of the dowel to get the effective area or projected area, i.e., .375 * .875 = .3281.

.3281 * 7474.445 = 2452.55 pounds. That figure seems way to high to me as well.

Please let me know if anyone sees what I did wrong

Thanks again,
John

Here is the input and output from the calculator...

SOLUTION TYPE:
Calculate Pressure Given Interference

INPUT PARAMETERS:

Material Properties:
Elastic Modulus Inner, E1 = 2.972501e+07
Poisson's Ratio Inner, Nu1 = 2.9e-01
Elastic Modulus Outer, E2 = 2.972501e+07
Poisson's Ratio Outer, Nu2 = 2.9e-01
Coefficient of Friction, mu = 1.5e-01

Applied Interference and Geometry:
Radial interference, d = .0001
Inner radius of inside part, ri = 0.0e+00 (zero for solid shaft)
Outer radius of outside part, ro = .785 (1.57 / 2)
Common radius, outer radius of inner part and inner radius of outer part, r = .375
Contact length, L = .875”

RESULTS:
Calculated Radial Pressure at r, p = 7.474445e+03

 

[GregLocock]

You need the surface area of the interface, multiply that by the contact pressure, to get the axial force. Working out the coning torque capacity is rather more difficult, I used calculus.

I think that you have made at least one mistake in the inputs to that program.

Cheers

Greg Locock

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