What Size Should the Shoulder of the Bar Be for a Press Fit?

[soundssupreme]

Hi,
I am new to this forum so please go easy on me :-)

I am making modifications to a hi-fi tonearm and I am making a new component for it and require help with a particular part of it. I have written to another engineering forum, but the replies are simply out of my depth to understand as the pointed me to sites that are simply gobbledygook to me !
Anyway, I have a aluminium ring about 35mm in diameter and 15mm high and 5mm thick.
I have drilled a hole through the edge of it (from od to id) with a 12mm drill. This made a hole of 12.03mm. To this I want to connect an aluminium bar that is 14mm thick and want to turn the end of the bar to a depth of 5mm so that it is a tight press fit (or I believe this is called an interference or force fit). What I mean is that it need to be tight enough that it is pressed in with a machine vice (very carefully !) but cannot easily be removed if at all. The reason fir this is that the size of the ring has to be fixed as it is for a commercial tonearm so the dimensions can't be changed to allow fixing with screws.

Can anyone tell me what size diameter to make the shoulder of the bar to be such a fit in the hole. The web sites I have been pointed to talk about H7 and s6 fits and such like and I am totally confused about it all.

If you can, could you point me to a website where this can be calculated by inputting the dimensions, or if not, just help me with this particular size I need.

Thank you in advance for any help you can give.


Soundssupreme.

 

[jw2007]

If I understand your question, you would like to know how much larger the dimension for the turned down end of the aluminum rod needs to be than your 12.03mm hole to get the proper interference fit. If that is correct, I will tell you that is most likely going to depend on what forces the fit will see. The items I deal with in my industry usually see between .003" and .005" or about .076mm-.127mm. This should probably be satisfactory if there is not much force exerted on the part. Something in this range we can mechanically press. If there is a requirement for a larger interference then cooling the part to shrink it and then inserting it and letting it warm to ambient temperature is another means to connect interference fits. Dry ice or liquid nitrogen are usually the cooling medium. A good reference for things mechanical is "Machinery's Handbook". Hope this provides some insight

KCF

 

[FredGarvin]

A rule of thumb (for those of us that still work in non-metric units) is that a heavy press is .001" per inch of diameter. Of course, this completely neglects all the stresses and deformations that occur because of it. A press this heavy will require heating the outer element with a heat gun (not a hair dryer) or chilling the inner member. You could use a press but it will gouge the two parts at the interface.

In regards to the info that has already been given to you, the H7s6 is a metric designation for the fit tolerances on both the female and the male part. H7 deals with the hole (hence the H designation) and s deals with the "shaft" part. The H7s6 falls into the medium press category. So when all is said and done, your inner part diameter must be a little bit larger than the outer or your math is wrong somewhere.

These fits are all tabulated and really only require knowing the general size to look up. For your instance, you have a hole that is 12 mm nominal. From the tables, the hole tolerance is H7 = 12.000-12.018. The shaft tolerance are 12.018-12.029. This will result in a fit that is 0.000 (what is referred to as line to line) to -.029.

You got lucky in that your drilled hole still falls within the tolerance band for the hole. In a lot of cases the hole needs to be reamed out to ensure a unified diameter the entire length of the hole and to ensure a close tolerance to the actual diameter machined.

 

[jw2007]

below is a link to a table regarding interference fits. Mr. Garvin made a good point I neglected by way of oversight. We use extremely hard alloys under extreme forces. About .001" would certainly be a heavy fit for aluminum. Look at this table and the web site. It contains good information.


http://www.engineersedge.com/class_v.htm

KCF

 

[Q_Goest]

Seems there are many answer-ers to this question:
http://www.eng-tips.com/viewthread.cfm?qid=202210&page=1

 

[FredGarvin]

Seems there are many answer-ers to this question:
http://www.eng-tips.com/viewthread.cfm?qid=202210&page=1

I figured a thread like that would get a lot of attention over there.