Hey everybody, I'm sure this is machinist 101, but I'm still pretty new to the whole thing. I'm attempting an engine build and there's one friction fit part, axle to a pulley. The plans call for a 2.9mm hole, however my metric drill set is .5mm increments and I'm curious as to how I get that size precisely? Is this a case where I need to make my own reamer? Or should I just modify the plans a bit and turn the axle down to fit a 2.5mm hole?
Jeremy -- At the most basic level, a ø2.9 mm drill
may be "standard" in one country or another, but there are no "standard" drill sizes with metric tooling. I mean by that there isn't a "standard set" such as we are used to working with here in the U.S. ø2.9 mm drill bits
are available. That's one answer set.
Another point to realize is that, in most materials, twist drills create holes slightly larger than their nominal size. (Titanium, some nickel alloys, and certain ceramics will compress a steel drill bit such that you end up with a hole
smaller than the nominal size.) Sequence drilling in three steps (usually 70% of finished size, 90% of finished size, and then finished size) will improve the size of a twist drilled hole. There are other techniques not normally seen in a "home shop" that can be employed. Reaming is generally the best way to get an "on size" hole. That's another answer set.
The thing is that you are going for a "friction fit" between a pulley and an axle. The pulley is going to apply some torque to the axle, right? If you look in
Machinery's Handbook, you will find a section on calculating the interference required to create a specific load level of "grip" between a bore and a shaft. These are called
press fits or
shrink fits.
Press fits are more commonly used to provide an accurate positioning between parts than a real "retention" as the exact amount of frictive force you end up with is fairly variable. However, so long is the resulting friction force is greater than the required clamping force, it does work -- especially for shaft/bore relationships too small for keys, pins, or setscrews (i.e. anything smaller than about 4.75 mm or .188 inch).
The point of this is that there is nothing "magical" about ø2.9 mm. That may be a dimension resulting from other "constraints" such as fitting through another hole elsewhere, but that is a "design choice" and
not some specific imperative. If you have some other set of drills or reamers that give you a good fit, use them. As you are listed as being in the U.S., I would note that ø2.9 mm = ø.1142 inches and that a #32 drill bit will produce a hole (with care) in the .118/.1165 inch range (.1160 nominal) in most common materials. If your shaft is truly ø2.9 mm then you would have a radial gap in the .002/.0015 range which is nearly perfect for "stud locker" type adhesives which will give you a total bond strength on the order of 2000 lb/in². If I assume that your "pulley" is only 1.5X the bore diameter in width (not an uncommon ratio), then you have .062 in² of bond area which gives you a bond strength of (2000 lb/in² * .062 in² =) 122 lbs. Acting on a ø.115 average bore, that is a maximum torque capability of 7 lb-in (0.79 N-m).
Not knowing anything about your application, I cannot say if that would be sufficient or not. However, these are the kinds of "options" available to you in working around the limitations of your tooling. There is an old mountain climber's adage that you either choose the route for the party or you choose the party for the route. I have re-written this as a design rule of choosing the design for the shop or the shop for the design. There is
always another way to do things.
