Simple design guide: Alignment dowels

Engineering design is a subject so vast that it is pretty much inevitable that new engineering graduates come out of school with gaping holes in their knowledge. It was certainly true for me. To address that issue, simple design guides, to help get through common problems, are probably quite helpful to those new to the field.

A little while ago, I wrote an article including some tips for the advanced beginner when designing parts to be CNCed. You can find it here:

More recently, I helped someone with a design problem similar to the one I will discuss in this article, and I felt the process would be useful for others as well, as it is a common situation. Before anyone jumps on me, I know that I am simplifying the issue. For example, I discuss neither the materials involved, nor statistical analysis of tolerances, which could bear on the discussion. This isn’t meant to be exhaustive, it’s a guide to get started.

The design problem we’re going to tackle is using two dowels to orient two parts that we want to mate together. Say we want the dowels to permanently stay in one part. We’ll need two holes in the first part to hold the dowels, and two ‘holes’ (quotations explained a bit further down) in the second to receive the dowels when aligning the parts.

Let’s say that we have determined that M5 dowels make sense for our design, and we’ll start by defining the tolerances on the holes that will hold the dowels in the first part. Not wanting to re-invent the wheel, take a look at the ISO system of fits. There are three basic classes of fit. A clearance fit is when the tolerances of the shaft and hole are such that the hole will always be bigger than the shaft (e.g. the hole is 4 +0.002/+0.008, and the shaft is 4 -.006/0, so that the biggest shaft will be 4mm exactly, and the smallest hole will be 4.002). Conversely, an interference fit is when the smallest shaft will be bigger than the largest hole. A transition fit is anything in between, when, depending on where the hole and shaft fall out within their respective acceptable tolerance ranges, the actual fit could be a clearance or an interference fit.

A ‘fit’ consists of two tolerance zones, one for the hole, and one for the shaft. One without the other is incomplete, because how things fit together depends on both sides of the equation. The ISO system has a selection of standard tolerance zones, each designated by a letter and a number. Tolerance zones for the hole side of things have capital letters, while tolerance zones for the shaft side of things have lower case letters. The letter tells you what kind of fit it is. For example, an ‘H’ tolerance means that your hole’s minimum size will be the nominal size, and the maximum size will be some amount larger than that. The number designates how tight the tolerance is, with lower numbers being tighter. For example, for a 5mm hole, an H7 hole would have a minimum size of 5.000mm, and a maximum size of 5.012mm, whereas the same hole with an H10 tolerance would still have a 5.000mm minimum, but the tolerance would be looser, with a 5.048mm upper bound.

We want the pins to stay in the holes, so let’s try for a light press fit. Looking at the description of preferred fits on the first page of that link above, a ‘Locational Interference’ sounds right. Ok, so our choices are H7/p6 and P7/h6. Problem: Assuming we’re using a standard metric dowel, our dowel has an m6 tolerance, not a p6 or h6. Neither H7/m6 or P7/m6 would give us the fit we want. Let’s take a look at what we’ve got. The actual m6 tolerance that we’re dealing with is +.004/+.012. a p6 tolerance for our 5mm dowel is +.012/+.020. An h6 tolerance is -.008/0. We’re fortunate, in that we are dealing with size ‘6’ tolerances all around, and the total tolerance range is .008mm in all cases.

Since we’re dealing with the same fit, we could really pick either of the given combinations to work from for our next step, but we’re going to do both, to check our work.

First let’s look at H7/p6. The conversion from the p6 tolerance given and the m6 tolerance we have in reality is simply to subtract 0.008mm from both ends of the tolerance zone (The lower end of p6 is +.012, which is 0.008 larger than the +.004 lower end of m6, and the higher end of p6 is +.020, which is 0.008mm larger than the +.012mm of m6). In order to get the fit we want, we simply have to subtract the same 0.008mm from the H7 tolerance (in other words, the pin we have is 0.008mm smaller than the pin described in the given tolerance, so the hole we use should also be 0.008mm smaller than the hole described in the given tolerance). The H7 tolerance gives 0/+.012, so the actual tolerance for our hole should be -.008/+.004.

Let’s test that value. First, is it actually a press-fit? Our largest hole is 5.004mm. Our smallest pin is also 5.004mm. Line-to-line at the extreme case counts. Check. Now, is it actually the same fit as we are looking for? H7/p6, using the smallest pin and largest hole gives a 5.012mm pin, and a 5.012mm hole. Line-to-line, just like our fit. Check. Median hole size for H7 is 5.006mm. Median pin size is 5.016mm. The interference is 0.01mm. The median hole size using our selected tolerance is 4.998mm. The median pin size is 5.008mm. The interference is 0.01mm. Check. Last test. The smallest hole for H7 is 5.000mm. The largest pin for p6 is 5.020mm, so the maximum interference for the fit is 0.02mm. Using our fit, the smallest hole is 4.992mm, and the largest pin is 5.012, also giving a maximum interference of 0.02mm. Looks like we got what we wanted.

Ok, so to be absolutely sure, let’s quickly take a look at the other option ISO suggests for this fit, P7/h6. To get from h6 to m6, you add +0.012. The tolerance for P7 is -.020/-.008. Adding 0.012 gives -0.008/+.004, which is the tolerance that we came out with before. Looks good. 

Let’s proceed to do the same for the other part. In this case, we want a slip fit. Let’s choose a ‘locational clearance’ from the ISO descriptions. We’re looking at an H7/h6 fit. Again, we simply convert from h6 to the m6 of our actual pin. As we noted one paragraph up, this entails adding +.012 to our H7 hole tolerance. H7 for a 5mm hole is 0/+.012, so our hole tolerance for our second part becomes +.012/+.024.

We run through the same checks we did with the first part, and determine that this tolerance makes sense, but we’re not done. What if our two holes aren’t exactly where they are supposed to be? What if the space between them is slightly larger or smaller? The pins won’t fit into the holes in the second part at the same time! Fortunately, there is a common design trick for just this situation (and now I will explain why, way back at the beginning of the article, I put ‘holes’ for the second part in quotations). Instead of two holes, we will put a single hole, and a slot in the second part. The axis of the slot will be along the line between the two hole positions. Make sure that the slot is long enough to capture the pin, no matter where the hole distances fall within the tolerances zones of both parts.

I was going to end this with a quick sketch of what the parts might look like, but my connection to Onshape, which is what I was going to use to punch it out doesn’t seem to be working. So, I’ll leave it open instead. There may be parts that could be explained better, perhaps with graphics. If you have any questions, feel free to put them in the comments, and I’ll try to address them.

Cheers!