# Bushing and Shaft Alignment

I"m working on a design and having difficulty calculating possible misalignment.

I have a shaft (diameter D, tolerance t) that is running through two bushings. The two bushings are separated by a distance L. The two bushings are of width W1 and W2 (measured along the axis parallel to the shaft), with internal diameter of D1 and D2 with tolerances of T1 and T2. (see image below)

How do I go about calculating the potential misalignment of the shaft? When I started this I though it was going to be a quick 5 minute challenge but the more I work on it the more difficult it's seeming.

• How are the bushings seated, are they in some kind of tube or machined cavity. The alignment of the bushings can only be as good as the accuracy of the method of generating their relative positions, obviously ;-) – William Hird Aug 22 '17 at 19:24
• The bushings are press fit into a tube using a hydraulic press, both simultaneously with the same insertion system. Their mounting 'depth' and the concentricity of their OD to the mounting tube is assumed to be perfect for what I'm working on right now. Obviously that's overly optimistic, but, I'm reasonably sure that the dimensions on what I listed is more critical to my alignment mainly given the tolerances I'm working with. I'll be working on THAT problem next after I nail this one down. – Diesel Aug 22 '17 at 19:31
• Let me be bold and suggest a new foolproof way of building the assembly. Instead of press fitting the bushings into the tube, get a bushing that will slip fit into the tube with shaft inside both bushings to align themselves perfectly and use Loctite bearing retaining compound to secure the bushings into the host tube. When the Loctite cures, just slide the shaft out. Now the bushings will be perfectly aligned with each other. Finis. – William Hird Aug 22 '17 at 19:43
• The assembly method is a good idea, however the tolerance stack up is still the primary question. I'm trying to calculate how much alignment I can expect given manufacturability tolerances of the bushings and the shaft. location and concentricity of the bushings to the outer tube isn't really a concern at this point. Also, just as an aside for the application adhesives are VERY not preferable due to the fact that they require a curing time, whereas a press fit operation can be done faster in a single step. – Diesel Aug 22 '17 at 20:12
• OK then , if you have to have a press fit then you can basically eliminate the tolerance stack up concern by selecting standard size shafts and bushings with standard tolerances guaranteed by the manufacturers and just ream out the bushings with a reamer that you will select by trial and error until the shaft slips into the assembly. You will want to ream both bushings with one shot so you will have to fixture this operation on a milling machine or drill press. – William Hird Aug 22 '17 at 20:46

I would like to point out that your CAD application can most likely do this for you. Just make a parametric drawing and sweep the values you need. This may be more practical as you go forward since the cad application can effortlessly make changes based on geometric constraints and you don't need to calculate this.

Image 1: Parametric sweep of one value. No harder than typing 3 values. Note the drawing s intentionally exaggerated.

Making the sweep by script is not much harder. This opens up statistical analysis by montecarlo simulation. Doing it this way reduces chance of error made during the mathematical analysis. Of course, this is both a good thing and a bad thing. Analytical analysis can give insights into global minima, but as the complexity of your analysis grows this might not be feasible so a local minima may suffice.

But yeah you can calculate this by hand too, the problem is that one you start adding the things you now consider trivial this may no longer be a trivial extension.

Image 2: Vector expression

You can express this as a equation of vector expressions

$$\vec a + \vec b + \vec c = 0$$

where $\vec a$ is known, magnitude of $\vec b$ is known and direction of $\vec c$ is linked to vector $\vec b$. So vector $\vec b$ is $D \cdot \{sin(\theta), cos(\theta)\}$ therefore $\vec c$ is $x \cdot\{-cos(\theta), sin(\theta)\}$. Since you have 2 directions and 2 unknowns this is solvable. But i wont take this further.

• Thanks! I've built a rough resolution table based on cad, very similar to the method you've mentioned. I've been trying to solve the problem analytically though which is where I've gotten stuck at. I like having the equations for my stack ups in areas specifically so that you can show danger zones with tolerance outliers. Brute forcing the problem will work it just takes more time. – Diesel Aug 23 '17 at 10:29
• @Diesel what computer algebra system do you use? Mathematica, Mathcad, Sympy?? – joojaa Aug 23 '17 at 10:55
• I've been doing all algebra by hand. Python (Sympy) would be my go to though. I've tried a few symbolic solvers on this problem but I've ended up with horrendous systems, I was hoping for a second set of eyes on the problem that might find a more elegant solution. – Diesel Aug 23 '17 at 11:49
• The 'best' solution I've come up with is to solve the problem from both sides (i.e. solve for theta), one unknown, two equations and then iterate for a matched solution. It's not what I was hoping for but seems to be the most useful. And, easy enough to automate when solving the vector loop for the total stack-up. – Diesel Sep 8 '17 at 16:18
• @Diesel Newton-Raphson root finding? – joojaa Sep 8 '17 at 16:21