# Stress in rotating ring segment

by matthewbanar
Tags: ring, rotating, segment, stress
 P: 5 Hi guys. I was wondering if any of you happened to know the equation for the radial and tangential stress developed across a segment of a ring in rotation. I have been looking through countless machine element design books, and all I can come across is one for a solid ring (like a flywheel). I am looking, moreover, for the stress developed at an element within a segment of a ring (like analyzing the chuck on a lathe) due to its inertia alone. I was asked to do an analysis on the stress developed (and test it versus the shear strength of the bolts) of a sector of a cylinder bolted in two equally spaced locations. Anyway. I bet this sounded horribly confusing. -matt
 Sci Advisor P: 2,340 If I understand the question correctly ("find the stress tensor within a hoop of large radius R and small radius r, R >> r, rotating about its symmetry axis with angular frequency $\omega$, and with given density and elastic constants"?), then oddly enough, this was the Newtonian background for a lengthy discussion in another PF forum, which led to this thread http://www.physicsforums.com/showthr...=171079&page=3 Do any surface forces act on this ring? Body forces? Or should we think of an isolated ring far from any matter rotating about its symmetry axis? (]i]Segment[/i] of a ring? How would that rotate?) By "solid ring" I presume you mean a "disk"? Like a pizza pie, rather than a wedding ring?
 P: 5 Well, somewhat. I know how to find the magnitudal radial and tangential stress over a sold ring in rotation about its centroid already. That's given by: $$\sigma_{r}=\rho\omega^{2}(\frac{3+v}{8})(r_{i}^{2}+r_{o}^{2}+\frac{r_{i }^{2}r_{o}^{2}}{r^{2}}-r^{2})$$ $$\sigma_{t}=\rho\omega^{2}(\frac{3+v}{8})(r_{i}^{2}+r_{o}^{2}+\frac{r_{i }^{2}r_{o}^{2}}{r^{2}}-\frac{1+3v}{3+v}r^{2})$$ where $$\rho$$ is the mass density of the ring/disk, $$\omega$$ is the angular velocity of the ring/disk, $$v$$ is the Poisson's ratio for the material, $$r_{i}$$ is the radius of the inner curvature, $$r_{o}$$ is the radius of the outer curvature, and $$r$$ is the radius of the element under consideration. This approximation only holds if the outside radius of the ring, or disk, is large compared with the thickness (at least by a factor of 10), the thickness of the ring is constant, and the stresses are constant over the thickness. ["3-15 Stresses in Rotating Rings". Shigley's Mechanical Engineering Design, Eighth Edition. p.110] That having been said, I need the analysis of only a segment of the ring, as I have crudely drawn and attached. Intuitively, I know that this stress is decidedly more than that of a ring. Here, let me think of a ficticious example. Let's say I have the world's most amazing Merry-Go-Round. It's built of a hyperalloy that cannot yield (even if I crank it up to 8000 rpm). I strip everything off it until it looks like a rotating, round platform of 15 feet in diameter. Now, let's say I have a chunk of mild steel that matches 30 degrees of the outer diameter of the Merry-Go-Round, but also has an concentric inner diameter of 12 feet. Let's also assume I bolt the 30-degree/15ft OD/12ft ID to the hyperalloy Merry-Go-Round such that it is concentric to it with four SAE Grade 1 Steel 1/2-13 bolts along a bolt circle of 13.5 feet. How fast can I crank the Merry-Go-Round before the bolts fail due to shear? If it was a solid ring bolted to the Merry-Go-Round, I could simply compare my radial stress and tangential stress to the shear strength of my bolt. However, it's not a solid ring. It's a ring segment. It has to be calculated differently, because a ring must actually yield to leave the platform due to symmetry, whereas this segment is only being held to the platform by bolts and not symmetry. I hope that clarifies my predicament. -matt Attached Thumbnails
 Sci Advisor P: 5,095 Stress in rotating ring segment Isn't this going to be more along the lines of a dynamics/rigid body type of problem? Granted you will develop some stresses in the wedge part, but they should be negligible compared to the stresses around the bolt holes due to the shear load loading on the bolts. You could also over come the bolt limitations by providing a proper clamping force on the wedges so that the bolts are not the primary shear load carriers.
 P: 5 Well, I'm analyzing a near-miss at work on a CNC machine. Six 10.5 lb wedges forming a circle around a part (i.e. each 60 degree segments and touching) flew off into the walls of the CNC machine after shearing through two 3/8-16 bolts (in each wedge) that were holding them in place. They were on a 15"RI 18"RO circle, and it was cranked to 800 rpm when the bolt failure occurred. Even at that circle and at 800 rpm, sae grade 1 3/8-16 bolts (located at about 16"R) should have definitely hold the wedges in place, unless they were not tightened down. The only way I know how to approach it is approximating it is as a centrifugal acceleration problem: $$F=mv^{2}/r=mrw^{2}$$ $$F=(10.5/32.2)(16/12)(800*2\pi/60)^{2}$$ which gives about 3050 lbs of force, distributed across two bolts. From that force, I can calculate stress and compare the stress to the shear strength of the bolt. Is this the wrong way to approach this? -matt