# Twisting of thick ring due to internal pressure

I have an axisymmetrical part that is subject to an internal (non-constant along the inner border, but constant in time) pressure (among other loads, but this is the one I am interested about) and I have to analytically evaluate the stresses across its cross section.

Due to the geometry of the part and the distribution of the load, I assumed symmetry about the central axis. I also chose to simplify the cross section like this:

The pressure can be decomposed into radial and axial pressure. I know how to deal with the hoop stresses generated by the radial pressure, I can use the thick wall model and get a good estimate from that. The axial pressure will compress the ring, this too looks like a simple calculation.

The problem comes when I want to evaluate the effects of the torque induced by the pressure about the centroid of the section.

I did some research and found this case covered in the book "Strength of Materials - Part 2" by Timoshenko, pages 138-143, but there are still some things I don't understand, because this topic is covered very briefly, and only for rectangle cross sections, so I still have some questions that I can't figure out. I've been stuck on this problem for some days, and I can't seem to find any other source that covers this particular topic in depth.

I also checked Roark for tables of various cross sections, but didn't find anything about this particular case.

I am also uncertain about how to exactly calculate the resultant forces and torques due to pressure, because the book calls this problem "Twisting of a circular ring by couples uniformly distributed along its center line", but I have the doubt that since the cross section rotates around the neutral fiber (similarly to how curved beams bend around the neutral axis) shouldn't the forces and moments be calculated about the neutral fiber ,and not about the centroid?

I also have no reason to think the centroid and the neutral fiber coincide because I am working in radial coordinates (which could generate radial offset, like in curved beams), and there's no symmetry about the horizontal plane in this case (which could generate vertical offset), like in the following picture, where G is the centroid, N is the neutral fiber:

My questions are:

• Should I decompose the pressure about the centroid or the neutral fiber?

• Are there manuals/handbooks/tables about this particular load that contain properties about various cross sections? I am in the process of finding these properties manually for a trapezoidal cross section, but it's quite a slow process and it's easy to miscalculate.

I could as well have misunderstood something, these topics about axial symmetry are something I am trying to study by myself, we didn't cover this at uni for now, but I need it for a student project. I am still trying to wrap my head around this topic and may have missed something obvious.

The ring is bolted to an external thin walled cylinder, which is also subject to internal pressure in the radial direction (this is a combustion chamber):

The cylinder provides the vertical equilibrium, and it is long enough so that what happens on the other side (top) of the cylinder won't affect what happens to the ring, apart for vertical forces that make the vertical equilibrium of the ring possible.

I chose to ignore the screw holes (8) at first to simplify the problem, I'll include their effects later after I'll manage to get a hold of the simpler case first. I also want to assume that the screws provide a distributed, rather than concentrated, reaction force. Again, I'll try to see if I can remove this simplification later in the analysis.

I am also thinking that since the ring is visibly stiffer (thicker) than the thin cylinder, I can assume that the radial component of the pressure is unloaded on the ring only, at least in this zone, so what I mean to say is that the cylinder doesn't help the ring much in resisting tangential elongation. The same can be said for the rotation that I am investigating I think, I can assume that the only thing resisting the rotation of the ring are only its fibers and not those of the thin cylinder. This also puts me in a safety condition, as the cylinder will surely help a bit somewhere.

## EDIT

Alright thanks to everybody who commented, I think I am getting a hang of this now. I had to go back to basics and follow carefully the derivation of the stress formula in the case of bending of a curved beam, to better understand the logic behind choosing the centroid as the point of reference. For this case the logic is similar, I am not forced to choose the centroid, but it's useful to choose it as reference point when deriving formulas and decomposing loads because there will be other loads that will cause other deformations in this part, so it's better to use the centroid as it's a point that is easy to utilize for other loading conditions too. If say I decomposed the loads with respect to the neutral fiber rather than the centroid, I would easily calculate the rotation of this ring, but the choice of the point of reference would make it harder to calculate stresses for the other components of the given load.

• Are there any other loads or supports beside the pressure? According to your second figure, the ring would not be in equilibrium in axial direction. Commented Feb 15 at 17:26
• By the way, the "neutral axis" here is horizontal, so the ring is not really curved with respect to it. Commented Feb 15 at 18:28
• Have you tried to consider the simpler, but maybe helpful case of a straight beam instead of a ring with the same load case as you have drawn in the second figure. On your question of how to decompose, on first impression I'd say it doesn't matter which point you pick as long as you calculate the moment of inertia relative to the same point.
– RJDB
Commented Feb 15 at 18:48
• Consider a beam fully supported (all DOFs constrained) on one end. This can be seen as approximating a piece of a ring with infinite diameter. Then with your load case, a part of the load will be bending the beam (hoop stress in a ring) and a part will be twisting the beam.
– RJDB
Commented Feb 15 at 19:02
• Adding the upward vertical force on the right side of the cross-section of the ring makes it clear a anti-clockwise torque is working on the ring
– RJDB
Commented Feb 15 at 19:11

# Solution using ring

Using figure from the question:

## Ring rotation stiffness

Section $$\Gamma$$ rotates by angle $$\varphi$$ around point $$S$$ located at height $$s$$ from the bottom. Axial coordinate $$z$$ starts at the bottom and inner boundary of the section can be expressed as: $$r_i(z) = R+(B-b)/h\cdot z_i$$ with $$k=(B-b)/h$$: $$r_i(z) = R+k\cdot z_i$$

### Hoop strain and stress

Suppose, that the section will just rotate around a point as reaction to moment. The points above the line will get closer to axis and points below the line will get further from axis: $$\epsilon_t = \frac{2\pi \cdot(r - (z-s)\cdot \varphi- 2\pi\cdot r)}{2\pi r} = -(z-s)\cdot\frac{\varphi}{r}$$

The hoop stress is then: $$\sigma_t = E\cdot \epsilon_t = -(z-s)\cdot E\cdot\frac{\varphi}{r}$$

### Relation between moment and stress

To understand relation between moment and stresses, you need to look at the situation from above. Imagine small angle $$d\vartheta$$ part of the ring. The hoop stress has a radial component with respect to central section: $$dF_r = -2\sigma_t\cdot dA\cdot \sin(d\vartheta/2)$$ which for small angle $$d\vartheta$$ is: $$dF_r = -\sigma_t\cdot dA\cdot d\vartheta$$

#### Position of the "neutral axis"

The moment should of course not result in net radial force on the section $$\Gamma$$, so it must be true, that: $$0 = \int\limits_{\Gamma} dF_r = \int\limits_{0}^h\left(\int\limits_{R+k\cdot z}^{R+B} \left((z-s)\cdot E\cdot\frac{\varphi}{r}\right)dr\right)dz = E\cdot\varphi\cdot\int\limits_{0}^h(z-s)\left(\int\limits_{R+k\cdot z}^{R+B} \frac{1}{r}dr\right)dz$$

In case of rectangular section, integral of $$1/r$$ is very simple and does not affect the second integral, positioning neutral axis at the center of mass. In your case however, it would be quite complicated. Luckily, the term $$1/r$$ can be simplified as a constant using a mean radius value when (radial) dimensions of the ring are much smaller then its distance from the axis. I will use weighted mean radius here: $$r_m \cdot A = \int\limits_{0}^h\left(\int\limits_{R+k\cdot z}^{R+B} r dr\right)dz = \frac{1}{2}\int\limits_{0}^h\left((R+B)^2-(R+k\cdot z)^2\right)dz$$ $$r_m\cdot \frac{B+b}{2}\cdot h = \frac{1}{2}\left(\left(2RB+B^2\right)h-Rkh^2-\frac{k^2 h^3}{3}\right)$$ finally: $$r_m = \frac{\left(2RB+B^2\right)-Rkh-\frac{k^2 h^2}{3}}{B+b}$$

##### Back to the neutral axis

With the mean radius simplification, condition for neutral axis position has following form: $$0\cdot r_m = \int\limits_{0}^h(z-s)\left(\int\limits_{R+k\cdot z}^{R+B} dr\right)dz$$ From this ($$s$$ is basically at the centroid): $$s = \frac{3B-2kh}{6B-3kh}h$$

#### Moment due internal stress from ring rotation

Whole moment acting on the $$\vartheta$$ part of the ring is sum of forces $$dF_r$$ acting on lever arms $$z-s$$: $$dM = \int_\gamma dF_r\cdot (z-s) = \int\limits_{0}^h\left(\int\limits_{R+k\cdot z}^{R+B} \left((z-s)^2\cdot E\cdot\frac{\varphi}{r}\right)dr\right)dz$$

Using the mean diameter $$r_m$$, this can be simplified to: $$dM = \frac{E}{r_m}\cdot \varphi\cdot \int\limits_{0}^h\left((z-s)^2\int\limits_{R+k\cdot z}^{R+B} dr\right)dz$$ Finally: $$dM = \frac{Eh}{r_m}\cdot \left(Bs^2-\frac{2Bs+ks^2}{2}h+\frac{B+2ks}{3}h^2-\frac{k}{4}h^3\right)\cdot \varphi$$

Full moment acting on the whole ring is sum of all the moments on $$d\vartheta$$ parts: $$M = 2\pi\cdot dM$$

## Moment caused by the external loads

First, all of the loads acting on the ring are required. The pressure indicated in the figure will be in axial equilibrium with a distributed axial force from the connected cylindrical shell: $$F_s = p\cdot \pi\cdot \left((R+B-b)^2-R^2\right)$$

For a full moment calculation, focusing just on a small angle $$d\vartheta$$ part of the ring might be clearer than using whole ring. Contribution from the force of attached shell (with thickness $$t$$) is straightforward: $$dM_F = F_s\cdot \frac{d\vartheta}{2\pi} \cdot (R+B+t/2-r_m)$$

Contribution of the pressure is trickier, because it acts on a trapezoidal section and local lever arm should be expressed with respect to center of rotation $$[r_m, s]$$.

# Solution using equivalent shell

From the first figure you have shown, it seems, you might be able to use a cylindrical shell with equivalent thickness (somewhere between $$B$$ and $$b$$) or even multiple such shells to capture bevelled geometry more accurately. This I think is at the edge of feasibility of available analytical solutions.

• Thank you for the answer. The mean radius is a nice approach I didn't think of, and that particular integral gave me a hard time. In the end I tried integrating by dz first and then by dr to get a logarithm only in the end, but I had to split the section into a triangle and a rectangle and it's been painfully slow. As for the torque due to the pressure, can this be solved by maybe considering half a ring and looking at the projected area in the plane of section for the torque due to horizontal component of pressure, and looking at projected area from above for torque due to vertical component? Commented Feb 18 at 11:00
• To further connect the dots, am I right if I think that this formula for the stresses can be used for curved beams subjected to a torque that bends them in a direction perpendicular to the plane of curvature? By intuition it looks like they're two different phenomena, but in the end both reduce to a torque in the cross section that has the same verse. Thank you again. Commented Feb 18 at 11:03
• If you struggle with an integral, it might help to calculate it numerically first. I think the moment due to pressure is best seen on a small angle part (like a piece of pie). Moment with respect to a point in general is a cross product of lever from the point to a point of applied force with that force $\overrightarrow{M}=\overrightarrow{r}\times \overrightarrow{F}$, so you could utilise this. $\overrightarrow{r}$ would go from center of rotation $[r_m, s]$ to location of force due to pressure $\overrightarrow{F} = \overrightarrow{p}\cdot dA$. Commented Feb 18 at 18:33
• You are right that a curved beam would respond to a local torque differently. The ring in your case has basically distributed torque applied to every cross section. Closest situation to the ring is when the curved beam is enclosed in a pipe, which is also bend and is much stiffer, in which case applying torque to the beam may cause it to rotate seemingly in place. This case was I think also of some historical relevance, because deformation of originally straight beams can be described using function for deformed centerline, but that might not be enough if the beam was originally curved. Commented Feb 18 at 18:44