---
title: "Thin closed ring under diametral compression"
subject: "Mechanics of Solids"
chapter: 6
key: "semester_5/mechanics_of_solids:closed-ring-diametral-load"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/closed-ring-diametral-load
site: BCT Engineering
---

# Thin closed ring under diametral compression

Mechanics of Solids · Chapter 6: Curved Beams

- Asked in: year not given, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/closed-ring-diametral-load
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-6

## Question

A thin circular ring of mean radius 150 mm has a rectangular section 25 mm wide (parallel to the axis) and 12 mm thick (radial). It is compressed by two equal and opposite forces of 1 kN acting along a diameter. Using Castigliano's theorem determine the bending moments at the load points and at the section at $90^\circ$ to the load line, the maximum bending stress, and the decrease in the loaded diameter. Take $E = 200$ GPa. Neglect the effect of axial and shear forces.

## Short answer

By Castigliano’s theorem, bending moments at load points and at 90° are 0.318PR (47.7 kN·mm) and 0.182PR (27.3 kN·mm). Maximum bending stress is 79.6 MPa at the load points. The loaded diameter decreases by 0.70 mm.

## Answer

#### Data

$R = 150$ mm, $P = 1000$ N, $b = 25$ mm, $t = 12$ mm, $E = 200\times10^3$ MPa.

$$
I = \frac{bt^3}{12} = \frac{25\times12^3}{12} = 3600\ \text{mm}^4
$$

#### Statically indeterminate moment

The ring is symmetric about both diameters. Cut it at the horizontal section: by symmetry the shear is zero, the normal force is $P/2$ (compression) and the moment $M_0$ is unknown. The slope of this section does not change, so $\partial U/\partial M_0 = 0$.

```text
          P
          |
        .-v-.
      /       \
  M0 |-- ring --| M0     (cut sections with P/2 each)
      \       /
        '-^-'
          |
          P
```

For a section at angle $\theta$ from the horizontal cut (towards the load):

$$
M_\theta = M_0 - \frac{P}{2}R(1 - \cos\theta),\qquad \frac{\partial M_\theta}{\partial M_0} = 1
$$

$$
\begin{aligned}
\frac{\partial U}{\partial M_0} &= \frac{4R}{EI}\int_0^{\pi/2}\left[M_0 - \frac{PR}{2}(1-\cos\theta)\right]d\theta = 0 \\
M_0\,\frac{\pi}{2} &= \frac{PR}{2}\left(\frac{\pi}{2} - 1\right) \\
M_0 &= PR\left(\frac12 - \frac1\pi\right) = 0.1817\,PR
\end{aligned}
$$

#### Bending moments

- At the horizontal section ($\theta = 0$, $90^\circ$ from the load): $M_0 = 0.1817\times1000\times150 = 27254$ N mm, tension on the **outer** surface.
- At the load point ($\theta = \pi/2$): $M = M_0 - PR/2 = -PR/\pi$, magnitude $0.3183\times1000\times150 = 47746$ N mm, tension on the **inner** surface (opposite sign to $M_0$).

More simply, $M_\theta = PR\left(\dfrac{\cos\theta}{2} - \dfrac1\pi\right)$.

#### Maximum bending stress

The maximum moment is at the load points:

$$
\sigma_{max} = \frac{M c}{I} = \frac{47746\times6}{3600} = 79.6\ \text{MPa}
$$

At the horizontal section: $\sigma = 27254\times6/3600 = 45.4$ MPa. (The direct stress $P/2A = 500/300 = 1.7$ MPa is small and is neglected, as stated.)

#### Change in loaded diameter

$$
\begin{aligned}
\delta_V &= \frac{\partial U}{\partial P} = \frac{4R}{EI}\int_0^{\pi/2} M_\theta\,\frac{\partial M_\theta}{\partial P}\,d\theta
= \frac{4PR^3}{EI}\int_0^{\pi/2}\left(\frac{\cos\theta}{2} - \frac1\pi\right)^2 d\theta \\
&= \frac{4PR^3}{EI}\left(\frac{\pi}{16} - \frac{1}{2\pi}\right) = \frac{PR^3}{EI}\left(\frac\pi4 - \frac2\pi\right) = 0.1488\,\frac{PR^3}{EI}
\end{aligned}
$$

$$
\delta_V = 0.1488\times\frac{1000\times150^3}{200\times10^3\times3600} = 0.697\ \text{mm}
$$

The loaded diameter decreases by this amount, and the horizontal diameter increases by about $0.137\,PR^3/EI$.

**Answer:** $M_{load} = 47746$ N mm (0.318 $PR$), $M_{90^\circ} = 27254$ N mm (0.182 $PR$, opposite sign), $\sigma_{max} = 79.6$ MPa, decrease in loaded diameter $= 0.70$ mm.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the unknown moment M0 at the horizontal cut of the ring found with Castigliano's theorem?
   A. Set dU/dP = 0
   B. Set dU/dM0 = 0 since the section does not rotate
   C. Set the shear force to zero at the load point
   D. Set the total energy equal to PR
   Answer: B. By symmetry the cut section does not change slope, so the partial derivative of strain energy with respect to M0 is zero. (Key point: Condition for indeterminate moment)

2. What is the bending moment magnitude at the load points of the ring?
   A. 0.182 PR
   B. 0.318 PR
   C. 0.5 PR
   D. 0.149 PR
   Answer: B. M = M0 - PR/2 = -PR/pi, magnitude 0.3183 PR = 47746 N mm. (Key point: Moment at load point)

3. The moment at the section 90 degrees to the load line is
   A. 0.182 PR with opposite sign to the load-point moment
   B. 0.318 PR with the same sign as the load-point moment
   C. Zero
   D. PR/2
   Answer: A. M0 = PR(1/2 - 1/pi) = 0.1817 PR = 27254 N mm, opposite in sign to the moment at the load points. (Key point: Moment at 90 degrees)

4. Where does the maximum bending stress occur and what is it (I = 3600 mm^4, c = 6 mm)?
   A. At the load points, 79.6 MPa
   B. At the horizontal section, 45.4 MPa
   C. At the load points, 45.4 MPa
   D. At the horizontal section, 79.6 MPa
   Answer: A. The largest moment is at the load points, so sigma = 47746 x 6 / 3600 = 79.6 MPa. (Key point: Maximum stress)

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