Thin closed ring under diametral compression

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 $\partial U/\partial M_0 = 0$.

          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.