How will it deform when I compress an elliptical ring along its long axis? For simplicity we could consider that the ring is homogeneous, elastic, has no internal stress at the beginning, and its cross section is small enough compared to its radius.

I think I should be able to find the normal stress and shear stress on any cross section if the ring were rigid. I feel that the next step should be establishing the relationship between stress and strain, but I don't know exactly what to do.


Follow these steps:

  1. Draw a nice free body diagram


  1. Find the internal moment inside the ring $$M(\theta) = M_A + F b \sin \theta$$

  2. Find the total strain energy $U$ when the ring deforms under the force $F$ $$ U = \int \frac{ M^2 }{2 E I} \, {\rm d}s = \int \frac{ (M_A + F b \sin \theta)^2 }{2 E I} \sqrt{a^2 - (a^2-b^2)\cos^2 \theta} \, {\rm d}\theta$$ where $M_A$ is the unknown reaction moment at point A, $E$ is the elastic modulus and $I$ is area moment of the section.

  3. Find the reaction moment by solving $$ \frac{{\rm d} U}{{\rm d}M_A} = 0 $$ for $$M_A \approx \frac{F a (5 \epsilon^2-12)}{6 \pi} $$ with the eccentricity parameter $\epsilon^2 = 1- \left(\frac{b}{a}\right)^2$. The expressions can only be calculated for small values of $\epsilon$ because $U(\epsilon)$ contains elliptical integrals.

  4. Get the total strain energy (approximation) $$ U \approx \frac{F^2 a^3}{E I} \left( \frac{208-27 \pi^2}{192 \pi} \epsilon^2 + \frac{\pi^2-8}{8\pi} \right)$$

  5. Calculate the deflection from $U = \int F {\rm d} \delta$ or $$\delta = \frac{{\rm d}U}{{\rm d}F} \approx \frac{2 F a^3}{E I}\left( \frac{208-27 \pi^2}{192 \pi} \epsilon^2 + \frac{\pi^2-8}{8\pi} \right)$$

  6. The internal stress comes from the bending $$ \sigma(\theta) = \frac{|M| y}{I} \approx \frac{y ( F a \left( \frac{2}{\pi}-\frac{5 \epsilon^2}{6\pi} \right) + F b \sin \theta)}{I}$$ where $y$ is the half width of the section.

  7. Maximum stress is at $\theta=\frac{\pi}{2}$ with $$\sigma_{max} \approx \frac{F\, y}{I} \left( \frac{b (6 \pi+13)}{6 \pi} - \frac{b^3}{6 \pi a^2} \right)$$

The above is only valid for small values of eccentricity, like $\epsilon < 0.5$. I estimate the approximation error at 10% for $\epsilon=0.78$ and 2% for $\epsilon=0.58$

For further details on this method see Roark formulas for Stress, Chapter 9 (p267).


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