2
$\begingroup$

Any physics textbook chapter on stress-strain curves will generally mention that stress is force acting upon an area, and when a shape is three-dimensional, that area is the cross-sectional area. However, every example I have ever seen describing this relationship used a prism of some kind, which has the same cross-sectional area along its entire length.

But then I considered a sphere under compression. At first, I assumed the cross-sectional area upon which the force is acting upon would be the equatorial cross-section, but when I went to double-check that assumption, I couldn't find any information on the topic. Perhaps it is the area of contact which would be more accurate? If that is the case, then would differently sized spheres (made of an extremely stiff material) be very similar in terms of resistance to permanent deformation via compression?

To phrase it in terms of a math question: You have a perfect sphere which is 5cm in diameter, made from a stiff material, which is placed between two flat surfaces. When 200 g of compression force is applied to the sphere, it undergoes catastrophic failure. If you have a sphere of the exact same material in the exact same situation, only it is 50cm in diameter, at what amount of force would it likewise break?

$\endgroup$
1
  • 1
    $\begingroup$ This is a mathematically difficult problem. You have to solve the for the stress and strain tensors in a complicated geometry. I doubt that there is any simple solution. $\endgroup$
    – mike stone
    Oct 21, 2021 at 12:48

1 Answer 1

0
$\begingroup$

The reason you see constant-cross-section geometries in introductory classes is that their stiffness (i.e., the load required to obtain a given displacement) is constant for small displacements. This isn't the case for a sphere or any geometry whose contact area shrinks to zero for small loads. For a gentle touch, these geometries have near-infinite compliance! This behavior is studied in the field of contact mechanics (see, for instance, Johnson's Contact Mechanics and Fischer-Cripps's Nanoindentation). You are right that the area of contact is an important parameter in this context.

The small-deformation case for a sphere contacting a flat surface has an exact solution for a single side (Case 2 here): the deformation $\alpha$ (retaining the nomenclature used in the link) is $$\alpha=\frac{(3\pi)^{2/3}}{2}P^{2/3}\left(\frac{1-\sigma^2}{\pi E}\right)^{2/3}D^{-1/3}=\frac{1}{2D^{1/3}}\left(\frac{3P}{ E^\star}\right)^{2/3},$$

where $P$ is the applied force, $\sigma$ is the sphere Poisson's ratio, $E$ is the sphere Young's modulus, $D$ is the sphere diameter, and $E^\star=\frac{E}{1-\sigma^2}$ is the reduced Young's modulus.

(You could calculate an effective contact area, analogous to a prismatic solid, by evaluating $\frac{PD}{\alpha E}$, but this isn't too useful, to my knowledge. Instead, the actual contact area is $\left(\frac{3PD}{8E^\star}\right)^{1/3}$, which may be useful in your similarity comparison of spheres of different sizes.)

To add a second side (the other side), we apply symmetry to obtain Case 3 in the link.

Regarding yielding, Johnson gives the yield load $P_\mathrm{Y}$ when applying the von Mises criterion as $$P_\mathrm{Y}=\frac{\pi^3D^2}{24E^{\star 2}}(1.60Y)^3.$$ where $Y$ is the yield stress. In comparison, $P_\mathrm{Y}=\frac{\pi D^2}{4}Y$ when applying an axial load on a cylinder of diameter $D$. Johnson thus notes that when designing for strength for curved contact, a low Young's modulus is desirable (the Young's modulus isn't a factor for yielding in the case of flat contact at the end of prismatic shape). This makes intuitive sense because the load is spread out over the larger contact area of a more compliant material.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.