Hertz contact stress with prescribed displacement

[FEAnalyst]

TL;DR

How to obtain the formulas for Hertz contact stress when force is replaced with prescribed displacement?

Hi,
one can easily find formulas for Hertz contact stress in various cases (two spheres, two cylinders and so on) when force is applied to one of the bodies. But how to get the equivalents of those formulas when prescribed displacement is used instead of stress ?

For example, I’ve found this equation for the case of two spheres made of the same material and having the same radii: $$\sigma_{C}=- \frac{E}{\pi (1- \nu^{2})} \sqrt{\frac{2h}{R}}$$
where: $E$ - Young’s modulus, $\nu$ - Poisson’s ratio, $h$ - prescribed displacement (one sphere moving towards the other), $R$ - radius.
For comparison, here’s the common formula for the same case but with force applied: $$\sigma_{C}=0.578304 \sqrt[3]{\frac{E^{2}F}{R^{2} (1- \nu^{2})^{2}}}$$
where: $F$ - applied force.
I am interested in other cases, such as a sphere on a flat plate. For this case, the formula with force is: $$\sigma_{C}=0.364309 \sqrt[3]{\frac{E^{2}F}{R^{2} (1- \nu^{2})^{2}}}$$
How can I convert it to get its equivalent for prescribed displacement ?

 

[FEAnalyst]

It seems that I've found a solution. I'm not 100% sure that it's correct but the results look good when compared with FEA.

In Roark's Formulas for Stress and Strain there are equations not only for the maximum stress but also for the displacement under a given force in various cases of Hertz contact. So I thought that maybe it would be enough to just rearrange the latter formula in such a way that it gives the force under applied displacement. Then it's just a matter of substituting this equation as force in the formula for stress. I wasn't sure if my reasoning makes sense so I tested it on the case with 2 spheres first. And the result from the formula obtained this way is the same as the one obtained using the first equation in my previous post. Thus, I used the same method for the case with a sphere and flat plate. Here's what I got:
- stress under given force:
$$\sigma=0.364309 \sqrt[3]{\frac{E^2 F}{R^2 (1- \nu^2)^2}}$$
- displacement under given force:
$$y=1.31032 \sqrt[3]{\frac{F^2 (1- \nu^2)^2}{E^2 R}}$$
- rearrange this to get the formula for force depending on the displacement:
$$F \approx \frac{0.666705 \cdot E \cdot \sqrt{R} \cdot y^{\frac{3}{2}}}{\sqrt{v^4 - 2v^2 + 1}}$$
- and finally:
$$\sigma=0.318259 \cdot \sqrt[3]{\frac{E^3 \cdot \sqrt{(v^2 - 1)^2} \cdot y^{\frac{3}{2}}}{R^{\frac{3}{2}} \cdot (v^2 -1)^4}}$$
Maybe this formula could be further simplified but the most important thing is that it gives meaningful results.

 

[Dr.D]

Sounds like a plausible approach to me.