classical mechanics - What is the stiffness of a crushed rod or cylinder?

If you are crushing a uniform rod between two plates with a known force, how do I estimate the deflection (and hence the stiffness) of the rod? I am interested in the overall deflection, including the effects of the contact and in the rest of the (blue) volume below.

Heuristically I see that stiffness $k = \frac{F}{\delta}$ should be inversely proportional to the diameter and linear to the length

$$k \propto \frac{\ell}{d}$$

I wonder if there is an analytical expression that shows the dependency on diameter, length and force applied.

Answer

Overall deflection

Considering that there is a small region of contact, and we can use the Hertzian model it seems that there is an analytical solution 1 (although I would not call this crushing)

$$2\delta = \frac{P}{L} (V_1 + V_2)\left[1 + \log\left\{\frac{2L^3}{(V_1 +V_2) P d}\right\}\right]$$

where $V_i = (1 - \nu_i^2)/(\pi E)$. If we assume that the planes are infinitely rigid compared to the cylinder we obtain

$$2\delta = \frac{P V_1}{L} \left[1 + \log\left\{\frac{2L^3}{V_1 P d}\right\}\right]$$

or

$$2 \delta = \frac{P}{\pi E_1 L} \left(1 - \nu_1^2\right) \log{\left[\frac{2 \pi E_1 L^3}{d P \left(1 - \nu_1^2\right)} \right]}$$

This equation can be inverted to obtain

$$P = \frac{2 \pi E_1 L^3}{d \left(1 -\nu_1^2\right)} e^{\operatorname{LambertW}{\left (- \frac{d \delta}{L^2} \right )}}$$

Stress at the interior

We can model the cylinder as a 2D problem: a disk with radial forces in the poles. The stress function for a disk of diameter $d$ with center in the origin, and radial inward and opposite forces $P$ placed at $(0, d/2)$ and $(0, -d/2)$ is given by

$$\phi = x\arctan\left[\frac{x}{d/2 - y}\right] + x\arctan\left[\frac{x}{d/2 + y}\right] + \frac{P}{\pi d}(x^2 + y^2)$$

We know that the stresses are given by

$$\begin{align} \sigma_{xx} = \frac{\partial^2 \phi}{\partial x^2}\\ \sigma_{yy} = \frac{\partial^2 \phi}{\partial y^2}\\ \sigma_{xy} = -\frac{\partial^2 \phi}{\partial x \partial y} \end{align}$$

that gives

$$\sigma_{xx} = 2 \left[\frac{P}{\pi d} - \frac{32 x^{4}}{\left(d + 2 y\right)^{5} \left(\frac{4 x^{2}}{\left(d + 2 y\right)^{2}} + 1\right)^{2}} - \frac{32 x^{4}}{\left(d - 2 y\right)^{5} \left(\frac{4 x^{2}}{\left(d - 2 y\right)^{2}} + 1\right)^{2}} + \frac{8 x^{2}}{\left(d + 2 y\right)^{3} \left(\frac{4 x^{2}}{\left(d + 2 y\right)^{2}} + 1\right)} + \frac{8 x^{2}}{\left(d - 2 y\right)^{3} \left(\frac{4 x^{2}}{\left(d - 2 y\right)^{2}} + 1\right)}\right]$$ $$\sigma_{yy} = 2 \left[\frac{P}{\pi d} - \frac{8 x^{2}}{\left(d + 2 y\right)^{3} \left(\frac{4 x^{2}}{\left(d + 2 y\right)^{2}} + 1\right)^{2}} - \frac{8 x^{2}}{\left(d - 2 y\right)^{3} \left(\frac{4 x^{2}}{\left(d - 2 y\right)^{2}} + 1\right)^{2}} + \frac{2}{\left(d + 2 y\right) \left(\frac{4 x^{2}}{\left(d + 2 y\right)^{2}} + 1\right)} + \frac{2}{\left(d - 2 y\right) \left(\frac{4 x^{2}}{\left(d - 2 y\right)^{2}} + 1\right)}\right]$$ $$\sigma_{xy} = - 8 x \left[\frac{4 x^{2}}{\left(d + 2 y\right)^{4} \left(\frac{4 x^{2}}{\left(d + 2 y\right)^{2}} + 1\right)^{2}} - \frac{4 x^{2}}{\left(d - 2 y\right)^{4} \left(\frac{4 x^{2}}{\left(d - 2 y\right)^{2}} + 1\right)^{2}} - \frac{1}{\left(d + 2 y\right)^{2} \left(\frac{4 x^{2}}{\left(d + 2 y\right)^{2}} + 1\right)} + \frac{1}{\left(d - 2 y\right)^{2} \left(\frac{4 x^{2}}{\left(d - 2 y\right)^{2}} + 1\right)}\right]$$

and for strains

$$\begin{align} \epsilon_{xx} &= \frac{1}{E}(\sigma_{xx} - \nu \sigma_{yy})\\ \epsilon_{yy} &= \frac{1}{E}(\sigma_{yy} - \nu \sigma_{xx})\\ \epsilon_{xy} &= \frac{\sigma_{xy}}{G} \, . \end{align}$$

For displacements, there are two options that come to my mind.

1. Rewrite the stress function in polar coordinates, and then use the Mitchell solution for displacements. The stress function should look something like

$$\phi(r,\theta) = r\theta \sin\theta + \frac{2P}{\pi d}r^2$$

2. Integrate the strains

$$\begin{align} u_x = \int\epsilon_{xx} dx + f(y)\\ u_y = \int\epsilon_{yy} dy + g(x) \end{align}$$

with $2\epsilon_{xy} = \partial u_x/\partial x + \partial u_y/\partial y$, differentiate this equation w.r.t $y$ and $x$ and solve for $f$ and $g$.

References

1. Puttock, M. J., & Thwaite, E. G. (1969). Elastic compression of spheres and cylinders at point and line contact. Melbourne, Australia: Commonwealth Scientific and Industrial Research Organization.