# 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.

• I have looked at this problem using numerical methods (FEA) and I was curious if there is an analytical approach out there. – ja72 Jan 31 '17 at 19:31
• One caveat about this problem is the nonlinearity that appears due to the contact (the boundary conditions are changing). – nicoguaro Feb 3 '17 at 16:32
• We can assume a line contact and known force applied. The contact width can be assumed to be zero as a first pass. This is because I can get the contact stiffness between two parallel rods from Hertz. – ja72 Feb 3 '17 at 19:03
• The stresses near the two points of contact (or, rather, the two lines of contact) are of course going to be much higher than the stresses throughout the rest of the blue rod. So relatively early in the compression you're going to exceed the yield strength of the rod material in those localized regions, and once that happens there really isn't much hope of finding an analytical solution to the problem. (Unless the rod is made out of some material like jello or rubber which doesn't hit a yield point until fairly high strains.) – Samuel Weir Feb 4 '17 at 1:07
• But in a Hertzian sense the line contact has to have a width to spread the force making the contact pressure finite and the deflections calculatable. – ja72 Feb 4 '17 at 18:42

## 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

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

1. 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.
• I understand I can calculate the contact deflection as you shown above, what I am after is the deflections (strains) of the rod away from the contacts. I am hoping there is an analytical stress field out there to describe this scenario. – ja72 Feb 3 '17 at 20:31
• @ja72, well, you asked about the stiffness: What is the stiffness of a crushed rod?. – nicoguaro Feb 3 '17 at 20:33
• Exactly. I think what is described above is the contact stiffness (twice) if the rod was a half elastic plane, not a finite volume. The reference states it clearly -> at the point of contact. I am interested away from the point of contact. – ja72 Feb 3 '17 at 20:42
• @ja72, not really. The calculation consider a region of contact of $L$ by $2b= [2PR/V]^{1/2}$. Now, if you are interested in the stresses/strains far away from the contact you should really restate your question to make it clear. – nicoguaro Feb 3 '17 at 20:49
• @ja72, the solution for a disk with point load diametrally opposite might be helpful.: nbviewer.jupyter.org/github/nicoguaro/notebooks_examples/blob/… – nicoguaro Feb 3 '17 at 21:48