Consider a plate $P$ of thickness $d$ (for example a plate of wood) laying on a hard surface $S$ (for example on concrete floor). Suppose you have a maximum value of pressure $p_{\mathrm{max}}$ $S$ could tolerate before it fails in some way. Suppose further you know a pressure distribution acting on plate $P$ downwards (for example a given force $F$ acting on an area $A$ on $P$).

Then I want to calculate the pressure distribution on $S$ to see if $p_{\mathrm{max}}$ is respected everywhere. What would the idea to do so and what are the relevant formulas? What parameter of the material of $P$ determines of how the force acting on $P$ is spread out over a larger area resulting in lower maximal forces acting on $S$?

The original motivation for this question was to get a better understanding the physics of a olympic weightlifting platform and how to numerically model the difference between rubber and wood as a material for the platform in terms of reducing the impact force on the concrete floor and thus protecting the concrete. More specifically this quote was the starting point of this question to understand it deeper:

A good sheet of 3/4″ rubber is good in that it absorbs some of the force, but it only spreads out the force a little bit. That’s where a platform comes in. Wood is rigid and will spread the force out.

(from https://www.tworepcave.com/4543/why-olympic-lifting-platform/)

So for example consider a barbell that falls from height $h$ directly on the concrete floor resulting in a pressure distribution $p_0$ on the concrete floor.

Then lay a plate of rubber on the concrete floor and drop the barbell again on the rubber surface. Since the rubber deforms by a distance $s$, it enlarges the stopping distance and thus reduces the force on the concrete floor. Additionally it "spreads" the force a little bit over the area of the plate resulting in a pressure distribution (acting on the concrete floor) $p_{\mathrm{rubber}} < p_{0}$.

Now do the same with the wooden plate (same dimensions as the rubber plate). In this case it deforms much less, but "spreads" the force much better because it is stiffer than the rubber, the combined effect seems to be much better than in the rubber case (but I don't see to estimate the order of magnitude of difference between the materials), resulting in pressure distribution $p_{\mathrm{wood}}$ with $p_{\mathrm{wood}} \ll p_{\mathrm{rubber}} < p_0$.

I didn't find anything in the web so far discussing this or a similar problem more in depth (either theoretically nor experimentally). So if you have further references, it would be great to include it in your answer.


For the case of the barbell and the rubber tile, I would say that there for protection of the floor the rubber layer seems to have three main effects:

  • A "Cushioning effect": Since the barbell plate sinks a bit into the rubber tile, the contact area is increased (compared to the contact with the bare concrete), thus the force is spread over a specific area on the top of the rubber tile.
  • Increasing the "breaking distance": Since the rubber tile is softer than the concrete, it enlarges the breaking distance and thus reduces the total impact force significantly.
  • "Spreading the force over an even larger area" on the bottom of the rubber tile. This effect is usually utilized in weightlifting platforms where one uses relatively stiff wooden plates to "spread the force over a large area". Since the rubber is much less stiff than wood this effect will be much smaller than in the case of a wooden plate, but I guess that it also plays an important role when you use high density rubber tiles. However I am not able to quantify this effect.

I am not sure of how large the three effects are compared to each other. Maybe one can show that one effect is negligible.

  • $\begingroup$ Interesting question. Regarding similar problems: Consider for example the physics of helmets or other protection equipment. Here is a paper discussing the case of a mouthguard for example. $\endgroup$
    – student
    Commented Dec 27, 2020 at 11:32
  • $\begingroup$ If you have access to multiphysics software you could try to simulate it (something like cosmol multiphysics or solidworks etc.) Here is an example of such a simulation. However it doesn't match exactly your question since it doesn't consider the "force transmission" and "spereading" to another plate (in your case the concrete floor). $\endgroup$
    – student
    Commented Dec 27, 2020 at 11:42
  • $\begingroup$ Another similar context is the analysis of pavement design, see here or here for example $\endgroup$
    – student
    Commented Jan 4, 2021 at 21:32

2 Answers 2


There is a huge amount of parameters to that problem that make it impossible to derive general formulae:

  • shape of the barbell (especially how sharp the edges are) and moments of inertia
  • impact angles of the barbell on the floor
  • dynamic (velocity dependent/viscous) stiffness/damping of the platform materials
  • nonlinearity of the platform materials (especially progression in the case of rubber, but also plasticity for other materials)
  • anisotropy of the materials (especially wood)
  • boundary conditions of the platform
  • random effects like grain size distribution of the aggregate (for concrete) or permanent stresses/fracture mechanics (think of a platform made out of glass as an illustrating extreme example)

The maximum stress (or even the stress distribution) during impact very sensitively depends on all of these parameters.

For calculating such events, one usually uses finite element software, namely explicit time-dependent solvers (crash simulation in the automotive industry, or bullet impact in defense). A commercial product dedicated to this is LS-Dyna (I am not paid for mentioning that, and I don't particularly like the product either, just had to work with it in the past). There may be other products, which anyone may feel free to mention in the comments.

Trying to control the problem by deriving closed formulae is a waste of lifetime, because you will always have to focus on a specific view on the problem, and neglect others. For the special case of the static problem (which you should be cautious about substituting for the dynamic problem!), however, Hertzian contact theory might be of some insight with respect to general load distribution.

  • $\begingroup$ Do you have a good simplified model to get a better feeling for some parameters of the model? $\endgroup$
    – Sarah
    Commented Jul 7, 2021 at 10:23
  • $\begingroup$ @Sarah: as mentioned, a simplified model will neglect almost every aspect of the impact event. Besides that I have none. But if I had to develop a simplified model at all cost, I would start with the Hertzian contact theory. Especially the influence of contact surface curvature is of foremost importance as well as the nonlinear behavior of the elastic forces. $\endgroup$
    – oliver
    Commented Jul 7, 2021 at 17:04
  • $\begingroup$ I just posted a follow up question which focuses on a particular aspect of the problem, maybe there is a simple model to explain that particular aspect. $\endgroup$
    – Sarah
    Commented Aug 18, 2021 at 19:58

When a barbell is slowly placed on the plate, both are elastic deformed according to the elastic constants of the materials, the barbell weight, and the geometry of the contact.

But when it falls, the impact is not transmitted at once to all volume of the bodies to generate the same deformed geometry. The small region of the contact on the plate is forced to accelerate downwards, and that movement is communicated to the surroundings in the form of an elastic wave with a finite speed. The mathematical description is outlined in this answer.

The tensions are proportional to the deformations, as shown in https://physics.stackexchange.com/a/558821/195949, for the simplified case of uniaxial elastic waves. The wave speed is $v = \sqrt{\frac{E}{\rho}}$ in this case, where $E$ is the modulus of elasticity, and $\rho$ the density.

The 3D situation is more complex, but anyway the greater the stiffness of the plate, the greater is the wave speed. And the wave speed is a measure of how fast the impact is spread to the plate as a whole.

As the stiffness of rubber is much smaller than wood, much of the impact will be transmitted to the ground behind. Steel would be even stiffer, but I'm afraid that the barbell could be damaged by the impact.

The sound produced during the impact is a measure of the energy of the elastic waves. In the case of wood, part of the energy is also converted in heat, associated to a permanent deformation in the place of contact.

  • $\begingroup$ Thanks. You explain the difference between the static and dynamic case, but how do you get for example the pressure distribution to see if $p_{\mathrm{max}}$ is respected everywhere? $\endgroup$
    – Sarah
    Commented Dec 27, 2020 at 11:28
  • $\begingroup$ If even the static pressure produced by the weight of the barbell is comparable to the strenght of the ground, the analysis must be different. But I presume that it is not the problem, and the ground can support easily the pressure (if the weight is slowly placed there). The problem is the energy of the impact, that is related to waves and its dissipation.. $\endgroup$ Commented Dec 28, 2020 at 21:11
  • $\begingroup$ If you drop a barbell (say with steel plates) on a bare concrete floor, the floor is likely to break since the impact force will be extremly high (tiny stopping distance, small contact area). I.e. the max admissible pressure of the concrete surface will be exceeded. Let the impact force be $F_i$. Now consider a barbell with (the extremly high) mass $m := \frac{F_i}{g}$ just resting on the floor. Then this resting barbell would equivalently exceed the max pressure of the floor. This way I tried to connect the dynamic problem to a static problem with an "equivalent" mass of a static barbell. $\endgroup$
    – Sarah
    Commented Dec 28, 2020 at 21:47
  • $\begingroup$ @Sarah: even if you could derive a rough estimate of the stopping distance, the problem is that there can be genuinely dynamic effects. Think of viscosity of fluids transferred to solids as an illustrating example for these effects. Just as you cannot treat a viscous fluid flow like a static fluid, you cannot treat a solid with those dynamic effects as a statically loaded solid. Sometimes these effects might be tiny, at other times they may be significant. $\endgroup$
    – oliver
    Commented Jul 4, 2021 at 8:21

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