The exact deflection result is a function of the shelving's structural system. In the OP comments you mention that the two options would be either a wooden skin with metal edging or a wooden lattice skinned with wood.
Let's start with the wooden skin with metal edging. Structurally, this is equivalent to a slab with multiple supports. The sort of support depends on... the sort of support. If the shelf is fixed to a wall, then that shelf-face will have a fixed support (restricts vertical displacements and rotations). If the shelf is not fixed to a wall, then it must either rest on metal edging (which restricts the vertical displacements of the shelf but would allow for small rotations) which transmits the load to a column or there is no metal edging at all and the shelf itself needs to transmit the entirety of the load to the column.
For a rectangular shelf, you will have up to four supports, one on each face. Considering only more plausible configurations, you have the following possibilities:
- One face fixed on the wall*
- One face fixed on the wall, lateral faces with metal edging
- One face fixed on the wall, other faces with metal edging
- Two faces fixed on the wall (corner-shelf), other lateral face with metal edging
- Two faces fixed on the wall, other faces with metal edging
- All faces with metal edging
- All but one face with metal edging
- Both long faces with metal edging*
- Both short faces with metal edging*
- No edging, supported directly on columns.
As you can see, this is a long list and I've removed a few which I found to be implausible. For this reason, there is no simple equation that can be adopted. This is made worse by the fact that the behavior of slabs is defined by the following multivariate differential equation:
$$\dfrac{\partial^4w}{\partial x^4} + 2\dfrac{\partial^4w}{\partial x^2\partial y^2} + \dfrac{\partial^4w}{\partial y^4} = \dfrac{p}{D}$$
where $D = \dfrac{Eh^3}{12\left(1-\nu^2\right)}$. This is a pain to solve, so no one does.
There are two ways of getting around these equations. I found this article which shows a lot of the necessary math in each of the support configurations, but in practice one doesn't go to such lengths. I can't find any resources for the case of wooden slabs, but in reinforced concrete slabs there are different approaches obtained by smarter men that I (Marcus, Czerny, Nadai, etc) which result in dozens of tables of coefficients. These coefficients can then be used to figure out how much load is transmitted in the x-axis and how much in the y-axis. Knowing that, you can calculate the slab as two perpendicular beams, which should result in similar deflections. The biggest problem translating this to a wooden slab is wood's anisotropy.
The other way of getting around that ugly differential and the one I recommend is a finite element model. This is also probably the only one that can handle the "no edging, supported directly on columns" case and it can take into consideration the flexibility of the supports (in the case of the metal edging). It may seem like overkill but it's the best way to get a good estimate of what your actual deflections will be.
You'll notice that some of the options on the list are marked with an asterisk*. These cases are exceptions and can be quite trivially calculated, since they behave in a manner identical to that of beams. Also, long slabs (where one side is more than twice the other), can usually be assumed to be "infinite", meaning that they too can be considered equal to beams in the perpendicular direction (if $x \geq 2y$, then the slab works like beams with span $L=y$). For a beam or slab fixed on a wall with all other faces free, the maximum deflection is
$$\begin{align}
\delta = \dfrac{PL^3}{3EI} && \delta = \dfrac{qL^4}{8EI}
\end{align}$$
for a concentrated load $P$ and a distributed load $q$, respectively, where $L$ is the slab's span from fixed to free face.
A slab supported at two opposing faces is equivalent to a simply supported beam of equal span.
$$\begin{align}
\delta = \dfrac{PL^3}{48EI} && \delta = \dfrac{5qL^4}{584EI}
\end{align}$$
for a concentrated load $P$ at midspan and a distributed load $q$, respectively.
Now onto the lattice method. If the lattice is very dense, it will behave very similarly to the slab above. If it is less dense, then the behavior may differ significantly. What the threshold is, I can't say.
Assuming the lattice doesn't act like a slab, it will behave like a grid. Once again, this can't be trivially solved and the best solution is an analysis tool, though in this case the tool can be much simpler (grids can be solved analytically without the need of FEM).
