Adding to @grfrazee's answer, there is an intermediate method, less rigorous than Rourke but more precise than simply dividing by the perimeter.
That is the trapezoidal method, where one divides the load being applied to each supporting member. The only relevant information is the boundary conditions of the plate: whether it is free, pinned or fixed. Usually, slabs are said to fix each other, so if you have a beam supporting two slabs, one to either side, then both slabs are considered fixed on that face (the only exception is if one of the slabs is a cantilever, in which case the cantilevered slab is considered fixed but the other is pinned, since a cantilevered slab won't resist rotations due to loading on the other slab). On the other hand, if a beam is supporting only one slab, then that slab-face is considered pinned. If the edge of a slab is not supported by a beam, then it is considered free, of course. This method doesn't work for slabs supported directly by columns. It also assumes the slab is under uniform loads which cover the entire slab and is not strictly valid for other loading conditions (concentrated loads or variable loading within a slab).
Now, if two adjacent and perpendicular faces of a slab are equally supported (both are pinned or both are fixed), then one draws a 45° line starting from their shared vertex. If, however, one face is pinned and the other is fixed, then the line must be 60° (from the fixed face). These lines describe the load division to each supporting member. (After a quick googling, I've discovered some people simply consider 45° in all cases. Do as you wish, but I think giving more load to the stiffer member makes sense. That being said, I can't prove it, nor can I demonstrate why 60° is a good angle, I simply learned this in college).
Extending these lines until they fully divide the slab area, you now know the fraction of load that will go to each face.
This can be easily understood with an example such as this one:
Here we have three slabs A, B and C. Slab A is supported by beams B2.1, B3.1 and B4.1 and has a free end (the dotted line). Slab B is supported by beams B2.2, B3.2, B4.1 and B5.1. Slab C is supported by beams B1, B2.2, B4.2 and B5.2. In this example, lets just look at the load distribution of Slabs A and B.
Slab A is free on one face, pinned on faces B2.1 and B3.1 and fixed on B4.1. This is because the stiffness of slab B on the other side of B4.1 impedes (or at least limits) the rotation around face B4.1, while the torsional rigidity of beams B2.1 and B3.1 is considered insufficient to influence the slab's rotation. Given the rules described above, lines are drawn from the vertices between the faces. In this case, since B4.1 is fixed and B2.1 and B3.1 are pinned, the lines must be at 60° from B4.1. The remaining load is distributed between the pinned faces. The load distribution is therefore:
For the same reasons as described above, Slab B is fixed on faces B2.2 and B4.1 (due to the slabs on the other side of the beams) and pinned on faces B3.2 and B5.1. Since both B2.2 and B4.1 are fixed, the line between them is at 45°. Since both B3.2 and B5.1 are pinned, the line between them is also 45°. Since B4.1 and B2.2 are fixed and B3.2 and B5.1 are pinned, the lines between them are at 60° from the fixed faces. The load distribution is therefore:
Now one can calculate how much load goes to each beam by multiplying the "tributary area" of the beam by the applied load. In the case of beam B3, for example, the applied load is:
$$ q_{B_3} = \begin{cases}
\dfrac{A_3 \cdot q_A}{L_{B_{3.1}}} \text{ along B3.1} \\
\dfrac{B_3 \cdot q_B}{L_{B_{3.2}}} \text{ along B3.2}
\end{cases}$$
where $q_{B_3}$ is the uniform load on beam B3, $A_3$ and $B_3$ are the tributary areas found in the figures and $q_A$ and $q_B$ are the loads applied in each slab (in force-per-area units). The loading of beam B3 is therefore
In the case of beam B4, the loads would be
$$ q_{B_4} = \begin{cases}
\dfrac{A_4 \cdot q_A + B_4 \cdot q_B}{L_{B_{4.1}}} \text{ along B4.1} \\
\dfrac{C_4 \cdot q_C}{L_{B_{4.1}}} \text{ along B4.2}
\end{cases}$$
(C4 is the tributary area of beam B4 in slab C, which hasn't been drawn). Along B4.1 one must obviously add the loads applied by both slabs supported by the beam.
These equations are somewhat simplified, since they consider the loading along the beam to be uniform, which isn't really the case. The loading should actually be trapezoidal according to the tributary area. To calculate this more precise form, the loading on the beam is found by
$$q(x) = p \cdot h(x)$$
where $q(x)$ is the value of the distributed load at point $x$, $p$ is the area-load on the slab and $h(x)$ is the height of the tributary area perpendicular to the beam's axis (displayed as $h_{3.1}(x)$ in the second figure above). The loading therefore becomes:
