To compute the expected number of intersections, use the fact that expectation is additive. The expected number of intersections is just $\binom{N}{2}p$ where $p$ is the probability of an intersection. 

To compute the probability of an intersection you can make your life easier (with essentially no cost to the accuracy) by assuming your surface/line wraps around. This means there are no special places on your surface. Now you need to compute the probability that two rods intersect. 

In one dimension imagine the first rod is placed somewhere. What is the probability that the second rod intersects it? $2L/A$ (either the second rod is placed so its left end lies inside the first rod; or that its right end lies inside the first rod).

In two dimensions, fix the position of the first rod. Then suppose the second rod is inclined at angle $\theta$ to the first rod (we can assume that $0<\theta<\pi/2$). We now need the area of the set of positions of the "top left" vertex of the second rod such that there is an intersection with the first rod. This set of positions is an octagon (whose area I would expect to be able to compute using Mathematica). It should be of the form $C+D\cos\theta+E\sin\theta$ for constants $C$, $D$ and $E$ depending on $W$ and $L$. (In the case where $W\ll L$ it reduces to a rhombus for which the area is just $L^2\sin\theta$). The intersection probability is then just $(2/\pi)$ times the integral of the area over $\theta$ divided by $AB$.