Q:

What component of the payload appears on the cross member?

I have a set of DIY steps; I've modelled it as the steps themselves "s", a short vertical support "a", a long vertical support "b" and a cross-member "c". Here's a picture:

Just to be sure I modelled it correctly, here is also a photo of the actual structure (it was built separately from the tree fort, please assume it's not connected to it).

I know that if you have a step-ladder in the shape of an 'A' then the cross piece of the 'A' shape carries a part of the load as the sides try to splay apart. But what of this case, where the supports A and B are vertical? If someone is standing on the steps, what component of their weight would find itself on the cross piece 'C'? How would you calculate it from the angles and lengths?

A:

TL;DR: We don't have enough information to say, but almost certainly less than 50% of the load placed on the ladder.
A simple model:
We can model the setup as a triangle:

Each length (beam) is made of wood and I'll assume they have a Young's modulus of $E=15$ GPa (typical of wood) and cross-sectional area of around $A_0=0.001$ m$^2$, which looks the right order of magnitude.
I'll also assume that the beams are connected together in a way that the angles can all be approximated as being fixed by harmonic potentials with a spring constant $k_\alpha$.
The initial (equilibrium) dimensions are $l_1^0=1.5$ m and $l_2^0=3.0$ and $\alpha_2^0=90^\circ$. And a weight $W$ is suspended from a length $rl_3$ along the ladder.
The potential energy of this system is then:
$$ U=Wrl_3\cos\alpha_1+\frac{EA_0}{2}\sum_i\frac{1}{l_i^0}(l_i-l_i^0)^2+\frac{k_\alpha}{2}\sum_i(\alpha_i-\alpha_i^0)^2 $$
The stable configuration is found by minimising $U$ with respect to the lengths $l_1,l_2,l_3$ (note that the three angles are uniquely determined by these lengths). And the resultant force experienced by the cross beam, expressed as a fraction of the load $W$, is
$$ f=\frac{1}{W}\frac{EA_0|l_1-l_1^0|}{l_1^0} $$
The results:
I numerically solved this equation to find the dependence of $f$ on $k_\alpha$ (how stiff the angle connections are) half-way up the ladder ($r=0.5$) and at the very top ($r=1.0$):

It's interesting to see that a maximum peak is obtained. Here's why: when the stiffness $k_\alpha$ is very small, there are no moments acting and therefore no weight is transferred to the cross beam. As you increase the angle stiffness, more weight is transferred to the cross beam. However, if the angle stiffness is extremely large, then the angle connections absorb more and more of the strain and therefore the force experienced by the cross beam decreases.
The result, according to this model, is that less than 50% of the load is transferred to the cross beam, and quite plausibly <1%, depending on the angle stiffness.
Caveats:
The stiffness of the individual angles will be different to each other in which case it may be possible to distribute more than 50% of the load to the cross beam (e.g. if $\alpha_3$ is very stiff but $\alpha_1$ and $\alpha_2$ are not). But this seems unlikely.
Also, some of the load will be absorbed by the wood bending but I've neglected this from my model. It wouldn't be difficult to modify the equations to account for this.

