And I am interested in seeing Heiwa's rebuttal.
First you must decide what you are going to model, e.g. a steel tower structure where part C one way crushes part A, when C is dropped on A by gravity; C = 1/10 A, A carried C before.
How to go about it? And does size/scale matter? Let's do it full scale!
First select a suitable building standard to get the design stresses and the redundancy right. Let's also simplify and that only static design stresses due to gravity loads are considered (in this example).
Let's start with a very simple tower 100 floors high. It has a core of 8 vertical columns (100 floors high) in a square; three columns each side; 4 corner core columns and 4 intermediate columns. At every floor the core columns are interconnected with horizontal/slooping beams for redundancy reasons. The core is like a big mast of 8 columns. It is evidently self-supporting due to the beams between the core columns.
The core is surrounded by perimeter wall columns; say 7 each side, thus total 24! At every floor the perimeter core columns are interconnected with horizontal spandrels for redundancy reasons.
All the columns are primary, load bearing elements. At present no loads (except own weights) are applied to the columns. The horizontal beams and spandrels are secondary elements installed for redundancy reasons and act as support to keep critical stresses (buckling) low in the columns (as per the standard).
Now we connect the core columns with the perimeter columns with horizontal floor trusses! 5 perimeter columns at each wall corner are connected to one core corner column (thus 5 trusses) and the intermediate core columns are connected by one truss to the remaining, intermediate perimeter columns.
Thus there are 24 floor trusses between perimeter columns and core; 4 x 5 trusses connected to the core corner columns, 4 just between remaining columns. You follow? The trusses are just bolted to the columns. The trusses are secondary elements.
This can be done in any scale, size.
Now put a thin steel plate over all floor trusses and add your loads on the floor trusses.
These loads are evidently transmitted to the columns that compress the columns. Now ensure that the static stress in each column is according to standard. Evidently the corner core columns carry more load than any other column but ... the stresses are the same. The columns get heavier lower down.
Now check the structure for redundancy! Say that the criteria according to standard is that you can remove any
one column between any floor and that the spandrels and the horizontal beams shall be able to transmit the column load around the 'failure' (the removed column bit) to adjacent coumns, that will not fail. It means, e.g. that the intermediate core columns become quite strong.
OK, your structure is ready. In any size, scale. The static stresses and the redundancy are as per standard.
Now we are going to remove all columns, one after the other, between parts C and A (so we can drop C on A).
Let's say we first remove one centre, the 4th middle one, perimeter wall column in one wall. No problem - there is redundancy enough.
Then we remove the two adjacent perimeter wall columns. What happens? Well, it is a possiblity that the 4th middle perimeter wall column above (and its local loads) will drop down to ground, the spandrels
above connected to the 4th middle perimeter column shear off, unless the floor trusses (and the hat truss!) above can pull it (and the load!) in position.
Regardless, it seems that part C will be severely locally stressed and possibly damaged when we start removing columns between parts C and A, thus before C is even dropped on A.
This is another reason why part C cannot one way crush down part A. C is simply less strong than A and subject to bigger local loads due to local failures between C and A. There are many others. And none of them has to with scale or modelling. It is just simple structural damage analysis that is required and it is the same for any structure.
Anyone suggesting that you can remove all columns between parts C and A and then drop C on A is a fool. C is locally damaged before that, parts of C and loads should drop off an the remaining parts of C cannot crush down A.
Read my papers to get a better feel for the problem.
