12 x 21.5/144 = 1.792 cubic feet.

In total there are 240 such columns, so the volume of steel so far is

240 x 1.792 = 430 cubic feet.

Now lets deal with the volume of steel in the column end plates. Each end plate is 14 inches wide by 11.75 inches deep and 1.375 inches thick, giving a volume of

14 x 11.75 x 1.375 = 226.2 cubic inches = 226.2/1728 = 0.130896 cubic feet.

Since, on each floor, one third of the columns are joined, and each join involves two end plates, the per floor volume of steel in the end plates is

2 x 0.130896 x 240/3 = 20.9433 cubic feet.

The spandrel plates are large, being 52 inches high and 3/8 inches thick. Each floor has the equivalent of one spandrel beam that stretches 4 x 207 = 828 feet right around the building. The volume is easily calculated to be

828 x 12 x 52 x 3/8 = 193752 cubic inches = 193752/1728 = 112.125 cubic feet.

So the overall per floor volume of steel in the perimeter wall is

430 + 21 + 112 = 563 cubic feet.

Now, we wish to calculate the per floor volume of steel in the core section of the building. To do this, we first need to calculate the volume of steel in each of the core columns. This is complicated by the fact that the dimensions of the columns reduced in size with increasing height. For example, at the base of the WTC some of these columns were 36 inches wide by 16 inches deep and 4 inches thick, whereas at the top, these box columns had transitioned to H-sections (I-sections) fabricated from 3/4 inch steel (the transition to H-sections occurred at floor 85). We will ignore the reduction in width and breadth of the columns, and only take into account the reduction in column thickness by assuming an average thickness of 2 inches (this roughly corresponds to a reduction in thickness of one quarter of an inch, every seven floors, up to floor 85). In reality, the column width and breadth decreased quite considerably and we only make this very generous assumption as the actual reductions in the width and breadth are unknown. So, we assume each core column has the following cross-section:

The cross-sectional area is (36 + 12 + 36 + 12) x 2 = 192 square inches = 192/144 = 1.333 square feet. Since each floor is 12 foot high, the per floor volume of steel in one such column is 12 x 1.333 = 16 cubic feet. Reports as to the number of core columns vary from 44 to 47. Once again, we will be generous in our assumptions and choose the higher figure of 47. Thus, the total volume of steel (per floor) in the core columns is

47 x 16 = 752 cubic feet.

On each floor, the core columns were bound together by a rectangular grid of beams. As the dimensions of these beams are not known we will assume they were, 14 inch by 14 inch box sections fabricated from 3/4 inch steel. Again, this is a very generous assumption. The cross-sectional area of such a box section is:

( 2 x 14 x 0.75 ) + ( 2 x 12.5 x 0.75 ) = 39.75 square inches = 39.75/144 = 0.276 square feet.

The core section is 137 feet wide x 87 feet deep. Hence, our rectangular grid comprises six 137 foot sections and eight 87 foot sections, for a total length of 822 + 696 = 1518 feet. Additionally, the outer two 137 foot sections have to extend to the perimeter wall (to give support for the trusses). Actually, the «official» version has a much smaller U shaped beam, but as I have mentioned above, we are being very generous. This adds another 140 feet to the length. The volume of the 1518 + 140 = 1658 feet of box section is:

1658 x 0.276 = 458 cubic feet.

Thus the overall volume of steel in the core section is:

752 + 458 = 1210 cubic feet.

We now turn our attention to the floor support system.

The floor slab was poured on 1.5 inch corrugated 22-gauge steel decking. Now, 22-gauge steel is 0.0336 of an inch thick. The corrugations lead to 1.25 square feet of steel decking for every square feet of floor slab. Hence, the volume of steel involved is:

207 x 207 x 1.25 x 0.0336/12 = 150 cubic feet.

To complete our calculations, we need to calculate the volume of steel used in the system of trusses which supposedly supported the concrete floor slabs. The following graphic illustrates the truss system. The double trusses (of which, in this graphic, we only have an end view) ran perpendicular to the transverse trusses, and were essentially two transverse trusses bound together.

Consider one of the 3 foot four inch (40 inch) sections illustrated in the above graphic. The diagonal rod has a diameter of 1.09 inches (radius 0.545 inches) and a length of twice the square root of 20 squared plus 30 squared, that is, a length of

2 x srt( 20^2 + 30^2 ) = 2 x srt( 1300 ) = 72 inches.

Here, srt stands for the square root.

The cross-sectional area of the rod is 3.14 x 0.545 x 0.545 = 0.933 square inches. Hence the volume of rod in this segment is 72 x 0.933 = 67.2 cubic inches.