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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.

This gives a volume of 67.2 x 12/40 = 20.16 cubic inches per foot of truss.

Pictured above, is the connection of one of the double trusses to the perimeter wall. The cross section marked X—X in this graphic, is pictured below. Note that the original graphic from the WTC-report was so out of scale, that it was necessary to stretch it somewhat.

The first image below is apparently the real life version of the above graphic (supposedly obtained from the WTC wreckage). The second image shows the gusset plate and seat connection.

The dimensions quoted in the following section were made by taking measurements from these two photos. Standard adjustments for perspective had to be made for measurements from the second photo.

The gusset plate is 4 x 2 x 3/8 and has a volume of 3 cubic inches. The seat angle has a volume of roughly 2 x ((9 + 4) x 14.5 x 3/8) = 141 cubic inches and the

«stiffeners» add another 9 x 1.5 x 3/8 = 5 cubic inches. Since there were (at most) 120 gusset plates and seat angles, these add in 120 x 149 = 17880 cubic inches. The 76 horizontal diagonal brace plates add in another 76 x 90 x 3/2 x 1/2 = 5130 cubic inches for an addition of (17880 + 5130)/1728 = 13.3 cubic feet of steel to our total.

The upper chord (top section) of one of the double trusses consisted of four pieces of 1/8 inch thick angle iron, as illustrated below (it is circled in red).

Below, is a more detailed view of the cross section of the top chord of a transverse truss (left) and double truss (right).

So, the upper chord has a cross sectional area of

((2 + 1.25) + (1.25 + 2))/8 = 0.8125 square inches for a transverse truss and,

((2 + 1.25) + (1.25 + 7 + 1.25) + (1.25 + 2))/8 = 2 square inches for a double truss.

Since we have no information concerning the lower chord (and the «official» pictures are inconsistent and nowhere near to scale) we will assume it has the same dimensions as the upper chord.

Now summing the volume of steel in the top and bottom chords and diagonal rods, we have the following per foot volumes:

2 x 0.8125 x 12 + 20.16 = 39.7 cubic inches per foot for the transverse trusses, and

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