11 сентября 2001 - Мейссан Тьерри - Страница 49

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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

2 x 2 x 12 + 2 x 20.16 = 88.3 cubic inches per foot for the double trusses.

Now we need to calculate the total length of double and transverse trussing. There were apparently, 60 double trusses spanning the 60 feet from the perimeter wall to (a beam attached to) the core and 24 double trusses spanning the 35 feet from the perimeter wall to the core. They are pictured in the following graphic:

The overall length of double trussing was thus 60 x 60 + 24 x 35 = 4440 feet. Transverse trusses ran perpendicular to the double trusses as illustrated:

The overall length of transverse trussing was thus 8 x 207 + 4 x 87 = 2004 feet.

There was also a lesser supporting feature called «intermediate support angle». Since all we know about the intermediate support angle, is that its support capabilities were inferior to the double and transverse trusses, we shall be generous and assume that it was similar in nature to the transverse trusses. This adds another 1764 feet, to give a total of 2004 + 1764 = 3768 feet of transverse trussing.

Hence, the volume of steel in the double trusses was 4440 x 88.3/1728 = 227 cubic feet.

And the volume of steel in the transverse trusses was 3768 x 39.7/1728 = 86.6 cubic feet.

So the overall per floor volume of steel in the floor support system was

150 + 13.3 + 227 + 86.6 = 477 cubic feet.

The total per floor volume of steel, is now the sum of that in the perimeter wall, the central core section and the floor system. This is 563 + 1210 + 477 = 2250 cubic feet.

So why have we gone to all this trouble to calculate the per floor volume of steel? Well, we know that 96,000 tons of steel was used in the construction of each of the WTC towers. The WTC towers had 117 floors (110 above and 7 below the Plaza level) so an average floor contained 96,000/117 = 820 tons of steel. Since the density of steel is 490 pounds per cubic foot, we see that each floor contained about 820 x 2000/490 = 3347 cubic feet of steel.

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