Тьерри Мейссан - 11 сентября 2001 Страница 42

The «Truss theory» is ludicrous

The truss theory is the absurd belief that the only support (between the central core and the perimeter wall) for the concrete floor slabs, was lightweight trusses. It was invented to explain away what were obviously demolitions and has become the «official» dogma. The central core, perimeter wall and the mythical trusses are all introduced in the next section. There you will find out their dimensions, their numbers and their supposed usage. After reading the rest of this article you should return to this section and (with improved understanding) read it again.

According to the «official» story, there is no significant lateral support for the walls (against wind loading) between the ground and top floors. This is like a bridge with a 1,300 foot span between supports. Even though the tube structure of the perimeter wall was designed for maximum rigidity (within the given weight specifications) the 1,300 foot span between supporting pillars, meant that even this very rigid design would sag in the midsection under wind loading, just like a bridge with such a span. In a typical steel framed building the span between pillars is only 12 feet (one floor) and such a problem does not arise.

The World Trade Center towers were like huge sails in the wind. These sails had to be able to resist the 140 mile per hour winds of a hurricane. Such hurricane force winds exerted a large (some 6000 tons) lateral force on the building. This lateral force is called the wind loading (or force of the wind) on the building. According to the «official» story, the only possible intermediate support comes from the flimsy trusses and the lightweight concrete floors. The WTC was designed to survive a 45 pounds per square foot, wind loading. This translates to a 12 x 207 x 45/2000 = 56 ton force on each of the floor segments. What this 56 ton force on each floor segment means, is that if one was to lay the World Trade Center on its side and use the pull of gravity as a substitute for the push of the wind, then each of the 110 floors would need to be loaded with a 56 ton block of steel (so the entire wall would have to support 110 such blocks of steel, that is, 110 x 56 = 6160 tons in total).

The fact that the tubular structure of the walls is very rigid, does not stop the central core from needing to bend when the walls bend. This means that the walls have to transmit the full force of the wind to the core, so that the core will flex to the same extent as the walls (this is obvious, otherwise if the walls flex while the core does not, the floor slabs would, by definition, be crushed). Again, it is important to note that the rigidity of the walls does not protect the central core from the full force of the wind, what it does, is it limits the distance that the walls (and hence the whole structure) can bend. The more rigid the design the less it tilts in the wind.

In strong winds the midsection of the windward wall will be pushed several feet towards the core. In a typical steel framed building of WTC type design, heavy steel beams transmit the wind loading to the core, which then bends together with the walls. However, in the WTC (as described in the «truss theory») the trusses and floor slabs are too weak to transmit this force to the core without buckling, so the core will stay in its original position as the wall advances to it. This will crush the trusses and floor slabs, leading to the collapse of many floors. Since this did not occur during the 30 years in which the buildings stood, we must assume that the «official» story is false. To see how utterly ridiculous the «official» story is, lets calculate the lateral loading (wind loading) that each one of these trusses was expected to resist. Consider, a one floor segment. Here, we have 30 trusses and a slab of concrete supporting 56 tons. That is about 2 tons per truss and piece of slab. If you balanced a 2 ton block of steel on top of one of these flimsy 60 foot long trusses and (a 60 foot long by 6 foot 8 inches wide by 4 inches thick) slab of concrete, we all know what would happen – the truss and slab would buckle and collapse.

Another point to consider, is that if the walls alone handle lateral loading, then the pressure on the windward wall must be transmitted via the corners to the remaining walls (this transmission of loading to the other walls is what gave the WTC its rigidity) but corners are far too weak to handle this task alone.

Although the «truss theory» is ludicrous, it has been pushed by many «experts». It should be noted that it is inconceivable that these experts did not know that it was false.

Where is the steel?

Since the trusses are incapable of resisting the wind loading, we know that the «official» explanation of the WTC collapse is false. If the floor joists (supports) were not the claimed trusses, then what were they? They had to be strong enough to support the floor slab and stiff enough to resist the wind loading. In fact, they had to be large steel beams. This is not to say that trusses were not used at all in the construction, but just that (contrary to the «official» line) the main floor joists were steel beams and not trusses.

The above argument using wind loading is certainly enough to tell one that trusses were not really used as the floor joists, but there are also other ways to determine this. Another approach is adopted in this section. We will:

• Calculate the weight of steel theoretically used in the construction of one of the towers assuming that the floor joists were trusses.

• Compare the result of this calculation to the 96,000 tons of steel known to have been used in the construction of each of the towers.

• Note that the calculated weight of steel is only 67 percent of the required 96,000 tons.

• Conclude that the 32,000 tons of steel unaccounted for, is due to the fact the the floor joists were actually weighty steel beams and not flimsy trusses (and thus that the official story is a lie spun to explain away what were obviously demolitions).

• Calculate a rough cross-sectional area for the steel beams that did serve as floor joists.

Since a cubic foot of steel weights 490 pounds, it is enough to deal with volumes rather than weights. We will calculate the volume of steel on a per floor basis.

To calculate the per floor volume of steel used in the construction of the twin towers, we will divide the calculation into three parts, namely, the volume of steel in the perimeter wall, the volume in the central core and the volume used in the floor support system.

The perimeter wall was comprised of box columns welded to large spandrel plates. Two typical prefabricated sections are illustrated below. Each consists of three spandrel plates welded to three box columns and each is three floors high.

The first figure below shows the cross section of one of the perimeter box columns and its surrounds. The second and third figures detail the dimensions of two actual perimeter columns that were salvaged from the rubble.

The numbers in the figure denote:

• 36 – the steel column

• 38 and 39 – fire resistant plaster

• 40 – aluminum facade

• 42 – window glass

• 43 – the window frame.

To obtain an estimate of the «typical» perimeter column, the dimensions of the perimeter columns listed in the WTC Steel Data Collection documentation were averaged. Whether this accurately reflects the true distribution of perimeter column thickness, is unclear, but it is all one has to go on (till those who hold the architectural details release them).

So, our «average» perimeter column has dimensions:

d = 13.4, t_w = 0.48, b_f = 12.9, t_(tf) = 0.32 and t_(bf) = 0.32.

and cross-sectional area:

2 x (13.4 x 0.48) + (12.9 x 0.32) + (14 x 0.32) = 21.5 square inches,

The parameters d, t_w, b_f, t_(tf) and t_(bf) are as in the following diagram from Appendix D which is part of the report found at http://www.house.gov/science/hot/wtc/wtcreport.htm.

For the time being we will ignore the column end plates and the spandrel beams. Since each floor is 12 feet high, the per floor volume of steel in an average perimeter box column is:

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.

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