24-08-2013, 10:22 AM
With apologies to Dr. Seuss:
Jeffrey Orling at 515:
When the core columns at floors 93-5 lost their ability to carry load the columns above would be hangin (in tension) from the hatt truss. This meant that instead of the loads bearning DOWN.. they were being HUNG from the hat truss.
Tony Szamboti at 523:
Jeffrey, when the core load is applied to one side of the hat truss over the arm of the A-frames that generates a bending moment in the A-frames which they could not take. They would fail as soon as that load was applied. The connection at the perimeter side could not take the shear load either.
Regarding the location and extent of damage, Tony Szamboti:
The failure was not over three stories it was at the 98th floor and went across the building on that floor in a 250 millisecond time frame. In my opinion, it would be nothing short of magic if this was a natural event.
Regarding the strength of the hat truss in relation to the role ascribed to it by Jeffrey Orling, Tony Szamboti submits calculations above:
The 12 story upper section of the North Tower weighed about 73 million lbs. and the core load would have been about half of that at 36.5 million lbs.. So if we have 16 outrigger A-frames you are proposing that they could take 2,281,250 lbs each and the longer distance 60 foot span outriggers would have a moment of 2,281,250 lbs. x 60 feet = 136,875,000 ft-lbs. or 1,642,500,000 in-lbs. of torque applied to them trying to bend them. That is 1.6425 billion in-lbs. of torque in case you don't quite follow.
The maximum yield stress of the medium grade steel which would have been used is about 50,000 psi. Since bending stress = MC/I, lets see how deep a 2 foot wide solid rectangular beam we would need to be to take the load you are saying the outriggers could take
50,000 psi = (1,642,500,000 in-lbs. x depth/2) / (1/12 x 24 x depth ^3) and we can get depth by itself as
50,000 psi = (1,642.500,00 in-lbs. x depth/2) /(2 x depth^3) = (821,250,000 x depth) / (2 x depth^3) = 410,625,000/depth^2 so
depth = sq. root [410,625,000/50,000] = sq. root [8212.5] = 90 inches.
That is a 7.5 foot deep x 2 foot wide solid beam and you would need 16 of these to transfer the core load to the perimeter. The beams themselves would weigh nearly half a million lbs., and I didn't consider the self weight in the calculation. The A-frame outriggers were 3 stories tall (36 feet) but they were far from solid and were never meant for the kind of bending stress dumping the core on them would apply. Those outriggers could not take the bending you want to impose on them and in reality they failed before they ever transferred the core load to the perimeter, as we can see with the antenna coming down before the roofline. The roofline then came down because the core pulled the perimeter inward where it was falling at the 98th floor. Your theory doesn't explain why the perimeters would fail at the 98th floor either.
With a 4 inch thick web I-beam (which would be an enormously thick web) the depth would need to be 222 inches or 18.5 feet deep. A 3 inch thick web I-beam would need to be 256 inches or 21.3 feet deep to take the bending stress you want to put on them and that is just barely taking it with no margin. If you had a 1.5 margin the 4 inch web would need to be 272 inches or 22.67 feet deep and a 3 inch web with a 1.5 margin would need to be 314 inches or 26.17 feet deep. These are solid I-beams which the outriggers certainly were not. The outriggers could not have been able to take even half the stress imposed by the core load on them with their fulcrum at the perimeter. Additionally, these beams I am talking about would be full height or depth across the full span. The maximum stress would have been at the perimeter side and the A-frames tapered to about 2 foot deep beams at that point. The bending resistance is a function of the depth cubed so a 2 foot deep beam is about 64 times weaker in bending than an 8 foot deep beam. The outriggers were designed to transmit antenna wind loads out to the perimeter and would have worked fine in that capacity.
The video supports this 98th floor failure in rapid order, while the calculations eliminate the participation of the hat truss in a 93-95 floor core failure--itself not explained: as the required heat was not present to weaken the steel, and the few elements of the aircraft are not of adequate size or number to suffice.
Hence, the continuing probability of demolition charges supplants the theory of a mythically strong hat truss reacting to mythic damage at lower cores.
That Jeffrey responds with repeated accusations of Tony lying or making up emails to the contrary notwithstanding.
Controlled demolition persists as the likely cause of collapse in view of absence of heat and demonstration by calculation.
The official explanation is served by Jeffrey arguing over the years against controlled demolition while failing to provide an alternative.
Jeffrey Orling at 515:
When the core columns at floors 93-5 lost their ability to carry load the columns above would be hangin (in tension) from the hatt truss. This meant that instead of the loads bearning DOWN.. they were being HUNG from the hat truss.
Tony Szamboti at 523:
Jeffrey, when the core load is applied to one side of the hat truss over the arm of the A-frames that generates a bending moment in the A-frames which they could not take. They would fail as soon as that load was applied. The connection at the perimeter side could not take the shear load either.
Regarding the location and extent of damage, Tony Szamboti:
The failure was not over three stories it was at the 98th floor and went across the building on that floor in a 250 millisecond time frame. In my opinion, it would be nothing short of magic if this was a natural event.
Regarding the strength of the hat truss in relation to the role ascribed to it by Jeffrey Orling, Tony Szamboti submits calculations above:
The 12 story upper section of the North Tower weighed about 73 million lbs. and the core load would have been about half of that at 36.5 million lbs.. So if we have 16 outrigger A-frames you are proposing that they could take 2,281,250 lbs each and the longer distance 60 foot span outriggers would have a moment of 2,281,250 lbs. x 60 feet = 136,875,000 ft-lbs. or 1,642,500,000 in-lbs. of torque applied to them trying to bend them. That is 1.6425 billion in-lbs. of torque in case you don't quite follow.
The maximum yield stress of the medium grade steel which would have been used is about 50,000 psi. Since bending stress = MC/I, lets see how deep a 2 foot wide solid rectangular beam we would need to be to take the load you are saying the outriggers could take
50,000 psi = (1,642,500,000 in-lbs. x depth/2) / (1/12 x 24 x depth ^3) and we can get depth by itself as
50,000 psi = (1,642.500,00 in-lbs. x depth/2) /(2 x depth^3) = (821,250,000 x depth) / (2 x depth^3) = 410,625,000/depth^2 so
depth = sq. root [410,625,000/50,000] = sq. root [8212.5] = 90 inches.
That is a 7.5 foot deep x 2 foot wide solid beam and you would need 16 of these to transfer the core load to the perimeter. The beams themselves would weigh nearly half a million lbs., and I didn't consider the self weight in the calculation. The A-frame outriggers were 3 stories tall (36 feet) but they were far from solid and were never meant for the kind of bending stress dumping the core on them would apply. Those outriggers could not take the bending you want to impose on them and in reality they failed before they ever transferred the core load to the perimeter, as we can see with the antenna coming down before the roofline. The roofline then came down because the core pulled the perimeter inward where it was falling at the 98th floor. Your theory doesn't explain why the perimeters would fail at the 98th floor either.
With a 4 inch thick web I-beam (which would be an enormously thick web) the depth would need to be 222 inches or 18.5 feet deep. A 3 inch thick web I-beam would need to be 256 inches or 21.3 feet deep to take the bending stress you want to put on them and that is just barely taking it with no margin. If you had a 1.5 margin the 4 inch web would need to be 272 inches or 22.67 feet deep and a 3 inch web with a 1.5 margin would need to be 314 inches or 26.17 feet deep. These are solid I-beams which the outriggers certainly were not. The outriggers could not have been able to take even half the stress imposed by the core load on them with their fulcrum at the perimeter. Additionally, these beams I am talking about would be full height or depth across the full span. The maximum stress would have been at the perimeter side and the A-frames tapered to about 2 foot deep beams at that point. The bending resistance is a function of the depth cubed so a 2 foot deep beam is about 64 times weaker in bending than an 8 foot deep beam. The outriggers were designed to transmit antenna wind loads out to the perimeter and would have worked fine in that capacity.
The video supports this 98th floor failure in rapid order, while the calculations eliminate the participation of the hat truss in a 93-95 floor core failure--itself not explained: as the required heat was not present to weaken the steel, and the few elements of the aircraft are not of adequate size or number to suffice.
Hence, the continuing probability of demolition charges supplants the theory of a mythically strong hat truss reacting to mythic damage at lower cores.
That Jeffrey responds with repeated accusations of Tony lying or making up emails to the contrary notwithstanding.
Controlled demolition persists as the likely cause of collapse in view of absence of heat and demonstration by calculation.
The official explanation is served by Jeffrey arguing over the years against controlled demolition while failing to provide an alternative.