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If 50 minutes of relatively mild fire can bow the exterior wall to the point of collapse, why I can't see any sign of deformation in this photo?
According to NIST - both of these areas should be stripped of the fireproofing ( east part of floors 78-84 and west part of the floors 98,97.
So, 56 minutes of mild fire acting upon "naked" trusses should cause extensive bowing to the point of failure, but 75 minutes of severe fire acting on another naked trusses did not result in one/two decimeters of bowing?
In addition to the 42 inch sagging of the main floor trusses allowed to occur due to the removal of the bridging trusses from between the core and perimeter in their model, the NIST added artificial 5 kip lateral loads on the south wall perimeter columns of WTC 1, as the model apparently did not produce the inward bowing with the sagging trusses. I only read 5 kips per column not 5 kips per floor per column, but I'll have to check again. Interestingly, they then say the maximum inward bowing they generated was 31 inches.
I just calculated the concentrated force required to generate a center span deflection of 55 inches in a 14 inch square steel beam supported at both ends that was five stories in length and with a .289" wall thickness like those at the 98th floor of WTC 1 would have had. I used a story height of 149 inches so the beam is 745 inches long. The equation is
deflection = (force x length^3)/(48 x modulus of elasticity x moment of inertia)
modulus of elasticity for steel = 29 x 10^6 psi
moment of inertia for a 14 inch square beam with .289" wall thickness = 496.8 in^4
length = 745 inches
to find force just manipulate the equation with the deflection being 55 inches
force = (deflection x 48 x modulus of elasticity x moment of inertia)/(length^3)
The answer I get is about 91,989 lbs. or 92 kip. The perimeter columns in the towers would actually require a little more as they would have help from the spandrels.
For 31 inches of inward bowing the lateral force required for five unsupported stories is 51,848 lbs. or 52 kip. Did the NIST make an order of magnitude mistake or did they have 25 kips of catenary force with the trusses and added 5 kips per floor per column to get another 25,000 lbs. per column artificially? I know Dr. Bazant was off by a factor of ten with his calculation of the axial stiffness of the columns in the towers. He shows 71 GN/m and we calculated 7.1 GN/m.
The average weight per floor of a 1000 row column at the 98th floor was about 4,200 lbs.. The floors outside of the core were about 31,000 square feet and weighed about 2,500,000 lbs giving 80 psf. The south wall side between the core and perimeter was 137 foot long x 60 foot wide giving an area of 8,220 square feet and at 80 psf would weigh about 657,600 lbs.
If a five story section of the 1000 row columns was cut and dropped about 1 story, generating an angle of 12 degrees with the floor, then the lateral load on the perimeter columns directly in front of the core would be about 4.7 times that of the vertical load due to catenary action. So given the 4,200 lbs./column per story (33,600 for eight columns) and 657,000 lbs. floor weight between the core and perimeter and five stories of floor you get 690,600 lbs. x 4.7 x 5 = 16,229,100 lbs.
This would then be applied to 40 perimeter columns which are directly across from the core. So you would get about 400,000 lbs./column of lateral load. This could definitely do the job.
So it could be done with cutting the 1000 row columns and leaving them hang and you would only need a couple stories of 1000 row columns completely hanging to get 55 inches. With the spandrels involved the bowing would also be parabolic to some degree with the worst being in the center.
These calculations also show a very large force could be applied to pull in the perimeter columns with a couple of stories of the entire central core and all of the floors going down.
If perimeter column inward bowing did occur on the south wall of WTC 1 for minutes before collapse I believe 1000 row cutting to be the only viable explanation. There is no way that the hat truss could have applied the extraordinarily high vertical load necessary to do it and it again raises the issue of why any inward bowing caused by it wouldn't occur on the 106th floor. The fact that the NIST needed to artificially add lateral loads shows the floor truss sagging could not do it either.
The load would predominantly transfer to the east and west perimeter walls through the spandrels and the NIST analysis itself shows the additional load on the southern columns of the east and west walls is only 69% on columns which can take 500% of their design load. Additionally, a rigorous analysis would also show that the central core columns wouldn't be affected much by a south wall failure.
COLUMN RESPONSE TO SHORTENING
To find answers we need to know how columns react to being overloaded. A force a typical column can transmit is shown as a function of column shortening below.
The elastic region is in yellow. The plastic region in pink. Columns have strong spring-back abilities when compressed. In order to buckle one from overload it is necessary to compress it past it's natural ability to spring back. The high spike in the graph between the yellow and pink regions shows the minimum downward force necessary to buckle the column (Fo). It is quite high relative to the normal load the column handles (Fc), meaning we need substantial additional force (2x to 5x Fc) applied downwards to buckle a column.
Dr Bazant explains his graph (from the paper):
"Let u denote the vertical displacement of the top floor relative to the floor below
(Figs. 3 and 4), and F(u) the corresponding vertical load that all
the columns of the floor transmit. To analyze progressive collapse,
the complete load-displacement diagram F(u) must be
known (Figs. 3 and 4 top left). It begins by elastic shortening and,
after the peak load F0, curve F(u) steeply declines with u due to
plastic buckling, combined with fracturing (for columns heated
above approximately 450°C, the buckling is viscoplastic). For
single column buckling, the inelastic deformation localizes into
three plastic (or softening) hinges (Sec. 8.6 in Bažant and Cedolin
2003; see Figs. 2b,c and 5b in Bažant and Zhou 2002a). For
multistory buckling, the load-deflection diagram has a similar
shape but the ordinates can be reduced by an order of magnitude;
in that case, the framed tube wall is likely to buckle as a plate,
which requires four hinges to form on some columns lines and
three on others (see Fig. 2c of Bažant and Zhou)."
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