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As we have seen, the plane would have to crash into the 2nd floor from the roof so no progressive global collapse ensues.
Originally posted by -PLB-
Why stop at 4.2GJ? If the plane crashed into the 31st floor from the roof (similar to the other tower), you end up with a kinetic energy of 4.34GJ at the moment the top impacts of the next floor. And you will have progressive collapse again.
Of course it would, because you're back to arguing that once one floor fails, global collapse ensues. It doesn't when the FoS is greater than 1, because the impacting mass could pick up that speed only once and each floor it crushes would decelerate it because each floor would be designed to withstand ten times the force of the mass in a static situation. It would dissipate the kinetic energy until none is left, just like my towers do and just like everything else that can be stacked and have a part of the stack dropped on the rest.
I clearly understand that under these conditions, global collapse is inevitable. My question to you is: how did it remain aloft then in the first place? Collapse wouldn't even have to be initiated by anything, it would have to be hanging by a rope to keep its potential energy.
This comes down to dynamic load vs static load. For the top to get into motion, it first has to overcome the counter force that the columns are offering. In a static situation, this force is only that of the gravitational pull on the mass above it. So Fs=ma=mg=10m.
For the dynamic situation, the same formula counts. However, "a" is no longer "g'. "a" has now become the deceleration of the mass as result of the impact with the lower floor. Lets for example assume it has 20cm to slow down before the supports fail (displacement of the supports result in failure). We can calculate v at the moment of impact:
v²=2gs=2*10*3.7=74 -> (v=sqrt(74)=8.6m/s
To decelerate something that goes 8.6m/s in a distance of 20cm you get (reverse of above)
a=v²/2s=74/2*0.2=185m/s²
so Fdyn=ma=185m.
Now compare it to the static situation, you get a force that is Fdyn/Fs=185/10=18.5 times larger. Even a FoS of 10 isn't going to help you here.
Therefor, the Twins had an FoS of 0.4 and 0.25.
by exponent
the acceleration of collapse is 0.6g for WTC2 and 0.75g for WTC1 (iirc, might have those backwards)
Originally posted by Akareyon
if you stack 6 floors of upright Jenga blocks and paper, the whole structure warps around its vertical axis and topples.
Originally posted by Akareyon
As we have seen, the plane would have to crash into the 2nd floor from the roof so no progressive global collapse ensues.
Of course it would, because you're back to arguing that once one floor fails, global collapse ensues. It doesn't when the FoS is greater than 1, because the impacting mass could pick up that speed only once and each floor it crushes would decelerate it because each floor would be designed to withstand ten times the force of the mass in a static situation. It would dissipate the kinetic energy until none is left, just like my towers do and just like everything else that can be stacked and have a part of the stack dropped on the rest.
Only with an FoS < 1 global progressive collapse can happen, because then each floor wouldn't be able to withstand even the static load, so when a dynamic situation occurs, the fall is only decelerated at the rate of 9,81*(1-FoS) m/s².
I did.
Originally posted by -PLB-
It seems to me that you are arguing that once the columns fail, they still offer resistance, and still slow down the mass. This is incorrect.
The correct way to counter my post is to show where my calculations are wrong.
Originally posted by Akareyon
This load-displacement diagram shows what happens when the force m*g is greater than the force needed to crush a column.
The mass crushes the column and gets decelerated in the process, decreasing its force. When u, the displacement, reaches the point where the m*g force of the mass is greater than the force with which the columns can resist to their deformation because they have developed plastic hinges from being crushed (u_c), the mass picks up velocity through gravity again, gaining enough force to do the same to the next story.
This is the scenario where m*g is greater than F_c (also called the maxwell line) and also greater than F_0. F_c is the average force of the F(u) curve, F_0 is the maximum force with which the column resists. This is the point shortly before your ruler snaps.
There is another scenario where m*g is not greater than F_c. In that case, the columns may be displaced a little, up to u_0. This is the dead load scenario. It does not bend very much and easily holds everything above it aloft with ease.
In the first scenario, you can have F_0 > m*g > F_c (FoS < 1) without the columns buckling yet.
Originally posted by -PLB-
No it doesn't. The force required to crush a column is F0 (or in other words, the load capacity of the column). This force is clearly larger than m*g in that figure, about factor 3.
Originally posted by Akareyon
This load-displacement diagram shows what happens when the force m*g is greater than the force needed to crush a column.
I never said otherwise. Except for m*g.
m*g, Fc and F0 are constant values. They do not change. That is why they are horizontal lines in that figure.
No, I was confusing m*g with the m*a (or F(u) after a free fall of m) I mentioned a few sentences later, and m*a is greater than F_0. It was a mistake, a wiser man would have noticed and corrected it, but it doesn't invalidate the point of my argument at all.
mg or Fc is nowhere greater than F0. F0 is the peak value. You are confusing it with F(u).
Which definition?
There is another scenario where m*g is not greater than F_c. In that case, the columns may be displaced a little, up to u_0. This is the dead load scenario. It does not bend very much and easily holds everything above it aloft with ease.
In the first scenario, you can have F_0 > m*g > F_c (FoS < 1) without the columns buckling yet.
If F0 > m*g, then by definition FoS > 1. As FoS=F0/m*g.
Originally posted by Akareyon
No, I was confusing m*g with the m*a (or F(u) after a free fall of m) I mentioned a few sentences later, and m*a is greater than F_0. It was a mistake, a wiser man would have noticed and corrected it, but it doesn't invalidate the point of my argument at all.
This is the scenario where F(u) is greater than F_c (also called the maxwell line) and also greater than F_0
Which definition?
Is there one I missed which says you are allowed to strain a structure with a dead load way beyond its maxwell line?
You don't have to, you know it already. If you insist however, I can correct my mistake and repost my last post to clear up any potantial for misunderstandings. But I understand why you are so happy I finally made a mistake you can pick on, we're all humans, after all ;-)
Originally posted by -PLB-
And no, I am not going to guess what you actually meant to say.
Obviously more like this one:
This one: "absolute strength (structural capacity) to actual applied load".
F0 = absolute strength (structural capacity).
m*g = actual applied load
Which one else did you have in mind?
Originally posted by Akareyon
You don't have to, you know it already. If you insist however, I can correct my mistake and repost my last post to clear up any potantial for misunderstandings. But I understand why you are so happy I finally made a mistake you can pick on, we're all humans, after all ;-)
"absolute strength (structural capacity) to actual applied load"
F_c = maxwell line = absolute strength (structural capacity)
m*g = actual applied load
I would find it logical that engineers would know about this phenomenon and use F_c, the mean crushing force of a column, for their designs and calculations instead of an arbitrary peak value of the load-displacement curve. What was your reasoning?
Originally posted by -PLB-
Correct, but the energy lost to crush a single floor is (according to bazant) 0.5GJ. The energy gained when the mass falls a height of 1 floor is 2.1GJ. So after each floor that is crushed there is a net (kinetic) energy leftover of 1.6GJ.
0.5jg to break all the connections of a floor
Want me to be honest or polite? :-)
Originally posted by -PLB-
Do I sound happy?
I'm always glad if I can make someone happy.
You would make me happy if you repost your argument with all mistakes corrected
I, on the other hand, wouldn't design a building hoping nothing ever triggers the collapse of a single floor because when it does, utter mayhem ensues. I would want it to behave like other buildings.
I don't find it logical, because you don't design a building based on the idea that it is collapsing in a way the building would never collapse anyhow.
That's not the point of our little sparring, in my opinion. After we have taken the practical approach and experimented with dish washer tabs and Jenga blocks, we discuss the implications, look for explanations, try to form a theory and square it with existing physical theories and universal realities. Bazants theory is just a crutch, not the bible or something. It is easy enough to understand when you know that E=m*g*h=1/2*m*v² and all that stuff and why that is, how it all correlates - and what an integral is. I think as a very, very basic model for the overall phenomenon - the structure falling through, into and out of itself - it helps a layman more than some fancy FEM could to understand what gerenally happened - that the weight of the structure had to be greater than its maxwell line, in other words, it had to be in an unstable equilibrium (again, thanks to exponent or whoever it was for throwing the ball, I was looking for that word).
You do understand that the figure we are discussing is a hypothetical situation that never happened in the WTC buildings right? Column to column impacts was not the failure mode.
Originally posted by Another_Nut
reply to post by -PLB-
did plb just admit the towers built were in an unstable equilibrium?
That some intelligence could have done that on purpose?
Originally posted by Another_Nut
did plb just admit the towers built were in an unstable equilibrium?
Only if it is not in place already :-)
Originally posted by -PLB-
reply to post by Akareyon
So to summerize, you are pleading for a law that says that buildings have to be designed so that Fc>mg.
That's what those truthers say usually....
*In practice, global progressive collapse almost never happens.
...and much better even to design your building so that if one floor fails, not everything goes down the drain!
*It seems to me its better to design your building so that a floor never fails unintentionally in the first place.
Indeed. Surely Bazant has been looking at single column buckling and ideal columns crashing end-to-end, the "most optimistic" scenario, one that never happened. The real event includes the floor slabs, trusses, core, perimeter, furniture, office cubicles, kerose, plane parts, fireproofing, heat and so many unknown variables for each impact and bending angle, force and counterforce, compression, tension, shear, torsion, weldseam, bolt, nut, rod and rivet that we would have a hard time making sense of it all. It's a very chaotic process.
*In an actual collapse, the value for Fc can be completely different than in theory, and is rather unpredictable. Kind of hard to design for it.
Firstly there is a problem over the use of models. A 1/10th scale model has linear dimensions that are 1/10th of the original, but physics dictates that its area dimensions will be 1/100 of the original, and that it will have only 1/1000 of the volume. This means the model only gives useful results if the variables are of the same dimension. For instance, models in wind tunnels work because the factors involved are distance and velocity so everything is linear. Burning buildings (or airships) don't work because the fuel is volume while the burned surface is area – for a given length there is a hundred times too much fuel compared to the real thing! The fire brigade have found that to see how fire behaves in buildings they have to build full-size buildings in one of the Cardington airship hangars, and burn them: models don't work.
Galileo proposed a new science, the study of the strength of materials, that considered how the size and shape of structural members affects their ability to carry and transmit loads. He discovered that as the length of a beam increases, its strength decreases, unless you increase the thickness and breadth at an even greater rate. You cannot, therefore, simply double or triple the dimensions of a beam, and expect it to carry double or triple the load
Who does not know that a horse falling from a height of three or four cubits will break his bones, while a dog falling from the same height or a cat from a height of eight or ten cubits will suffer no injury? Equally harmless would be the fall of a grasshopper from a tower or the fall of an ant from the distance of the moon.
The point Galileo is making here is that small things are sturdier in proportion to their size.
Originally posted by Akareyon
...and much better even to design your building so that if one floor fails, not everything goes down the drain!
However, the overall picture would not change much if we'd derive very rough mean values for F(u) and m*g of each floor: when all forces are summed up, m*g would still have to be above the maxwell line to make global progressive collapse possible. As you said yourself, in practice, this almost never happens because usually, even if a floor fails, the impact is dampened by the rest of the structure.
If twice the office space and half the cost means the whole thing comes down and flattens everything in it instead of one or two floors getting crushed, thousands of lives are saved and objects of value can be retrieved, an economic evaluation would most surely come get my skyscraper, not yours. No offense intended :-)
Originally posted by -PLB-
If "better" means half the office space and twice the cost, I don't think everyone will agree.
Oh, FEMA, NIST and those "debunkers" went to great lengths (examples abound in this thread, as in the parallel discussion between ANOK and GenRadek) to prove the opposite. A simple office fire could have made the trusses of the floor slabs sag, pulling the outer columns inward (analogous to a built-in, heat-driven "screw clamp", so to speak).It almost never happens because the support columns on a floor almost never fail.
However, the overall picture would not change much if we'd derive very rough mean values for F(u) and m*g of each floor: when all forces are summed up, m*g would still have to be above the maxwell line to make global progressive collapse possible. As you said yourself, in practice, this almost never happens because usually, even if a floor fails, the impact is dampened by the rest of the structure.