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# Video Nullfies Pancake/CD Theory

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posted on May, 26 2013 @ 08:44 AM

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.
As we have seen, the plane would have to crash into the 2nd floor from the roof so no progressive global collapse ensues.

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.
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.

by exponent
the acceleration of collapse is 0.6g for WTC2 and 0.75g for WTC1 (iirc, might have those backwards)
Therefor, the Twins had an FoS of 0.4 and 0.25.

posted on May, 26 2013 @ 09:10 AM

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.

Your Jenga building is missing the spandrel plates that the WTC Towers had. That is why it is having a torsional failure. You need to attach something to at least two of the blocks on each level to resist torsional load,

posted on May, 26 2013 @ 09:24 AM

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.

If we are talking about deliberate attacks (about the only scenario where disconnect of top from lower structure happens) it doesn't take too much planning to aim a bit lower. So does it make sense to make a building almost 10 times stronger to be able to handle a scenario which you dislike, while the terrorists would just aim lower and still get global progressive collapse?

For what exactly are you designing it then? How does it become safer?

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 just explained in detail why this is not true. I showed you how in my hypothetical situation the forces on the support columns are 18.5 times higher when the top section impacts the support columns below, so that even a FoS of 10 would not safe the building. These calculations are completely generic, and count for any structure.

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. Just iterating what you believe to be correct is not the right way to counter this.

edit on 26-5-2013 by -PLB- because: (no reason given)

posted on May, 26 2013 @ 10:51 AM

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.
I did.

Open this document, go to page 3 of it and zoom in on Fig. 3. What you see is a load-displacement diagram.

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. They are under extreme strain, but m*g is smaller than F_0 and doesn't quite manage to overcome the maximum force with which the column resists. But if you increase m*g to m*a, for example by dropping m from a certain height, it easily exceeds F_0 as well - it climbs the hill of F_0, deplaces the column to the point where m*g is greater than F(u) (that's where u_c is) and here picks up the force it needs to overcome the next F_0 (W_c < W_b). Total global progressive collapse ensues.

In the second scenario, you have F_0 > F_c > m*g, that is, a FoS > 1. The maxwell line is above F_c now. Again, you can increase m*g to m*a by dropping m from a certain height. It easily exceeds F_0, climbing its hill, deplaces the column to the point where m*g is greater than F(u). But this happens much later now than in the first scenario (second scenario's u_c is greater than the u_c of the first scenario), so more kinetic energy is dissipated, and less potential energy can be converted to kinetic energy. If it is large enough, it may well be enough to overcome the next floor's F_0, but again, less potential energy can be converted to kinetic energy than kinetic energy was dissipated in the displacement of the columns (W_c > W_b). So at some point, the force is diminished so it doesn't manage to climb the F_0 hill of the n-th floor anymore and the collapse comes to a screeching halt.

The first scenario is what happened on September 11th, 2001, to the Twins, this is what happens in a vérinage at the moment the hydrauilic cylinders spring into action, in Rube Goldberg machines, nuclear chain reactions, -PLB-s dishwasher tablet tower and so on.

The second scenario is what happens in card houses, stacks of banana crates, tubes full of eggs or lightbulbs, buildings where the demolition goes wrong and collapse arrests because F_c has not been properly diminished.

eda ²waypastvne:

I know, thank you : ) Indeed, that would introduce the "stabilizing triangles" I mentioned a few pages back, but in turn would make the tower way too strong and that's what must be avoided at all costs to ensure a progressive global collapse :-)
edit on 26-5-2013 by Akareyon because: (no reason given)

posted on May, 26 2013 @ 11:34 AM

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.

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.

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.

m*g, Fc and F0 are constant values. They do not change. That is why they are horizontal lines in that figure.

mg or Fc is nowhere greater than F0. F0 is the peak value. You are confusing it with F(u).

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.

I think you should reconsider your interpretation of the figure. When you have, we can discuss it further.

posted on May, 26 2013 @ 12:26 PM

Originally posted by -PLB-

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.
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.

Yes it does. The average force required to crush the column is F_c (or in other words, the maxwell line), because when F_0 is overcome, only very little force is needed because the column doesn't resist anymore.

m*g, Fc and F0 are constant values. They do not change. That is why they are horizontal lines in that figure.
I never said otherwise. Except for m*g.

mg or Fc is nowhere greater than F0. F0 is the peak value. You are confusing it with F(u).
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.

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.
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?

posted on May, 26 2013 @ 12:43 PM

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.

A wise person isn't going to guess what someone means when what he writes doesn't make sense, but confronts the person with it. Why didn't you just call it F(u)? It is called like that in the figure too.

Anyway, when by m*g you actually meant m*a which is F(u), you wrote:

This is the scenario where F(u) is greater than F_c (also called the maxwell line) and also greater than F_0

You say that F(u) is greater than F0. That is logically impossible, as F0 is the peak value of F(u). So it still does not make sense. And no, I am not going to guess what you actually meant to say.

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?

This one: "absolute strength (structural capacity) to actual applied load".

F0 = absolute strength (structural capacity).

Which one else did you have in mind?
edit on 26-5-2013 by -PLB- because: (no reason given)

posted on May, 26 2013 @ 01:57 PM

Originally posted by -PLB-
And no, I am not going to guess what you actually meant to say.
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 ;-)

This one: "absolute strength (structural capacity) to actual applied load".

F0 = absolute strength (structural capacity).

Which one else did you have in mind?
Obviously more like this one:

"absolute strength (structural capacity) to actual applied load"

F_c = maxwell line = absolute strength (structural capacity)

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?

posted on May, 26 2013 @ 02:07 PM

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 ;-)

Do I sound happy? You would make me happy if you repost your argument with all mistakes corrected, so that I do not have to solve a couple of puzzles in order to know what you are trying to say.

"absolute strength (structural capacity) to actual applied load"

F_c = maxwell line = absolute strength (structural capacity)

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?

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. 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.

I do not believe you will find your definition of FoS anywhere in any literature. But I can be wrong so feel free to prove me wrong.

posted on May, 26 2013 @ 02:47 PM

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.

How did Bazant come up with those figures? 0.5gj sounds a little low to me. I think he just made them up to fit his hypothesis. Do you really believe that 15 floors would crush 95 to their foundations? The towers did not collapse in blocks, you can quite clearly see the top and bottom collapsed independently of each other. You can clearly see the top floors crushing before the bottom floors even start to collapse. So you are already losing mass before the collapse of the bottom even started. The top floors would have all been gone long before they could crush anything.

And this is what Bazant claimed happened....

You can't seriously think that is what happened?

Equal and opposite reaction, conservation of momentum, and resistance PLB, try adding those to your hypothesis.

Again yours and Bazants hypothesis relies on the dropping floors losing no mass, yet mass was being ejected, and floors were blowing out ahead of the bottom "block" collapsing.

Also if it only took 0.5jg to break all the connections of a floor, then there is no way those connections would have survived NIST's pull-in hypothesis.

You can't have it both ways. Your hypothesis is contradictory, and doesn't fit the visual evidence.

edit on 5/26/2013 by ANOK because: (no reason given)

posted on May, 26 2013 @ 03:13 PM

You really have no idea what is being discussed, nor what we are talking about. Just one example:

0.5jg to break all the connections of a floor

No, 0.5GJ to make all columns buckle.

I could dissect your entire post but it will be ignored.

posted on May, 26 2013 @ 04:56 PM

Originally posted by -PLB-
Do I sound happy?
Want me to be honest or polite? :-)

You would make me happy if you repost your argument with all mistakes corrected
I'm always glad if I can make someone happy.

Please open Metaphysics, errrm, Mechanics of Progressive Collapse, by Zdeněk Pavel Bažant and Matthieu Verdure, on page 3 and zoom in on Fig. 3. What you see is a load-displacement diagram.

This load-displacement diagram shows how an "ideal" steel column reacts when you apply a compressive force to it. This diagram is based on very old knowledge and a firm theoretical and experimental basis.

At first, the column resists with increasing force to its compression (like a plastic ruler or a spoon), but at a certain point (F_0) it starts to buckle. It forms "plastic hinges" (which decreases its tensile strength because of Law of the Lever and other complicated stuff going on macro- and microscopically) so its resistance to the compressive force is diminished the farther it is compressed.

A compressive force, for example the live load of the floors above (m*g) resting on them in a static scenario, would just have to be smaller than F_0 to remain aloft. If this column would very, very slowly and carefully be compressed with an additional force, for example by means of a giant screw clamp, 1mm beyond u_c, m*g would suddenly be greater than F(u_c+1mm) and snap! the weight on top of the column would do the rest of the "work" and compress the column all the way down and gain so much kinetic energy in the process that it would impact the next floor with a force so great it would crush its columns too, all the way to u'_c, and thus set off a global progressive collapse all the way to the basement.

This is the scenario where m*g is greater than F_c (also called the maxwell line).

There is another scenario where m*g is not greater than F_c. Again, you can use a giant screw clamp to compress the columns very, very slowly to the point where m*g is greater than F(u). This time however, u_c is greater than in the first scenario because F(u) intersects the m*g line much later. So you have a little more screwing to do in this scenario. When finally you have reached u_c (the floor has a really low ceiling now), and give the screw clamp a last little turn, F(u_c+1mm) is smaller than the weight above which does the rest of the work. It gains kinetic energy in the process, but, alas, it is not enough to crush the next floor's columns, they just get bent a little maybe. Collapse is arrested. Boooo.

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.
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.

Like those that are meant to be demolished and get their base blown away but then the explosive charges in the upper floors don't go off for some reason and so the rest of the building freefalls one story and... remains standing.

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.
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).

Truly, still the question is not settled if the engineers effed up big time or all tall structures are made like that or a mechanism was built in on purpose or if that beautiful chain reaction is the result of dark magic, black tech or an Aether hiccup (you can imagineI still find the first two options the least credible ;-)

I'd be happy to walk the E=p*V path with someone, too.

But we know what to do now. Look out for stuff where m*g > maxwell line, check why and compare it with WTC.

posted on May, 27 2013 @ 01:27 AM

So to summerize, you are pleading for a law that says that buildings have to be designed so that Fc>mg. Maybe its a good idea, maybe it is not. Again, I am not a building safety expert and I haven't studied this. Just some points that come to mind:

*In order to increase Fc you have to add weight, and you automatically also increase mg. Seems to me it can have rather huge implications, especially for high rise buildings.
*In practice, global progressive collapse almost never happens.
*It seems to me its better to design your building so that a floor never fails unintentionally in the first place.
*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.

Anyhow, I think this discussion is more interesting among people who do building safety for a living, and have studied the subject in detail. I am not planning to become an expert in this area.

ps. I now do mostly agree with your interpretation of that figure you pointed to
. See, I am much more happy now.
edit on 27-5-2013 by -PLB- because: (no reason given)

posted on May, 27 2013 @ 02:55 AM

did plb just admit the towers built were in an unstable equilibrium?

That some intelligence could have done that on purpose?

And to the possibility of a built in cd sans explosions?

Therefore making bothsides of the argument null and void.

Showing we were all chasing shadows . and each other in circles.

And who likes to play with dominos?

posted on May, 27 2013 @ 03:16 AM

Originally posted by Another_Nut

did plb just admit the towers built were in an unstable equilibrium?

That some intelligence could have done that on purpose?

Any time you raise a weight above the ground you put in energy, energy that can be used to destroy things under it.

This is an unavoidable consequence.

Akareyon I'll do my best to come up with a reply to you ASAP, it's quite tricky reading through someone elses discussion and PLB seems as confused as I!

posted on May, 27 2013 @ 03:17 AM

Originally posted by Another_Nut
did plb just admit the towers built were in an unstable equilibrium?

I don't agree the towers were in an unstable equilibrium. That would mean that even a person leaning against the building would make it collapse. I do agree that the towers were in a metastable state though. For instance, I talk about it here:

www.abovetopsecret.com...

posted on May, 27 2013 @ 03:58 AM

Originally posted by -PLB-

So to summerize, you are pleading for a law that says that buildings have to be designed so that Fc>mg.
Only if it is not in place already :-)

*In practice, global progressive collapse almost never happens.
That's what those truthers say usually....

*It seems to me its better to design your building so that a floor never fails unintentionally in the first place.
...and much better even to design your building so that if one floor fails, not everything goes down the drain!

*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.
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.

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.

Especially in a steel frame, since steel has a high mechanical velocity of propagation so the shock wave of the impact force would run to the basement and back a few times long before the first column is displaced beyond its crushing point. But that's a different topic alltogether ;-)
edit on 27-5-2013 by Akareyon because: (no reason given)

posted on May, 27 2013 @ 07:48 AM

Models,Models,Models I will say this once again all the models on both sides of the arguement are FLAWED! I had this arguement with psikeyhackr as well, the models in NO way represent the WTC towers or the true physics of that day.

The materials are not a scalable match, the dimensions the relative strengths and forces are not a match , the floors were suspended between the walls NONE of the models are, the relative areas ,volume ,mass & strengths of the components are not the same and the poles through psikeyhackr's and your model with the records represents NOTHING in the structure of the towers.

I will give you an example and a couple of links to read.

If I made a 100th scale model of one of the towers and wanted to film an object/ part of the construction falling for a special effect in a film I would even have to scale TIME !!! as of course anything falling under gravity will be at 9.81m/sec/sec the same as if it fell off the real building only problem is it's 1/100th the distance so to create realism the drop has to be slowed down that's why if not done well models are 100% obvious in films.

Models suffer from the square cube law to find out what really happens you need FULL size!

As I said before aerodynamics for car design can be achived with models , crash test data is always full size WHY do you think, do you not think any of the manufacturers really expenzive cars would use models if the results were accurate why crash a £/\$200,000 chassis when you could crash a model for a fraction of the cost.

Here is a comment from another site made about a recent documentary shown on Tv about the Hindenburg Disaster.

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.

With the Cardington Fire test a full scale model of part of a steel frame structure using real steel sections was built to show the effect of an office fire on steel, it also showed from the test data areas of the steel got to above 700c(1292f) in just over 20 mins and some areas got to over 900c(1652f) well within the time frame of the 9/11 events.

Here are a couple of links re models/scale

Scaling Laws

The link below has references to biology but also how Galileo looked at this problem.

Scaling & Estimation

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

The opposite is true that smaller objects appear sturdier!

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.

Models are flawed!

wmd

posted on May, 27 2013 @ 11:22 AM

Originally posted by Akareyon
...and much better even to design your building so that if one floor fails, not everything goes down the drain!

That remains to be seen. "Better" is kind of subjective. If "better" means half the office space and twice the cost, I don't think everyone will agree.

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.

It almost never happens because the support columns on a floor almost never fail. And even in situations where it does happen (I can only think of controlled demolition), I don't know of any situation where collapse is arrest after several floors being crushed. The situations I know of in controlled demolition is where the peak (F0) has not been overcome, so you see immediate arrest. That does not necessarily mean that Fc>mg.

posted on May, 28 2013 @ 03:46 AM

Originally posted by -PLB-
If "better" means half the office space and twice the cost, I don't think everyone will agree.
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 :-)

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.
It almost never happens because the support columns on a floor almost never fail.
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).

At least that's what they say, not that you or I would buy that explanation for the initiation of the collapse sequence ;-)

Increasing F_c without increasing m proportionally is not all that hard, by the way. wmd_2008 raises a valuable point. If you look at a long steel column suspended horizontally by a crane, you can see how it bends visibly under its own weight - as if it were made of clay or paper. So what you want to do -- and has been done for centuries -- is to increase F_c not by making the columns stronger and much heavier, but by simply keeping them short. The shorter, the better. Because the strength of a column is inversely proportional to its length: the smaller u is, the greater F_c gets.

But you can't raise a high building then, you say, and just stacking a lot of short columns on top of each other doesn't really solve the problem?

You're completely right, and that's why you put horizontal beams in between them. The house I grew up in for example was made like that, just with oak beams, not steel beams (and, btw, it didn't have diagonal bracings in each rectangle between beams and columns either, just here and there). The horziontal beams make buckling for the columns harder, thus raise F_c -- at relatively little cost in m. The result is a steel grid called framework. Keeping F_c greater than m*g without much compromise in cost and office space is no problem anymore.

It's also somewhat easier to design then, because F_0 is hardly ever taken into consideration in engineering. The formula to calculate the critical force or maximum load (where F(u_c)=m*g, the point you never want to reach because a small nudge would be enough to bring it down) is

F_critical=(π²*E*I)/(K*L)²

where E is the modulus of elasticity (in N/mm² or psi, the "stiffness" depending on the material, in engineering given for the columns you're about to buy (for example 29,000,000 psi in ASTM-A36 steel)),

I is the second moment of inertia (in meters^4, depending on the shape of the cross section, in engineering given for the columns you're about to buy (for example 12.19 m^4 in a HE 160 A I-Beam)),

L is the length of the column and

K is the "effective length factor" ranging from 1/2 to 2 depending on whether one or both ends of the columns are fixed or free to rotate.

Check units: N=1 * N/mm² * m^4/ (1 * m)², that*s N*(1000*mm)²/m², that's N... yup, correct (Notice how Young's modulus is four-dimensional, therefor time and velocity come into the equation and a glimpse of the fabric of the universe is allowed? I love it!).

So, this is the F_critical you want to keep as high as possible, 2 or 4 times greater than m*g (so that 1 < FoS = F_critical/m*g). Since E and I are given by the stuff you're using, except if you want to use diamonds or adamantium to increase E or use solid columns to increase I, you have only K*L to play with, so you keep it small to increase F_critical by keeping the column short and fixing both ends with bolts and rivets and weldseams (thus halving the effective length). From there you go and design a stable and cost-effective framework with plenty of office space, without ever knowing or worrying what F_0 might be :-)
edit on 28-5-2013 by Akareyon because: format

edit on 28-5-2013 by Akareyon because: typos

edit on 28-5-2013 by Akareyon because: got m*g/F_critical wrong way round, sorry!

edit on 28-5-2013 by Akareyon because: K*L and so on.

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