Claim: Jim Hoffman's "9/11 progressive collapse challenge" can't be met

Jeffrey Orling

Senior Member
according to Euler... the limit is a slenderness ratio of 66 for aluminum... the height limit would be 66 times the base diameter. This doesn't mean that it would crush... it means it would buckle from instability.
 

econ41

Senior Member
according to Euler... the limit is a slenderness ratio of 66 for aluminum... the height limit would be 66 times the base diameter. This doesn't mean that it would crush... it means it would buckle from instability.
For what cross section of column does that 66 apply Jeffrey? Can you quote and link the reference please?
 

Jeffrey Orling

Senior Member
For what cross section of column does that 66 apply Jeffrey? Can you quote and link the reference please?
Look at the graphic I posted previously which I sniped from the www years ago... it was related to Euler instability for columns and I believe the material's modulus of elasticity is a factor. Shorter access governs.
 

Mick West

Administrator
Staff member
I just want to be clear that I understand this, and what (I think) we're talking about is whether the exterior shell, without any of the loading of the floors, would buckle itself, thus being squashed by its own weight or would fail by "not being perfectly vertically aligned".

The can is a good example of what I'm thinking of. How tall would the can have to be to crush or buckle itself? How would it fail if you added no additional loads?

[Edit: or how many cans would you need to glue on top of each other before it's likely that, without any other loading, including laterally, they would collapse?]
Depends somewhat on the glue. To be analogous to the walls they would have to be very loosely attached, like with two small pieces of Scotch tape.

The exterior shell, without any bracing (meaning no floors, roof, or hat truss, would be very unstable. If there's zero wind and you are very lucky, then it might stand. But probably just differential heating from sunlight would push that luck.

It's not a super relevant point, because the walls collapsed in sections from the top-down, because of the combined effects of floor stripping and collisions.
 

econ41

Senior Member
My internet has been down for 48 hrs so I can't search for a reference... sorry
I was just hinting that the Euler Buckling limit isn't directly relevant to hollow cans where wall thickness dominates the form of buckling. Buckling of a structural column involves buckling of the whole structural element - macro level, Buckling of an Al can hollow tube will involve more micro aspects - more localised collapse of the section. So it will be different to "normal" Euler buckling. The critical length will be less and I'm not convinced that Euler is strictly relevant. Yes it is still a useful concept. At the "longer things buckle and bend easier" level of reasoning. But risky to put specific numbers on outcomes. Hence my several disclaimers in my posts and a couple of PMs. "We" are trying to explain complex matters in simplified terms. It is a risky game. ;)
 

deirdre

Senior Member.
just fyi... you do need to reinforce the bottom of paper towers . mine (with no floors) is buckling on itself, and twisting a bit.
it was rolled of course, but i put the "roll" side inside (so i think it should be concaving if anything...so that's kinda weird). each side 12" and about 76inches high. 20210328_220143.jpg

20210328_220128.jpg
 

econ41

Senior Member
just fyi... you do need to reinforce the bottom of paper towers . mine (with no floors) is buckling on itself, and twisting a bit.
Please excuse me if I comment - but your base buckling of your "tower" illustrates the issue I was identifying for Jeffrey Orling. Your model Deirdre is buckling at that lower level BUT it is a "micro" scale buckle - a local effect caused by local stress concentration. It is NOT Euler Buckling which would affect your tower overall - one big buckle near the middle of thetower.

We cannot conclude that your tower has an Euler limit of 6.33 because your 76" high building has a 12" square base. The buckling is a local effect. NOT buckling of the whole tower.

And it is the same sort of local effect which would cause result in "denting" or local crushing of beer cans which also is NOT Euler Buckling.
 
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Thomas B

Active Member
The exterior shell, without any bracing (meaning no floors, roof, or hat truss, would be very unstable. If there's zero wind and you are very lucky, then it might stand. But probably just differential heating from sunlight would push that luck.

It's not a super relevant point, because the walls collapsed in sections from the top-down, because of the combined effects of floor stripping and collisions.
My understanding is that the progressive stripping of the floors (and the chaotic collisions) destroyed both the exterior walls and the core. So it's relevant to know how strong these structures were before the floors began to collapse and do all this destructive work. It's important to understand how much work they were doing to hold up the buildings on an ordinary day. Maybe "know" is too strong a word; it's relevant to imagine or have a "feel" for the strength of the structures that were being broken into sections.

Let me work through my intuitions here.

First, I think it would be fair to count the hat truss as part of the exterior shell, but it's not an essential point.

We imagine a WTC tower on a windless sunny day before 9/11. My thought experiment proceeds in stages. First, we magically remove all floors except the "mechanical floors". We will leave the core in place. This will, of course, remove the lateral bracing the floors provided (I don't deny that they did provide such bracing) but it will also remove their gravity load. While the effective length of the columns (for purposes of doing a Euler buckling calculation) will be longer, they are also under considerably less strain.

My gut tells me that the combination of the (effective) lengthening of the columns and reduction of gravity loads puts the tower in an overall stronger position, not a weaker one. In this configuration it would be better able to handle a hurricane or an earthquake. The structure is doing much less work than it was designed to.

Next, remove the mechanical floors. This again removes some load and some lateral support. At this point I might begin to imagine that the tower could "twist" ... but that would require me to forget the lateral bracing of the spandrel plates that tie the columns together around the perimeter. Also, there's still the hat truss at the top. Also, the core is still connected to the hat truss.
IMG_0229.JPG
Remove the core. This has the consequence of removing the support directly under the gravity load of the hat truss. But the way that truss works is of course to distribute that load to the perimeter. So now the perimeter columns are doing a little more work again. But the truss itself doesn't weigh very much (and we removed the radio antenna before we started, let's say). And the truss ties the top together in a way that I mainly imagine makes the structure stronger rather than weaker.

Finally, remove the the hat truss. You now have what I imagine is an empty tube, similar to my paper model. It is designed to carry million of pounds of gravity loads (My paper tower can carry at least ten AA batteries on its top), and resist hurricane force winds and earthquakes, but it is carrying no weight, and there is no wind or ground vibrations. Like Mick says, it's just being gently warmed in the sun.

I find it very difficult to imagine that it will soon simply crumble and fall to ground.

I really don't know what to think. Like I say, the idea that the outer shell wasn't self-supporting is completely new to me, and I'm surprised to hear anyone suggest it. I'm just stating what I've been thinking all these years while trying to understand the arguments on both sides. It never occurred to me that either truthers or debunkers believed that the outer shells couldn't stand up on their own.
 

FatPhil

Active Member
I realize you're making an empirical statement so it's going to come down to an experiment. But doesn't that seem wrong to you intuitively? Imagine 7 stacks of ten empty beer cans standing side by side -- seven 4-foot aluminum towers. Look at them with your mind's eye. Feel their weight in your mind. Now imagine putting them all on top of each other. One big tower of 70 beer cans glued together, 28 feet tall. Do you expect them to collapse? With no additional loading, whether vertical or lateral?

[Edit: the tower would weigh just over 2 pounds. But an aluminum can can "easily" "support a 100lb person".]

The problem isn't the load, it's the torque. This should become clear once you notice that the height features at a power higher in the Euler equation than it would do if it was just to do with the load.

Try crushing a long piece of wire between finger and thumb - it's trivial to get it to bend under a compressive load. Halve its length, still easy. Halve it again - it starts to get apparently stiffer, even though the modulus of the material hasn't changed. Halve it again, and again. Eventually, you'll just end up piercing your fingertips, the wire's stronger than your skin. The reason you were bending it successfully early on was because you were using leverage and torque, not just raw compressive load.

If you want to make it a better analogue to a tower, hold one end rigidly in a clamp, and repeat the experiment. The ability for the wire to pivot around its point of contact with your skin increases its effective length, and makes it quicker to fail. The difference (squared, IIRC), should be detectable by hand. (Perhaps use thin paper on a soft block like a sponge as an analogue to your skin and see at what lengths the paper gets pierced in these two scenarios rather than just using perception of force on fingertip as a measurement device - they really ought to be noticeably different.)
 

Thomas B

Active Member
Try crushing a long piece of wire between finger and thumb - it's trivial to get it to bend under a compressive load.
This keeps coming up, so let me stress again that I'm asking about a completely unloaded column. The analogy would be holding the wire between your finger and thumb at one end and pointing it straight up (90 degrees to the ground). If the wire is so long that it can't remain straight, then "pin" it with your other hand so it remains upright, but don't exert any force up or down.

Under those conditions, would a single, laterally unsupported (except for being "fixed" at the base and "pinned" at the top) WTC column collapse under its own weight?
 

Thomas B

Active Member
you do need to reinforce the bottom of paper towers . mine (with no floors) is buckling on itself, and twisting a bit.
I really like your Tower and I think it's a better answer to @Gamolon than I will be able to muster. Econ is right that
The buckling is a local effect. NOT buckling of the whole tower.
It's just because of small imperfections in these paper models we're building. They become even more pronounced under loads. There's a real art to making the folds straight and clean, and closing the seam at the open corner. That seam must be as strong and straight as the folds. I'm getting better and better at making these things, but it's never perfect. It's quite satisfying to see my craftsmanship improving though. And I think -- one "artist" to another -- your tower is really impressive overall.

As @Mick West points out, ultimately the structure will find one or more weak points to fail at. The trick to meeting the challenge is to get it to stand up first and then collapse from the top down. So I guess it will have to have "weak points" distributed through the structure that will fail only under the dynamic load of the downward moving mass. If it just fails at a lower corner and falls over it will not really be doing what we're trying to do, but I suspect it would also fail my stress tests it's "normal" state.
 

Oystein

Senior Member
IMG_0202.JPG
Thanks. I've already done this, albeit in a slightly simpler form. I've tested the tube with floors and without, in both cases putting the entire load on the top. The point you're making is indeed confirmed. But that's not something I needed to learn.

What I'm looking for is a stable configuration of floors, equally loaded with batteries, such that it's possible to destroy the whole tower solely by weakening either the paper, the cardboard floors, or the connections between them, in the top 20% of the structure. [Edit: a crucial point here is that this also means that at least 80% of the mass will be bellow the level of the weakening.]

I'm expecting that this will require a significantly bigger structure (i.e., that it has to be "scaled up" to leverage the necessary gravitational forces) but, since I will be using the same materials, I hope I will not need to build a 1400' x 200' paper tower with tons and tons of batteries on each floor!
You are on a fool's errand to think it will be educational to mimic the WTC in arbitrarily many, arbitrarily picked characteristics at once.

Good that you understood that a tube in tube design gets critically weakened by removing internal bracing. That was the entire point of my proposed model. If you understand that, and understand why floors must have failed by impacting masses from above, you'll also understand the ROOSD part of the actual collapse sequences and why perimeter collapse must have followed immediately.

Then all that's left is understanding why the core can not remain standing. Two main reasons: i. It wasn't designed to stand on its own at is too slender to resist normal lateral loads (wind...) without bracing, ii. It got massives kicks in the knees by the enormous falling and shifting masses at its foot.

It doesn't matter at all if there are 80 or 1 or 0 or a million more floors below the part of the model that helps understand the vulnerability of the tube to removal of internal bracing.
 

Thomas B

Active Member
Under those conditions, would a single, laterally unsupported (except for being "fixed" at the base and "pinned" at the top) WTC column collapse under its own weight?
I can't quite get my mind around it. They're box columns, right? And tappered to be much stronger at the bottom than at the top, right? If we imagine building a single column from the ground up (actually cantilevered into the ground, i.e., fixed at the bottom), and always pinning the top so it remains straight, every length we add would be lighter as we increase the height. When does it buckle? (At what height does it need its first lateral bracing?)
 

Gamolon

Member
What I'm looking for is a stable configuration of floors, equally loaded with batteries, such that it's possible to destroy the whole tower solely by weakening either the paper, the cardboard floors, or the connections between them, in the top 20% of the structure. [Edit: a crucial point here is that this also means that at least 80% of the mass will be bellow the level of the weakening.]
Let's talk about the flooring structural subsystem first. Do you think the top floor of the remaining 80% of the building was designed to withstand the dynamic load of the upper 20% falling on it? Of that flooring subsystem, what structural components do you think would fail upon that initial impact?
 

Thomas B

Active Member
What reasons do you think that tower is a better answer? Just curious.
Instead of messing around with tape, as I will have to, it just uses a big piece of paper. Very elegant. It says, "Here's how it would look if the seams were perfectly made." (i.e., if the sheets of paper were joined seamlessly.)
 

Gamolon

Member
Instead of messing around with tape, as I will have to, it just uses a big piece of paper. Very elegant. It says, "Here's how it would look if the seams were perfectly made." (i.e., if the sheets of paper were joined seamlessly.)
I guess I'm asking you why Dierdre's model better replicates the WTC design of an unbraced perimeter facade (connections, materials, etc.)? Why seamless joints versus my two pieces of Scotch tape per joined edge for example?
 

FatPhil

Active Member
Yes. Of course. That's why we're talking about buckling "under its own weight".

But you're failing to undertand the failure mode. This is a completely different mode from an inability to support the downward force in a downward direction. This is an inability to provide a nett restorative force in an orthogonal direction to counter any perturbations. WIthout that, perturbations lead to higher torques which lead to higher perturbations, ad finitem.
 

Thomas B

Active Member
But you're failing to undertand the failure mode.
This isn't about the failure mode, it's about the structural system of the towers when it is not collapsing. It's about why the buildings stood up, not why they fell down. While discussing this, Mick mentioned that he didn't think the facades could stand up without the lateral bracing of the floors even in the absence of a downward force. I still don't think that can be right. And that's what we're talking about. We're not talking about how the buildings failed.
 

Thomas B

Active Member
Why seamless joints versus my two pieces of Scotch tape per joined edge for example?
Your two pieces of tape create an obvious weakness along the length of the tower. The structural strength of the sheet of paper (which let's the original 4-sheet version stand up) is literally "broken" at the joints, so it's not surprising that the whole collapsed neatly the way it did. (Do note, however, that the individual sheets of paper were largely undamaged.) If the paper is better joined (more tape, a little overlap of paper) then the structure will (I predict) respond more like @deirdre 's tower.
 

Gamolon

Member
Your two pieces of tape create an obvious weakness along the length of the tower. The structural strength of the sheet of paper (which let's the original 4-sheet version stand up) is literally "broken" at the joints, so it's not surprising that the whole collapsed neatly the way it did. (Do note, however, that the individual sheets of paper were largely undamaged.) If the paper is better joined (more tape, a little overlap of paper) then the structure will (I predict) respond more like @deirdre 's tower.
Thomas B.

I'll ask again. Why does Deidre's tower more closely replicate the unbraced WTC perimeter façade design in that you accept her model as a better replication than mine? For example, what reasoning to you think her seamless connections more closely match the connections in the façade more than my "two pieces of tape"?

So far I've created a model that collapses and you and Deidre created a model that stands.
 
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Thomas B

Active Member
You are on a fool's errand to think it will be educational to mimic the WTC in arbitrarily many, arbitrarily picked characteristics at once.
The goal is for the model not to be arbitrary, and I've mentioned a few senses in which I'm hoping to achieve that. It might be useful to make a list, and you can tell me how it could be improved:

1. The load will be distributed evenly throughout the building: each floor will have the same load, and the floors will be evenly spaced.
2. This means that the collapse will be initiated with the downward movement of about 20% of the mass of the building.

3. The building will be strong enough to carry twice as much as weight as it will be loaded with when it is destroyed. (I don't know if that's the right safety factor; I'm not sure it has to be exactly right, just that there has to be a significant one. The building should not be right on the cusp of collapse before the collapse is initiated.)
4. The building will be strong enough to survive an "earthquake", i.e., a quick back-and-forth shifting of the foundations about 10% of the width of the footprint. Preferably a few times.
5. The building will be strong enough to survive a bit of local damage, i.e., poking some holes in the paper facade. This will simulate both the severing of columns and the force of the airplane impacts (which the towers of course survived, a testament to their strength).

6. The floors will be connected to the exterior (they will not be resting on top of column sections).
7. The floors will be connected weakly enough to allow them to be stripped away, but
8. strongly enough to break or buckle the exterior wall when then they are impacted by the upper floors.

Finally, there's the one we're discussing at the moment, which I haven't decided what I think about yet.

9. Without the floors (and their loads) in place, the exterior wall (the folded sheet of paper) should behave as the WTC's exterior shell would have behaved without the lateral support of the floor trusses. The model's shell should stand or fall as the WTC's shell would.

Some of these may not be as reasonable as I think they are, but I don't think this set of specifications is "arbitrary". If I can get a paper model like this to top-down collapse, I'll definitely have educated myself about the physics of the collapses. I may be a fool to try it, but I've already learned a lot. And it's been fun.
 

Thomas B

Active Member
you accept her model as a better replication than mine
Sorry, you misunderstood that comment. I said her tower is already better than I know mine will be. Whether yours or mine is a better model of the Twin Towers is another matter.
what reasoning to you think her seamless connections more closely match the connections in the façade more than my "two pieces of tape"?
I do think the facades were composed of what were intended to be continuous, unbroken columns. I think I've even heard engineers describe them as essentially single sheets of steel. There weren't four horizontal seams that they could be neatly folded along. (Even the prefabricated sections were assembled in a staggered pattern to avoid this. See @Jeffrey Orling 's drawings.)
 

Gamolon

Member
Whether yours or mine is a better model of the Twin Towers is another matter.
It's the ONLY matter is it not? With my model, I showed the façade would have collapsed. Your model (and Deidre's) show that the façade would have stood.

Now what?
 

Thomas B

Active Member
Now what?
I suppose one next move would be for you to show that putting floors in would keep it standing. And then to put a serious load on those floors.

My models -- and I suspect @deirdre 's too -- can stand with floors that are significantly loaded (many, many times the weight of the paper.)
 

deirdre

Senior Member.
So it's relevant to know how strong these structures were before the floors began to collapse and do all this destructive work.
it isnt really. relevant.
the SECTIONS of the exterior mesh broke due to being pulled and pushed and knocked about by forces even i can't comprehend.
 

benthamitemetric

Senior Member
I do think the facades were composed of what were intended to be continuous, unbroken columns. I think I've even heard engineers describe them as essentially single sheets of steel. There weren't four horizontal seams that they could be neatly folded along. (Even the prefabricated sections were assembled in a staggered pattern to avoid this. See @Jeffrey Orling 's drawings.)
Pure and utter nonsense. You "think you've heard" engineers saying something completely nonsensical? Really? You cannot possibly believe anyone actually thinks the column connections were as strong as the midpoints on the columns. (Just look at how, in reality, the columns came apart at their connections rather than being torn in half.) To beat the deadhorse again, you are going about this question entirely wrong and should start with actually learning (and honestly reckoning with) first principles.
 

Gamolon

Member
I suppose one next move would be for you to show that putting floors in would keep it standing. And then to put a serious load on those floors.

My models -- and I suspect @deirdre 's too -- can stand with floors that are significantly loaded (many, many times the weight of the paper.)
Why? The argument here is whether an unbraced square tube can stand on it's own. Mine collapsed under it's own weight. Yours did not.
 

deirdre

Senior Member.
My models -- and I suspect @deirdre 's too -- can stand with floors that are significantly loaded (many, many times the weight of the paper.)
doubtful. if that were the case then they wouldn't have had to build such a strong core piece and hat truss.

(and windows arent steel columns, so just going on record that i reject your premise of the mesh as a "sheet of steel".)
 

Jeffrey Orling

Senior Member
I can't quite get my mind around it. They're box columns, right? And tappered to be much stronger at the bottom than at the top, right? If we imagine building a single column from the ground up (actually cantilevered into the ground, i.e., fixed at the bottom), and always pinning the top so it remains straight, every length we add would be lighter as we increase the height. When does it buckle? (At what height does it need its first lateral bracing?)
the perimeter column panels were bearing on 3 story tree columns...
But for you theoretical question you need to consider that a perimeter column's cross sectional represent ALL the loads and that included a part of the floor loads. And the columns "stepped back" in area as you move up.

Theoretically an un-braced steel column would be limited by the slenderness ratio... See this summary:

euler buckling - Copy_page1.jpg

So the limited would be that the elastic limited is EXCEEDED after the S/L ratio is greater than 150.... You need to calculated the radius of gyration... Mathematically the radius of gyration is the root mean square distance of the object's parts from either its center of mass or a given axis, depending on the relevant application... So the column shape will impact the calculation

The axis are the same - 14"... so it seems that a steel column of 14"x 14" x 150 is the limit for stability is about 175' or about 14 stories tall. I am assuming that this would be for a solid column... One that was made from thinner plates might fail at a short length...
 
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Gamolon

Member
What I'm looking for is a stable configuration of floors, equally loaded with batteries, such that it's possible to destroy the whole tower solely by weakening either the paper, the cardboard floors, or the connections between them, in the top 20% of the structure. [Edit: a crucial point here is that this also means that at least 80% of the mass will be bellow the level of the weakening.]
Couple of questions.

1. What does 80% of the mass of the towers being below the level of weakening have to do with anything?
2. Do you think the top floor of the remaining 80% of the structure was designed to withstand the impact of the upper 20%
3. Do cardboard floors effectively replicate how concrete floors would behave when impacted?
 

Thomas B

Active Member
Pure and utter nonsense. You "think you've heard" engineers saying something completely nonsensical?
If we think of "flanges" as the sheets of steel out of which box columns are made then Nordenson is essentially saying that the faces operate "like" single sheets of steel.
You cannot possibly believe anyone actually thinks the column connections were as strong as the midpoints on the columns.
Remember that were many dozen joints along each column (the prefabricated panels were three stories high, I think). The joints were of course the weakest points on the columns. But not all the joints failed. That is, while you can predict that if a column will break, it will break at a joint, you cannot predict which joint. So the joints are nothing like the taped connections in Gamolon's model -- which are its exactly predictable failure points. You could certainly not draw four horizontal lines along which the faces of the WTC would naturally break into four sheets if you remove the horizontal support. That's really all I meant in my response to @Gamolon.
 
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