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reasonwhy
This is a story by a reporter that toured the WTC basement before the evidence was destroyed.
QUOTE

By James Glanz
New York Times News Service
Published November 16, 2001

In the days after the collapse of the towers two months ago, the tangled steel was still so hot that it glowed like charcoal briquettes in the unlighted basement, O'Connor said, adding, "For seven weeks it was surreal down here."



A 3-foot stalagmite of steel, which looks for all the world like a drip candle, sits next to one of the immense steel columns that held up the north face of the tower. The column has a sort of compound fracture--the top has been pushed a foot south of the piece it is resting on.


http://web.archive.org/web/20011118112358/...0084nov16.story

This would have been before the torches were used. biggrin.gif

Move along folks, nothing to see here.

This is the debris of a perfectly normal gravity driven collapse.

reasonwhy
Double post
tikay
QUOTE (Grumpy+Feb 2 2007, 07:44 AM)
Never underestimate the ability of the general public to be misinformed.

laugh.gif
~so true!
einsteen
DBB, Although the wtc stories were no springs they had the same behavior in first order, i.e. force linear in the beginning with distance. The nice thing is that if you use those theoretical functions and set up your differential equations then nobody on earth can solve that, but on the other hand you could also substitute the collapse function and check if it is a possible solution...
Chainsaw,
QUOTE (reasonwhy+Jun 27 2007, 04:39 AM)
This is a story by a reporter that toured the WTC basement before the evidence was destroyed.


http://web.archive.org/web/20011118112358/...0084nov16.story

This would have been before the torches were used. biggrin.gif

Move along folks, nothing to see here.

This is the debris of a perfectly normal gravity driven collapse.

Unfortunately that is only evidence of heated steel not evidence of a CD. All you need are the right chemical conditions and steel itself will Oxidize, and a gravity driven collapse can form those conditions on its own.
It does however point out that it was not thermite thank you for that.
NEU-FONZE
I have a couple of questions about the heating of WTC steel with and without SFRM:

First consider some NIST data from NCSTAR 1-5B:

1. "Ceiling temperatures", that is AIR temperatures, from exposure to a 3 MW fire were generally ~ 800 deg C.

2. Uninsulated steel reached 600 deg C in 15 minutes.

3. Steel insulated with about a 2 cm thickness of SFRM reached 200 deg C in 15 minutes.

4. Steel with about a 2 cm thickness of SFRM reached 680 deg C in 50 minutes.

My questions are:

1. What temperature would UNINSULATED steel reach in 50 minutes?

2. After 15 min in a 3MW fire, what would be the temperature of a piece of steel with equal lengths that were insulated and uninsulated?
adoucette
QUOTE (reasonwhy+Jun 26 2007, 11:39 PM)
This is a story by a reporter that toured the WTC basement before the evidence was destroyed.


http://web.archive.org/web/20011118112358/...0084nov16.story

This would have been before the torches were used. biggrin.gif

Move along folks, nothing to see here.

This is the debris of a perfectly normal gravity driven collapse.

Bull.

This trip was TWO MONTHS after 9/11, they had been cutting in the pile since day 2 or 3.

Still it does have some interesting detail:

After a walk southward down the truck ramps and a dogleg right, to the west, the dancing flashlights illuminate the edge of the debris that fell nearly straight down through the north tower. At first the mind simply refuses to accept what the eyes see--the recognizable traces of 20 floors, much like geologic strata, over a 10-foot vertical span.
In one place, the steel decks of a half-dozen floors protrude like tattered wallpaper, almost touching where they are bent downward at the edge. "You're looking at roughly 60 feet of the building, smashed into about 3 feet," O'Connell said.

So NO MORE about the tower floors not PANCAKING.

Arthur


Chainsaw,
QUOTE (NEU-FONZE+Jun 27 2007, 12:29 PM)
I have a couple of questions about the heating of WTC steel with and without SFRM:

First consider some NIST data from NCSTAR 1-5B:

1. "Ceiling temperatures", that is AIR temperatures, from exposure to a 3 MW fire were generally ~ 800 deg C.

2. Uninsulated steel reached 600 deg C in 15 minutes.

3. Steel insulated with about a 2 cm thickness of SFRM reached 200 deg C in 15 minutes.

4. Steel with about a 2 cm thickness of SFRM reached 680 deg C in 50 minutes.

My questions are:

1. What temperature would UNINSULATED steel reach in 50 minutes?

2. After 15 min in a 3MW fire, what would be the temperature of a piece of steel with equal lengths that were insulated and uninsulated?

Do to the conductivity of steel the heat should transfer to the insulated part until the steel reached an even temperature unless air currents cooled it.
The heat transfer would be slower than on unheated steel but the cooling effect of air would not be as likely to have taken place on the insulated end.
The steel should reach in time close to the temperature that it is exposed to, unless another reaction is taking place.
As I recently explained to a Cter it is air that cools the end of the rod in a black smiths forge that is what keeps the end cool enough for him to handle. Even though the end is glowing.

Insulation removes the ability of the environment to both heat and cool the steel.

Right now I have been researching some very interesting stuff, about Fe304,

http://72.14.205.104/search?q=cache:8qfXvJ...clnk&cd=1&gl=us

http://www.freepatentsonline.com/4141759.html

http://www.mpi-magdeburg.mpg.de/research/p...riodicReforming

Right now I am working on an experiment to confirm my suspicions.

reasonwhy
QUOTE (adoucette+Jun 27 2007, 05:13 AM)
Bull.

This trip was TWO MONTHS after 9/11, they had been cutting in the pile since day 2 or 3.

Still it does have some interesting detail:

After a walk southward down the truck ramps and a dogleg right, to the west, the dancing flashlights illuminate the edge of the debris that fell nearly straight down through the north tower. At first the mind simply refuses to accept what the eyes see--the recognizable traces of 20 floors, much like geologic strata, over a 10-foot vertical span.
In one place, the steel decks of a half-dozen floors protrude like tattered wallpaper, almost touching where they are bent downward at the edge. "You're looking at roughly 60 feet of the building, smashed into about 3 feet," O'Connell said.

So NO MORE about the tower floors not PANCAKING.

Arthur

Nope,

He is taking a tour of the basement with flashlights.

Only a Idiot would be cutting the columns in the basement at the same time they are working above.

You go into an enclosed area with cutting torches and things blow-up.

QUOTE
The trade center's basement was once a six-level shopping center, parking ramp and underground train terminal spreading over more than 2 million square feet. Now it has become a place where the horror of the aboveground devastation is amplified by the gloom of the debris-strewn, claustrophobic space--a hazy darkness pierced only by flashlights and an occasional crater that lets dim sunlight filter through from above.


A rare journey to the bottom of the trade center's basement this week revealed a few places with only superficial damage, like the Commuter's Cafe, five levels below the trade center's plaza, where dust-encrusted bottles of liquor sit on the shelves.




But thanks for sharing your ignorance with us (even though you obviously read the article)!
laugh.gif laugh.gif laugh.gif
adoucette
QUOTE
Only a Idiot would be cutting the columns in the basement at the same time they are working above.


Who said the cutting was done in the basement?

Molten steel from cutting above could go quite a way down.

But here's the point you REALLY miss.

The "stalagmite" sits NEXT to the column. Which makes sense if its dripping from above.

He does not describe the column itself however as being damaged by anything molten:

QUOTE (->
QUOTE
Only a Idiot would be cutting the columns in the basement at the same time they are working above.


Who said the cutting was done in the basement?

Molten steel from cutting above could go quite a way down.

But here's the point you REALLY miss.

The "stalagmite" sits NEXT to the column. Which makes sense if its dripping from above.

He does not describe the column itself however as being damaged by anything molten:

The column has a sort of compound fracture--the top has been pushed a foot south of the piece it is resting on.


Which is what STRESS from ABOVE would do based on a gravitational collapse.

Oh, and about "before the evidence was destroyed"

But they thought they would give a NYTimes reporter a GUIDED TOUR of said evidence BEFORE they destroyed it?

laugh.gif

ROTFLMAO

Arthur
metamars
Theory for a "wire telegraph" is developed at http://www.du.edu/~jcalvert/waves/strtel.htm

QUOTE


Now suppose that at z = 0 two wires of different characteristics are joined. The displacement ξ must be continuous, and the tension in one wire must equal the tension in the other. Let a wave of amplitude A be incident from the left in wire 1, a wave of amplitude B be reflected and travel to the left in wire 1, and a wave of amplitude C propagated to the right in wire 2. The continuity of displacement gives the condition A + B = C, while the equality of forces gives (A - B )ρcS = Cρ'c'S', where primes refer to wire 2, and no primes to wire 1. These equations are easily solved for the unknowns B and C.




B = [(ρcS - ρ'c'S')/(ρcS + ρ'c'S')] A and C = [2ρcS/(ρcS + ρ'c'S')] A.* Let's look at the two limiting cases when the impedances are very different. First, if ρcS is much less than ρ'c'S', C is about zero, and B = -A. This is the case of a rigid wall, or rare-to-dense reflection. The incident wave is reversed in phase, and all the energy is reflected from the boundary.

If ρ'c'S' = ρcS, or the impedances are equal, then B = 0, and all the energy is transmitted.





In the limit that S' goes to infinity, C -> 0 => no transmittance. We know that there was tranmittance, so obviously this theory is not very illuminating, even if it is suggestive.


Not sure, but I think the earlier post re Zoeppritz equations are developed for infinite media. ( http://forum.physorg.com/index.php?showtopic=12383&st=4920# ), while the telegraph wire equations are developed for finite media.

I would think that theory has been developed for a situation closer to WTC/bedrock scenario, which would be finite (width) media for z < 0, infinite (width) media for z > 0. Anybody?


* I corrected the web page formula for C, which contains a typo. (They had ρ for ρ')
metamars
QUOTE (metamars+Jun 27 2007, 02:38 PM)



In the limit that S' goes to infinity, C -> 0 => no transmittance. We know that there was tranmittance, so obviously this theory is not very illuminating, even if it is suggestive.


Intuitively, as S' goes to infinity, the waves will proceed deeper into the bedrock long before waves 'going off to the side' will reach the end of S'. Thus, I wouldn't expect these equations to be valid for media where S' >> S.
wcelliott
QUOTE
As I recently explained to a Cter it is air that cools the end of the rod in a black smiths forge that is what keeps the end cool enough for him to handle. Even though the end is glowing


What you're saying is true, but may mislead some people who read it too literally.

The other factor at-work here is that steel has a relatively low thermal conductivity. It isn't like copper or aluminum, which under similar circumstances, would burn the blacksmith's hand despite the cooling effect of air.

Steel at 800C would sag and droop like taffy.
reasonwhy
QUOTE (adoucette+Jun 27 2007, 06:30 AM)

Who said the cutting was done in the basement?

Molten steel from cutting above could go quite a way down.

But here's the point you REALLY miss.

The "stalagmite" sits NEXT to the column. Which makes sense if its dripping from above.

He does not describe the column itself however as being damaged by anything molten:


A three foot pile of slag from above AFTER the column is out of alignment? laugh.gif

QUOTE (adoucette+Jun 27 2007, 06:30 AM)


Which is what STRESS from ABOVE would do based on a gravitational collapse.


Gravity now pushes columns horizontally ? biggrin.gif

QUOTE (adoucette+Jun 27 2007, 06:30 AM)

Oh, and about "before the evidence was destroyed"

But they thought they would give a NYTimes reporter a GUIDED TOUR of said evidence BEFORE they destroyed it?

ROTFLMAO

Arthur


They obviously were not worried about a REAL investigation by NIST or anyone else (no molten steel according to NIST). biggrin.gif
adoucette
Where do you get from that article that the "stalagmite" had ANYTHING to do with that COLUMN?

The article says it was NEXT to the column, it does not say it was MELTED from the column.

The term "NEXT" could be anywhere from one inch to several feet.

Besides, this is after TWO MONTHS of STEEL CUTTING OPERATIONS in the pile.

Do you expect that there WOULDN'T be these "stalagmites" from all that cutting that is going on up above?

FRIGGIN AMAZING.


Go sideways?

Yes, below this column is resting on an even STRONGER footing that is then resting on BEDROCK. When a force SUFFICIENT to break the column in a "Compound fracture" is applied, then YES, I would not be surprised that when the column fails it will ALSO get displaced sideways.

What about that surprises you?

Arthur
sirfiroth
Ever wonder why the world is in such disaray?

With an endless parade of experts nothing is ever resolved, it only fades away.

There is no black and white, only infinite shades of grey to consume and conceal.

Insipid discussion
David B. Benson
QUOTE (einsteen+Jun 27 2007, 09:28 AM)
DBB, Although the wtc stories were no springs they had the same behavior in first order, i.e. force linear in the beginning with distance.

The nice thing is that if you use those theoretical functions and set up your differential equations then nobody on earth can solve that, but on the other hand you could also substitute the collapse function and check if it is a possible solution...

Yes, so long as the elastic limit is not surpassed, the core columns would have behaved approximately like ideal mass-spring systems for small amplitudes.

The equations just for the forty masses and springs are easy to set up and solve numerically.

The crush-down equation will not be a solution for the equation for forty masses and springs.

But anyway, I've thought of another approach to describe the first 0.2 seconds after the walls buckled in WTC 1. I'll report it later.

metamars --- Yes, coupling the ends of core columns to the concrete is, in detail, more complex than I described. But since I am only interested in the reflection coefficient, I believe saying it is about -(2/3) is reasonable.
David B. Benson
Fundamental period of a core column in WTC 1 --- The core columns were 437 meters tall from the lowest sub-basement to the top. I will assume these columns all massed 660 tonnes, distributed evenly along the entire length, since the answer is the same if the columns were sections of a right cone or even a square pyramid.

When the three walls buckled, the excess load of the top portion of the south half of the tower was carried onto the core columns by the hat truss. From NCSTAR1-6D, Table 4--7, the excess load was 7717.6 tonnes. From Figures 4--71 and 4--81 there were 27 surviving columns in the south half, but only 7 were unstrained. If the sudden excess load is divided evenly over all 27 surviving columns, the average sudden load was 293.2 tonnes. If over only the 7 unstrained columns, the average sudden load was 1131.1 tonnes. I will assume that some of the strained columns were still capable of bearing a load to assume the sudden load was 660 tonnes.

I also assume that the reflection coefficient at the bottom of the columns was exactly -1.0.

From the Fox & Mahanty 1970 paper in Amer. J. Physics previously cited, the multiplier for the effective mass of the spring (core column) in this case is 0.35. So the total mass to be used in the mass-spring equations is M = (1.35)(660) tonnes. The density of mild steel is d=7848.3 and we take Young's modulus o be E=200x10^9. The inverse speed is

1/c = sqrt(Md/E) = 0.0.005913 s/m

and the period is

T = 2(pi)/c = 0.037 seconds.

Hence, at T/4 = 0.0093 seconds, compression is at its maximum, assuming elastic behavior, and rarefaction begins. At T/2 = 0.019 seconds, the mass-spring system has returned to its original position, again assuming elastic behavior. At (3/4)T = 0.0283 seconds, the ideal mass-spring system is at its greatest extension. At T = 0.037 seconds the ideal mass-spring system is again at its original position.

However, from NEU-FONZE's measurements, at t = 0.02 seconds the center of the top of the tower had descended about 0.1--0.14 meters and at t = 0.04 seconds it had descended about 0.4--0.55 meters. This indicates that already inelastic behavior, or at least out-of-plane bending, had occurred.
David B. Benson
QUOTE (David B. Benson+Jun 27 2007, 11:37 PM)
Fundamental period of a core column in WTC 1 --- The core columns were 437 meters tall from the lowest sub-basement to the top. I will assume these columns all massed 660 tonnes, distributed evenly along the entire length, since the answer is the same if the columns were sections of a right cone or even a square pyramid.

When the three walls buckled, the excess load of the top portion of the south half of the tower was carried onto the core columns by the hat truss. From NCSTAR1-6D, Table 4--7, the excess load was 7717.6 tonnes. From Figures 4--71 and 4--81 there were 27 surviving columns in the south half, but only 7 were unstrained. If the sudden excess load is divided evenly over all 27 surviving columns, the average sudden load was 293.2 tonnes. If over only the 7 unstrained columns, the average sudden load was 1131.1 tonnes. I will assume that some of the strained columns were still capable of bearing a load to assume the sudden load was 660 tonnes.

I also assume that the reflection coefficient at the bottom of the columns was exactly -1.0.

From the Fox & Mahanty 1970 paper in Amer. J. Physics previously cited, the multiplier for the effective mass of the spring (core column) in this case is 0.35. So the total mass to be used in the mass-spring equations is M = (1.35)(660) tonnes. The density of mild steel is d=7848.3 and we take Young's modulus o be E=200x10^9. The inverse speed is

1/c = sqrt(Md/E) = 0.0.005913 s/m

and the period is

T = 2(pi)/c = 0.037 seconds.

Hence, at T/4 = 0.0093 seconds, compression is at its maximum, assuming elastic behavior, and rarefaction begins. At T/2 = 0.019 seconds, the mass-spring system has returned to its original position, again assuming elastic behavior. At (3/4)T =  0.0283 seconds, the ideal mass-spring system is at its greatest extension. At T = 0.037 seconds the ideal mass-spring system is again at its original position.

However, from NEU-FONZE's measurements, at t = 0.02 seconds the center of the top of the tower had descended about 0.1--0.14 meters and at t = 0.04 seconds it had descended about 0.4--0.55 meters. This indicates that already inelastic behavior, or at least out-of-plane bending, had occurred.

I made three mistakes by being in a hurry, but didn't discover this until too late to edit.

The spring constant is given by k = E(A/L) where E is Young's modulus, A is the cross-sectional area, and L=437 meters, the height of the core columns. Assuming a fixed cross-sectional area A = 0.1 m^2, k = 0.04577x10^9. Then

1/c = sqrt(M.k) = 0.1395

and

T = 2(pi)/c = 0.8767 seconds,
T/2 = 0.438 seconds, and
T/4 = 0.219 seconds.

Now at T/4 the elastic compression is at a maximum which might be taken as agreeing with the observed 0.1--0.14 meter descent at t=0.2 seconds. But at both T/2 and T the elastic behavior implies a return to the original position, in comparison with the measurements and the calculations which give, at t=0.4 seconds, a drop of 0.4--0.55 meters, at t=0.8 seconds, a drop of 2.0--2.18 meters, and at t=1.0 seconds, a drop of 4.9--3.39 meters.
David B. Benson
Blast! Another correction:

1/c = sqrt(M/k)
einsteen
Isn't this in fact solving mx''=-kx?, that would indeed give a sinus shaped collapse, which also is approximately a parabolic shape. The bizar thing is that this is now applied stepwise for each 3.8 meter. Wouldn't that give an effective k which is 110 times higher ? The collapse could physically only be a total bypass, but that cannot explain the 0.5 GJ in the beginning.
einsteen
Oops should add mg of course... mx''=-kx+mg

ps. but in the beginning there is of course already strain and the translation x -> x+mg/k will reduce to the above...assuming we are still linear.
David B. Benson
QUOTE (einsteen+Jun 28 2007, 07:57 AM)
Isn't this in fact solving mx''=-kx?, ...

Yes. But what the posts, suitably corrected, show is that an intact core column has a quite short period in the vertical direction.

Indeed, thanks to shagster, I have a better estimate of the cross-sectional area, an average of 0.203 m^2, so the corrected period is

T = 0.6215 seconds

and

T/4 = 0.1563 seconds

is the first time of maximum compression. This compares well with my previous estimated time for a over-pressure front to travel to the bottom and return as a canceling under-pressure front, 0.146 seconds for the full round trip.

The point of these posts is that the elastic limit of core columns was already overcome in the first 0.2 seconds of the collapse, following wall buckling.
David B. Benson
Did the weakest part, near the top, of the core of WTC 1 fail before lower down?

The south and immediately thereafter, the east and west walls of WTC 1 buckled. This means that the load of the upper portions, above the buckling, are supported by the hat truss, which transfers these loads onto the core. We treat the south one-half only, using the loads of the entire south wall and the south half of each of the east and west walls as the imposed additional load. The data is from NCSTAR1-6, Tables 8--2 and 8--3.

From Table 8--2 the additional imposed load was 9850 tonnes for a total load on the south half of the core at floor 105 of 13,196 tonnes. Assuming an average DCR of 0.5 before damage, as NIST reports, the resulting average DCR on all columns of the south half, including those with destroyed or damaged members around floor 98, was 1.429.

However, that is the quasi-static DCR. To account for the fact that the imposed load was a sudden load of infinite duration, the general dynamical considerations of Simitses & Hodges, op. cite, require multiplying the above figure by two. So the average dynamical DCR is then

ave. DDCR = 2.9858

which is a stupendous overload.

It seems quite likely, from this analysis, that some (or all) of the upper members of core columns failed by buckling followed by connection failure to facilitate the tilting of the upper block, hinged for about one second on the north wall.

Unfortunately, it seems that NIST has no core columns from these elevations in its steel inventory.
Possibly re-analysis of the videos (by others, I can't do it) might develop evidence which would tend to support this conjecture.
RealityCheck
.
Damn!

Sorry NEU-FONZE!

Something went wrong at my end and I posted prematurely and when I removed it I lost the 'saved' copy in my word-processor file as well AT MY END!

All my fault, pressed the wrong button! I'm still not fully able to view screens properly....the recovery is still 'iffy', hehehe.

How are you, NF?

RC.
NEU-FONZE
RealityCheck:

Hi, long time no see... nice post!

I have believed for quite a while that tilting of a "top heavy" structure, causing "sagging" (compression) and "stretching" (tension), was the key to collapse initiation ..... gravity did the rest.
David B. Benson
QUOTE (RealityCheck+Jun 28 2007, 11:12 PM)
Something went wrong at my end and I posted prematurely and when I removed it I lost the 'saved' copy in my word-processor file as well AT MY END!

huh.gif Looked fine to me and I was about to answer when your post went away. Premature posting is always remediated via the edit button, if you do so within an hour. But I suppose you knew that long before I did...
RealityCheck
QUOTE (David B. Benson+Jun 28 2007, 11:58 PM)
huh.gif  Looked fine to me and I was about to answer when your post went away. Premature posting is always remediated via the edit button, if you do so within an hour. But I suppose you knew that long before I did...



Hehehe. Yeah, DBB, I lnew. That's what I was doing before I pressed the wrong button and exited from Physorg and was left with my word-processor file....which I TOO promply deleted by pressing the wrong button!

I knew people had days like tis! hehehe. I suppose I can blame my recovering eyesight for not taking care....although lack of sleep for the last few nights due to 'nerve discomfort' might have something to do with my laxness today. Luck I'm not in charge of heavy machinery, heh! hehehe.


Anyhow, I'm glad you and NF caught 'the gist' from my initial premature post!

I think I'll go close my eyes for a while, they seem to be more 'dangerous' than useful today!

Cheers!

RC.
.
NEU-FONZE
RealityCheck:

I'm fine, some of the time at least, and I hope you are too!

Don't you just love computers!

Some of my best posts have been "lost in cyberspace" too.....
Pierre-Normand
QUOTE (Grumpy+Jun 22 2007, 08:45 PM)
Godzilla=impossible, we're talking about New York, not Tokyo.

Couldn't this be construed to be coprolitic evidence for the Godzilla hypothesis?

User posted image
Pierre-Normand
QUOTE (David B. Benson+Jun 23 2007, 12:00 AM)
Grumpy --- Thanks for the useful link! An I'll agree to the banana peel analogy if you'll include some Japanese cooks to immediate slice up the peel parts which represent the east and south walls of WTC 1.

Who ordered Japanese skyscraper cooks?

http://www.cartoonstock.com/newscartoons/c...es/ksmn849l.jpg
wcelliott
I think this proves my Invisible-Godzilla hypothesis.
Pierre-Normand
QUOTE (David B. Benson+Jun 23 2007, 07:20 PM)
metamars --- By my calculations, the energy consumed per undamaged, truss floored, story for the first few seconds was 510 megajoules. Of this stupendous consumption, so far only about 20 megajoules is accounted for in breaking the truss seats. Nobody seems ready to supply figures for breaking connections in the core or breaking the bolts holding wall sections together. Nobody seems ready to estimate the energy required to crush the trusses. But qualitatively only, I'm having difficulty finding ways to expend the remaining 490 megajoules. Some small portion goes to air movement, but for the first few seconds this is negligible. The initial speeds seem a bit low for much concrete comminutiion.

I just thought of a possible mechanism for extra energy dissipation of E1 or equivalently of F_c * h. I will have to do some calculations to see if it is significant. I'll report back on this later today. It does not involve Godzilla this time.

Well... Thinking about it in a time-reversed fashion: The mechanism might have to do with rockets and launch pads! Let me just think this through.
lozenge124
QUOTE (David B. Benson+Jun 26 2007, 07:55 PM)
lozenge124 --- Yesterday I computed an approximation to the slenderness ratio of a hypothetical building consisting just of the exterior walls. I compute 31.2, probably an underestimate. It might be as high as 60 or even 70. However, the critical slenderness ratio for Euler buckling is 99.3.

So, provided the wind does not blow hard, this hypothetical building ought to stand up on its own.

To be fair to RealityCheck's hypothesis, he had the building coming apart at the seams, as it were, due to oscillations & resonance between the wall components (each consisting of 3 column sections and 3 spandrel plate sections welded together) - at least that's how I understood it.

But absent winds and with the building in a non critical state, per your calculation, it's hard to see why this would happen. Thanks for posting your result.
wcelliott
QUOTE
But absent winds and with the building in a non critical state, per your calculation, it's hard to see why this would happen. 


Unfortunately for this theory, there were gale-force winds coming out of the floors being crushed and the building *was* in a critical state as it was collapsing.

Aside from that, Mrs. Lincoln, how was the opera?
lozenge124
QUOTE (wcelliott+Jun 29 2007, 02:11 PM)

Unfortunately for this theory, there were gale-force winds coming out of the floors being crushed and the building *was* in a critical state as it was collapsing.

Aside from that, Mrs. Lincoln, how was the opera?

DBB's calculation refers to an undamaged building, genius.

The question was, would the perimeter walls hold up on their own if the core and floors were removed. This was actually originally prompted by your post that claimed they would just fall over - thanks for paying attention!
metamars
QUOTE (Pierre-Normand+Jun 29 2007, 07:30 AM)
Couldn't this be construed to be coprolitic evidence for the Godzilla hypothesis?

User posted image

Now you're being silly. Since when does Godzilla eat rebar?
Grumpy
Pierre-Normand

QUOTE
Couldn't this be construed to be coprolitic evidence for the Godzilla hypothesis?



Yes, but BIG G is getting entirely too much iron in his diet.

Grumpy cool.gif
Grumpy
lozenge124

If an undamaged outer frame could be constructed it would only last as long as the windspeed remained near zero. Any oscillation would soon bring it down.

Think of it as a reverse Slinky. As long as the forces remain vertical(as with floor bracing) it can remain . Any undamped horizontal force is multiplied by the unbraced length and would generate forces far in excess of those the bolted connections could sustain and it would fall like a Jenga tower. Just the available bending of the steel columns themselves would be enough to allow non vertical forces to reach the point of collapse. 1000 feet of steel beam flexes quite a lot, look up the Tacoma Narrows bridge, for example.

Grumpy cool.gif
wcelliott
QUOTE
The question was, would the perimeter walls hold up on their own if the core and floors were removed. This was actually originally prompted by your post that claimed they would just fall over


In the idealized case, a pencil balanced perfectly on its point will remain standing, too.

But in the real world, it falls over.

Does the idealized model of the perimeter wall include any tolerances for alignment at the joints, or are you assuming that steelworkers' efforts are uniformly perfect?
reasonwhy
QUOTE (wcelliott+Jun 29 2007, 07:41 AM)

In the idealized case, a pencil balanced perfectly on its point will remain standing, too.

But in the real world, it falls over.

Does the idealized model of the perimeter wall include any tolerances for alignment at the joints, or are you assuming that steelworkers' efforts are uniformly perfect?

What does a pencil balanced perfectly on its point and a square tube with a foundation going down to bedrock have in common?

Grumpy
reasonwhy

QUOTE
What does a pencil balanced perfectly on its point and a square tube with a foundation going down to bedrock have in common?


Sensitivity to perturbation. A 1000 foot high steel structure(which, incedently has the thickness/length ratio close to tin foil for the beams) will multiply any sideways force in a infinite positive feedback loop to failure unless specifically braced to damp such forces(IE core and floor bracing). Steel is nowhere near stiff enough to resist such forces(carbon fiber MIGHT be).

Grumpy cool.gif
David B. Benson
QUOTE (Grumpy+Jun 29 2007, 05:26 PM)
Sensitivity to perturbation. A 1000 foot high steel structure(which, incidentally has the thickness/length ratio close to tin foil for the beams) will multiply any sideways force in a infinite positive feedback loop to failure unless specifically braced to damp such forces(IE core and floor bracing). Steel is nowhere near stiff enough to resist such forces(carbon fiber MIGHT be).

The towers had an aspect ratio of 6.5:1. That's a rather stubby pencil. I treated the exterior wall, standing alone, as a single column to obtain a slenderness ratio well within acceptable parameters.
That means the walls alone would not undergo buckling unless there is some lateral forces. I did not compute the resistance to overturning, which is a different calculation.

The resulting safety factor was higher than actually used in constructing the complete tower. Finally, the resulting allowable load seemed too small.

The whole question is entirely hypothetical, an attempt an building physical intuition about tall columns. But yes, the actual tower design required the dampers to remove lateral vibrations.
wcelliott
QUOTE
The towers had an aspect ratio of 6.5:1. That's a rather stubby pencil.


My point wasn't that the tower-shell would itself topple over, my point was that the perimeter walls wouldn't be stiff enough to be considered a *unit*, but that the inherent flexibility of such thin walls would allow flexing off-vertical, and that off-vertical flexing would quickly exceed the strength of the bonds holding the individual perimeter trusses together.

Leading to them separating at broken connections, leading to collapse of the structure piecemeal.

In the idealized state, where the perimeter walls are all precisely vertical, there would be no side forces (as in the case of a perfectly-balanced pencil standing on-point), but in the real-world case, small perturbations from vertical would exaggerate themselves without something (like a floor) restraining their off-vertical leaning, the same way that when an almost-perfectly balanced pencil starts to tip over, the pencil's Cg moves in the same direction as the initial perturbation leading to positive feedback, accelerating it towards an unsustainable state (i.e., leaning outwards). It doesn't take much of an outward-leaning angle to create forces at the joints capable of breaking those connections (welds/brackets), due to the leverage of a long truss in-compression on the one side and in tension on the other. The outer edge forms the fulcrum of a lever that's as long as the perimeter assembly, the joint/bracket/weld on the other side of the column is at the business-end of the lever that's as long as the column (36feet?) and the leverage is the ratio of the length to the thickness of the column (12"? 16"?). The force on the bracket would be the weight of the column times the distance to the Cg (36'/2) times the sine of the lean-angle divided by the thickness of the column. If you assume that those columns act as a unit all the way up, the distance to the Cg ends up being half the height of the WTC tower, and the leverage is the ratio of the height to the Cg divided by the thickness of the perimeter column. It doesn't take much force with that sort of leverage to start snapping brackets/bolts/welds.

At which point the wall collapses piecemeal.
David B. Benson
QUOTE (wcelliott+Jun 29 2007, 07:44 PM)
My point wasn't that the tower-shell would itself topple over, my point was that the perimeter walls wouldn't be stiff enough to be considered a *unit*, but that the inherent flexibility of such thin walls would allow flexing off-vertical, and that off-vertical flexing would quickly exceed the strength of the bonds holding the individual perimeter trusses together.

You may have a point. Treated as a single column, the outer dimension is 63.7 meters and the inner dimension is only 0.1 meters less, above floor 8, which is the design I considered. That's a column, assumed to be constructed the way ordinary steel columns are made, that is very thin. The thinness ratio for the box column is 637:1, and ordinary steel box columns aren't made that thin, AFAIK.

Nonetheless, the efuda column buckling calculator only cares about the radius of gyration, which is quite insensitive to the thickness of thinness of the box column.

So strictly speaking, such a column would not buckle, but rather crumple...

Edited to add: Experimentalists can try this in their gym. Suppose corrugated cardboard is 4 mm thick. Construct a square (large) box of this cardboard which is 2.548 meters on a side to see if you can build one 16.562 meters tall. biggrin.gif

Edited again to add: Corrugated cardboard doesn't have the same Young's modulus. Try using sheet steel, as is found in duct work. Assuming this is 1 mm thick you'll need four pieces 0.637 meters by 4.16 meters. Punch holes representing windows. Assemble and observe. smile.gif
Pierre-Normand
QUOTE (metamars+Jun 29 2007, 02:36 PM)
Now you're being silly. Since when does Godzilla eat rebar?

QUOTE (metamars+Jun 29 2007, 02:36 PM)
Now you're being silly. Since when does Godzilla eat rebar?

He does not really eat it. He uses it to floss, and might've swallowed some by accident. This is a behavor Godzilla acquired from observing King Kong.

User posted imageUser posted image
Daru
The debat is over.

Game over.

WTC 1,2 & 7 could not have collapsed because of fire.

It has already been proven.

Most likely explosives were used.

The only question now is; What are the US people (and in fact all the world) going to do about it. What is the next step. About that is the debat now. (or should be)
David B. Benson
And returning to rationality (after the two, equally amusing, previous posts):

The hat truss was amazingly strong. From pages 306--308 and also page 170 of NCSTAR1-6D one can determine the ability of the hat truss outriggers to hold up the top portion of the east and west walls after the walls buckle. Here, again, we just consider the south 1/2 of these walls. For this portion there were two outrigger connections. Under tension these have a combined yield load limit of 3,415.6 tonnes and an ultimate load limit of 4,445.3 tonnes. But above the break, taken as floor 98, the south half of the east wall load to be supported was 2,778.07 tonnes, easily transfered to the core.

Interestingly, this seems to indicate that for WTC 2 the outriggers would fail, being unable to transfer all the load of the walls onto the core as there were about 1.65 times the mass in the top block for that gravity-driven collapse.
Pierre-Normand
QUOTE (David B. Benson+Jun 30 2007, 12:49 AM)
Interestingly, this seems to indicate that for WTC 2 the outriggers would fail, being unable to transfer all the load of the walls onto the core as there were about 1.65 times the mass in the top block for that gravity-driven collapse.

Unless the core failed before [(1/1.65)*100] percent of the perimeter weight had been transferred to the core, right? And why would it not have failed by then? Were the DCRs in the WTC2 core much lower than those of WTC1 just prior to the buckling of perimeter walls? If they weren't, then only a partial load transfer would have been achieved before ultimate failure lead to much unloading of the upper block core and thus to much relief of the tension in the outriggers' perimeter connections.
Grumpy
Daru

QUOTE
The debat is over.


No, the DEBATE is not over, there are still people like you who display their ignorance on this forum. Until you stop repeating such drivel the DEBATE will never be over.

QUOTE (->
QUOTE
The debat is over.


No, the DEBATE is not over, there are still people like you who display their ignorance on this forum. Until you stop repeating such drivel the DEBATE will never be over.

WTC 1,2 & 7 could not have collapsed because of fire.

It has already been proven.


By whom. All I see from the real scientists is that aircraft impacts and fires were the ONLY cause of those collapses. Have you or anyone else published peer reviewed papers falsifying NIST??? No??? I didn't think so. So, No, no one has proven any such thing and it is an incredibly stupid statement to make. Any other stupid statements you wish to make at this time???

QUOTE
Most likely explosives were used.


Evidence??? None??? Then this is just another incredibly stupid remark.

QUOTE (->
QUOTE
Most likely explosives were used.


Evidence??? None??? Then this is just another incredibly stupid remark.

The only question now is; What are the US people (and in fact all the world) going to do about it. What is the next step. About that is the debat now. (or should be)


Well, there is no DEBATE about who posts the stupidest, most moronic and ignorant statements on this forum, at least since dad1 left.

Grumpy cool.gif
David B. Benson
QUOTE (Pierre-Normand+Jun 30 2007, 01:22 AM)
Unless the core failed before [(1/1.65)*100] percent of the perimeter weight had been transferred to the core, right? And why would it not have failed by then? Were the DCRs in the WTC2 core much lower than those of WTC1 just prior to the buckling of perimeter walls? If they weren't, then only a partial load transfer would have been achieved before ultimate failure lead to much unloading of the upper block core and thus to much relief of the tension in the outriggers' perimeter connections.

Yes, this is only suggestive. However, NIST calculates that one of the outriggers failed in tension in WTC 2 before collapse commenced via wall buckling. I'm offering the opinion that more failed in tension just as collapse initiated.

How many and where might be partially resolved by reading the appropriate parts of NCSTAR1-6 and 1-6D. I'll not attempt it...
wcelliott
QUOTE
QUOTE 
The only question now is; What are the US people (and in fact all the world) going to do about it. What is the next step. About that is the debat now. (or should be) 

Well, there is no DEBATE about who posts the stupidest, most moronic and ignorant statements on this forum,


I'm having trouble deciding which of Daru's statements is stupidest.

They're all pretty close to being as stupid as humanly possible.

But since the technical questions have been addressed ad nauseam, let's examine his "What are the US people going to do about it?" question.

What would you suggest, Daru? Revolution?

Let's think about that for a moment.

I just checked, and there are just under 14 million US government employees.

There are 5.5 million people on Welfare.

There are 26.4 million veterans who reside in the United States; this is a ratio of about 1-in-8 (13 percent) of U.S. civilians 18 and over.

In 2007, nearly 50 million Americans will receive over $602 billion in Social Security benefits.

And I haven't even started discussing the number of people who work for companies whose main customer is the US government (like me, for instance).

There are about 280 million Americans, and it seems at least 100 million of those have a vested stake in keeping the US government in power, not counting people who are merely patriotic and prefer living in a democracy to changing to a different form of government, assuming there is a better one out there, and I can't think of any that aren't worse.

And let's count how many of you feel we should overthrow the government? I'm guessing that the number is a lot less than 100million. I'm guessing that it's a lot less than 100 thousand. So you're outnumbered by at least 1000 to 1 odds.

And the US government has the most sophisticated weapons systems on planet Earth (You're welcome!), so you don't have a technical advantage, either.

Hmmm...

Outnumbered at least 1000:1, not as well armed as the adversary, and the best you have in terms of a rallying cry is "Jet fuel can't melt steel!" (which it can, of course, that's why we don't use steel in the hot sections of jet engines).

And we're all supposed to decide to kill 100 million Americans to put people like *you* in power, I suppose?

You're wasting your time.

Bugger-off.
David B. Benson
Minor correction: I forgot to take into account, properly, the corner outriggers. Using 1/2 of its load carrying ability for the east wall, plus the (correct) load carrying abilities for the two outriggers along the south half of the east wall, there was a yield load of 3,755.8 tonnes and an ultimate load of 4,889.8 tonnes.

Also, I now have doubts about the one-hundredth scale model to actually be meaningful. The problem has to do with Young's modulus, which maybe also has to be scaled down. If so, build it out of sheets of 1 mm thick plastic.
wcelliott
QUOTE
The problem has to do with Young's modulus, which maybe also has to be scaled down. If so, build it out of sheets of 1 mm thick plastic.


I'd be more concerned with the assumption that you can model an assembly with a single sheet of anything, as it's the joints that failed, both in real life and in my analysis. Leverage of a squarish-cross-section beam leaning out will place the outside bottom of the beam in compression, and if it leans far enough, the inside of the beam (bracket/weld) will be in tension sufficient to break there at the joint, whether it's welded or bolted. It's a simple free-body diagram. I don't know how to post diagrams here, yet.

Maybe someone with more experience with this blog interface could help out?
David B. Benson
QUOTE (wcelliott+Jun 30 2007, 09:04 PM)
I'd be more concerned with the assumption that you can model an assembly with a single sheet of anything, as it's the joints that failed, both in real life and in my analysis. Leverage of a squarish-cross-section beam leaning out will place the outside bottom of the beam in compression, and if it leans far enough, the inside of the beam (bracket/weld) will be in tension sufficient to break there at the joint, whether it's welded or bolted.

Yes, this assumes that the through-welded plates and the bolts are as strong as the rest of the wall.

With regard to the failure mode you have proposed, I'll opine that was, in fact, the design. The observed failure mode is stripping walls of support so that (at least two of the) walls essentially fell over. One of these appears to have come apart of the way down.
wcelliott
DBB - Do you know the thickness of the perimeter column off the top of your head?

(I'm guessing you do.)

By "thickness", I mean outside dimensions, rather than material thickness.

I'm assuming the individual perimeter trusses are 36' long.

If you could tell me how much they weighed, I'll try to do the calculation of tension on the inside bracket/weld (which?) versus lean angle.

Would the Cg be at the geometric center of the perimeter truss?
David B. Benson
QUOTE (wcelliott+Jun 30 2007, 10:07 PM)
DBB - Do you know the thickness of the perimeter column off the top of your head? By "thickness", I mean outside dimensions, rather than material thickness.

I'm assuming the individual perimeter trusses are 36' long.

If you could tell me how much they weighed, I'll try to do the calculation of tension on the inside bracket/weld (which?) versus lean angle.

Would the Cg be at the geometric center of the perimeter truss?

(1) About 0.1 meter outside dimension according to either NCSTAR1--2 or 1--2A.

(2) Those are called perimeter (or exterior) columns. Yes, 36 feet high.

(3) I'll guess 1 tonne for the ones at the top, but I'm not sure (yet).

(4) For the perimeter columns , the center of gravity is significantly affected by the spandrels. See NCSTAR1--2A. However, the center of gravity of the entire hypothetical tower-walls-as box-column is, of course, in the center of the hypothetical structure.

Edited to add: NCSTAR1-2A, section 3.2.6, page 38 (72 ordinal) and following, discusses the exterior wall columns and spandrels. While nominally 14 inches square, the text points out the spandrels contributed significantly to stiffness in several coordinates. I recommend reading it before doing any calculations.

P.S. I'll be away from computers for the rest of the day. smile.gif
shagster
From a post by Adoucette in another thread:


The Perimeter Columns around the impact zone were mainly type 120.

The columns were made up of four welded plates

In column 120 there were 12 of these plates and they were all 0.25 INCHES thick ~14" wide (plate 2 starts out at 13.5" and shrinks by twice the increase in thickness of plate 1) and 36 Foot long.

There were two spandrels that were 0.375 inches thick that were 52" X 13 foot long (they amounted to ~ 1/4 of the weight of a tree).

This works out to (at 490 lb per cubic ft for steel) 14.2 cubic ft of steel and thus they were made from 6,975 lbs of steel.

Round that to 7,000 lbs per tree or 3.5 Tons.

There were 80 trees around the perimeter or 280 tons.

But these trees spanned 3 floors, so the FLOOR weight for the exterior columns was but 93 tons per floor (upper floors).

Which, in comparison to the ~ 1,500 tons that each floor weighed is relatively insignificant.

By the way, the core columns, were bigger individually, but there were many fewer and so their weight on a floor by floor basis would have been about the same.

Oh and though the exterior columns got heavier as they went down, the amount they increased was relatively small:

121 series column trees weighed in at 3.9 tons or 104 tons per floor
122 series column trees weighed in at 4.4 tons or 116 tons per floor
123 series column trees weighed in at 4.8 tons or 128 tons per floor
124 series column trees weighed in at 5.3 tons or 140 tons per floor
125 series column trees weighed in at 5.7 tons or 153 tons per floor
126 series column trees weighed in at 6.2 tons or 165 tons per floor
127 series column trees weighed in at 6.6 tons or 178 tons per floor
128 series column trees weighed in at 7.1 tons or 190 tons per floor
129 series column trees weighed in at 7.8 tons or 208 tons per floor
130 series column trees weighed in at 8.3 tons or 220 tons per floor
131 series column trees weighed in at 8.7 tons or 232 tons per floor
132 series column trees weighed in at 9.5 tons or 252 tons per floor

PS its NOT quite as simple as this as each floor had multiple different column series making up the floor, thus these top floors which were mainly series 120, also had a few 121, 122s and 123s mixed in and for this total I've ignored the twin beam corner columns.

Arthur

wcelliott
OK, it's official. I'm confused.

Are the perimeter columns 0.1m thick or 14" thick?

By "thick", what I'm looking for is if, standing at a theoretically-open window, you put a tape measure straight out from inside to outside of the window and measured the perimeter column's outside-diameter, inside surface to outside surface.

The 0.1m seems too small, the 14" seems too big. I had assumed it was about 8", give or take.

Please advise.

Thanks for the mass info!

Pierre-Normand
QUOTE (wcelliott+Jul 1 2007, 12:48 AM)
OK, it's official.  I'm confused.

Are the perimeter columns 0.1m thick or 14" thick?

By "thick", what I'm looking for is if, standing at a theoretically-open window, you put a tape measure straight out from inside to outside of the window and measured the perimeter column's outside-diameter, inside surface to outside surface.

The 0.1m seems too small, the 14" seems too big.  I had assumed it was about 8", give or take.

Please advise.

Thanks for the mass info!

The perimeter columns are exactly 14 inches wide (~distance between windows) according to both NCSTAR1-2A p. 40 and NCSTAR1-6C p. 7. The side plates are 13.5 inches large (~wall thickness). This excludes the thickness of the spandrel plate.

0.1 meter is roughly 4 inches. That would be the thickness of the floor slabs.

http://i185.photobucket.com/albums/x261/ho...n/Perimeter.jpg
wcelliott
QUOTE
The side plates are 13.5 inches large (~wall thickness).


Merci bien, Pierre!

Just what I was looking for.

Making the assumption/approximation that the Cg of a 36' high 14" thick column is at the geometric center, 18' up and 7" in, we can say the following:

When perfectly vertical, half the weight of the column will be on the inside edge, the other half will be on the outside edge. For the given weight of 7000 pounds, the inner and outer forces are 3500 pounds, each.

When the column/tree is leaning outward by 1.86 degrees. the Cg of the column is directly above the outer edge, so all the weight is resting on that edge, with zero weight on the inner edge. (Please, someone check my numbers.)

So far, what I've got is that the WT = F1 + F2
where:
WT = 7000#
F1 = the load on the outer edge of the perimeter truss
F2 = the tension on the inner edge of the perimeter truss

AND:

F2 = 7000#(18’-7”)sinA/14”

(if I've done the free-body diagram correctly.)

I'll enter this into Excel and print the results.
Pierre-Normand
QUOTE (wcelliott+Jul 1 2007, 05:57 AM)

Merci bien, Pierre!

Just what I was looking for.

Making the assumption/approximation that the Cg of a 36' high 14" thick column is at the geometric center, 18' up and 7" in, we can say the following:

When perfectly vertical, half the weight of the column will be on the inside edge, the other half will be on the outside edge.  For the given weight of 7000 pounds, the inner and outer forces are 3500 pounds, each.

When the column/tree is leaning outward by 1.86 degrees. the Cg of the column is directly above the outer edge, so all the weight is resting on that edge, with zero weight on the inner edge.  (Please, someone check my numbers.)

So far, what I've got is that the WT = F1 + F2
where:
WT = 7000#
F1 = the load on the outer edge of the perimeter truss
F2 = the tension on the inner edge of the perimeter truss

AND:

F2 = 7000#(18’-7”)sinA/14”

(if I've done the free-body diagram correctly.)

I'll enter this into Excel and print the results.

Are you trying to establish how much one column tree that is still being connected at the base can lean before the bolted connections fail? I don't know what the tensile yield strength of the bolts is. What you might do is compute the moment of force at the base of the column tree caused by gravity. This will be mgl/2*sin(A). This will be equilibrated by the tensile moment of the inner 3 pairs of bolts relative to the outer column webs. The outer bolts being much closer to the outer webs will not be stretched nearly as much and will be guarantied to fail when the inner bollts do. I am unsure how far from the plates the bolts are located. Let say the effective leverage is 12 inches (=d). Then the moments are compensated when:

mgl/2*sin(A) = d*y

where y is the combined yield strength of the 6 inner bolts. This gives you A, assuming there is some solution A (< 90 degree) to this equation, which there might not be if the self weight on one single tree held horizontally isn't enough to snap the bolts.
Pierre-Normand
I have a hard time trying to quantify in a principled manner what I thought could be some way to account for the extra 510MJ - 20MJ = 490MJ dissipated in the collapse process. The inverse-rocket analogy did not prove enlightening for this particular issue. I'll leave this problem aside for a little while.

User posted image

http://bitrot.net/blog/wp-content/uploads/2006/10/rocket.gif
adoucette
QUOTE (Pierre-Normand+Jul 1 2007, 04:23 AM)
Are you trying to establish how much one column tree that is still being connected at the base can lean before the bolted connections fail? I don't know what the tensile yield strength of the bolts is. What you might do is compute the moment of force at the base of the column tree caused by gravity. This will be mgl/2*sin(A). This will be equilibrated by the tensile moment of the inner 3 pairs of bolts relative to the outer column webs. The outer bolts being much closer to the outer webs will not be stretched nearly as much and will be guarantied to fail when the inner bollts do. I am unsure how far from the plates the bolts are located. Let say the effective leverage is 12 inches (=d). Then the moments are compensated when:

mgl/2*sin(A) = d*y

where y is the combined yield strength of the 6 inner bolts. This gives you A, assuming there is some solution A (< 90 degree) to this equation, which there might not be if the self weight on one single tree held horizontally isn't enough to snap the bolts.

The column bolts were ASTM 325

They had a ksi rating of 125 for up to 1 inch and 105 above 1 inch.

The columns had 4 bolts per column, spaced 6" apart (at least in the upper floors).

The butt plates were all 50 ksi steel.

Their thickness and the size of bolts varied based on height:

At the 96-99th floor the butt plates were 1.375" and used four .875" bolts

At the 92-96th floor the butt plates were 1.625" and used four .875" bolts

At the 91st floor the butt plates were 1.875" and used four 1.0" bolts

etc

The bolted column would fail at .18 inch beyond the ultimate capacity of the bolts.



The spandrels were all on the inside of the column, so a "stand alone" tower wall would have more weight on the INSIDE of the column center line.

Arthur

Palpatane
this thread has really devolved into the pursuit of the trivial.

What difference does it make how far the column tree can lean before the bolts fail. The point is, the bolts will eventually fail.
David B. Benson
QUOTE (Palpatane+Jul 1 2007, 02:39 PM)
this thread has really devolved into the pursuit of the trivial.

What difference does it make how far the column tree can lean before the bolts fail. The point is, the bolts will eventually fail.

Actually no.

The energy required to destroy the blots has not be established.

Arthur --- By the way, NIST states that only three bolts per column was the standard construction practice.

Assuming the exterior columns are simply connected at the spandrels, the slenderness ratio is 27. Assuming the exterior columns are fixed at the spandrels, the slenderness ratio is 13.5. Neither assumption accurately describes the action of the spandrels upon the columns and vice versa. However, the entire structure is a V. truss and any small. local perturbation spreads throughout the entire structure within 0.06 seconds. The vibrations reaching the concrete in the lowest sub-basement are partially damped there by being transmitted into the concrete. In a short time the vibrations damp to insignificance.

However, several local perturbations arising in different portions of the structure might combine into a so-called rogue wave in some column. This is certainly the case in the oceans, from whence the term arises.

My apologies to, especially, wcelliot, for the incorrect depth of the columns. I meant to write 0.3 meters, but indeed 14 inches is correct.
wcelliott
QUOTE
The bolted column would fail at .18 inch beyond the ultimate capacity of the bolts.


I'm not quite happy with the equation I derived before.

I'm trying to quantify the forces on the inner and outer side of the perimeter column/tree at the base as a function of the lean angle. The first cut at the equations indicate that the force on the inner edge is zero when the angle is zero, which I know is wrong, but I'm not sure where my error was. (A lot of distractions during the derivation last night.)

Anyway, when the angle is zero, the forces, which I'm calling F1 and F2 should be equal, and should sum to 7000#.

At the angle where the Cg is directly over the outer edge, F2 should be zero, and F1 should be 7000#.

As it tips further out, the net torque imparted by the Cg being outside the outer support must be countered by the torque of the difference between F1 and F2, with the continuing constraint that F1+F2=7000#, but recognize that when the Cg is outside the outer edge, the force at the inner edge is negative (in tension), so the force at the outer edge ends up being greater than the weight of the truss. With relatively small angles, the moment arm of the Cg ends up requiring the inner tension to be much larger than the weight of the truss itself, and the compression on the outer edge is that same force + 7000#.

This is for a single tree, but the forces get much greater when more than one tree is added to the "unit", double the weight for two trees bolted together, and double the moment arm. As the number of trusses taken as a unit increases, the forces rapidly exceed the strength of the brackets/bolts, which is why they snapped apart as the walls leaned out.

I agree in principle that this is kind of a waste of effort, but there have been CDiots here who seem to think that the perimeter columns could only have broken apart if explosives had been used. I thought it'd be worth calculating the forces involved. (Not that hard facts ever persuaded a CDiot that they were wrong...)
David B. Benson
NIST's terminology --- From NCSTAR1-2A, pages 36 and 38, the exterior units are called exterior wall trees only from floors 4 through 9. From floors 9 through 106, these units are called exterior wall panels, consisting (usually) of three columns and three spandrels sections. These latter are the units involved in the damaged and buckled portions of the towers.
adoucette
QUOTE (David B. Benson+Jul 1 2007, 03:27 PM)
Arthur --- By the way, NIST states that only three bolts per column was the standard construction practice.


Really?

I missed that, do you remember what report because NIST NCSTAR 1-3 (Sec 4.2.4 pg 23) says that 4 bolts were used for the upper floors and 6 bolts for the lower floors.

Arthur
David B. Benson
QUOTE (adoucette+Jul 1 2007, 11:50 PM)
I missed that, do you remember what report because NIST NCSTAR 1-3 (Sec 4.2.4 pg 23) says that 4 bolts were used for the upper floors and 6 bolts for the lower floors.

NCSTAR1-3B?

(Not in Prof. Astaneh's report)

In any case, this is in agreement with Ground Zero evidence, where there is no damage to one of the four bolt holes in every photograph...

Edited to add: Page 23 of NCSTAR1-3 is, for me, ambiguous.

I've never seen a photograph of one with six bolt holes, but Prof. Asteneh says there were some, although he also offers no photographic evidence. Perhaps the perimeter trees had six bolt holes while the exterior panels above the trees had only four bolt holes?
adoucette
QUOTE (David B. Benson+Jul 1 2007, 07:01 PM)
NCSTAR1-3B?

(Not in Prof. Astaneh's report)

In any case, this is in agreement with Ground Zero evidence, where there is no damage to one of the four bolt holes in every photograph...

Edited to add: Page 23 of NCSTAR1-3 is, for me, ambiguous.

I've never seen a photograph of one with six bolt holes, but Prof. Asteneh says there were some, although he also offers no photographic evidence. Perhaps the perimeter trees had six bolt holes while the exterior panels above the trees had only four bolt holes?

I found the reference in NIST NCSTAR 1-3C Sec 3.4.2 pg 112

But the three bolt practice was ONLY for the mechanical floors since those column splices were welded and thus the bolts were merely a construction aid (like the floor truss bolts).

Arthur

David B. Benson
QUOTE (adoucette+Jul 2 2007, 05:09 AM)
I found the reference in NIST NCSTAR 1-3C Sec 3.4.2 pg 112

But the three bolt practice was ONLY for the mechanical floors since those column splices were welded and thus the bolts were merely a construction aid.

Thanks. That is indeed the reference which I did not remember correctly.
metamars
From my post on p. 273

QUOTE (metamars+Jun 8 2007, 07:09 AM)
If we can get an accurate value of power output from the seismic record, should this just equal, at any given time before "crush up", KE(t) - Wd??

If so, then we have a way to test the Bazant, Le, Greening, Benson theory, using a quantifiable observable other than just speed of collapse.

.
.
.
QUOTE

a] would require that the energy needed to collapse each floor decrease as the collapse progressed during "crush down" (i.e., was less at lower floors), which is quite the opposite of reality. If b] is true, then we can roughly estimate that the energy needed to collapse each floor during "crush down" will be at most  {3% of KE during "crush up" / (84 floors) }, which is roughly (using Greening's figures)*:

{3% x (10^^12 J)} / 84 = 

357,142,857 J ~.4GJ

This is roughly in accord with  Greening's  .6GJ esitmate. 

However, by the time the collapse front reaches ground level, and "crush up" commences, the speed of the collapse front is more than an order of magnitude greater. In other words, in the same unit of time, we are collapsing (say) 10x as many floors, and yet the rate at which energy is dissipated to accomplish this is roughly constant. Consequently, we get an upper bound of .04GJ per floor, which is completely inconsistent with Greening's .6GJ.



Implications of BLGB for Crushing Power
=======================================

Taking the height z to increase downwards,

z = 1/2 at^^2

taking a = 2/3g, we see that

z(t) = (1/2)(2/3)gt2

so z(1) – z(0) = (1/3)(9.8)(1)
= 3.3m
or not even 1 storey height

However,
z(10) – z(9) =
3.3(10^^2 – 9^^2)
= 62.7m
16 storey heights

So, before “crush-up”, the rate of floor crushings has increased by a factor of 16x.
The lowest column sections were about 16x stronger than the topmost ones. Call the difference, for the purpose of this calculation, 10x.

Therefore, before “crush-up”, crushing energy is being applied at a rate about 160x during 9s < t < 10s vs. 0s < t < 1s.

I expect such a wide disparity to show up in seismic records, somehow.



If we assume that ‘crushing energy in’ is proportional to ‘seismic energy out’, then if the seismic records should reveal that only 3% of KE dissipated as ‘storey crushing energy’ during “crush down”, and that this seismic energy is created at a constant rate, we can figure out what crushing energy per floor should be as an implicit function of height (and explicit function of time). I solve for the scenario of ignoring column strength differential, assuming that "crush down" takes 10 seconds.


Call the total energy expended (by dissipating KE) in crushing the bottom of a WTC tower, as a function of height z (z positive downward) "E_crush(z)"


z(t) = (1/3)g t^^2

therefore d(E_crush(z)) / dt = d(E_crush(z))/dz * dz/dt = E' * { (2/3)g t } ( just using shorthand E' = d(E_crush(z))/dz )

Interpreting the seismic record as noted above implies that d(E_crush(z)) / dt = k, where k is some constant to be determined

therefore E' * { (2/3)gt } = k
E' = 3k/(2gt)


Now, recall Greening's estimate of total KE or 10^^12J, as well as my measurements implying 3% dissipation into seismic energy during "crush-down" vs. 97%

dissipation into seismic energy during "crush-up" *.

(More carefully, looking at "Energy Transfer in the WTC Collapse" by Greening, he estimates total KE as 4.6 x 10^^11 J, and assumes 75% "available" to crush concrete. So, we can consider maximum KE available to create seismic waves to be 25% of 4.6 x 10^^11 J, or 1.5 x 10^^11 J. Not sure if 10^^12 was a revised
estimate. I'll use 10^^12 J for the purposes of this calculation.)

3% x (10^^12 J)} = 3 x 10 ^^10 J dissipated during "crush down"


Therefore, to find k, we multiply both sides of
d(E_crush(z)) / dt = E' * { (2/3)g t } = 3k/(2gt) { (2/3)g t } = k
by dt and integrate

note that integrating d(E_crush(z)) from t = 0 to t = 10 gives 3 x 10^^10 J (as per Greening estimate) and thus
E_crush(z(10 seconds)) - E_crush(z(0 seconds)) = k (10 - 0)

So 3 x 10^^10 J - 0 J = k(10)

therefore

k = 3 x 10^^9 J/s


=> E'(z(t)) = 3k/(2gt) = (1/t) (3/2) (3 x 10^^9)

therefore

d(E_crush(z)) / dt = k
=>

E_crush(z) = (3 x 10^^9 J/s) t + K2

At t = 0, E_crush(z) = 0, therefore K2 = 0


///////////////////////////////////////
// E_crush(z(t)) = (3 x 10^^9 J/s) t //
///////////////////////////////////////


By calculating t values corresponding to a given floor, we can determine the energy necessary to crush that floor. Using

z(t) = (.5)((2/3)g)t^^2 = 3.27 t^^2

t(z) = sqrt ( z / 3.27 )


For the first crushed storey
-----------------------------
after a fall of h,


t = sqrt ( 3.87 / 3.27 )
= 1.09 s


So E_crush(1st crushed storey) = (3 x 10^^9 J/s) ( 1.09 s) = 3.27 GJ


I'm frankly not sure whether to declare this in good agreement with Greening, et.al. estimate of .6GJ. I would think not, from the point of view that the earliest stage of the collapse yields the crispest data. Thus, a 5-fold difference here seems particularly bad.

OTOH, poor agreement here may be an artefact of the homogenization process, and, in fact, I expect BLGB assumptions to have more (but still questionable :-) ) validity for a collapse after it had progressed for a while.

BLGB is also in trouble with lower floors involved with "crush down" ....


For the last crushed story during "crush down"
-----------------------------------------------

after a fall of 83 h

t = sqrt ( 83 * 3.87 / 3.27 )
= 9.91 seconds


after a fall of 84 h


t = sqrt ( 84 * 3.87 / 3.27 )
= 9.97

So E_crush(84th crushed storey) = (3 x 10^^9 J/s) (9.97s - 9.91s) = .18 GJ


E_crush(1st crushed storey) / E_crush(84th crushed storey) = 18 x

This is very bad, considering the correct figure should be something like
E_crush(1st crushed storey) / E_crush(84th crushed storey) = .1 x



* recalls Benson's complaints about my drawing inferences from the graph in BLGB! I still have not looked into these complaints.


edited:

Changed the line "However, BLGB is in far more serious trouble with lower floors involved with "crush down" .... "

to

BLGB is also in trouble with lower floors involved with "crush down" ....

since being off by a factor of about 5 over seems worse than being off by a factor of about 3 under. It is the ratio of values of upper and lower storeys that may indicate "serious trouble", more than anything else.
David B. Benson
QUOTE (metamars+Jul 2 2007, 07:23 PM)
The lowest column sections were about 16x stronger than the topmost ones.

I expect such a wide disparity to show up in seismic records, somehow.

If we assume that ‘crushing energy in’ is proportional to ‘seismic energy out’, ...

(1) The only part which matters is the strength of the construction splice welds in the core. These were indeed stronger as the members were more massive. BLGB takes this into account.

(2) I don't. The modest sized amplitudes nearer the beginning of the trace are surely due to exterior wall panels breaking off to fall freely in advance of the progressive collapse. Also, the energy escaping seismically can only be about (1/9)th of the elastic energies in core columns (maybe similarly in exterior wall columns).

(3) That is a reasonable assumption, but the total energy consumed eventually becomes far larger due to (i) concrete comminution and (ii) air movement. Neither of these will leak into seismic energies.

(4) The rate of total energy consumption, or even just the crushing energy consumption, is not constant. Attempting to determine such a rate ignores the fact that the downward speed of the crushing front continues to increase. Even the average rate is of no interest.

(5) If at first you don't succeed,...
NIST
[removed: no sci-tech]
frater plecticus
[removed: no sci-tech]
NIST
[removed: no sci-tech]
adoucette
QUOTE (punk+)
The militia movement has millions of members. And when the civil war will go down we will also be backed by the crips and the bloods and many other gangs


ROTFLMAO

But I did like your recruitment slogan, "Join or Die"

laugh.gif

Arthur
wcelliott
QUOTE
I'm guessing that it's a lot less than 100 thousand" The militia movement has millions of members. And when the civil war will go down we will also be backed by the crips and the bloods and many other gangs,


I'm still certain that your numbers are far below 100,000, and that the Crips/Bloods will be killing more militia members than government agents, especially once they realize that militias are mostly motivated by racism. The biggest militant group is White Supremists, and the Bloods/Crips have no allegiance to people who refer to them using the "n-word".

zoktoberfest
[removed: no sci-tech]
newton
[removed: no sci-tech]
wcelliott
Since this is supposed to be about Physics, rather than politics, I'll invite you to consider the following:

What is the stability of world peace?

In other words, how likely is the state where the entire world is at peace, no wars anywhere?

The naive among us seem to assume that war is something that can be avoided everywhere in all cases, that diplomacy will keep us at peace, if only we were more evolved. The jaded seem to think we're looking for wars to fight just to spend tax dollars on arms.

Here's the real story.

International relations is inherently chaotic, as in the "Chaos Theory" use of the word. Small differences in initial conditions result in macroscopic outcomes. A flock of butterflies in China can create a war with Cuba, that sort of *chaotic*.

You can try all you care to to avoid conflicts and resolve differences diplomatically, but the mere fact that a country avoids conflict, itself, invites provocation, and if this provocation is itself ignored, that country appears too weak to defend itself and that invites war.

War is inevitable in a chaotic environment, and so long as the nature of international relations is chaotic, wars will happen regardless of our best intentions.

Those who are able to convince themselves that every random event is an omen from God are also able to interpret every random international event as being part of a Grand Conspiracy by the NWO.

It's a sign of schizophrenia.

Even if you assume there is a conspiracy by a group of people called the NWO, it's a mistake to believe that any group, no matter how powerful, is actually *running* anything. World Politics isn't an aircraft, it doesn't have wings and it doesn't have controls, and even if it did, that doesn't mean anyone would know how to fly it. Even perfectly functional, well-designed aircraft sometimes get into modes that a good pilot can't fly his way out of, and this world is either in a flat spin or it's in a death spiral. I doubt any group of rich people would know how to exert any control over it, with Nature and Overpopulation and Diseases and Bonehead Political Activists and Terrorists and Ignorance and Stupidity and Religious Zealots all trying to crash the system.

War is inevitable. Deal with it.
NEU-FONZE
wcelliott:

As we all know, Switzerland has had five hundred years of democracy and peace... and the Swiss also invented the cuckoo clock!

Is that so bad?
wcelliott
QUOTE
As we all know, Switzerland has had five hundred years of democracy and peace... and the Swiss also invented the cuckoo clock!

Is that so bad?


Switzerland, as we all know, is where the NWO has its headquarters.

(just joking)

Neutrality isn't an available option for the US.
metamars
QUOTE (David B. Benson+Jul 2 2007, 08:42 PM)
(1) The only part which matters is the strength of the construction splice welds in the core. These were indeed stronger as the members were more massive. BLGB takes this into account.


(In the below, I refer to the June 22 version of BLGB)

And, what are the details? The word "splice" does not appear in your paper. Please explain.

Furthermore, looking at BLGB, figure 5a), South Tower, we see that after floor 64, until "crush up", Fb, Fs, and Fa are close to flat. Since I don't expect 4 inches of concrete to be any harder to crush at floor 64 than 4 inches of concrete would be at floor "F" (i.e., end of "crush-down"), I have to assume that a graph of absolute magnitude of Fs vs. floor # will also be flat between floor 64 and floor F.


From this, it follows that graphs of absolute magnitued of Fb and Fa vs. floor # would also be flat.

So, while your mathematics may have allowed the possibility of an increasing Fb, I am not seeing that in your graphs. Instead, it looks like your model is telling us that Fb is relatively constant for a large number of floors.

However, that is contrary to reality, even ignoring any considerations of what the seismic graphs tell us.


QUOTE

2) I don't. The modest sized amplitudes nearer the beginning of the trace are surely due to exterior wall panels breaking off to fall freely in advance of the progressive collapse.


And what percentage of Fb does this represent?


QUOTE (->
QUOTE

2) I don't. The modest sized amplitudes nearer the beginning of the trace are surely due to exterior wall panels breaking off to fall freely in advance of the progressive collapse.


And what percentage of Fb does this represent?



Also, the energy escaping seismically can only be about (1/9)th of the elastic energies in core columns (maybe similarly in exterior wall columns).

I don't think this has been nailed down. Are you referring to a former post of yours?



QUOTE


(3) That is a reasonable assumption, but the total energy consumed eventually becomes far larger due to (i) concrete comminution and (ii) air movement. Neither of these will leak into seismic energies.


I probably should have been clearer. I was only thinking of Fb, not Fc or Fa. As per my comments above, your graphs don't seem to agree with what you write here. We are talking about energy consumed (via Fb) per storey, correct?

QUOTE (->
QUOTE


(3) That is a reasonable assumption, but the total energy consumed eventually becomes far larger due to (i) concrete comminution and (ii) air movement. Neither of these will leak into seismic energies.


I probably should have been clearer. I was only thinking of Fb, not Fc or Fa. As per my comments above, your graphs don't seem to agree with what you write here. We are talking about energy consumed (via Fb) per storey, correct?


(4) The rate of total energy consumption, or even just the crushing energy consumption, is not constant. Attempting to determine such a rate ignores the fact that the downward speed of the crushing front continues to increase. Even the average rate is of no interest.


You're confusing me. Why cannot a rough meaure of power consumption be calculated, if, indeed, we make the observations I note above, and use reasonable estimates for acceleration during crush down? Actually, why don't you calculate that, for floors 64 to F for the South Tower?

QUOTE

(5) If at first you don't succeed,...

Good advice, which I hope you take. I think it's a good thing that you are making revisions to your paper. However, I still doubt that it is salvageable. I think the WTC towers are too complex for simple models to not quickly develop contradictions either with reasonable extrapolations, or ("obviously") with serious and reasonable questions regarding how realistic their assumptions are.

NIST
[removed: no sci-tech]
David B. Benson
QUOTE (metamars+Jul 3 2007, 04:02 PM)
And, what are the details? The word "splice" does not appear in your paper.

Furthermore, looking at BLGB, figure 5a), ...

Are you referring to a former post of yours?

(1) The paper assumes that the mass increases downwards.

(2) These are percentages of the total F_c(t), which is not constant.

(3) Yes.

It is clear that you still fail to understand BLGB. Ask questions and I'll attempt to clarify.
David B. Benson
QUOTE (metamars+Jul 3 2007, 04:02 PM)
And what percentage of Fb does this represent?

The energy required to break the bolts and shove an exterior wall panel out is quite small in comparison to the total or even the total per story.

The panel then falls freely to the ground. Upon impact, the ground shakes. Entirely due to the PE --> KE conversion during the fall.
zoktoberfest
[removed: no sci-tech]
frater plecticus
[removed: no sci-tech]
David B. Benson
QUOTE (ap-bot+Feb 1 2007, 10:22 PM)

[continued from 9/11 Events - New thread ]

Previous two threads :
Basic Physics, Correct Analysis of WTC Towers Collapse
911 WTC - Evidence of an Nuclear explosion?

Go to the first link. Read what administrator ap-bot states in the first post.

This is simply the continuation of that thread, necessary, I believe, for technical reasons.

Pay heed to what the administrator says there. ph34r.gif
wcelliott
QUOTE
below 100,000 . You will be surprised.


The surprise will be all yours when you discover just how many of your "friends" turn out to be undercover government agents.

QUOTE (->
QUOTE
below 100,000 . You will be surprised.


The surprise will be all yours when you discover just how many of your "friends" turn out to be undercover government agents.

Obviously, you didn't open the link to "Zeitgeist", nor is there any indication that you have ever read Orwell. You have exposed yourself, as an empty vessel, in the realm of geo/political dynamics. Even people, far to the right of your vapid outlook would readily acknowledge that.

If you believe that randomness is the engine driving earthly affairs


Not since Malmo was ejected has a post so long contained nothing but errors.

I did open the link to Zeitgeist, I did read "1984", I do participate in geo-political dynamics (that's the ultimate result of my 9-5 job), and I said that international relations were *chaotic*, not random, and if you don't know the difference between those two, maybe you should go learn something and get back to us.
newton
[removed]
wcelliott
[removed: no sci-tech]
newton
[removed: no sci-tech]
wcelliott
[removed: no sci-tech]
frater plecticus
QUOTE (David B. Benson+Jul 3 2007, 11:21 PM)
Go to the first link. Read what administrator ap-bot states in the first post.

This is simply the continuation of that thread, necessary, I believe, for technical reasons.

Pay heed to what the administrator says there. ph34r.gif

Hey David, I got a great Idea...



Why don't you start your own debate out of the "off topic" section of the forums, then you won't have to moan about what other people write here..
RealityCheck
QUOTE (frater plecticus to David B Benson+Jul 4 2007, 06:55 AM)
Hey David, I got a great Idea...



Why don't you start your own debate out of the "off topic" section of the forums, then you won't have to moan about what other people write here..



Hehehe.

Hi frater!

Why don't YOU try to start an 'associated' '9/11 POLITICS' thread where the 'political/religious off-shoots' from THIS thread can be discussed without upsetting anyone interested ONLY in the science/physics of 9/11 events?

Good luck, matey!

RC.
.
einsteen
Physics folks....
frater plecticus
[removed]
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