WTC: Rate of Fall (rate of crush)

Tony Szamboti

Active Member
No it is not. I think your analysis is quite lacking.
I showed my calculations for the composite beam (truss and slab) vertical deflection at temperature, and they are based on the beam deflection equation for simply supported end conditions with a distributed load, which is

deflection = (5 x total load x length^3) / (384 x modulus of elasticity x moment of inertia)

Where are your calculations saying any different?
 
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slenderbeam

Member
@slenderbeam
It seems from my reading that @Tony Szamboti is correct that a 7.4m SHS (as above) is stout. Do you agree?
There's no arguing they were robust built up sections, and could have been close to stout - wouldn't rule it out. The whole premise of this argument is farcical. Columns are hard enough to align when erecting let alone dropping through the air during a collapse.

Maybe if timoshenko was god this would be possible.
 
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qed

Senior Member
@Tony Szamboti

In your paper you do take as initial condition an already buckled stout SHS (buckled by a lateral force) and then consider the response to compression.
  • Which of your references analyze the compression response of an a priori buckled stout column?
 

slenderbeam

Member
I showed my calculations for the composite beam (truss and slab) vertical deflection at temperature, and they are based on the beam deflection equation for simply supported end conditions with a distributed load, which is

deflection = (5 x total load x length^3) / (384 x modulus of elasticity x moment of inertia)

Where are your calculations saying any different?
Elastic beam equations (the one you showed above) assumes a beam with relatively small shear deflection. A truss has substantial shear deflection because the web diagonals extend/shorten, as do the chords. The truss deflection is related more to axial elongation of the members because every member (ideally) is highly stressed and therefore contributing substantial elongation when compared to a beam web which has areas of high and low stress.

Compare a simple model truss (pin jointed) to a beam and you will see the difference.
 

slenderbeam

Member
@Tony Szamboti

In your paper you do take as initial condition an already buckled stout SHS (buckled by a lateral force) and then consider the response to compression.
  • Which of your references analyze the compression response of an a priori buckled stout column?
Why bother with this argument at all, the premise is that columns somehow fall 2 storeys or something perfectly onto the ones below. It is ludicrous.
 

slenderbeam

Member
We haven't seen any structural analysis from you, so you aren't teaching anyone anything.
I just told you why 5wL4/384EI is not appropriate - you chose to ignore it. I don't need to do calculations to prove this. I am asking you to open your mind and consider you've made a mistake.
 

Tony Szamboti

Active Member
I just told you why 5wL4/384EI is not appropriate - you chose to ignore it. I don't need to do calculations to prove this. I am asking you to open your mind and consider you've made a mistake.
It is appropriate as long as the diagonals in the truss do not buckle. The diagonals are sized not to buckle.
 

slenderbeam

Member
You are picking on nits and that is why you can't show the results of my analysis would be any different. You really should stop it.
I don't carry analysis software on me everywhere i go but I will do some numbers when I get a chance. What drawings did you use for the typical trusses? If you can send me the links I'll get around to it.
 

gerrycan

Banned
Banned
Columns are hard enough to align when erecting let alone dropping through the air during a collapse.
Interesting thread to sit and watch, but every so often I see something that is just so plain wrong that it requires me to log in and reply. @slenderbeam ....Do you have any understanding at all about how the transitional connections on these columns were designed and formed? Because the above quote would suggest that you have very little understanding of how such elements and connections function, or indeed of design in general. In reality, it would be difficult to position these columns at the construction phase wrongly in terms of alignment.
 

Tony Szamboti

Active Member
I don't carry analysis software on me everywhere i go but I will do some numbers when I get a chance. What drawings did you use for the typical trusses? If you can send me the links I'll get around to it.
The dimensions for the trusses can be found in the NIST WTC report. The actual drawings have never been released.

My calculations were done in Excel using standard engineering equations. There is no need for special analysis software here.

Just for you and others here, please look at pdf pages 19 and 20 of the attached, which is about composite slab and truss beams, and you will see they used the same simply supported distributed load deflection equation I did in my calculations.
 

Attachments

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OneWhiteEye

Senior Member
The reality is that the bottom of the North Tower's upper section disintegrated for four stories or more before contacting the lower section of the building. I don't know of a natural explanation for this. The only natural forces available to cause this would be contact with the lower section. This explains the lack of deceleration in the early stages.
A simple natural explanation is that column ends were not aligned.

You can have a misalignment after four stories of collapse and I would agree that at that point there is some visual evidence of it.
But you don't agree there's visual evidence for it after two stories? As I said, in order for it to be barely detectable in grainy video, it's all ready way out of line. Scrubbing forward in that video shows a steady movement to the severe misalignment (many feet) I showed earlier.

I also showed how the lowest portion of the upper north wall can be seen slipping outside the lower wall from the very beginning, and how it moves downward with the roofline as a rigid body over at least six stories. So it is plainly apparent there was misalignment straight away. With that wall slipping outside, what's going on inside? The lower wall is impacting floor assemblies coming down from above. What's that going to do to the deceleration on the north roofline? The upper section was (partially) disintegrating in the first few stories because it had to, given that geometry.

Revisiting your comment:


... before contacting the lower section of the building.
In the case of the leading tip of the north wall, it never contacted the lower section.

If you adjust your kinematic illustration posted earlier to account for the observed, factual offset of the north wall, what happens to the alignment of all the other columns? If the upper section were truly rigid, they'd all be off by the same amount. All of them miss and the south upper wall goes inside the lower. I suppose considering the upper section to be rigid might be asking a bit much. So, what happens to the structure itself if south face columns are aligned but the north face is off by two or three feet to the outside? Can the upper section actually be stretched 1-2+% horizontally and be considered intact? Maybe that's why it disintegrated.
 

slenderbeam

Member
Interesting thread to sit and watch, but every so often I see something that is just so plain wrong that it requires me to log in and reply. @slenderbeam ....Do you have any understanding at all about how the transitional connections on these columns were designed and formed? Because the above quote would suggest that you have very little understanding of how such elements and connections function, or indeed of design in general. In reality, it would be difficult to position these columns at the construction phase wrongly in terms of alignment.
It generally takes 1 crane operator, 2-3 very skilled labourers to position a single panel in place, correctly aligned.

However in your's and Tony's opinion you would have the entire building (while twisting and rotating), falling through the air and somehow land every column in alignment.

And somehow my posts inspire incredulity?
 
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slenderbeam

Member
The dimensions for the trusses can be found in the NIST WTC report. The actual drawings have never been released.

My calculations were done in Excel using standard engineering equations. There is no need for special analysis software here.

Just for you and others here, please look at pdf pages 19 and 20 of the attached, which is about composite slab and truss beams, and you will see they used the same simply supported distributed load deflection equation I did in my calculations.
Tony - as long as you used Section 5.5 steps 1-5 then I don't see a problem. This is an approximation to what I was saying.

Big difference to a beam.
 

Tony Szamboti

Active Member
I don't see why you are asking about the columns involved in the initiation here. It is the impacted columns below that would affect the propagation possibilities and for which the energy dissipation needs to be calculated.
 

qed

Senior Member
I don't see why you are asking about the columns involved in the initiation here. It is the impacted columns below that would affect the propagation possibilities and for which the energy dissipation needs to be calculated.
upload_2013-12-23_7-41-5.png

Are these not the columns whose response curve is in question in your theory of single floor arrest?
 

Tony Szamboti

Active Member
View attachment 5212

Are these not the columns whose response curve is in question in your theory of single floor arrest?
No, the arrest involves the columns of the impacted story below absorbing the energy from the one story fall. Of course, the energy absorbed over the first story is important so that you know what is left to impact the story below. It is that energy absorption subtracted from mgh which is what impacts the columns of the story below.
 

Tony Szamboti

Active Member
That is not what I seem to be reading here.
You either forgot to continue reading further down that same page where we say

One can also choose to be more conservative and degrade the average resistance by another 25%. Thus, the energy absorbed over the first story travel will be
Π = 0.75 Pav h = 0.75 × 325.4 × 106 N × 3.7 m = 903 × 106 N-m.
This time U > Π, therefore stopping is not complete over this story and further motion
must be considered. Noting that the gross kinetic energy available to overcome the resistance of the next story below is
Ek0 = 1,204 – 903 = 301 MN-m
Before further motion continues, there is an energy loss incurred due to the accretion of
the slab. When this is treated as a fully plastic collision, that loss, according to [5] is
(2)
where Mb = 33 × 106 kg is the mass of the descending part of the building and Ms = 2.74 ×
106 kg is the mass of the accreted slab and its tributaries. After substituting, one finds ΔEk =
23 MN-m. Consequently, at the outset of travelling down by a further story, the following status
develops. The moving part now has M = (33 + 2.74) × 106 kg = 35.74 × 106 kg and its potential
energy relative to the next floor is U = Mgh = 1,297.3 × 106 N-m. The energy available to
overcome the column resistance is now
Ek0 – ΔEk + U = (301– 23.0 + 1,297.3) × 106 = 1,575.3 × 106 N-m.



or you just want to be a nudge.

The paper is a critique of Zdenek Bazant's WTC analyses and shows that in contrast to what he said there was far more column energy absorption than there was kinetic energy to continue the collapse.

In reality, the bottom of the upper section of the North Tower disintegrated before ever contacting the lower section and that is why there was no deceleration observed.

You should really tell us what you think happened and as far as the paper is concerned tell us what you think is wrong. Go ahead be a man and take your full shot at it, instead of trying to nitpick ad infinitum with trivial nonsense you often aren't even right about. I am not here to be abused by a nonsensical nitpicker and nobody should blame me for ignoring one.
 
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qed

Senior Member
I was going to post this:

------------------------------------------------------------------

I am only requesting clarification with regard to single story arrest. I would love your argument to be sound.

Is this then correct?
Code:
┃
┠───────      c-1     ─
⎞                     ⥌  3.7m
⎬┄┄┄┄┄┄┄┄ →  c        ─
⎠
┠───────      c+1
┃
One story fall at "free fall" (near no resistance). This gives you U.
Then first collision.
Now begins first story travel from which ∏ is calculated?

  • Is first story travel from c+1 to c+2?
----------------------------------------------------------------------------------------------
but have just seen your edit!!!!!!!!
 
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qed

Senior Member
You either forgot to continue reading further down that same page where we say

One can also choose to be more conservative and degrade the average resistance by another 25%. Thus, the energy absorbed over the first story travel will be
Π = 0.75 Pav h = 0.75 × 325.4 × 106 N × 3.7 m = 903 × 106 N-m.
This time U > Π, therefore stopping is not complete over this story and further motion
must be considered
. Noting that the gross kinetic energy available to overcome the resistance of the next story below is
Ek0 = 1,204 – 903 = 301 MN-m
Before further motion continues, there is an energy loss incurred due to the accretion of
the slab. When this is treated as a fully plastic collision, that loss, according to [5] is
(2)
where Mb = 33 × 106 kg is the mass of the descending part of the building and Ms = 2.74 ×
106 kg is the mass of the accreted slab and its tributaries. After substituting, one finds ΔEk =
23 MN-m. Consequently, at the outset of travelling down by a further story, the following status
develops. The moving part now has M = (33 + 2.74) × 106 kg = 35.74 × 106 kg and its potential
energy relative to the next floor is U = Mgh = 1,297.3 × 106 N-m. The energy available to
overcome the column resistance is now
Ek0 – ΔEk + U = (301– 23.0 + 1,297.3) × 106 = 1,575.3 × 106 N-m.
  • No I did not forget.

That is your theory of two floor collapse, which I thought I had already beaten (given that you refused to respond when I pointed it out).
My argument then was that in the very next line you say,
and I pointed out that after the floor has been smashed, this column is damaged:
  1. it has been damaged as part of the above collapsing column,
  2. it has been damaged by by the lateral pull of the smashing floor,
  3. and, this column is now in fact only pinned at the bottom with the top free!
I am currently asking about your theory of single story collapse. This is entirely contained within the section of your paper that I quoted.
  • So then, in your theory of single story collapse, which column* yields ∏, c to c-1, c+1 to c, or, c+2 to c+1?
Code:
┃
┠───────      c-1     ─
⎞                     ⥌  3.7m
⎬┄┄┄┄┄┄┄┄ →  c        ─
⎠
┠───────      c+1
┃
[*Can you see my confusion? Since you refer to the next column as undamaged, I assumed the previous column (which yields ∏ above) was the initially buckled column. Is it or isn't it?]
 
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slenderbeam

Member
The dimensions for the trusses can be found in the NIST WTC report. The actual drawings have never been released.

My calculations were done in Excel using standard engineering equations. There is no need for special analysis software here.

Just for you and others here, please look at pdf pages 19 and 20 of the attached, which is about composite slab and truss beams, and you will see they used the same simply supported distributed load deflection equation I did in my calculations.
I had a look at your calc's and the Moment of Inertia (truss stiffness) wasn't reduced as per Section 5.5 of the guide which you provided. This is exactly the effect I was talking about.

Basically it means the deflection is probably around 40% more than what you report.
 

Tony Szamboti

Active Member
I had a look at your calc's and the Moment of Inertia (truss stiffness) wasn't reduced as per Section 5.5 of the guide which you provided. This is exactly the effect I was talking about.

Basically it means the deflection is probably around 40% more than what you report.
More like 35%, since the procedure is to

- calculate the composite "I"
- multiply the "I" by 0.15 and subtract this from the "I"
- divide the new value by (1.0 + 0.15 +0.15)

resulting in a reduced "I" value which would be 0.85 / 1.3 = 0.65 x original unreduced "I"

So with this reduced "I" value, instead of a main 60 foot double truss deflection of about 2.00 inches at 700 degrees C, it would be about 3.00 inches. This small amount of difference does not change my argument that the twin tower floor system outside of the core was robust, and it is now a quite conservative calculation.
 
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qed

Senior Member
Of course, the energy absorbed over the first story is important so that you know what is left to impact the story below. It is that energy absorption subtracted from mgh which is what impacts the columns of the story below.
So this is ∏ and that column is initially buckled.
  • Which of your references analyzes the compression response of an initially buckled stout column?
Can you now understand why I have been asking this question? I think you calculate ∏ incorrectly.
 
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Jeffrey Orling

Senior Member
So Charlie Thornton (of the NIST oversight committee) finally tells the "TRUTH" . . . the WTC Towers were cheap, flawed, and didn't meet minimal strength design for a high rise construction . . . the perfect storm for a chance demolition . . . well . . . I have to say it makes more sense than anything I have heard of since this whole debate began . . . write the hole thing off as people using the cheapest route to a project that should allow the victims to sue the Port Authority for incompetence and negligence . . .
More or less what I've stated for 2 years running.
 

Aref

New Member
Hi
I have a question about your interesting model. Actually I want to model a similar thing, as I attached. I need to model kind of bending deformation of a solid rod to a semi-circle using COMSOL. Could you please help me with the loading types and constraints that I need to use for this model. Also, I need to consider my elastic-perfectly plastic with different yield stress at compression and tension. Is there any option for this in COMSOL?
I really appreciate your help.
Thanks,




I was unnecessarily prickish in the last post. The points could be made without the jabs. It is difficult to ignore past history, and even more difficult to pretend that the current situation resembles a reasonable technical discussion.

Anyone who wants to believe that a 144" tall column of width 14-16" can turn two 90 degree bends and one 180 without fracture... I don't think I can fix that. I can try.

Even Szuladzinki's FEA shows fracture top and bottom, and an inability to keep the top segment plumb despite restraining forces, thus the sum of angles is not zero and it's more like two 70's and a 180. My point, very simply, was always that the fracture depicted was insufficient and did not reflect reality. If his sim has any credibility at all, it shows that fracture is indeed mandatory, even when you let the elements stretch to 5x their ductile limit and survive. No argument has been made as to why the 20% ductile limit does not apply to his FEA. If it did, the column would've broke long before the final deformation depicted. Therefore it necessarily exaggerates residual capacity, probably greatly.

I'd like to proceed with the foundation set down above. I can expound on it if desired, but I'm going to work with the expression r = 2.5w relating minimum radius of curvature to a solid steel section of width w. The simple first approximation is to use differential arc lengths based on radius of curvature. Obviously hollow sections and I or H configurations can deform in such a way as to change the inner and outer radii, and more esoteric factors like transverse shear and so on are ignored. This is planar analysis and ignores the other horizontal dimension of the member. Deformations are assumed elastic without work hardening, even though this is patently unrealistic anywhere near the ductile limit. This is the absolute simplest model possible for bending deformation.



The figure labeled A represents a solid prismatic steel column with height:width ratio of 9:1 with fixed endpoints. If it were 144 inches high, it would be 16 inches wide. Figure B shows the same column deformed into a circular arc (constant radius of curvature) of just a little over 200 degrees end rotation. Note that column bowing is NOT a circular shape, and bending would not assume such a shape unless the ends were constrained as shown. But, being bent into this circular shape by constraint, the relations I derived above hold true (more or less).

The color shading represents the distribution of tensile and compressive strain in the member, with blue being compression, red tension, and green nominal (no distortion). The neutral axis, where the strain crosses from tension to compression, is shown as a dotted line. The radius of this axis from the center of the arc is set according to the relation for maximum elongation at the edge without fracture, 2.5 times the column width. This axis divides the cross section into regions of tension and compression, and the circumference at the outer edge is 1.2 times longer (the ductile limit) than the circumference at the neutral axis, which is at nominal length. For a 144" nominal length, the outside edge is stretched to 172.8 inches, and is on the verge of tensile fracture.

This circular arc is the tightest bend that can be done in this model without fracture. The end rotation is over 200 degrees, no problem. If the column were taller, it could go all the way around, 360. BUT, if the radius of curvature is decreased (circle gets smaller) while keeping the same width, the outer edge length will exceed 1.2 times nominal and fracture. Likewise, if the radius were kept constant but the width increased, same thing - fracture. The column cannot be bent tighter and a wider column can't be bent this tight.

The geometry of this arrangement is independent of absolute size, it only depends on the proportion of width to radius of curvature. Therefore, this representation applies to a column or sheet of any size, theoretically. This leads to figure C, which is simply completing the circle. The purpose of this is to provide a template against which to judge the tightest bend possible. It's only necessary to scale the circle so that the width of the circular band matches the width of the column in question. In images which have roughly orthogonal perspective, the circle can overlay the column bend and show where the minimum radius of curvature is exceeded by a bend.

This model predicts fracture for members bending tighter than the curvature of the circle, and survival for those less tight. We'll see.
 

Attachments

Aref

New Member
upload_2015-5-19_15-4-8.png

I was unnecessarily prickish in the last post. The points could be made without the jabs. It is difficult to ignore past history, and even more difficult to pretend that the current situation resembles a reasonable technical discussion.

Anyone who wants to believe that a 144" tall column of width 14-16" can turn two 90 degree bends and one 180 without fracture... I don't think I can fix that. I can try.

Even Szuladzinki's FEA shows fracture top and bottom, and an inability to keep the top segment plumb despite restraining forces, thus the sum of angles is not zero and it's more like two 70's and a 180. My point, very simply, was always that the fracture depicted was insufficient and did not reflect reality. If his sim has any credibility at all, it shows that fracture is indeed mandatory, even when you let the elements stretch to 5x their ductile limit and survive. No argument has been made as to why the 20% ductile limit does not apply to his FEA. If it did, the column would've broke long before the final deformation depicted. Therefore it necessarily exaggerates residual capacity, probably greatly.

I'd like to proceed with the foundation set down above. I can expound on it if desired, but I'm going to work with the expression r = 2.5w relating minimum radius of curvature to a solid steel section of width w. The simple first approximation is to use differential arc lengths based on radius of curvature. Obviously hollow sections and I or H configurations can deform in such a way as to change the inner and outer radii, and more esoteric factors like transverse shear and so on are ignored. This is planar analysis and ignores the other horizontal dimension of the member. Deformations are assumed elastic without work hardening, even though this is patently unrealistic anywhere near the ductile limit. This is the absolute simplest model possible for bending deformation.



The figure labeled A represents a solid prismatic steel column with height:width ratio of 9:1 with fixed endpoints. If it were 144 inches high, it would be 16 inches wide. Figure B shows the same column deformed into a circular arc (constant radius of curvature) of just a little over 200 degrees end rotation. Note that column bowing is NOT a circular shape, and bending would not assume such a shape unless the ends were constrained as shown. But, being bent into this circular shape by constraint, the relations I derived above hold true (more or less).

The color shading represents the distribution of tensile and compressive strain in the member, with blue being compression, red tension, and green nominal (no distortion). The neutral axis, where the strain crosses from tension to compression, is shown as a dotted line. The radius of this axis from the center of the arc is set according to the relation for maximum elongation at the edge without fracture, 2.5 times the column width. This axis divides the cross section into regions of tension and compression, and the circumference at the outer edge is 1.2 times longer (the ductile limit) than the circumference at the neutral axis, which is at nominal length. For a 144" nominal length, the outside edge is stretched to 172.8 inches, and is on the verge of tensile fracture.

This circular arc is the tightest bend that can be done in this model without fracture. The end rotation is over 200 degrees, no problem. If the column were taller, it could go all the way around, 360. BUT, if the radius of curvature is decreased (circle gets smaller) while keeping the same width, the outer edge length will exceed 1.2 times nominal and fracture. Likewise, if the radius were kept constant but the width increased, same thing - fracture. The column cannot be bent tighter and a wider column can't be bent this tight.

The geometry of this arrangement is independent of absolute size, it only depends on the proportion of width to radius of curvature. Therefore, this representation applies to a column or sheet of any size, theoretically. This leads to figure C, which is simply completing the circle. The purpose of this is to provide a template against which to judge the tightest bend possible. It's only necessary to scale the circle so that the width of the circular band matches the width of the column in question. In images which have roughly orthogonal perspective, the circle can overlay the column bend and show where the minimum radius of curvature is exceeded by a bend.

This model predicts fracture for members bending tighter than the curvature of the circle, and survival for those less tight. We'll see.
 

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