Does the exclusion of stiffness from Nordenson's falling girder calculations demonstrate anything?

Jeffrey Orling

Senior Member
Oy thanks, understood... I think this is again splitting hairs.... and has to assume "idealized" conditions. This is not true or knowable for the freed dropping beam nor for the one it it dropped on. We don't even know that the collision is axial and not eccentric to the the beam below... Actually why would the beam drop perfectly aligned on the one below?

It seems as if Tony's argument is that progressive collapse would not take place from perhaps a single section of floor collapsing at that location. That is even if it managed to move off the seat... east or west... it simply couldn't mess up the beam below. There is a certain irony here because for the beam to drop... perfectly vertically... both seats at 79 and 44 would have to fail or one remain as a hinge with an axis bring the falling on dead onto the beam below.

Come on... let's be real... that ain't happen... The likely scenario is the falling beam shattered the concrete slab... eccentrically loaded the girder below and it was perhaps twisted free of its connections. the 10-12" impact is 1 D problem and seems a fantasy.
 

Oystein

Senior Member
Let me try to illustrate roughly what I mean:



I have drawn in the way the two girders would bend down (the gray one below less than the red falling one), and the forces acting on them. At the end of the falling girder, we have equal and opposite forces. Now of you switch frame of reference such that the co-ordinate system is fixed to the ends of the falling beam, then the beam experiences a gravity force (because of its inertia) that is F/2 (I drew a single arrow of half length at the mid-point, but don't mean to imply a point force there, but rather that gravity force is centered on the center of the beam). "D" is the location where I think deflection is measured
 

Oystein

Senior Member
Oy thanks, understood... I think this is again splitting hairs.... and has to assume "idealized" conditions. This is not true or knowable for the freed dropping beam nor for the one it it dropped on.
No, not splitting hairs. If you accept that Nordenson's Appendix B is a reasonable model, then, if Tony is right, his objection makes a large difference and is much more than splitting hairs - it changes the result by at leat an order of magnitude - or two, or three.
On the other hand, if you think that Nordenson's Appendix B is bunk because "this is not true or knowable", then say so, and trash Nordenson altogether. Then of course you should also stop your own investigation into the collapse of WTC7, for none of whatever you might come up with would be "knowable" or even "true".
We don't even know that the collision is axial and not eccentric to the the beam below... Actually why would the beam drop perfectly aligned on the one below?
I suggest you should read Nordenson's Appendix B before commenting on it, because then you'd know that he determines quite exactly how much the impact is misaligned, and he takes that into account.

...
Come on... let's be real... that ain't happen... The likely scenario is the falling beam shattered the concrete slab... eccentrically loaded the girder below and it was perhaps twisted free of its connections. the 10-12" impact is 1 D problem and seems a fantasy.
Nordenson's model is 3D. Read the Appendix B!
 

Jeffrey Orling

Senior Member
If I understand his geometry.... the impact is pulled by the girder rotating and the attached beams rotating both about the seats/hinge. Still this is idealized and it does lead to non axial impact.
 

Oystein

Senior Member
If I understand his geometry.... the impact is pulled by the girder rotating and the attached beams rotating both about the seats/hinge. Still this is idealized and it does lead to non axial impact.
Read the Nordenson Report (part 2), Appendix B. You didn't, did you? The Figure B5.1 that econ and I reproduced and annotated illustrate only one concept - why the impact occurs 10-12 inches away from the column. It is a SIMPLIFIED rotation diagram.
Section B3 explains the geometry in 3D. Figure B3.13 shows how far the impact is excentric to the axis. Figure B5.2 shows in 2D, looking from above, how the rotated girders overlaps the intact girder.
Nordenson takes this non-axial impact into account.

I suggest that you read Appendix B before commenting on it again.
 

Jeffrey Orling

Senior Member
Read the Nordenson Report (part 2), Appendix B. You didn't, did you? The Figure B5.1 that econ and I reproduced and annotated illustrate only one concept - why the impact occurs 10-12 inches away from the column. It is a SIMPLIFIED rotation diagram.
Section B3 explains the geometry in 3D. Figure B3.13 shows how far the impact is excentric to the axis. Figure B5.2 shows in 2D, looking from above, how the rotated girders overlaps the intact girder.
Nordenson takes this non-axial impact into account.

I suggest that you read Appendix B before commenting on it again.

read those sections... thanks
 
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Tony Szamboti

Active Member
*big sigh*

Neither of you has answered my questions in #26. My purpose was two-fold:
1. To check with you guys if my understanding of Nordenson's approach, and his steps of work, matches your understanding
2. To find out if you agree that you both agree that Nordenson's approach, his model assumptions and values are okay with you within reasonable bounds.

In the process, I identified one element where Tony explicitly deviates from Nordenson's model - I identified it several times, but it really is the same element each time: Tony wants to add one term - the stiffness of the falling girder.

So my question to you is:
econ: Do you think that the Nordenson model is a correct model to answer the question of whether or not the girder connection below would fail?
Tony:Do you think that the Nordenson model plus adding the stiffness of the falling girder is a correct model to answer the question of whether or not the girder connection below would fail?

If you both answer in the affirmative, then I'd suggest that
- contrary to econ's assertion that Tony's model is undefined, it is very well defined
- contrary to econ's assertion that Tony is playing mix and match, Tony has merely identified what he thinks is a flaw in Nordenson's model - a missing term - and added what he thinks is a proper amend.
- We need to agree on which model is the correct one (or that both are flawed), and work from there.

Now Nordenson has explained all the things he has included in his model, but he has not explained all the things he has not included, so looking at Nordenson's report to find an explanation why the falling girder's stiffness is not included yields nothing, for trivial reasons.

Tony needs to explain why he wants to include it.
Actually, I think I can provide an easy explanation, but I want to give Tony one chance to provide an explanation within the framework of Nordenson's logic.

Then econ would have to explain why Tony's explanation doesn't cut it.

But Step 1 is your answers to whether or not I have framed the models you agree with correctly.
Oy,

Yes, I agree with everything in Nordenson's analysis, except for his having left out the falling girder stiffness and deflection in the impact. It is not very stiff compared to the area 10 to 12 inches from the girder seat below. When the falling girder stiffness and deflection are included my calculations show the force applied to the girder seat below is much lower than what Nordenson predicted and that the girder seat below will not fail.
 

econ41

Senior Member
So my question to you is:
econ: Do you think that the Nordenson model is a correct model to answer the question of whether or not the girder connection below would fail?
I think the Nordenson model is a valid model. There could be other valid models. The issues in contention here include (a) Does the Nordenson explanation using his Model include all the relevant and significant energy sinks; AND (b) valid assignment of burden of proof to the person asserting that Nordenson is wrong.

- We need to agree on which model is the correct one (or that both are flawed), and work from there.
Yes - and even more fundamental - we need to agree the topic. It isn't "prove econ41 wrong" ;) "Prove T Sz wrong" is a valid topic given it is his OP - unless he wants to disown the OP which would make the discussion moot.

Tony needs to explain why he wants to include it.
Only if he wants to support the de-facto OP which is his claim:
Nordenson's failure mechanism for the northeast corner floors below could not have occurred. He errs in not including the much lower stiffness of the falling girder, which will limit the impact load to well below the shear capacity of the girder seat at column 79.
Welcome to the party with this:
..... I want to give Tony one chance to provide an explanation within the framework of Nordenson's logic.
Go for it. I've already given him multiple chances. He may respond to your offer.
Then econ would have to explain why Tony's explanation doesn't cut it.
"would have"? :) I would be under no compulsion. Reality is if Tony got it right I would congratulate him. If he presented reasoned argument and still got it wrong I would congratulate him for engaging in the reasoned argument and probably offer correcting advice. If he made another bare assertion with or without overlaid dishonesty I would near certain ignore the post.

But Step 1 is your answers to whether or not I have framed the models you agree with correctly.
That is a broader assertion that the scope of your question. I agree that Nordenson's "model" is a valid model and I described the "frame" of the model at Post #21 repeated at Post #29. It is in broad terms - a true "frame" onto which details can be added. Do you agree that "frame"?
 

econ41

Senior Member
The idea now in Nordenson's model is to compute how far the girder below will deflect under a point load 10 to 12 inches from the column such that its elastic strain energy is equal to the kinetic energy of the falling girder. That deflection is proportional to the applied force.
He then converts that force on a point 10 to 12 inches from the column to the resulting vertical force on the connection itself, using the girder's moment of inertia, and compares with the connection's capacity.
Near enough correct so I don't need to get pedantic.
Nordenson does not consider the bending/deflection of the falling girder which inhales strain energy as it bends down in the middle. This strain energy is not available to strain the girder below...
Both those points are:
1) In contention;
2) Not proven by Tony in accordance with his burden of proof for his claim. (Nor proven by you yet Oystein as you support Tony.)
3) I am not persuaded (a) that he needs to consider it OR (b) that all of it OR a significant portion of it is "lost".

IF some of that energy is lost AND it is large enough to warrant inclusion in the losses THEN include it in the losses. It is not legitimately dealt with by Tony's artifice of "combining" the flexibilities. Attempting to do so shows misunderstanding of the mathematical reasoning used by Nordenson.

Nordenson uses flexibility as a pivot device in his maths. He says in effect:
"This is the net energy landing on the lower beam"
"It would cause a deflection of that LOWER beam"
"That deflection could be caused by an effective force"
He works out that force and concludes
"The effective force is greater than the connector strength".

The flexibility of the lower beam as used by Nordenson in the relevant calculations is nothing more than a pivot for the calculations. And it is based on the NET ENERGY landing on that lower beam.

Tony suggests that the two flexibilities be combined. They cannot in the Nordenson model which is premised on what is the NET ENERGY that lands on the LOWER beam. Combining stiffnesses does not fit into the Nordenson maths. So incorporating the missing energy loss needs EITHER a different model of the maths using "combined stiffnesses" OR a dfifferent way of incorporating the missed energy factor legitimately into the Nordenson model. I would accept either way - provided we are explicitly clear what model we are following.

So - accepting for moot purposes that Nordenson missed a factor - I can identify two ways to incorporate that missing factor:
A) By validly modifying the Nordenson argument* - and that means that the "falling beam energy loss" needs to be quantified as an energy sink and added to the "losses" already recognised and incorporated. In effect using the same mechanism Nordenson already uses for the other energy sinks - reducing the net PE/KE input BUT still using lower beam stiffness as the pivot for the maths; OR
B) Someone - preferably the proponents, Tony or Oystein, formulate an alternate model which validly uses the two combined stiffnesses. The challenge to that path would be how to structure a legitimate mathematical model.

* Let me make the issue explicit. I am asserting that combined or "effective" stiffness CANNOT be substituted in the Nordenson method. Doing so "breaks" the maths. The combination of Nordenson Method AND "effective or combined stiffnesses" is not possible. I'm prepared to explain for any members who cannot see why that is so. I have no problem with the alternate method - quantify the loss of energy involved. (And prove that it is not already covered by Nordenson - whether explicitly or by his conservative margins.)
 
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Tony Szamboti

Active Member
Near enough correct so I don't need to get pedantic.
Both those points are:
1) In contention;
2) Not proven by Tony in accordance with his burden of proof for his claim. (Nor proven by you yet Oystein as you support Tony.)
3) I am not persuaded (a) that he needs to consider it OR (b) that all of it OR a significant portion of it is "lost".

IF some of that energy is lost AND it is large enough to warrant inclusion in the losses THEN include it in the losses. It is not legitimately dealt with by Tony's artifice of "combining" the flexibilities. Attempting to do so shows misunderstanding of the mathematical reasoning used by Nordenson.

Nordenson uses flexibility as a pivot device in his maths. He says in effect:
"This is the net energy landing on the lower beam"
"It would cause a deflection of that LOWER beam"
"That deflection could be caused by an effective force"
He works out that force and concludes
"The effective force is greater than the connector strength".

The flexibility of the lower beam as used by Nordenson in the relevant calculations is nothing more than a pivot for the calculations. And it is based on the NET ENERGY landing on that lower beam.

Tony suggests that the two flexibilities be combined. They cannot in the Nordenson model which is premised on what is the NET ENERGY that lands on the LOWER beam. Combining stiffnesses does not fit into the Nordenson maths. So incorporating the missing energy loss needs EITHER a different model of the maths using "combined stiffnesses" OR a dfifferent way of incorporating the missed energy factor legitimately into the Nordenson model. I would accept either way - provided we are explicitly clear what model we are following.

So - accepting for moot purposes that Nordenson missed a factor - I can identify two ways to incorporate that missing factor:
A) By validly modifying the Nordenson argument* - and that means that the "falling beam energy loss" needs to be quantified as an energy sink and added to the "losses" already recognised and incorporated. In effect using the same mechanism Nordenson already uses for the other energy sinks - reducing the net PE/KE input BUT still using lower beam stiffness as the pivot for the maths; OR
B) Someone - preferably the proponents, Tony or Oystein, formulate an alternate model which validly uses the two combined stiffnesses. The challenge to that path would be how to structure a legitimate mathematical model.

* Let me make the issue explicit. I am asserting that combined or "effective" stiffness CANNOT be substituted in the Nordenson method. Doing so "breaks" the maths. The combination of Nordenson Method AND "effective or combined stiffnesses" is not possible. I'm prepared to explain for any members who cannot see why that is so. I have no problem with the alternate method - quantify the loss of energy involved. (And prove that it is not already covered by Nordenson - whether explicitly or by his conservative margins.)
Using the effective stiffness does not change how the math is done whatsoever. Where do you even get this notion?
 

jaydeehess

Senior Member
Just to be clear, does Nordenson take into account any heating of the lower floor girder and connections, does he take into account added mass from the failed upper floor, loading the lower floor, IOW, does he assume the lower floor and supporting girder to be in as-built condition?
We already understand that the fire damage the ARUP report dealt with occurs on two floors in the NE corner of WTC 7. They conclude that the girder came off one on one of those two floors. Obviously they will assume some damage either above initial failure floor, or below it.
 

Jeffrey Orling

Senior Member
Just to be clear, does Nordenson take into account any heating of the lower floor girder and connections, does he take into account added mass from the failed upper floor, loading the lower floor, IOW, does he assume the lower floor and supporting girder to be in as-built condition?
We already understand that the fire damage the ARUP report dealt with occurs on two floors in the NE corner of WTC 7. They conclude that the girder came off one on one of those two floors. Obviously they will assume some damage either above initial failure floor, or below it.
I find these math studies a bit crazy because there is so much that is assumed... so many variables and the process is taking place in 3 dimensions over time... The margin of error must be large and so arguing over the math ... assumed values seems just a waste of time. Steel frames are "protected" for a couple of hrs and then bets are off. And with 7wtc there was no fire fighting. It looked like a CD falling because it failed low down in the structure and so it fell much like a CD... And of course it had all those transfers which could move the failures about the structure at the bottom rather efficiently.

We'll never know the details... but educated guesses with reasonable assumptions will have to do.
 

Tony Szamboti

Active Member
I find these math studies a bit crazy because there is so much that is assumed... so many variables and the process is taking place in 3 dimensions over time... The margin of error must be large and so arguing over the math ... assumed values seems just a waste of time. Steel frames are "protected" for a couple of hrs and then bets are off. And with 7wtc there was no fire fighting. It looked like a CD falling because it failed low down in the structure and so it fell much like a CD... And of course it had all those transfers which could move the failures about the structure at the bottom rather efficiently.

We'll never know the details... but educated guesses with reasonable assumptions will have to do.
Applying math is what keeps the assumptions reasonable and with a plausible basis in reality. The math (and I include FEA here) has shown that the assumptions made in the NIST WTC 7 report are impossible.
 

Jeffrey Orling

Senior Member
Applying math is what keeps the assumptions reasonable and with a plausible basis in reality. The math (and I include FEA here) has shown that the assumptions made in the NIST WTC 7 report are impossible.
Sure when designing a static structure.... load calcs and so on make perfect sense. In these buildings you had a very dynamic and chaotic situation... and so the sort of static calcs makes very little sense... to me.
 

jaydeehess

Senior Member
Sure when designing a static structure.... load calcs and so on make perfect sense. In these buildings you had a very dynamic and chaotic situation... and so the sort of static calcs makes very little sense... to me.
I don't know about it making "little sense". I believe it certainly does make sense to attempt to find a plausible sequence of failure.
However, all one can do is determine in many cases is if a mechanism comes close to failure in an FEA. Given the high number of assumptions and margins of error, close counts in determining plausibility, IMHO.

So do Tony's calcs remove the failing girder/floor section from being able to fail a lower floor section/girder?
I'm not sure he's demonstrated that at all.
 

econ41

Senior Member
Using the effective stiffness does not change how the math is done whatsoever.
Nonsense Tony. And there is no point in me explaining further given that I have already posted sufficient explanation and whilst ever you are locked in denial.

Where do you even get this notion?
I got tired of the nonsense being posted so decided to read through the docs to see where you were going wrong. Then the "Eureka" point when, with a bit of serious thinking, I realised what your fundamental errors were. (I knew you were wrong from the outset with the "effective stiffness" nonsense - couldn't satisfactorily explain why till yesterday. And it is "bleeding obvious" - once you see it - hence my reference to "Eureka".) The irony being that I really didn't need to read the paper in close detail - just exercise the "Little Grey Cells"*.


* Christie, A - multiple references 1922-1956.
 

econ41

Senior Member
Sure when designing a static structure.... load calcs and so on make perfect sense. In these buildings you had a very dynamic and chaotic situation... and so the sort of static calcs makes very little sense... to me.
Jeffrey I suggest that you don't fall for the trap of making too broad generalisations. The specific issue being discussed here is admittedly somewhat complex. Certainly it looks complex if you try to read the lengthy papers.

However there are really only two issues you need to understand about THIS SPECIFIC topic - which is the part of Nordenson's paper where he explains how ONE portion of ONE floor of WTC which disconnects from Col 79 drops and impacts on the next floor down. Nordenson correctly shows why that "bit of a single floor dropping at one corner" would have more than sufficient impact to shear the next lower floor. So that is WHAT we are discussing.

The two key issues then are:
A) Understanding Nordenson's explanation of how he reached his conclusion. I have summarised that explanation three times in previous posts in this thread.

B) T Szamboti has asserted that Nordenson is wrong. He has - in effect - asserted a straw-man which superficially looks reasonable. It isn't. And I outlined the reasons why in a recent post. If the brief explanation of the error is not clear enough I can give a bit more detail.
 

econ41

Senior Member
@jaydeehess Current discussion is about Nordenson's explanation as contained in Section B of his paper. So my answers are limited to the content of that section. And E&OE - I may have missed something.
Just to be clear, does Nordenson take into account any heating of the lower floor girder and connections,...
No - with one exception. His scenario addresses what happens when each floor slab for the range of affected floors loses support at Col 79. The floor area is the four sided section with Col 79 at one corner. And he describes how a triangular section of the corner at Col 79 fails and drops onto the next floor down close to Col 79. He assumes the falling floor section impacts at 1/12 span - relatively close to Col 79 - where the lower floor beam is "very stiff". The one exception and concession to heat is where he allows for heat distortion of the Floor 13 girder failure landing on Floor 12 and he assumes impact at the 1/5 span point - a conservative concession. He makes no general allowances for heat of the style I think you are alluding to. No allowances for heat == a conservative analysis. He is asserting - with sound reason - why ONE floor failing and impacting the next lower floor would without doubt in his explanation result in failure of that next lower floor. And once initiated the failures would continue sequentially down the building.
....does he take into account added mass from the failed upper floor, loading the lower floor,
If I reframe slightly - NO. He takes each floor - the failing floor and the next floor down that it impacts - as being original condition other than the failure damage resulting from loss of Col 79 support. AND he does NOT accumulate "levels of debris". So as floor N lands on N-1 only Floor N weight is used. And next step as floor N-1 lands on N-2 ONLY N-1 weight is considered - no accumulation - a deliberate conservative choice.
.... IOW, does he assume the lower floor and supporting girder to be in as-built condition?
Yes - AFAICS the only concession is the point of impact at 1/5th for Floor 13 landing on Floor 12.
...We already understand that the fire damage the ARUP report dealt with occurs on two floors in the NE corner of WTC 7. They conclude that the girder came off one on one of those two floors. Obviously they will assume some damage either above initial failure floor, or below it.
Those are broader issues not considered in the specific section we are discussing. The section is a tightly defined focus on one mechanism - single floor failing impacts next floor down and has far more than enough energy to fail that next floor. He then builds a cascading sequence of successive floor failures.
 

Oystein

Senior Member
econ,

I am convinced your personal "Eureka" moment is still ahead of you.
You may be right in a very pedantic sort of way that using combined stiffness "changes the math", but you will find in the end that it is actually the smart way to do the math. It will turn out to be equivalent to doing the math the way Nordenson did it, but simpler.

Let's put the math back on the shelve for a moment - I hope Tony is reading, too - and try to think clearly what it is we want to compute:

The falling girder is in motion as its lower end touched the girder below. At this time, it has a certain Kinetic Energy (KE), for which Nordenson provides a specific number. He derived this number from the masses and geometry of his SAP2000 model that gave him the Potential Energy differential (PE), and he subtracted from that the energy dissipated in several areas and modes of deformation (DE1 through DEn), all of which he discusses individually and provides numbers for. <- Do you agree that this is what Nordenson did in B1-B3? Do you accept, at least tentatively, that Nordenson's value for KE is in the right ballpark?

So this is the situation we are looking at - SIMPLIFIED (not to scale, and disregarding certain complications such as that the falling girder has beams attached to it and a deformed floor slab loading it, with which there may be some residual composite action; this all may affect the numbers we'll have to plug in in the next few steps of math, but we are not there yet)


I drew in a few things in blue and pink.
What Nordenson want to solve for if the unknown force F that the falling girder would exert on the girder below if all the KE (remaining at that moment - see above) got converted to elastically bending the girder below ("strain energy below" = SEb): KE = SEb <- Please indicate clearly if that is a sufficiently accurate rendering of what Nordenson attempts to do in B5!

Now comes the difference between the understandings of Nordenson and econ on one hand, and Tony and myself on the other:
The girder below isn't the only sink that the KE pours into!
At the same time that the girder below experiences a force down, the falling girder experiences an equal and opposite force up. As a result, it bends down. This bending down builds up strain energy in the falling girder (SEf). The source of SEf has to be the same KE that concurrently feeds SEb: KE = SEb + SEf.

So. I have now proven with reasoned arguments, using true premises, that Nordenson's approach is incomplete, resulting in an overestimate of the strain energy in the girder below, that he then goes on to convert to an equivalent force F. So he overestimates F, too!

econ, please
  • either agree that Nordenson should have included a term for SEf, which will reduce the resulting equivalent force
  • or point out why SEF ought not be considered (i.e. what is wrong with my proof). Citing your Eureka moment will not convince.
 

Jeffrey Orling

Senior Member
Oy:
This is not clear:

"At the same time that the girder below experiences a force down, the falling girder experiences an equal and opposite force up. As a result, it bends down. This bending down builds up strain energy in the falling girder (SEf). The source of SEf has to be the same KE that concurrently feeds SEb: KE = SEb + SEf."

It bends down... means what? Are you saying the the falling girder will bend upon impact? And this consumes some of the kinetic energy? Is "It" the fall girder?

This all seems to ignore the fact that the falling girder was connected to and composite with the beams and the slab including the metal fluted deck.. admittedly more stiff in one axis than the other... and all of this falling composite is acting as a concentrated load on the floor girder below and not axially aligned because of the rotation in two axes.
 

Tony Szamboti

Active Member
I don't know about it making "little sense". I believe it certainly does make sense to attempt to find a plausible sequence of failure.
However, all one can do is determine in many cases is if a mechanism comes close to failure in an FEA. Given the high number of assumptions and margins of error, close counts in determining plausibility, IMHO.

So do Tony's calcs remove the failing girder/floor section from being able to fail a lower floor section/girder?
I'm not sure he's demonstrated that at all.
You say you can't follow the math but then want to say it isn't valid. You really do need to follow the math first and show your points before making a proclamation about it. Otherwise, it can rightfully be said that you are just talking out of your hat.
 

Tony Szamboti

Active Member
Oy:
This is not clear:

"At the same time that the girder below experiences a force down, the falling girder experiences an equal and opposite force up. As a result, it bends down. This bending down builds up strain energy in the falling girder (SEf). The source of SEf has to be the same KE that concurrently feeds SEb: KE = SEb + SEf."

It bends down... means what? Are you saying the the falling girder will bend upon impact? And this consumes some of the kinetic energy? Is "It" the fall girder?

This all seems to ignore the fact that the falling girder was connected to and composite with the beams and the slab including the metal fluted deck.. admittedly more stiff in one axis than the other... and all of this falling composite is acting as a concentrated load on the floor girder below and not axially aligned because of the rotation in two axes.
The girder and the beams are said to have broken their shear studs in the ARUP and NIST analyses, so there is little to no composite action remaining. If the studs don't break the girder could not be pulled off its seat. If the girder doesn't fall off the seat then we don't even need to consider the next step where Nordenson gets involved.

The falling girder will deflect upon impact and consume some of the kinetic energy. This is how an automobile is designed to limit shock loading in an impact. It deforms to absorb the energy over a longer duration and limit the deceleration and thus the load experienced. Maybe you don't understand how shock loading works. The stiffer the items in an impact the less they deflect and the shorter the duration. The area under the G vs. time curve is the energy. The energy does not change but the force is dependent on deceleration or acceleration and if the impact happens over more time the forces are lower. If time is short there is higher deceleration of the impacting object and high acceleration of the impacted object. The resulting loads are determined by mass x deceleration or acceleration. Do you realize that Nordenson is not calculating the force using static load? He comes up with an equivalent to the dynamic load based on stiffness and deflection. The equivalent load is much higher than the static load, which could never even have a chance of shearing the seat below.

If you don't know enough about this you should sit back and watch and learn something instead of complaining that it is too complicated. It actually is not that hard to understand and you should be able to learn it fairly easily if you pay attention. Complaining won't solve the problem and it sure doesn't allow for you to make a proclamation that things just happened a certain way.
 
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Tony Szamboti

Active Member
It is obvious that Econ cannot support his proclamation that effective stiffness is not involved in a collision which determines the deceleration and acceleration of the items involved and ultimately the forces they experience.

The dynamic load applied by the falling girder to the girder below is a function of its deceleration which is a function of its stiffness and how much it deflects on impact. If one where to apply the load from the falling girder in multiple tests, in an increasingly slow manner each time, it ultimately would be just the static load and the static load has no chance of shearing the seat of the girder below.
 
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Jeffrey Orling

Senior Member
Well yea a dynamic load of mass X would be greater than a static one of mass X... that's intuitive.... the value may not be.

I am curious about the failure of the shear studs... were they sheared or remained attached to the steel and the concrete around them fractured?

Thank you for the explanation of crash zones.... I was aware of it.

I may not be the person to calculate the math in this scenario but I can see it is a dynamic and chaotic on and this seems to be perhaps an over simplification.
 

Tony Szamboti

Active Member
Well yea a dynamic load of mass X would be greater than a static one of mass X... that's intuitive.... the value may not be.

I am curious about the failure of the shear studs... were they sheared or remained attached to the steel and the concrete around them fractured?

Thank you for the explanation of crash zones.... I was aware of it.

I may not be the person to calculate the math in this scenario but I can see it is a dynamic and chaotic on and this seems to be perhaps an over simplification.
On what basis are you saying there is an over simplification as far as the math used to determine the load applied to the girder seat below?
 

econ41

Senior Member
econ,

I am convinced your personal "Eureka" moment is still ahead of you.
You may be right in a very pedantic sort of way that using combined stiffness "changes the math", but you will find in the end that it is actually the smart way to do the math. It will turn out to be equivalent to doing the math the way Nordenson did it, but simpler.
I'm sorry that you are not following and choose to disregard the substantive points in my several posts...however let's focus on ONE issue - the alleged "missing energy sink". I'll address it in your context:
Oystein said:
.... He derived this number from the masses and geometry of his SAP2000 model that gave him the Potential Energy differential (PE), and he subtracted from that the energy dissipated in several areas and modes of deformation (DE1 through DEn), all of which he discusses individually and provides numbers for. <- Do you agree that this is what Nordenson did in B1-B3?
Yes I agree - I have sought to assert understanding of Nordenson on several occasions.

Nordenson looks for the NET energy which flows through to the lower beam. There are two consequent issues - one is subject of your contention that he has NOT allowed for all the energy dissipation.

Now to me "dissipation" in this context means "doesn't flow forward in the net energy"; it is "lost"; it does not contribute to the impact on the lower beam.

All of the "energy sinks" which Nordenson identifies "use up" energy so it is not available to "flow forward" in the process.

What happens to the energy involved in the bending of the falling girder? Is it lost? Is any of it lost? How much is lost? How is it lost? AND - if "it" or part of it is not lost - where does it go?

I responded this way to one of your earlier comments:
....3) I am not persuaded (a) that he [Nordenson] needs to consider it OR (b) that all of it OR a significant portion of it is "lost"....
I continued by suggesting how it should be accommodated - and identifying the existence of a fundamental problem with the maths required by Tony's alternate suggestion:
...IF some of that energy is lost AND it is large enough to warrant inclusion in the losses THEN include it in the losses. It is not legitimately dealt with by Tony's artifice of "combining" the flexibilities. Attempting to do so shows misunderstanding of the mathematical reasoning used by Nordenson...
I will aquiese with your suggestion to leave the maths concerns "on the shelf" at this stage.

I'll pause there in accordance with my suggestion "one issue at a time"

What happens to the energy involved in falling beam flexure? Is it "lost"? If so - How?
 

Tony Szamboti

Active Member
For those having trouble following this discussion an example here might intuitively give them insight.

Imagine dropping a 20 lb. concrete or steel ball from 10 feet up onto a wooden deck made with 2 x 6 planks which are 5 feet between supports. After falling 10 feet the velocity of the ball is 25.4 feet/second. The kinetic energy is 200 ft-lbs. You can imagine the concrete or steel ball breaking the plank it lands on or its connection/seat.

Now drop a rubber ball with the same mass from the same height. It has the same energy (200 ft-lbs) but it will deflect much more than the concrete or steel ball and very likely will not break the plank or its connection/seat. This is because the actual load experienced is a function of the stiffness and deflection of both items and in this case the rubber ball deflected much more and absorbed the energy over a longer duration and reduced the deceleration and thus the force imparted to the plank and its connection/seat. The force can be calculated and stress analysis done to determine if the plank or its connection/seat would break.

The above shows that the stiffness and deflection of the falling object matters along with that of the impacted object. It is the combined stiffness and deflection which determine the deceleration in the impact and thus the actual dynamic load involved. This is what I have shown with the falling girder and girder seat below with the Nordenson analysis. The calculations show the falling girder cannot break the girder seat at column 79 when its stiffness and deflection are also considered in conjunction with that of the girder below at 10 to 12 inches from its seat. Nordenson erred when he did not also consider the stiffness and deflection of the falling girder. He considered the falling girder to be an infinitely stiff point load and that is erroneous and it makes a big difference.
 
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Jeffrey Orling

Senior Member
Thanks Tony for the example. Obviously the kinetic energy will either shatter the connection materials, the beam(s) and or DEFORM them.

I can't compute this. I suspect that the weak link is the connection of the beam to the column... and this will fail. I find it hard to conceptualize that the connection would restrain the beam and cause it to deform.

On the other hand there were tens of thousands of falling beams, and columns in these collapses. Most seemed to be remarkably intact with little to no deformation... at least what I saw. Sure there were some for sure.. But the debris was not MOSTLY mangled deformed heavy sections... And this leads me to the conclusion.... once again... that the forces present dismantled the structure by failing the connections... I suspect as well that deformation from expansion (and perhaps cooling likewise failed connections... and freed mass to collapse and bust apart the structure.

Has the connection been examined as a failure location rather than the beam / girder?
 

Tony Szamboti

Active Member
Thanks Tony for the example. Obviously the kinetic energy will either shatter the connection materials, the beam(s) and or DEFORM them.

I can't compute this. I suspect that the weak link is the connection of the beam to the column... and this will fail. I find it hard to conceptualize that the connection would restrain the beam and cause it to deform.

On the other hand there were tens of thousands of falling beams, and columns in these collapses. Most seemed to be remarkably intact with little to no deformation... at least what I saw. Sure there were some for sure.. But the debris was not MOSTLY mangled deformed heavy sections... And this leads me to the conclusion.... once again... that the forces present dismantled the structure by failing the connections... I suspect as well that deformation from expansion (and perhaps cooling likewise failed connections... and freed mass to collapse and bust apart the structure.

Has the connection been examined as a failure location rather than the beam / girder?
Jeffrey, in this thread we are talking about a collapse initiation, not what was happening once the collapse was underway. Try to stay focused on the initiating event in WTC 7.

As for what happens to the connection we know the shear capacity of the girder seat at column 79. It was fairly robust with a 14 inch deep x 18.875 inch wide plate welded to the column side plates, with 14 inch long 3/8" fillet welds on both sides, supporting the girder. The shear capacity was 632 kips. That is a lot of force required and the static weight of the girder and beam mass was only a small fraction of that.

Although it was very robust, the connection would take less force to fail than shearing the girder or bending it enough to fail at the seat. The girder had a 38.3 sq. inch cross section and a 50,000 psi tensile yield strength. The shear yield strength would be 57.7% of the tensile yield strength, so shearing the girder would take a load of about 1,105 kips.
 
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Jeffrey Orling

Senior Member
This discussion also seems to be about what happens once the upper girder has left its seat... and whther it destroys/fails the girder and slab surrounding it on the floor below... (which had intact shear studs or no?) and isn't that is... the collapse underway? Isn 't this the threshold case for a progressive collapse/ destruction of this line of girders?
 

Tony Szamboti

Active Member
This discussion also seems to be about what happens once the upper girder has left its seat... and whther it destroys/fails the girder and slab surrounding it on the floor below... (which had intact shear studs or no?) and isn't that is... the collapse underway? Isn 't this the threshold case for a progressive collapse/ destruction of this line of girders?
Yes, but you seemed to be getting into things that weren't germane, such as the chaotic nature once a lot of things start falling.

We are talking about whether or not the single failure of the girder coming off its seat could cause propagation by failing the next floor down. You should limit your discussion to this subject.

I don't think it matters whether there were intact shear studs on the floor slab and frame below if the girder seat was sheared below.
 

Oystein

Senior Member
Oy:
This is not clear:

"At the same time that the girder below experiences a force down, the falling girder experiences an equal and opposite force up. As a result, it bends down. This bending down builds up strain energy in the falling girder (SEf). The source of SEf has to be the same KE that concurrently feeds SEb: KE = SEb + SEf."

It bends down... means what? Are you saying the the falling girder will bend upon impact? And this consumes some of the kinetic energy? Is "It" the fall girder?

This all seems to ignore the fact that the falling girder was connected to and composite with the beams and the slab including the metal fluted deck.. admittedly more stiff in one axis than the other... and all of this falling composite is acting as a concentrated load on the floor girder below and not axially aligned because of the rotation in two axes.
Yes, the falling girder would bend down. In the Figure B5.1 with my hand-drawn additions, the pink curve below the falling girder shows im what direction it bends - down.

Yes, any part of the floor assembly connected to the falling girder potentially affects the stiffness - but that doesn't negate the principle that the falling girder (or floor) has a finite stiffness, and that its stiffness must be considered. We need to agree on this first, then think about how to compute that stiffness (I don't know how yet).
 

qed

Senior Member
Has anyone actually done the calculation with stiffness? Surely that will save much thread space?
 

Jeffrey Orling

Senior Member
OK

Stiffness depends upon material properties and geometry. The stiffness of a structural element of a given material is the product of the material's Young's modulus and the element's second moment of area. Stiffness is measured in force per unit length (newtons per millimetre or N/mm), and is equivalent to the 'force constant' in Hooke's Law.

The deflection of a structure under loading is dependent on its stiffness. The dynamic response of a structure to dynamic loads (the natural frequency of a structure) is also dependent on its stiffness.

In a structure made up of multiple structural elements where the surface distributing the forces to the elements is rigid, the elements will carry loads in proportion to their relative stiffness - the stiffer an element, the more load it will attract. This means that load/stiffness ratio, which is deflection, remains same in two connected (jointed) elements. In a structure where the surface distributing the forces to the elements is flexible (like a wood framed structure), the elements will carry loads in proportion to their relative tributary areas.

A structure is considered to fail the chosen serviceability criteria if it is insufficiently stiff to have acceptably small deflection or dynamic response under loading.

The inverse of stiffness is flexibility.

When a beam is not stiff enough it flexes or deflects under load. Is the core of Tony's argument that the beam defects, absorbs the energy and remains there bent? How much would it be bent?
 

Oystein

Senior Member
What happens to the energy involved in falling beam flexure? Is it "lost"? If so - How?
Already explained, by me and by Tony.

In B4, Nordenson computed the "Floor collapse potential energy". Here is the sentence in B4.1 that might contain the point that I am not getting: "The gravity load reaction at the former location of Column 79 under the 1.0DL + 1.0SDL + 0.25LL load combination was found to be 46 kips at Floors 13 through Floor 8" - I am not sure what to make of the phrase "gravity load reaction".
In B4.2 he constructs an idealized floor area that has "the equivalent 46 kip load as the corner reaction found in the SAP2000 model". Looking at the geometry of this floor area, he determined that the center of gravity of this floor area droped a height of 83 inches.
In B4.3 he simply multiplies 46 kip * 83 inches = 3818 kip-inches initial PE.
From this he subtracts several deformation energies (my "D1 to Dn"), listed in Table B6.1:
D1: South Bay Tensile Fracture Energy = 17 kip-in
D2: West Bay Tensile Fracture Energy = 14 kip-in
D3: Slab Boundary Rotations = 147 kip-in
D4: Slab Deformation = 70 kip-in
D5: Slab Shear = 14 kip-in
D6: Girder Impact Corner Yielding = 82 kip-in
That's Total Dissipated Energy Sum(D1..D6) = 345
And so
Remaining [kinetic] Energy at impact = 3818 kip-in - 345 kip-in = 3473 kip-in.
This is the NET energy in your post.

Please note that none of the terms D1..D6 takes care of how the falling girder bends!

Nordenson then assumes that all that NET Energy of 3473 kip-in must be dissipated by bending the girder below and ONLY the girder below! That's why he takes the stiffness of the girder below K = 7627 kips/in (derived in B5.2), and plugs into formulas in B5.3. Let me make those computations explicit. Nordenson writes:
"Using the girder impact stiffness values, the impact energy was converted to a static force via deformation using the formula:
PE = 1/2 K * D^2
where D is girder deflection and K is the girder spring stiffness
This equation can be rearranged and solved for deflection.
"​
Nordenson skips this, so I write it out - and note that he should write KE instead of PE:
D = sqrt(2PE / K) = sqrt (2 * 3473 kip in / 7627 kip/in) = sqrt(0,9107 in^2) = 0,9543 in
Nordenson continues:
"This deflection value can then be multiplied by the girder stiffness to find the equivalent static force using the equation:
F = K * D
"​
Let me to this quickly:
F = 7627 kip/in * 0,9543 in = 7279 kip
(Nordenson writes in the text that this "yields a static force of 4133 kips" - this is wrong, but Table B6.1 has the correct value)

So this proves that Nordenson wants to dissipate ALL of the KE established AFTER subtracting D1..D6 by bending the girder below ONLY.

Now look again at this drawing:



Your question: "What happens to the energy involved in falling beam flexure? Is it "lost"? If so - How?"

My answer:
At the same time as the girder below is bent down by the impact force of the end of the falling girder, the inertia of the falling girder makes the falling girder bend, since it must react to the impact force of the girder below, which is equal in size and opposite in direction.
This bending builds up strain energy in the falling girder at the same time that straim energy build up in the impacted girder.
There is only one source of energy to bend both girders at the same time: The NET Kinetic energy of 3473 kip-in.
Therefore Nordenson is WRONG to imply that ALL of that net KE flows into the girder below - some flows into the falling girder at the same time.

Nordenon's formula
KE = 1/2 K(below) * D(below)^2​
is wrong. It should be
KE = 1/2 K(below) * D(below)^2 + 1/2 K(falling) * D(falling)^2​

The equivalent static force is still
F = K(below) * D(below)​
but with a smaller D(below); and at the same time
F = K(falling) * D(falling)​

I will have to think hard about how to re-arrange these formulas, but my math intuition tells me already that Tony's solution/formula for springs in series will solve the system.

In the meantime, you need to wrap your mind around the fact that the falling girder bends and thus uses some of the energy which is then no longer available to the girder below, resulting in less deformation below, meaning less equivalent static force.

Unless you can show me that the bending-down of the falling girder is already accounted for somewhere.
You can try to deny that the falling girder would bend and thus use up some of the available energy, but then you simply aren't getting it.
 

benthamitemetric

Senior Member
Already explained, by me and by Tony.

In B4, Nordenson computed the "Floor collapse potential energy". Here is the sentence in B4.1 that might contain the point that I am not getting: "The gravity load reaction at the former location of Column 79 under the 1.0DL + 1.0SDL + 0.25LL load combination was found to be 46 kips at Floors 13 through Floor 8" - I am not sure what to make of the phrase "gravity load reaction".
In B4.2 he constructs an idealized floor area that has "the equivalent 46 kip load as the corner reaction found in the SAP2000 model". Looking at the geometry of this floor area, he determined that the center of gravity of this floor area droped a height of 83 inches.
In B4.3 he simply multiplies 46 kip * 83 inches = 3818 kip-inches initial PE.
From this he subtracts several deformation energies (my "D1 to Dn"), listed in Table B6.1:
D1: South Bay Tensile Fracture Energy = 17 kip-in
D2: West Bay Tensile Fracture Energy = 14 kip-in
D3: Slab Boundary Rotations = 147 kip-in
D4: Slab Deformation = 70 kip-in
D5: Slab Shear = 14 kip-in
D6: Girder Impact Corner Yielding = 82 kip-in
That's Total Dissipated Energy Sum(D1..D6) = 345
And so
Remaining [kinetic] Energy at impact = 3818 kip-in - 345 kip-in = 3473 kip-in.
This is the NET energy in your post.

Please note that none of the terms D1..D6 takes care of how the falling girder bends!

Nordenson then assumes that all that NET Energy of 3473 kip-in must be dissipated by bending the girder below and ONLY the girder below! That's why he takes the stiffness of the girder below K = 7627 kips/in (derived in B5.2), and plugs into formulas in B5.3. Let me make those computations explicit. Nordenson writes:
"Using the girder impact stiffness values, the impact energy was converted to a static force via deformation using the formula:
PE = 1/2 K * D^2
where D is girder deflection and K is the girder spring stiffness
This equation can be rearranged and solved for deflection.
"​
Nordenson skips this, so I write it out - and note that he should write KE instead of PE:
D = sqrt(2PE / K) = sqrt (2 * 3473 kip in / 7627 kip/in) = sqrt(0,9107 in^2) = 0,9543 in
Nordenson continues:
"This deflection value can then be multiplied by the girder stiffness to find the equivalent static force using the equation:
F = K * D
"​
Let me to this quickly:
F = 7627 kip/in * 0,9543 in = 7279 kip
(Nordenson writes in the text that this "yields a static force of 4133 kips" - this is wrong, but Table B6.1 has the correct value)

So this proves that Nordenson wants to dissipate ALL of the KE established AFTER subtracting D1..D6 by bending the girder below ONLY.

Now look again at this drawing:



Your question: "What happens to the energy involved in falling beam flexure? Is it "lost"? If so - How?"

My answer:
At the same time as the girder below is bent down by the impact force of the end of the falling girder, the inertia of the falling girder makes the falling girder bend, since it must react to the impact force of the girder below, which is equal in size and opposite in direction.
This bending builds up strain energy in the falling girder at the same time that straim energy build up in the impacted girder.
There is only one source of energy to bend both girders at the same time: The NET Kinetic energy of 3473 kip-in.
Therefore Nordenson is WRONG to imply that ALL of that net KE flows into the girder below - some flows into the falling girder at the same time.

Nordenon's formula
KE = 1/2 K(below) * D(below)^2​
is wrong. It should be
KE = 1/2 K(below) * D(below)^2 + 1/2 K(falling) * D(falling)^2​

The equivalent static force is still
F = K(below) * D(below)​
but with a smaller D(below); and at the same time
F = K(falling) * D(falling)​

I will have to think hard about how to re-arrange these formulas, but my math intuition tells me already that Tony's solution/formula for springs in series will solve the system.

In the meantime, you need to wrap your mind around the fact that the falling girder bends and thus uses some of the energy which is then no longer available to the girder below, resulting in less deformation below, meaning less equivalent static force.

Unless you can show me that the bending-down of the falling girder is already accounted for somewhere.
You can try to deny that the falling girder would bend and thus use up some of the available energy, but then you simply aren't getting it.
Isn't Nordenson saying that the corner yielding occurred at impact as a result of the energy transferred to the falling girder? I could see the point that he may be under-counting the energy transferred to and absorbed by the falling girder, but if you introduce the stiffness variable as something completely separate from that yielding, then I think you may be over-counting that energy, no?
 

Oystein

Senior Member
Has anyone actually done the calculation with stiffness? Surely that will save much thread space?
Early in this thread, Tony presented numbers for the stiffness of the falling girder, derived under the assumption that the falling beam is a cantilevered beam pinned at column 44. That value of stiffness was far lower than the stiffness of the girder below that Nordenson derived. Treating the two beams that collide as springs in series, you can compute their effective (joint) stiffness, re-do Nordenson's "equivalent static force" calculation, and the result is that the (average) impact force is far lower than the shear capacity of the girder connection.

econ doesn't accept that the falling girder's stiffness needs to be considered at all.
I do accept it needs to be considered, and am mainly trying to tutor econ such that he finally has his Eureka moment.
After we are all on the same page, we'll need to think a bit harder about what the stiffness of the falling girder really is (it is not a cantilever as the far end is not pinned; it's free to rotate). This may get complicated as there will be some free-body physics involved (the impact on the loose end constitutes a large angular momentum, the girder would want to rotate about a point 2/3 L away from the loose end, resulting in a force on the rolling end; I am currently getting knots in my brain trying to think what this means for moment loads along the length of the girder. I intuitively believe that a quasi-gravity load of F/2 along its entire length is a reasonable first approximation).
 

Oystein

Senior Member
Isn't Nordenson saying that the corner yielding occurred at impact as a result of the energy transferred to the falling girder? I could see the point that he may be under-counting the energy transferred to and absorbed by the falling girder, but if you introduce the stiffness variable as something completely separate from that yielding, then I think you may be over-counting that energy, no?
You mean the last sink considered by Nordenson - what I denoted as "D6"?
Hmm, you may have a point. There also would/might be some plastic deformation of the concrete floor below, and of the girder below.
Still, the falling girder will bend and thus consume some energy; and Tony's first back-of-the-handkerchief calculations suggests that this could be VERY significant.
 

jaydeehess

Senior Member
You say you can't follow the math but then want to say it isn't valid. You really do need to follow the math first and show your points before making a proclamation about it. Otherwise, it can rightfully be said that you are just talking out of your hat.
I don't know about it making "little sense". I believe it certainly does make sense to attempt to find a plausible sequence of failure.
However, all one can do is determine in many cases is if a mechanism comes close to failure in an FEA. Given the high number of assumptions and margins of error, close counts in determining plausibility, IMHO.

So do Tony's calcs remove the failing girder/floor section from being able to fail a lower floor section/girder?
I'm not sure he's demonstrated that at all.
I made a proclamation? Do proclamations usually contain the words " do"( in an interrogatory sense), and"not sure", in your world?
 

jaydeehess

Senior Member
I certainly understood what TSz is getting at. Confirmed by his steel vs. rubber ball analogy.
What happens to the component of energy from the KE that bends the falling girder? Does the girder spring back or remain bent? ( as JSO asked re the lack of deformed girders in the rubble)

Could other factors make up the difference by taking into account stiffness as per TSz?
As said many times, analysis is done in a simple representation of reality. TSz introduces one more level of complexity and that could be a good thing unless it begs introducing another equal or near equal component of the real situation. Perhaps Nordenson considered stiffness contributions to be equal to a (or combination of) other secondary contributions.
 
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