The 9/11 Forum

Okay, I'm reading along, and as usual, I understand about 50 to 60% of it

Don't want to interrupt your flow, interesting stuff.

In the mean time I can say I'm a reductionist, so I believe in atomizing the building into a computerized model, plugging in the equations, and pressing "run", then seeing what happens. But it's quite easy for me to talk about it in the hypothetical, whereas you seem to be going for the jugular.

No it's not easy... at all. First of all you have tens of thousands of elements with unique properties. Then you have the uneven distribution of the super imposed live loads. Then you have some asymmetrical damage from the planes. Then you have the unknown extent of the wakening/damage from the ensuing fires. Then you have the wind loads. If an FEA was so simple as plugging in a few numbers it would have been done.

If you reduce this in a reductionist approach, your model will fail to work and you are heading in the direction of the blocks.

It appears to me that the floors could be failed from over loading from descending debris. That's basic engineering. Now it might be more complex to describe HOW the floors failed and ultimately turned to crushed up grains and dust... and little more.

It's not unexpected that columns with unbraced lengths exceeding that for slender columns would be unstable and buckle - breaking apart at the weakest points.. the connections between the 36' sections. That's engineering too. Could the collapsing floors destroy the bracing? Seems perfectly plausible considering the forces eventually developed. Can a model show how brace X on the 84th floor came to be ripped off? Sure if the model was in some super computer that ran the simulation for a year or so (guess).

I do think that physics should be able to predict the speed of the collapsing floors more or less within a range considering the non uniformity of the actual live loads. And remember that no one has a precise measurement for the duration of the collapse. When did the clock start and more difficult... when was it over?

OneWhiteEye wrote:Of the coefficients r,s,a,b listed in:

v' = 9.81 - r*v/x - s*v^2/x - a*v - b*v^2
x' = v
r,s,a,b are constant coefficients

these are the effects, or very nearly so:

r => constant acceleration, increasing r is decreasing acceleration
terminal acceleration, increasing r is decreasing asymptotic value
s => terminal acceleration, increasing s is decreasing asymptotic value
a => terminal velocity, increasing a is decreasing time to max and increased max velocity
b => terminal velocity, increasing b is decreasing time to max and increased max velocity

Corrected. The non-constancy of acceleration is not apparent at small values of r, but it's there.

Thanks for the comments, guys. I know, why go back to playing in blocks world? Unfinished business, mostly. While I don't think this sort of analysis has much to do with any of the 9/11 collapses in their entirety, it might be pretty good at characterizing some of the verinages where there really is block vs. block. Also, I actually see the potential for describing an interior floor collapse, at least in the sense of fitting parameters to actual data.

Where this technique might work - should work - is any case where there is uniform engagement of colliding parts, as in floor slab sections dropping onto equal size slabs or whole building top dropping squarely on the bottom.

The other reason is I wouldn't mind proving Seffen wrong. He's the one oddball who says the acceleration converges towards g/2. If there's something wrong with his analysis, then I've got a publishable comment. Might not do it, but the option is nice.

Maybe that's silly. I don't put myself on that level, anyway.

Another use for this as a rough tool for estimating the conditions under which an interior floor collapse could outpace the global collapse, and for how long, given x amount of head start. More on that later.

OneWhiteEye wrote:1) a per-unit-length energy dissipation added at the beginning of the formulation
2) a corrective force added as a term in the final equation

Revised appraisal: Either way is okay up to the limits of what they're capable, but #2 is capable of more and is better; I'll say why later but, in a nutshell, it's a perfectly valid approach (in contradiction to some of my earlier comments) and is more flexible. I've also solved the imaginary conundrum of the momentum conservation going to the continuum limit - it's not a problem. And so there is at least some physical motivation for this form:

a = 9.81 - s*v^2/z - a*v - b*v^2

The term with s coefficient is the momentum transfer term and the two velocity dependent terms could be interpreted as viscous drag and turbulent drag, respectively. Sorry for switching around the displacement coordinate, now I show it as z, but I've been talking this up with Benson too and he objects to y for vertical. Which reminds me that Benson's avalanche theory, which is based on fitting drop data, indicates the term in v is not needed or counterproductive. So, no viscous drag. His model also indicates nothing in the way of structural resistance contributed, either.

Interpreting the velocity dependent sinks as drag may be semantics, if this form is correct, or it may not. The velocity dependent sinks like pulverization, which is more appropriate to the system, may have a slightly different form.

Honestly, before all is said and done, I'll probably re-derive Bazant's simple analysis from scratch and certainly be in a better place to have another look at what Seffen did. A complete rederivation of the basic mechanics will open up exploration of the parameter space, making it a living analysis instead of a snapshot as applied to the towers.

I always thought the 1D models of the type here would be good for characterizing verinage. After all this time spent dabbling, I look back on many arguments on this subject as being so primitive. Even at this forum, and from me, though less so than what you generally see.

Why does verinage decelerate but the towers do not? Look at the buildings!

1) Concrete slabs fracture to remain as fragments with shorter effective average length but also greater capacity
2) Concrete slabs have a different load-displacement response under compression and bending
3) Cellular geometry ensures higher cross-sectional load bearing surface during disorder of collapse; is relatively insensitive to misalignment
4) Cellular geometry has a greater tendency to trap air in the interior

Don't discount #4. Pressure builds quickly if the air doesn't escape. The compaction acts (quickly) to close off flow paths to the exterior yet the pressure increase wants to open the orifice wider, even if that means blowing chunks out. There will be a point of dynamic equilibrium on this alone. #4 is not something that will cause deceleration in itself, but it becomes a large factor as the collapse progresses. Since the acceleration is due to the sum of forces, a high magnitude in this category sharply limits the acceleration - rolls it off like the curves above.

#2 and #3 don't necessarily cause deceleration in themselves, either. They are reasons why the global load displacement response would be a greater magnitude in relation to a given static DCR.

It is #1 which has the greatest potential to produce a consistent average resistive force which exceeds the static demand. That favors propagation over compaction and means the crush front can reach ground faster, perhaps a lot faster. In a tall tower, that doesn't matter for a long time, if ever (because of mass lost outside of footprint).

In these short apartment buildings it can reach ground quickly and, at that point, the mode goes to exclusive crush-up (no matter what mix of up and down it was before) and the rules change a bit. Compaction continues to some degree but remaining uncompacted stories are favored to crush, if anything at all is going to crush. The driving mass is now much smaller.

Crush up exhibits a tail-off of velocity (theoretically, ideally) to zero. That means deceleration! The shorter the building is and the greater the effect of #1, the sooner this commences and the faster the rate.


OK, must attend to job.

David Benson gave me the answer on how the value of g/3 is determined from the equation of motion in the OP.

David B. Benson wrote:Respective of the physically correct formulation (where it certainly should not matter which approach one uses), consider the governing equation a = g - v^2/z to assume that for large t one has approximately v = kt with a = k. It follows that
z(t) = z0 + (1/2)kt^2
and for t sufficiently large z0 can be neglected. Then
k = g - (kt)^2/[(1/2)kt^2] = g - 2k;
k = g/3.

Have to use a different governing equation to obtain Seffan's result; indeed I believe he does derive something different.

Simple as that.

I don't know if the small string of equations from DBB is that simple.

The DDB quote is taken from which post?



I got most of the logic. Interesting. Maybe that DDB logic could be clarified a bit?

That post was at physforum.

http://www.physforum.com/index.php?showtopic=12383&view=findpost&p=510877

Here goes. If the value of acceleration converges on a constant, then for large t ( the independent variable) the relationship v = at is approximately correct. This is the familiar formula for determining velocity under constant acceleration. So, if the convergence actually occurs, we can work with this simplified formula to obtain a value for acceleration. Otherwise, it will lead to a contradiction (like z=z^2 or 3=5 some such).

Benson uses v = kt with k representing the constant acceleration we'll assume equates to a. To find the position as a function of time, v = kt can be integrated once to give:

z = z₀ + (1/2)kt²

also a familiar formula for position change under constant acceleration. The constant offset from integration z₀ is the initial position which can be taken to be well after the onset of constant acceleration (really, wherever you want) and, if you let t range MUCH larger than that, z₀ can be ignored because (1/2)kt² will be much larger. Now the value of z under a long period of constant acceleration is approximately:

z ~ (1/2)kt²

Even though we're considering the limit of very large t, the original equation of motion still holds as always, whether or not the assumption of constant acceleration is true, so that relation can be used. Plugging in the values of a, z and v from the large t approximation into the eq of motion should yield either a meaningful result or a contradiction.

eq of motion: a = g - v^2/z

Let a = k, v = kt, z = (1/2)kt²
=>
k = g - (kt)²/[(1/2)kt²]
=>
k = g - 2k²t²/kt²
=>
k = g - 2k²t²/kt²
=>
k = g - 2k
=>
3k = g
=>
k = g/3


I'm a little embarrassed I couldn't do this. I've only ever done this sort of thing in classes a long time ago, and it didn't come up that much. Still, it's pretty easy and I had a sense of what to do but couldn't remember where to start. Start by assuming the constant relation and obtain formulas for the dynamic variables; then plug into original equation and see what comes out.

The assumption of (near) constant acceleration at large t holds true for g/3.

Benson also said he thought Seffen was wrong. I'll be digging deeper into the why. Seffen deliberately appeals to a special formulation, claiming it's necessary under these circumstances. That formulation has acceleration converge to g/2. It's not a huge difference, but significant. Regardless of which formula is supposed to be correct, physics engine simulations indicate g/3.

From your OP:

a = g - v2/y


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


This is Cherepanov's equation of motion. Wow, that was easy! Now, how does one get asymptotic approach of acceleration to g/3 from this?

So your acceleration a is a function of velocity and y. Cool.

From the OP you ask if there is a simple proof that a converges to g/3?

Well, the position y will take the form y=y(t).

If a(v, x) converges on a constant then for large y, t.... v=kt. fine.


If so, than y takes the form of a parabola y= p + qt + 1/2 k t^2 where we don't care about p.


Then plug that back in for y in the equation a= a(v, y)? .....

k=g/3 ? I'll think about it.


How did that other guy get g/2?

.....................

You already ran this situation with your toy variety model. In which post?

Here is your small proof:

eq of motion: a = g - v^2/z

Let a = k, v = kt, z = (1/2)kt2
=>
k = g - (kt)2/[(1/2)kt2]
=>
k = g - 2k2t2/kt2
=>
k = g - 2k2t2/kt2
=>
k = g - 2k
=>
3k = g
=>
k = g/3

I would say, for very large t,

a goes to k
v goes to m +kt
z goes as p + qt + 1/2 k t^2.


If plugged into the a(v, z) term in the limit of very large t, k goes to g/3.

Same idea. Makes sense.

This is from your observation of terminal acceleration in the "toy" thread.

From this post

Similar situation? g/3 is your result, too?

From the "toy" thread. OWE:

The simulator is not junk. It's a little difficult to reconcile a figure of 0.6g with 0.3g, however. Maybe WTC1 starts at 0.6g and goes to 0g, whereas this starts at g and goes to 0.3g, and it all comes out in the wash. ???

???