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Building Collapses in Rio

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posted on Jan, 31 2012 @ 02:07 PM
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Originally posted by DrinkYourDrug
A column does not require an equal or higher psf rating than another material to exert its axial capacity on said material.

This doesn't address anything I've said, so I'll pass on it.


Do you:

-Disagree that floor slabs and debris would be compacted into a stack which is crushing the rest of the tower?


I do disagree with the notion that all slabs and debris would be compacted into a stack, and further disagree that it would crush the rest of the towers. Some debris was compacted, but a lot of it ended up in a pile around the base. What went down into the basement was probably largely compacted, and this is where the 'sedimentary' layers were found. A significant amount of debris was found outside the footprint. Direct observation of video shows much of the perimeter peeling and a significant core remnant, so these were not crushed by anything, let alone a nice neat stack of pancakes. Also directly observable is the extreme spatial inhomogeneity of the crush front(s), making impossible any neat stacks of entire floors.

Ultimately, no, there was no evidence of 110 floors stacked up at the bottom, so of course I disagree. I believe the layers discovered numbered a maximum of 20 floors, which would naturally be from near the bottom of the tower. It could be that a majority was compacted into layers transiently, and definitely possible that a sizable percentage ended up compacted, but you're taking it as a given that it all did. I won't do that because the evidence clearly indicates otherwise.


-Feel that a column could punch through a stack of several floor slabs (whist not even getting close to applying its axial capacity)?

I already gave you my take on that. One slab, no problem, Two, very likely. Three, probably still possible. By five, I was saying no. One is all I was EVER talking about, so extension of the context of my remarks is not my obligation to address, and I said why in an earlier post. To be even clearer, I had in mind initial impacts in the first few stories of descent, but I wasn't explicit. After that, momentum takes over, and things learned in statics classes become mostly irrelevant to the dynamics.


It's ok, I have a degree in structural engineering, I know how that stuff works.

Good, most don't.


Yes, the tiny area of the section of the column would obviously crush, but as it does so and the column is driven into the stack of floors the effective area the column applies its axial load to increases.

Agreed.


I can go into more detail here if you persist that a column can effortlessly drive through a stack of floors.

I've never even said it once, let alone persisted. Please direct your argument at my argument. I restated what was clearly stated the first time. I do NOT feel a column can effortlessly drive through a STACK of floors. I showed, however, that it likely can drive effortlessly through one.


Yes, but I believe the overwhelming difference would be in the unrestrained length of column.

To the degree that such lengths exist, true.




posted on Jan, 31 2012 @ 02:14 PM
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Originally posted by DrinkYourDrug
Agreed, and a lot of energy would be exerted in doing so.

A LOT of energy was available. A huge excess. I don't imagine structural engineers make use of potential energy calculations much.


Axial capacity of columns are determined with this in mind. In real conditions the column is not braced for its entire length and deformations under capacity axial load generate moments in addition to any other moments it may be designed to carry. The in-situ axial capacity is closely related to its moment capacity. For the column to fail in bending while its unrestrained length is close to original configuration will require almost the same amount of force as its design capacity.

For a given length, axial compression as failure mode ALWAYS dissipates more energy than bending. This says nothing about which failure mode will happen in a given circumstance but the lower energy mode will be favored if accessible. Without near perfect alignment, failure in bending is probable. If indeed, in a particular instance, the force required to induce bending is comparable to the force needed to compress, then we've already seen from my calculations that column wins against floor slab in that case, too.


The context was collapse progression which I always assumed meant undamaged structure being impacted.

OK.


By this point a compacted stack of floors and debris must have accumulated.

Maybe. Not in every circumstance, I'm quite sure.


How would loose rubble crush (and completely decimate) 80 or so undamaged levels of structure at ~2/3rds free fall? Uncompacted rubble can't apply much force without compacting itself.

Uncompacted debris cannot bear load, there's a difference. I'm not one of these people who'll claim a 10 pound bag of flour is the same as a 10 pound brick. Clearly, a rigid body of mass M imparts a greater peak impulse than a collection of particles with mass summing to M. But, the notion that it can't apply much force when in motion once again ignores the role of momentum. I suppose structural engineers don't calculate momentum too frequently, either.

But the matter of descent at 2/3g deserves further consideration. (Thanks for not claiming freefall) The figure of 2/3g is only applicable for the first few seconds. While it is not possible to directly measure the displacement of the upper section past three seconds and change, the leading interior crush wave of WTC1 has been measured over a substantial later interval and was found to propagate at a near constant velocity of just under 30 m/s. How does that fit with your remark?


I'm not saying it should necessarily arrest, just that it is suspicious the undamaged structure only applied an average force to the falling top section of about one third of the force it applied to it as it held it stationary.

You are a structural engineer. Have you done the calculations for the dynamics yourself, or followed anyone else's treatments (Bazant, Seffen, Greening, Cherepanov, etc * )? I recommend it. You may be surprised. I understand your incredulity. I had it, too, originally. Now I save my incredulity for more bizarre stuff than that, and there is plenty of bizarre to go around.

I'm not saying these simplified 1D models capture the collapse mechanisms well, because I firmly believe they don't. On the other hand, they do provide bounding cases and certainly get results within a small percent of actual, which is reason enough to give attention.


Listen, I'm not going to fight this one particular point forever to the nth detail. My claim was simple and solid, excessive parsing will not change that. A (isolated) floor slab CANNOT support the same load as a column, period. The calculations, while rough, show at least a three orders of magnitude difference in yield points. End of story.


* Edit: even Tony Szamboti, member of AE911Truth!

I repeat: As-built FOS is MEANINGLESS in a messy collapse. Even so, ALL of the cited authors demonstrate, through meticulous engineering calculations, that the average resistive force even in a perfect axial collapse is less than the static load.
edit on 31-1-2012 by IrishWristwatch because: (no reason given)



posted on Jan, 31 2012 @ 02:17 PM
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In general, I won't be responding to your tiresome and uniformly fallacious crap, Darkwing, but I'll make an exception early on to call you on your lying nature, so it's clear to everyone what you're about.


Originally posted by Darkwing01...the fact that Irish returns to the ROOSD hypothesis again and again ...


At the other forum, the only time I mentioned ROOSD was in arguments with you against your ignorant and unfounded criticisms. It's not my theory; I will defend it against feeble moronic attacks such as yours, but I don't promote it. I have not promoted it here, nor do I intend to start - ever. I've not mentioned it even once here (before now), so you are a liar. Proven. And, I might add, an unscrupulous liar.

I realize you apparently exist for the sole purpose of riding my ass, but all I have to do is scroll past your posts as I encounter them, which isn't difficult.

SnowCrash continues his trend of handing you your ass at every turn. You should stop embarrassing yourself, especially if your secret thrill is believing that I'm reading your tripe.


edit on 31-1-2012 by IrishWristwatch because: (no reason given)

edit on 31-1-2012 by IrishWristwatch because: (no reason given)



posted on Jan, 31 2012 @ 02:25 PM
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Originally posted by psikeyhackr
It is not just momentum that slows it down.

Agreed. Most people don't realize that it does, though.


That is all very nice but why don't we just demand that floor assemblies be tested in furnaces to see if that phenomenon can be produced in TWO HOURS of heating.

Sure, that's preferable to my rinky-dink model. Even so, the model puts bounds on the geometric possibilities.



posted on Jan, 31 2012 @ 03:25 PM
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Originally posted by DrinkYourDrug

A column does not require an equal or higher psf rating than another material to exert its axial capacity on said material.

I may not have understood your statement when I replied above. It seems you are responding to my "weakest link" claim that, in a collision or even (especially) quasi-static application of force, the weaker of the two objects establishes the peak force between them. If your comment is a rejection of this premise, then I certainly disagree.

When two bodies are in contact, the magnitude of forces applied by one body to the other are equal and directed oppositely, Newton's 3rd, I'm sure we're in agreement on that. What is the magnitude of the force? It is given by the load-displacement responses of the respective bodies. The peak force of that load displacement curve is the peak force the body in question is able to offer in resistance to deformation. If two bodies with dissimilar load-displacement response apply force to each other, the force between them can never exceed the minimum of the two response curves at the given points of deformational displacement in both bodies at a single point in time.

If you disagree with that, you'll have to explain how a material can afford resistance greater than its ultimate yield strength.

I'll give you the answer. My claim is true to a (very useful) first approximation. It holds true for small momentum changes due to deformation within the contacting bodies. The inertial effects exhibited by the deformed portions (which may yet be in motion) provide a resistive force according to the momentum these portions acquire, per Newton's second law in the form F = -d(mv)/dt.

Thus there is effectively no limit to the mutual force experienced by two bodies in collision. The high momentum case is dominated by these effects; at high enough momentum for any two bodies, the contribution of force from 'structural resistance' can be made negligible.

In the circumstance of the tower collapses, given the member masses, speeds, and typical load displacement response, it is best to assume my claim valid only for the first few seconds of collapse. When it matters. When there's even a 0.001% chance of arrest (IMO).

The weakest link, in those circumstances, dictates the outcome. Floor diaphragms cannot support even the static weight of the upper section when loaded through contact points of column ends. This is easily demonstrated, and I did so earlier by comparing capacities. My lowball estimate, derived from wildy overinflated floor specs, was a factor of 1000 difference. It's probably more like 10000. The columns did not have an FOS of 10^4, therefore they were designed for nominal stresses far in excess of the yield of the floors, therefore the floors cannot support their nominal load. QED.

I can't believe I have to say this to a degreed structural engineer.

The force in collision is dictated by the yield of the weaker member.



posted on Jan, 31 2012 @ 03:36 PM
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Originally posted by DrinkYourDrug


Why? Why can't it be loose (uncompacted) rubble?

How would loose rubble crush (and completely decimate) 80 or so undamaged levels of structure at ~2/3rds free fall? Uncompacted rubble can't apply much force without compacting itself.


For someone others are calling an expert that statement right there proves Irish is not.

How could a person who understands physics think rubble can crush solid floors, and the rubble not be crushed also in an equal opposite reaction, and conservation of momentum. It shows a lack of understanding of basic laws of motion. Physics 101.

Irish agrees though that sagging trusses can not pull in columns, which is the collapse mechanism according to NIST. Yet he still thinks rubble can cause solid undamaged floors to fail?

Newtons third law, and conservation of momentum is constantly ignored by OS supporters.


edit on 1/31/2012 by ANOK because: typo



posted on Jan, 31 2012 @ 03:40 PM
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Originally posted by IrishWristwatch

Agreed. Most people don't realize that it does, though.


OS supporters don't realise that.

They ignore equal opposite reaction, and conservation of momentum laws, and think momentum is all that matters.

Try reading some of the OSers posts.



posted on Jan, 31 2012 @ 03:42 PM
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reply to post by ANOK
 





Yet he still thinks rubble can cause solid undamaged floors to fail?

I don't see much difference between chunks of concrete and gravel.
So you are saying I can shovel as many tons of gravel as I want into my living room floor and not worry about it collapsing?
edit on 31-1-2012 by samkent because: (no reason given)



posted on Jan, 31 2012 @ 03:43 PM
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Originally posted by snowcrash911
What seems to be mainly at issue here is if the upper block (unfortunate term) is broken up, it becomes incapable of doing damage. Not true. We are talking about an avalanche of debris thundering downward, accumulating and accruing matter as it advances. The rubble is layered: the layer at the collision front isn't as fragmented as the layers on top of that which have already seen multiple collisions. How are the floors, whose connections have roughly the same capacity throughout the building, going to arrest collapse? We are not talking about blocks here, we are talking about a massive accumulation of loosely associated rubble impacting a flat, concrete surface of limited thickness supported by trusses designed to carry desks, chairs, cabinets, office furnishings and people, not a catastrophic collapse, a downwardly accelerating mass of heavy debris. Imaginary physics isn't going to change the hard fact that the floors were overloaded every collapse iteration, with mass reigning down over the 3.7 meter distance from ceiling to ground level.


Yep, this seems to be one of the main points of contention here.

I've asked ANOK & friends repeatedly, at what floor should the collapse have completely arrested. But I can't seem to get any sort of a straight answer; or even a monkey's guess.

Once the collapse initiated- what was supposed to stop that tremendous amount of debris and its downward force dead in its tracks I wonder? Was it the vertical steel columns' job to do this? Hmm..

Collapse initiation of the south tower- We're talking about dropping essentially a 30 story building from a height of 12 feet on to one story. Not 95 stories, as ANOK & friends keep asserting.

Each floor functioned independently of the one below it as it pertained to supporting a very sudden, and enormously dynamic and chaotic load- that of the dozens of floors coming straight down on top of it. It was beyond the design and scope of what the structure was meant to withstand, no matter what these truthers say.

In short, once the collapse initiated it arrested where it should have.... on the ground.



posted on Jan, 31 2012 @ 03:56 PM
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How would loose rubble crush (and completely decimate) 80 or so undamaged levels of structure at ~2/3rds free fall? Uncompacted rubble can't apply much force without compacting itself.

Loose rubble also doesn't accumulate on column ends, does it? By all rights, this alone stops your argument dead in its tracks. There are three places for rubble to go: the void in the core, the floor area, and outside the footprint. The portion impacting the negligible cross-section of the column ends is, of course, negligible.

So who cares what the FOS of the columns are in a rubble-driven collapse, if such a thing could happen? If it's NOT rubble, but rather a nice stack of intact pancakes too thick to punch through, then you can have a serious problem of dynamic overloading of the columns. The columns may then be the weakest link. Go ahead, invoke their full capacity, FOS and all, it won't matter above a certain threshold of MOMENTUM. A stack of pancakes can overload the columns via impulse. Easily, with sufficient momentum.


edit on 31-1-2012 by IrishWristwatch because: (no reason given)



posted on Jan, 31 2012 @ 04:07 PM
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Originally posted by PhotonEffect
I've asked ANOK & friends repeatedly, at what floor should the collapse have completely arrested. But I can't seem to get any sort of a straight answer; or even a monkey's guess.


You have, I don't recall.

But who knows? How can anyone answer that question, and why do you think it makes a difference?


Once the collapse initiated- what was supposed to stop that tremendous amount of debris and its downward force dead in its tracks I wonder? Was it the vertical steel columns' job to do this? Hmm..


The tremendous amount of undamaged floors? Why do you think the falling debris was more than what it was falling on? 15 floors fell on 95.


Collapse initiation of the south tower- We're talking about dropping essentially a 30 story building from a height of 12 feet on to one story. Not 95 stories, as ANOK & friends keep asserting.


No, you have it wrong. It is a block of 15 floors falling on block of 95 floors, you keep ignoring the mass of the 95 floors.

Think about this. Both blocks were the same construction, floors held by the trusses. When the first two floors impacted the force of the falling floors will effect BOTH impacting floors, equally, not just the one the top is impacting. Both floors would be damaged. Equal and opposite reaction. As the collapse continues the top floors would be destroyed before the bottom ones. Do the calculation, 95 - 15 = 80. To give you the benefit lets say the top floors only received 50% of the damage the bottom ones did, so to keep it simple I'll double the falling floors, 95 - 30 = 65. If two bottom floors were destroyed for every top one, 47.5 - 30 = 17.5. Even stretching the maths the collapse would not have been complete.



posted on Jan, 31 2012 @ 04:11 PM
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Originally posted by IrishWristwatch
So who cares what the FOS of the columns are in a rubble-driven collapse, if such a thing could happen?


Again you prove you are not an engineer.

If you don't know the FoS, how can you claim what force can cause it to fail?

If it has an FoS of 2, it would take less force to fail than if it had an FoS of 4.

I remember when we first started discussing FoS, and OSers thought it was safety equipment and such lol.


“ . . . the choice of an appropriate factor of safety is one of the most important decisions the designer must make. Since the penalty for choosing too small a factor of safety is obvious, the tendency is to make sure that the design is safe by using an arbitrarily large value and overdesigning the part.

blogs.nasa.gov...


edit on 1/31/2012 by ANOK because: typo



posted on Jan, 31 2012 @ 04:12 PM
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Originally posted by IrishWristwatch

Originally posted by psikeyhackr
It is not just momentum that slows it down.

Agreed. Most people don't realize that it does, though.


Just to get this straight, are you talking about inertia here? If so, isn't saying "momentum that slows it down" a bit misleading? If not, what is it you are talking about here?



posted on Jan, 31 2012 @ 04:13 PM
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Originally posted by ANOK

Originally posted by DrinkYourDrug


Why? Why can't it be loose (uncompacted) rubble?

How would loose rubble crush (and completely decimate) 80 or so undamaged levels of structure at ~2/3rds free fall? Uncompacted rubble can't apply much force without compacting itself.


For someone others are calling an expert that statement right there proves Irish is not.

To be clear, I've not referred to myself as an expert. I did question whether someone of obvious lay background should be condescendingly lecturing someone who is degreed in the subject, but each of our arguments stand on their own merit, naturally, as they should.

Having said that, where we disagree on the matters discussed, you are wrong. In each and every case, I've provided rigorous refutation straight from standard freshman texts. That you, and certain others with the commonality of not one whit of formal education on the subject, cannot recognize it - is not my problem. You have, in virtually every instance, left my objections unaddressed.


How could a person who understands physics think rubble can crush solid floors, and the rubble not be crushed also in an equal opposite reaction, and conservation of momentum. It shows a lack of understanding of basic laws of motion. Physics 101.


As for what follows next, let it be seen by ALL that I did NOT raise this subject. Is that clear Darkwing? I have every right to address the subject, especially as it is here associated with impugning my skills.

Rubble falling into air doesn't do anything, but rubble falling on floors counts. At the very least, it counts for the static mass of the accumulation to any given point. If the local distribution of mass exceeds the yield point of the floor, it will yield. If NIST's figures are to be believed, this occurs when the local accumulation exceeds 12 stories worth of (presumably live) load across the applicable region. That means the entire floor need not fail in order for local failure to occur.

Realistically, rubble originates from dynamic conditions. All momentum must be dissipated and bring the mass to rest in order for static accumulation to occur, by definition. Therefore, there is a(n unquantified) higher probability for failure in advance of excess static loading than not.

The issue of self sustaining continuation is clearly a boundary value problem and is an open question. In no way has such a system been numerically quantified and there is no reason except faith that such a progression can account for observables. There is but one way, however, to avoid the matter of continuation in the static case, once sufficient source rubble amount and density has been generated to cross global thresholds : mass loss across the boundaries of the floor regions.

Theoretically, rubble can drive collapse progression in these circumstances. For the feeble of mind (and you know who you are), all I'm doing here is addressing reality, not promoting a theory. I am not claiming the circumstances exist to arrive at such conditions, nor that the actual dynamics would be correctly described should those conditions arise. The feeble minded take any mention of the possibility as promotion of a theory when, in fact, the skilled and knowledgeable understand that to ignore the possibility (especially by means of ignorant hand-waving) is profoundly unscientific. To single-mindedly rail against the possibility borders on religious zealotry.

Now, back to this:


How could a person who understands physics think rubble can crush solid floors, and the rubble not be crushed also in an equal opposite reaction, and conservation of momentum. It shows a lack of understanding of basic laws of motion. Physics 101.

Once again, you (who hasn't taken Physics 101!) presume to lecture me on physics and, once again, you fall flat on your face. You're batting a thousand on that. But, I'll cut you some slack. Apparently, it's how you learn.


Irish agrees though that sagging trusses can not pull in columns, which is the collapse mechanism according to NIST. Yet he still thinks rubble can cause solid undamaged floors to fail?

Without a shred of doubt, given certain (not improbable) conditions.


Newtons third law, and conservation of momentum is constantly ignored by OS supporters.

Patently false. Your abuse of elementary physics is an abomination to the field and an affront to its diligent practitioners. You're in over your head, by a wide margin.
edit on 31-1-2012 by IrishWristwatch because: (no reason given)



posted on Jan, 31 2012 @ 04:18 PM
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Originally posted by IrishWristwatch

To be clear, I've not referred to myself as an expert.


I didn't say you did. Others are saying you are, but they are obvioulsy as confused as you are.



posted on Jan, 31 2012 @ 04:21 PM
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Originally posted by PhotonEffect
Yep, this seems to be one of the main points of contention here.

And pretty much everywhere collapses are discussed. It is one of the central hamster wheels in the debate, and it's been spinning (needlessly) for years. Devotees of the exercise obtained by running on said wheel would like nothing more than to entrain you into their world which, despite their incessant repitition of thoroughly discredited points, they fail to perceive as circular.

**** 'em. I'll poke at the runners on that wheel, but not indulge neuroses. Too much of that already. Blind repetition is not an argument; once refuted should be sufficient.


I've asked ANOK & friends repeatedly, at what floor should the collapse have completely arrested. But I can't seem to get any sort of a straight answer; or even a monkey's guess.

And you won't, because they can't. No one I've seen in this thread arguing against natural collapse can even set up and solve ANY equation of motion, which is the beginning of most dynamics. None of these people have solved a single physics problem in their entire life, and don't intend to start now.



posted on Jan, 31 2012 @ 04:30 PM
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Originally posted by ANOK

Originally posted by IrishWristwatch
So who cares what the FOS of the columns are in a rubble-driven collapse, if such a thing could happen?


Again you prove you are not an engineer.

Again, you prove that blowing smoke is your idea of an argument. Look above. Rubble bypasses columns. Care to calculate the ratio of floor area to column end area? A chunk of debris hitting a column end does not arrest any collapse, except perhaps the motion of that chunk. Few chunks hit column ends. Doesn't take a physicist or an engineer to divine that, should be obvious to the most casual observer.

Face it, this subject is not your cup of tea. Why do you keep trying to drink from it?


If you don't know the FoS, how can you claim what force can cause it to fail?

The FOS of columns bypassed, rubble or otherwise, is wholly irrelevant. By definition, not opinion.


If it has an FoS of 2, it would take less force to fail than if it had an FoS of 4.

Do I have to say 'Duh' to every correct triviality you utter, which also is irrelevant? Those columns bypassed do not contribute, and a goodly portion are directly observed to be bypassed in video. Those which are not impacted axially do not afford their full capacity. Rubble doesn't settle on column ends in any significant quantity. Momentum can overload even the strongest of columns.

But I've said all of this, some things many times in this thread. When are you going to learn repetition of refuted assertions does not increase your chances of winning an argument?


I remember when we first started discussing FoS, and OSers thought it was safety equipment and such lol.

Good for them. That has nothing to do with me, or the fact that you are consistently wrong by a country mile when it comes to all things physics.


edit on 31-1-2012 by IrishWristwatch because: (no reason given)

edit on 31-1-2012 by IrishWristwatch because: (no reason given)



posted on Jan, 31 2012 @ 04:36 PM
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Originally posted by ANOK

Originally posted by IrishWristwatch

To be clear, I've not referred to myself as an expert.


I didn't say you did. Others are saying you are, but they are obvioulsy as confused as you are.

I appreciate the positive comments, but it is not true that I am an expert in the subject of progressive collapse. Few people are, and I know it a whole lot better than most (who yammer endlessly about it). This is through the sweat of my brow.

It is true I'm light years ahead of you.



posted on Jan, 31 2012 @ 04:56 PM
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Originally posted by -PLB-

Originally posted by IrishWristwatch

Originally posted by psikeyhackr
It is not just momentum that slows it down.

Agreed. Most people don't realize that it does, though.


Just to get this straight, are you talking about inertia here? If so, isn't saying "momentum that slows it down" a bit misleading? If not, what is it you are talking about here?

I am talking about inertia, and I did mean momentum slows it down, although that's an inaccurate statement generally (i.e., so long as collapse is proceeding). What I should have said is, momentum transfer is responsible for a significant amount of resistive force in a collapse where accretion occurs.

In a medium of uniform density (not applicable to the towers, I know, just trying to make a point), non-conservative or inelastic accretion converges on an acceleration of g/3, where for conservative or non-impulsive accretion the convergence is on g/2. See Bazant and Seffen, respectively for each figure, for derivations in a collapse context. This result, interestingly enough, is independent of material properties. That is, it doesn't matter the residual capacity of a system, it will converge on constant acceleration in the limit, regardless. Greater capacity just means a longer time to converge.

Seriously, it's almost magical, but it isn't.



posted on Jan, 31 2012 @ 05:15 PM
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At the risk of being accused of being someone who DOES know how to set up equations of motion, I offer the following.

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

Uniform density ρ given total height H and mass M:
ρ = H/M

The change of mass accumulated with respect to time:
dm/dt = ρdy/dt = ρv

Conservation of momentum in incremental inelastic collision:
mv = (m + Δm)(v + Δv)

Multiplying out quantities in parentheses:
mv = mv + mΔv + vΔm + ΔmΔv

Eliminating original momentum from both sides:
0 = mΔv + vΔm + ΔmΔv

Dropping second order quantity:
mΔv = -vΔm

Dividing both sides by Δt, the increment of time:
mΔv/Δt = -vΔm/Δt

Allow increment to go to infinitesimal:
mdv/dt = -vdm/dt

Substitute expressions for instantaneous acceleration and mass accumulation rate:
ma = -ρv^2

Thus the resistive force due to inelastic accretion is:

Fr = -ρv^2

where the negative sign indicates the force opposes the direction of motion.

If the moving portion of the mass is subject to gravitational force, then the total force on the moving portion at a given time is

F = mg - ρv^2

where m, ρ, and v are all functions of time.

Acceleration is therefore:

a = g - (ρ/m)v^2

Since the instantaneous upper block mass m is:

m = ∫ρdy evaluated on [0,y] = ρy

the above relation a = g - (ρ/m)v^2 becomes:

a = g - v^2/y

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

This is the continuum equation of motion for inelastic accretion of a uniform material. The only resistance is from inertia of entrained material. The derivation is not unique by any stretch, and I confess to having perused it long before doing it all by myself, It also was not fresh in my mind!

I am pleased by this modest accomplishment, which directly shows the resistive force I was referring to in the previous post. I am less pleased that I could not immediately figure out how to show convergence to g/3, and that David Benson (of BLGB) had to show me how. But here is his take, verbatim:


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.


And, after tidying it up to be more explicit:

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)kt^2

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)kt^2 will be much larger. Now the value of z under a long period of constant acceleration is approximately:

z ~ (1/2)kt^2

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)kt^2
=>
k = g - (kt)^2/[(1/2)kt^2]
=>
k = g - 2k^2t^2/kt^2
=>
k = g - 2k^2t^2/kt^2
=>
k = g - 2k
=>
3k = g
=>
k = g/3


This does not address the case of structural resistance, but you will observe it is invariant with respect to material property, indeed entirely independent of such. Even MASS! Only depends on uniform density. The eqs get considerably harder with non-uniform density.

A picture is worth a thousand equations. Here is a numeric solution of the equation for position and velocity versus time.



See how the velocity goes to a near straight line (constant slope, constant acceleration)? It's g/3. From momentum only. That's what I was talking about in my remark to psikeyhackr.



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