Some help from those with Mathematical knowledge

[gumboot]

Hi,

I'm doing some simple equations regarding the WTC collapses. I just want to make sure I'm going in the right direction.

I'm wanting to calculate the force applied to the first intact floor, as spread over the combined footprint of the 47 core columns and 240 exterior columns.

The value for the footprint is 45m2

(This is based on some assumptions that are detailed in my calculations)

I calculated the ENERGY of the impact based on the calculated mass and instantaneous velocity of the upper masses at impact and came to 9.7GJ (WTC1) and 30.3 GJ (WTC2).

This gives us a distribution of 215 MJ per square metre (WTC1) and 673 MJ per square metre (WTC2).

The problem here is to calculate PRESSURE I need FORCE, not ENERGY.

I know I've gone off track somewhere.

Help? :o

-Gumboot

[CurtC]

Instead of energy, I think you should be using momentum. Momentum in an impact is mass times velocity, but it's also force times time.

You don't know the time of the impact here. An impact that takes a longer time involves smaller forces, and vice-versa.

[Newtons Bit]

CurtC is correct. You need to be using f = m*a or rather f = m * dv/dt. To do this you need to know how long each floor held up and how much of the velocity it reduced.

[Anti-sophist]

Calculating the forces involved in a collision is fairly non-trivial. This is complicated further because you don't really care about the 'average force' over some small delta-t but you what you really care about is the maximum instantaneous force. I don't know if there is even a satisfactory way too approach this problem in this case.

[The Silver Shadow]

ETA: Never mind...

[gumboot]

Just looking at this now...

this (http://www.jw-stumpel.nl/bounce.html) site has some stuff on impact forces.

Given that my investigation here is to determine whether the upper mass should stop falling, am I right to apply an elastic collision scenario?

Assuming maximum resistance from the lower mass, it seems the impact time would relate to how long it takes the shockwave to travel through the upper columns and back (at the speed of sound through steel).

I'm just trying to get my head around all this because I think it's an interesting exercises to look at what sort of impact forces we're dealing with based on a number of simplifying assumptions.

From my point of view, I can look at the collapse and understand that the dynamic load of the upper mass, after having fallen through the aircraft impact floors, has enormous energy which exceeds the capabilities of the lower structure.

But as an exercise in learning, I'm just trying to put a ballpark figure on what sort of force is in play here.

Part of my intention is to illustrate to CTers just how much difference the "solid vs structure" mistake can cause, for example in WTC1, treating the upper area as a single block, the mass's GPE at impact (9.7J) is spread across 3,986.65m2, resulting in 2.4MJ per square metre. In contrast, if you take into account that the upper mass is a structure and the columns will exert the initial impact mass only (assuming the columns are a solid mass rather than hollow) that 9.7GJ is spread over only 45m2 resulting in 217.1MJ per square metre - almost 100x as much.

-Gumboot

[gumboot]

Just saw everyone's posts.

Dagnammit.

Heh.

-Gumboot

[natureboy]

Gumboot, and friends

Let me just give you a heads up,

No matter how many irrelevant "calculations" you perform, There is no way (using newtons laws) to get those towers (all three) to fall as fast as they did (13-14 sec)

You can calculate all force,pressure,energy (in megajules) you want(till your blue in the brain) but...

The top 1/5 of the north tower....is only just that! only 1/5th the total mass of the tower....(no matter how much "pressure" there is).....DUH

No matter how fancy your calculations are, there is just no way to get 1/5 the mass of the world trade center to plumett through 4/5 (of the mass of the world trade center) in 12 -14 sec.

[defaultdotxbe]

No matter how fancy your calculations are, there is just no way to get 1/5 the mass of the world trade center to plumett through 4/5 (of the mass of the world trade center) in 12 -14 sec.
can i see your calculations? :D

[gumboot]

No matter how fancy your calculations are, there is just no way to get 1/5 the mass of the world trade center to plumett through 4/5 (of the mass of the world trade center) in 12 -14 sec.




Garbage.

If I dropped an asteroid the size of Texas on the WTC from space I guarantee you I could get the top 1/5 of the mass to plummet through the lower 4/5 in much less than 12-14 seconds.

-Gumboot

[defaultdotxbe]

Garbage.

If I dropped an asteroid the size of Texas on the WTC from space I guarantee you I could get the top 1/5 of the mass to plummet through the lower 4/5 in much less than 12-14 seconds.

-Gumboot
actualyl i have it on fairly good authority that an asteroid dropped onto the undamaged towers would just drive them intot he ground like a nail :)

[gumboot]

actualyl i have it on fairly good authority that an asteroid dropped onto the undamaged towers would just drive them intot he ground like a nail :)


:D

Very true.

-Gumboot

[westprog]

Gumboot, and friends

Let me just give you a heads up,

No matter how many irrelevant "calculations" you perform, There is no way (using newtons laws) to get those towers (all three) to fall as fast as they did (13-14 sec)

You can calculate all force,pressure,energy (in megajules) you want(till your blue in the brain) but...

The top 1/5 of the north tower....is only just that! only 1/5th the total mass of the tower....(no matter how much "pressure" there is).....DUH

No matter how fancy your calculations are, there is just no way to get 1/5 the mass of the world trade center to plumett through 4/5 (of the mass of the world trade center) in 12 -14 sec.

There are a number of obvious flaws here - but the main one is the assumption that the bottom 4/5ths will have no effect on the collapse.

Clearly, once the rest of the building starts to fall, it's potential energy will be released. How this might happen is extremely complex. If, for example, the initial impact of the top 1/5 falling one floor caused total structural failure of the rest of the building, then it would fall at close to free fall. If, as Judy Wood supposed, the gigantic impact of the top 1/5 was totally absorbed (an absurd scenario) then the building wouldn't fall down at all.

Exactly what force would make the lower 4/5 fail isn't obvious. The lower 4/5 was designed to withstand the stationary weight of the top 1/5. It wasn't designed to hold the top 1/5 once it was in motion.

[rwguinn]

Gumboot, and friends

Let me just give you a heads up,

No matter how many irrelevant "calculations" you perform, There is no way (using newtons laws) to get those towers (all three) to fall as fast as they did (13-14 sec)

You can calculate all force,pressure,energy (in megajules) you want(till your blue in the brain) but...

The top 1/5 of the north tower....is only just that! only 1/5th the total mass of the tower....(no matter how much "pressure" there is).....DUH

No matter how fancy your calculations are, there is just no way to get 1/5 the mass of the world trade center to plumett through 4/5 (of the mass of the world trade center) in 12 -14 sec.

Is Christophera back already?
This is totally unrealistice.

[Dan O.]

Has anyone done the simulation where you assume the entire mass of the buildings are in the floors and ignore the columns? The top of the building would begin falling at 1G and as the floors come together they would combine to form a common falling mass conserving the total momentum. At what height would the top of the building meet this common mass? How long does it take for all floors to reach ground level?

[natureboy]

Scuse me, I should have said,

"No matter how fancy your calculations are, there is just no way to get 1/5 the mass of the world trade center to plumett through 4/5 (of the mass of the world trade center) in 12 -14 sec." .....at sea level w/ the top 1/5 portion starting its descent from 1360 feet high, and hitting another 1/5 portion every 272 feet down, in sucession, in a vacuum, with no resistance offered by the other "portions" (exept) their own mass!

[AZCat]

actualyl i have it on fairly good authority that an asteroid dropped onto the undamaged towers would just drive them intot he ground like a nail :)

No! When the top of the asteroid [bangs table] hits the top of the building [bangs table] the NET FORCE IS ZERO! [bangs table repeatedly while waving his hand] It could crush some floors but it can't accelerate through the structurally sound portion of the tower!

[AZCat]

Has anyone done the simulation where you assume the entire mass of the buildings are in the floors and ignore the columns? The top of the building would begin falling at 1G and as the floors come together they would combine to form a common falling mass conserving the total momentum. At what height would the top of the building meet this common mass? How long does it take for all floors to reach ground level?

This is trivial and a number of people have done this. Judy Wood tried to, but totally screwed it up.

The following took me all of ten minutes in Excel:

http://img488.imageshack.us/img488/7540/billiardballsfixedjg6.jpg?pic

[Anti-sophist]

No matter how many irrelevant "calculations" you perform, There is no way (using newtons laws) to get those towers (all three) to fall as fast as they did (13-14 sec)


Prove it. Or rather, give us a scientifically compelling reason that this is true. Appealing to personal incredulity doesn't count.

[gumboot]

Scuse me, I should have said,

"No matter how fancy your calculations are, there is just no way to get 1/5 the mass of the world trade center to plumett through 4/5 (of the mass of the world trade center) in 12 -14 sec." .....at sea level w/ the top 1/5 portion starting its descent from 1360 feet high, and hitting another 1/5 portion every 272 feet down, in sucession, in a vacuum, with no resistance offered by the other "portions" (exept) their own mass!


Why not?

-Gumboot

[natureboy]

Prove it? OK

Ummm. lets see It seems I said ....."at sea level w/ the top 1/5 portion starting its descent from 1360 feet high, and hitting another 1/5 portion every 272 feet down, in sucession, in a vacuum, with no resistance offered by the other "portions" (exept) their own mass!"

(do you realize that I'm giving this to you on a silver platter in a vacuum with no resistance from the welded floor truss brackets)? ....just mass at rest as resistance!

Here goes!

One fifth of the mass of the world trade center begins falling... from a height of 1360 feet (above sea level) in a vacuum (on earth). It falls for 4.15 seconds,reaches a speed of almost 132.9 feet per second and has traveled 272 feet when BAM! ...It slams down on top of another 1/5 portion of the tower and its downward velocity has been reduced by HALF! to 66.45 fps.............


I'll stop there for a minute,.....check my work for me.(using newtons laws) I'll continue if...I havent made any mistakes so far. Ho hummmm...

[Furcifer]

Prove it? OK

Ummm. lets see It seems I said ....."at sea level w/ the top 1/5 portion starting its descent from 1360 feet high, and hitting another 1/5 portion every 272 feet down, in sucession, in a vacuum, with no resistance offered by the other "portions" (exept) their own mass!"

(do you realize that I'm giving this to you on a silver platter in a vacuum with no resistance from the welded floor truss brackets)? ....just mass at rest as resistance!

Here goes!

One fifth of the mass of the world trade center begins falling... from a height of 1360 feet (above sea level) in a vacuum (on earth). It falls for 4.15 seconds,reaches a speed of almost 132.9 feet per second and has traveled 272 feet when BAM! ...It slams down on top of another 1/5 portion of the tower and its downward velocity has been reduced by HALF! to 66.45 fps.............


I'll stop there for a minute,.....check my work for me.(using newtons laws) I'll continue if...I havent made any mistakes so far. Ho hummmm...

lol, it's very convenient you stopped there. i haven't even checked your figures, but assume there right. you're intial velocity was 0m/s, from rest. you have a new initial velocity of 66.45 fps (feet per second?, no self respecting scientist would use this figure, but whatever) so now you have the upper mass, plus a little extra, falling another 3.7M (free fall) until it hits the next floor. subtract the kinetic energy absorbed by the next floor before it collapses, 6.28ee8 J, from the kinetic energy of the new mass. i'm going to let you do this because you need to learn, and the rest of us need to see if you have any idea whats going on.

[gumboot]

What is with you Americans? Can't you use International Units like the rest of the planet? Ugh.

-Gumboot

[rwguinn]

What is with you Americans? Can't you use International Units like the rest of the planet? Ugh.

-Gumboot
Units is units

Now, is 1 mN a small force or a large torque?

[gumboot]

Prove it? OK

Ummm. lets see It seems I said ....."at sea level w/ the top 1/5 portion starting its descent from 1360 feet high, and hitting another 1/5 portion every 272 feet down, in sucession, in a vacuum, with no resistance offered by the other "portions" (exept) their own mass!"

(do you realize that I'm giving this to you on a silver platter in a vacuum with no resistance from the welded floor truss brackets)? ....just mass at rest as resistance!

Here goes!

One fifth of the mass of the world trade center begins falling... from a height of 1360 feet (above sea level) in a vacuum (on earth). It falls for 4.15 seconds,reaches a speed of almost 132.9 feet per second and has traveled 272 feet when BAM! ...It slams down on top of another 1/5 portion of the tower and its downward velocity has been reduced by HALF! to 66.45 fps.............


I'll stop there for a minute,.....check my work for me.(using newtons laws) I'll continue if...I havent made any mistakes so far. Ho hummmm...



This should be entertaining... continue.

-Gumboot

[gumboot]

Units is units

Now, is 1 mN a small force or a large torque?



Yes but our units are better, only stinky people use American units. I'm telling my mum on you.

-Gumboot

[gumboot]

Units is units

Now, is 1 mN a small force or a large torque?



You don't mean 1 N.m do you? (1 Newton metre of torque)

-Gumboot

[Newtons Bit]

Gumboot: The vast majority of the engineers on this forum work in inches and kips (thousands of pounds). We're old dogs. Can't teach us new tricks at all.

[Hokulele]

Unless you are measuring bridges. Those are done in smoots (http://en.wikipedia.org/wiki/Smoot).

[rwguinn]

You don't mean 1 N.m do you? (1 Newton metre of torque)

-Gumboot
see- The problem is torque is Distance (cross) Force.
You guys had to turn it backwards so that the units would work out. in* lb and lb*in are understandable either way! :D
that's what you get for naming every unit after dead guys.
And go tell mommy. I don't care. Wait till she hears you been baiting toofers and making people laught at them!:p

[Furcifer]

Gumboot: The vast majority of the engineers on this forum work in inches and kips (thousands of pounds). We're old dogs. Can't teach us new tricks at all.

one of my profs had a very funny story about the Mars Polar Lander probe he worked on. if you remember, it was lost 10 minutes before it was about to land. the reason it was lost? there was a small rounding error when they converted the work the US engineers did to SI. units.

[Newtons Bit]

see- The problem is torque is Distance (cross) Force.
You guys had to turn it backwards so that the units would work out. in* lb and lb*in are understandable either way! :D
that's what you get for naming every unit after dead guys.
And go tell mommy. I don't care. Wait till she hears you been baiting toofers and making people laught at them!:p

And when we talk mass, we use (lb * s^2)/in and we LIKE IT!

[gumboot]

Just so I make sure we're on the right page...

"foot" is the measurement from the current President's heel to his large toe... yes?

:boxedin:

-Gumboot

[Newtons Bit]

No, it is the precise measurement of some old english dead guy's foot. Whenever we need to calibrate our standards, we take out the old dried up decayed foot and whip out a calibration. For some reason, the distance of a foot has become smaller and smaller over the centuries. I wonder if it has anything to do with how well they perserved that foot...

[Furcifer]

Just so I make sure we're on the right page...

"foot" is the measurement from the current President's heel to his large toe... yes?

:boxedin:

-Gumboot

you're thinking:
"fool" is the measurement from the current President's head to his large toe...yes?

[rwguinn]

And when we talk mass, we use (lb * s^2)/in and we LIKE IT!
youbecha!
None of them slug things here. Feet are for walking on.
the towers were 1.632E4 inches tall.

[qarnos]

What is with you Americans? Can't you use International Units like the rest of the planet? Ugh.

-Gumboot

Haven't you heard The Stonecutters' song?

"Who keeps the metric system down? We do! We do!"

Hmmm... that song should be our NWO theme song. :D

[qarnos]

one of my profs had a very funny story about the Mars Polar Lander probe he worked on. if you remember, it was lost 10 minutes before it was about to land. the reason it was lost? there was a small rounding error when they converted the work the US engineers did to SI. units.

Are you sure you didn't mean Mars Climate Orbiter?.

The Polar Lander was lost because of a faulty sensor telling the computer it had already landed when, in fact, it was still falling through the air. Made for a kinda bumpy landing.

[Dan O.]

This is trivial and a number of people have done this. Judy Wood tried to, but totally screwed it up.

The following took me all of ten minutes in Excel:

http://img488.imageshack.us/img488/7540/billiardballsfixedjg6.jpg?pic

That's close but the collapse didn't start at the 110th floor. The collapse starts at the point of impact with the upper and lower structured both collapsing into the middle falling mass. The very top of the structure will free fall until it meets this middle mass. The total collapse time will be much closer to the free-fall time.

[chippy]

Don't know if anyone has mentioned this, but you're already halfway there to find the pressure since you already have the area. Pressure is force divided by area, and you said you already know the area is 45 m^2. So all you need is the force of the buildings. You know the mass of the buildings, so multiply that by the acceration due to gravity and you'll have your force (F = ma, or "mg" in this case). Divide that by area and you'll have your pressure.

That, however, is how you would define a static pressure. The falling of the building is a dynamic load that changes things a lot. I'll have to think about this some more.

[A W Smith]

actually the foot was derived from a multiple of inches which as i understand it was a measure of three grains of rice placed end to end. Why anyone would want to measure the length of rices instead of simply weighing is is beyond me.

Interesting story (or maybe not) a few decades ago maybe even three they built an entirely metric demonstration house as in Canada. Everything went well until they got to the gypsum board. Since it would have been prohibitively expensive to make a short run of gypsum board in anything but a 48 inch width module, they were stuck with a foot module in a metric house. Run horizontally which is typical for gypsum board in a residential application they were short of the ceiling. They had to cut a filler strip for the void between the bottom and top course of sheets. THOSE BUTCHERS!!

[gumboot]

Somehow I ended up talking about this in about three threads... :boxedin:

In this (http://forums.randi.org/showthread.php?t=81444&page=2) thread Myriad provided some useful information, and further down Gravy provided links to some studies which I will attempt to understand... :p

-Gumboot

[Furcifer]

Are you sure you didn't mean Mars Climate Orbiter?.

The Polar Lander was lost because of a faulty sensor telling the computer it had already landed when, in fact, it was still falling through the air. Made for a kinda bumpy landing.

oops, you're absolutely right. can't blame em all on the engineers. :)

[e^n]

One fifth of the mass of the world trade center begins falling... from a height of 1360 feet (above sea level) in a vacuum (on earth). It falls for 4.15 seconds,reaches a speed of almost 132.9 feet per second and has traveled 272 feet when BAM! ...It slams down on top of another 1/5 portion of the tower and its downward velocity has been reduced by HALF! to 66.45 fps.............

Ok fine, your example can be abstracted to 5 flat plates suspended at 1/5th increments, when the first plate collides with the second its speed is (according to you) 132.9fps. As the second plate is identical, p=mv means when you double the mass you half the velocity. Fine, now the second plate and first plate begin to fall, as they hit the third plate their velocity is now 199.35fps using your math. Now the impact only adds 1/3 of the mass so conserving momentum the resultant speed is 132.9fps, in the time it takes to hit the fourth plate, your initial mass is now at twice the speed it hit the first plate at.

Would you like me to calculate the total collapse time or would you like to revise your model?

PS. Not an engineer, probably wrong 8)

[Dan O.]

This is trivial and a number of people have done this. Judy Wood tried to, but totally screwed it up.

The following took me all of ten minutes in Excel:

http://img488.imageshack.us/img488/7540/billiardballsfixedjg6.jpg?pic

Ok, I did the spreadsheet thing for the upper section free falling into the collapsing pancake which is collecting floors from both the fixed structure below and the falling structure above. With the collapse starting from the top, the total collapse time is about 14.8s. Starting on about the 95th floor the total collapse time is about 12s. And starting from about the 80th floor the collapse time is about 10.3s. These are approximate because I may be off by 1 counting floors up or down.

Starting anywhere below the 66th floor and the tower collapses is about 9.2 second and is limited by the free fall time of the top floor.

[AZCat]

Ok, I did the spreadsheet thing for the upper section free falling into the collapsing pancake which is collecting floors from both the fixed structure below and the falling structure above. With the collapse starting from the top, the total collapse time is about 14.8s. Starting on about the 95th floor the total collapse time is about 12s. And starting from about the 80th floor the collapse time is about 10.3s. These are approximate because I may be off by 1 counting floors up or down.

Starting anywhere below the 66th floor and the tower collapses is about 9.2 second and is limited by the free fall time of the top floor.

Cool. When I originally did the graph, it was in response to a Judy Wood graph that started at floor 110, so I wanted to keep that consistant between the two (even though we all know the collapse didn't start there).

As you probably figured out, it's not a hard calculation at all.

[Dan O.]

Cool. When I originally did the graph, it was in response to a Judy Wood graph that started at floor 110, so I wanted to keep that consistant between the two (even though we all know the collapse didn't start there).

As you probably figured out, it's not a hard calculation at all.

It just took me a while to remember how to use the old quadratic formula to calculate the collision time for each impact.

I still need to check my numbers. I thought I saw somewhere that free fall from the top was 8.2s but my calculation came up with about 9.2s. Maybe if a UFO were flying overhead it would exert a negative gravity pushing the building down faster. :)

[R.Mackey]

I'm doing some simple equations regarding the WTC collapses. I just want to make sure I'm going in the right direction.

I'm wanting to calculate the force applied to the first intact floor, as spread over the combined footprint of the 47 core columns and 240 exterior columns.

The value for the footprint is 45m2

(This is based on some assumptions that are detailed in my calculations)

I calculated the ENERGY of the impact based on the calculated mass and instantaneous velocity of the upper masses at impact and came to 9.7GJ (WTC1) and 30.3 GJ (WTC2).

This gives us a distribution of 215 MJ per square metre (WTC1) and 673 MJ per square metre (WTC2).

The problem here is to calculate PRESSURE I need FORCE, not ENERGY.


You can approximate this as an impulse problem. Impulse I = F dt = m dv.

You know the mass (m) and velocity after falling that is all dissipated (dv) of the upper block. This gives you the total impulse the upper block exerts on the lower block, and vice versa.

To get force, you then need to estimate what the time dt is for the impact to be dissipated. This involves some hand waving.

Assuming both blocks stayed intact, the time dt would be approximately the length of time it took the pressure wave to travel through the upper block -- treating the impact as an extremely large phonon phenomenon. The time is equal (roughly) to the height of the block times the sound speed of the material, or 5100 meters per second for steel. Not a very long time.

From this you can estimate the total force F while the momentum is being dissipated, added of course to the static load, and from that and the contact footprint you can estimate the pressure.

However, take this derivation for what it is, namely an extremely simplified model. In the real case, we don't expect the two blocks to hit each other squarely, so the force exerted on any given contact point will vary wildly. I would not base any firm conclusions on so coarse an estimate.

Also, in computing the likelihood or not of collapse, what's more relevant is the ability of the structure to absorb the impact in terms of total strain energy, which is related to the integral of force, although both force and impulse are relevant especially for fast phenomena. NIST discusses this in its Simplified Model, chapter 10 of NIST NCSTAR1-2B, which would probably be a good thing to review.

[AZCat]

Assuming both blocks stayed intact, the time dt would be approximately the length of time it took the pressure wave to travel through the upper block -- treating the impact as an extremely large phonon phenomenon. The time is equal (roughly) to the height of the block times the sound speed of the material, or 5100 meters per second for steel. Not a very long time.
(emphasis mine)

I may be nit-picking, but shouldn't your second-to-last sentence be "The time is equal (roughly) to the height of the block divided by the sound speed of the material, or 5100 meters per second for steel." Multiplying height by speed would give units of length^2/time, whereas dividing gives units of time.

[R.Mackey]

I may be nit-picking, but shouldn't your second-to-last sentence be "The time is equal (roughly) to the height of the block divided by the sound speed of the material, or 5100 meters per second for steel." Multiplying height by speed would give units of length^2/time, whereas dividing gives units of time.

Yup, you're right.

This is why we carry our units and show our work!

[AZCat]

Yup, you're right.

This is why we carry our units and show our work!

Cool - thanks. I like it when things make sense.

[Anti-sophist]

Also, in computing the likelihood or not of collapse, what's more relevant is the ability of the structure to absorb the impact in terms of total strain energy, which is related to the integral of force....

Since we've basically reached the limit of my knowledge of this type of physics, it's question&answer time.

In the simplest mental experiment, two objects impacting squarely, and determining whether one "breaks": my initial "guess" at modeling this scenario involved modeling "breaking" as the maximum force a particular object could withstand. The sentence quoted above seems to imply that is not the right approach. Is my simplistic approach flawed? What are you integrating force over? Time? That is a momentum quantity, then? Distance, maybe?

Under the presumption that maximum force is a valid criteria for "breaking", the problem becomes even more difficult because an impulse calculation would result in an average force, not a maximum force. So that presents a new problem of determining the max-F from the average-F. Unless, of course, we can express F(t) in some analytic fashion? If we model the collision as a shock-wave, is it linear? How much does elasticity (or lack there of) effect such a guesstimate?

[Newtons Bit]

Gah, can't think. Suffice to say, the resisting force of the member is variable with the displacement of that member.

[R.Mackey]

In the simplest mental experiment, two objects impacting squarely, and determining whether one "breaks": my initial "guess" at modeling this scenario involved modeling "breaking" as the maximum force a particular object could withstand. The sentence quoted above seems to imply that is not the right approach. Is my simplistic approach flawed? What are you integrating force over? Time? That is a momentum quantity, then? Distance, maybe?

I was integrating over the distance, i.e. the amount something was squashed.

When you start talking about solid mechanics, namely treating the interior of materials rather than their boundaries only, the notion of "force" is less important than its replacement, known as "stress." Stress (http://en.wikipedia.org/wiki/Stress_%28physics%29) is a volumetric quantity that describes how much force there is at any point within a solid, and it can vary depending on where you are in the solid. Stress is defined in terms of force per area, much like a pressure, although unlike pressure the direction of the stress vector is not always normal to the boundary.

To take an extremely simple example, consider a homogeneous rectangular solid of cross-sectional area A, lying at rest beside an immovable object. If you press on the other side of the block with force F, the block doesn't move, because the immovable object has a reactive force also equal to F. If you look inside the block, the stress at any point in the block is equal to F / A, because if you were to cut into the block, no matter where you do so, you will have opposing forces of strength F at the boundary. Thus we say the stress σ = F / A.

In the example above, the stress is a "normal stress," because the stress is normal (viz. perpendicular) to the boundaries of the solid. You can also have shear stresses, caused for example by friction, but many analyses can be simplified to one dimension where only normal stress is used.

As my old solid mechanics professor once explained, "stress is momentum flow." This is more correct than it may sound. If you draw a loop through any solid, the total flux of stress through the loop is equal to the total force across that boundary, and force F = dp / dt according to Newton.

Under the presumption that maximum force is a valid criteria for "breaking", the problem becomes even more difficult because an impulse calculation would result in an average force, not a maximum force. So that presents a new problem of determining the max-F from the average-F. Unless, of course, we can express F(t) in some analytic fashion? If we model the collision as a shock-wave, is it linear? How much does elasticity (or lack there of) effect such a guesstimate?

Good intuition. Usually we don't think of there being a maximum force, but rather a maximum strain. Strain (http://en.wikipedia.org/wiki/Strain_%28materials_science%29) refers to how much a material under stress stretches or squishes in response. Strain is a dimensionless quantity, since it refers to (change in length) / (length).

Strain and stress, in the one-dimensional case, are related (http://en.wikipedia.org/wiki/Stress-strain_curve) by a constant called the Young's Modulus (http://en.wikipedia.org/wiki/Young%27s_Modulus). In the simple example above, we write σ = E ε, where E is the Young's Modulus and ε is the strain. This equation looks like the classical spring equation named after Hooke (http://en.wikipedia.org/wiki/Hooke_law), i.e. F = k x, and for good reason -- ordinary materials in one-dimensional strain are springy, at least for a while. Apply a force and they compress; release the force, and they spring back to their original shape, until the force is so large that the solid is no longer elastic and enters its plastic (http://en.wikipedia.org/wiki/Plasticity_%28physics%29) phase.

So now it's time to answer your question. For any given material, we know how much strain it can handle before it breaks. We can then work backwards to find the stress resulting in this much strain, known as the Ultimate Strength (http://en.wikipedia.org/wiki/Ultimate_strength), beyond which a solid will flow until it comes apart unless the force is removed very quickly. Multiply the ultimate strength by the cross-sectional area (or integrate over area if you must), and you have the total amount of force a solid can take before it fails. Similarly, you can think of a certain "strain energy," equal to the product of stress along the strain direction (force times distance when summed over the whole body) that the solid can contain before it ruptures.

Impact modeling, however, introduces a new wrinkle -- rather than gently loading a solid, you're now creating stress waves (http://en.wikipedia.org/wiki/Seismic_wave). This can lead to strains varying wildly within the solid, and thus pieces of it can fail independent of others. How important this is to the fracture mechanics depends on how fast the impact occurs and much more detailed understanding of the solid's behavior. This is a very difficult thing to treat in general, and you need to choose your simplifications carefully, depending on the subject.

[Anti-sophist]

Actually, that all makes a good deal of sense and is surprisingly intuitive. It has always amazed me how "right" the right physics feels.

Suffice to say, from my point of view, that my original statement holds. Attempting, in any way, to analytically and confidently answer the "should the building have held" question seems, shall we say, non-trivial.

[westprog]

Since we've basically reached the limit of my knowledge of this type of physics, it's question&answer time.

In the simplest mental experiment, two objects impacting squarely, and determining whether one "breaks": my initial "guess" at modeling this scenario involved modeling "breaking" as the maximum force a particular object could withstand. The sentence quoted above seems to imply that is not the right approach. Is my simplistic approach flawed? What are you integrating force over? Time? That is a momentum quantity, then? Distance, maybe?

Under the presumption that maximum force is a valid criteria for "breaking", the problem becomes even more difficult because an impulse calculation would result in an average force, not a maximum force. So that presents a new problem of determining the max-F from the average-F. Unless, of course, we can express F(t) in some analytic fashion? If we model the collision as a shock-wave, is it linear? How much does elasticity (or lack there of) effect such a guesstimate?


Part of the problem - imagine that each floor is simply resting on a number of little shelves, with gaps between them. The shelves have enormous strength. If the descending mass is going exactly straight down, it will be held. If it is even slightly skewed, it will dislodge the floor which will move into the gaps between the shelves, and start falling.

Obviously the WTC wasn't designed like this, but as a thought experiment, it shows the difficulty in calculating the forces required to dislodge a floor. Any calculation will involve a number of assumptions, which may or may not be true. The Judy Wood approach of making up an arbitrary number and working from there is quite tempting.