The presentation was supposed to take place on August 7
at 10.15 am - 12.15 pm at 206 Arch Room of the Norris Center
of the Northwestern University (Session 5.7).
The top part C of any structure is
originally carried by four parts A1, 2, 3 and 4
below one above the other of similar structure; even
if lowest part A4 is four times stronger than
A1 and five times stronger than top part C.
A1 is just double as strong as C. Then top
part C is given downward motion due to gravity (some
supports between parts C and top A1 are
buckled) and top part C applies, when impacting
A1, released potential energy on the top part
A1 that is crushed down into rubble B -
POUFF - as top A1 cannot absorb the energy
applied by C! The part A1 becomes smoke and
dust or rubble B or whatever. It cannot really be
seen due to the smoke and dust. Then the rubble B
or the smoke (?) is accelerated by the top part C
and gravity and more released potential energy is applied
on the second part A2 that is crushed into more
rubble B - another POUFF - as second part
A2 cannot absorb the energy applied! The second
part A2 becomes more smoke and dust or more
B or whatever. It cannot be seen either. Smoke,
you know! The crushing down will be repeated as many
times as is required to crush parts A3 and
A4 into more rubble B or smoke and dust. In this
example the third and fourth parts A3 and
A4 thus become smoke and dust or even more
B or whatever. Then all A parts are rubble
B because all A parts could not absorb the
energy applied by top part C and rubble B!
Then the rubble B crushes up top part C
into more rubble B - the second final
POUFF! The last POUFF is that B
becomes a pyramid shaped heap of rubble. Bažant, Z. P has
confirmed above nonsense several times in peer reviewed
scientific papers! It cannot be demonstrated in a
laboratory or on a building site - so you have to believe
Bažant, Z. P.
It is however very easy to demonstrate theoretically,
numerically, in a laboratory and/or in real scale that
any man- made structure A cannot globally collapse
by itself from top down, when a small top part C
of A drops on and impacts remainder of A by
gravity from above. Reason is simply that such a man-made
structure is always stronger at bottom than top.
The hazard, risk or danger of global collapse from top
down does not exist in the real world!
This is good news. It will assist US authorities and
learned societies like ASCE and EMI, Engineering
Mechanics Institute, to improve safety against
progressive collapse of structures
everywhere without too much effort.
The risk does not exist.
I will explain it using a very simple model.
A simple 2D model of any
A structure is evidently three dimensional, 3D, but it
can be represented in two dimensions, 2D, by a line or
chain of connected elements, e.g. springs, at nodes that
in turn are subject to forces acting on a mass of the
node. The springs compress longitudinally and
transversely, 2D. The findings evidently also apply in
Let us therefore consider a tower structure of
N floors (nodes), each of mass m,
labeled from top (i=1) to bottom (i
· Adjacent floors (nodes) are separated
· Floor n is supported by n
· The spring ends are fixed to the floors.
Let's assume the spring itself fails in overload prior
to its end joint connections.
All springs are weightless, have rest length
L (m), and are perfectly elastic with
constant stiffness k (N/m) until compressed
longitudinally and laterally, 2D, a critical longitudinal
displacement , when they compress plastically another
displacement, after which they finally fail (and the
length becomes 0).
The springs represent the load carrying elements of
the tower structure, e.g. external walls, pillars and
The tower structure is placed in a vacuum and fixed to
a very large inertial mass, i.e. ground and subjected to
a uniform gravitational field directed downwards.
g = 9.82 m/s².
The structure is in static equilibrium if
Fn = nkx = nmg (N), x = mg/k.
In static equilibrium the displacement x is
everywhere the same and the same is true for the strain
e =x/L and stress s
(N/m² or Pa) in the spring.
The n springs below floor n can each
support a load of FC = kxC
before plastic deformation starts. If the actual design
load of a spring is F, then the safety factor
S = FC/F.
Elastic and plastic strain
The work performed in compressing a spring elastically
is E = kx²/2 (J).
With x = 0.01L, which is typical for a
pillar in a tower structure, a spring can elastically
absorb ES =
If a spring is overloaded elastically and starts to
deform plastically, the spring will develop plastic
hinges which absorb more energy EP and
which takes a certain time tP - and its
effective length finally becomes 0 - it has broken!
Let's assume that EP =
ES. Actual values are easy to establish in
Assume that the tower has N = 110
floors, each with mass m = 3.6 Mkg (i.e. 3 600
tons) and that L = 3.7 m.
It means that the tower is 407 m tall, like WTC 1 or
The total number of springs in a tower of
N floor is
Thus a tower with 110 floors consists of 6 105
springs. The top floor (or roof) is supported by 1
spring and the bottom floor is supported by 110
The assembly of the 13 uppermost floors contains 91
springs, while the assembly of 97 lowest floors contains
6 014 springs.
Let's call the assembly of the 13 uppermost floors
top part C and let's remove the 13 bottom springs
Top part C then consists of 13 m
connected by 78 springs (1 between top two m, 12
between lowest two m). The 78 springs in top
part C can together absorb
EC=78ES. Top part
C weighs 48 800 tons.
Let's call the assembly of the 97 lowest floors bottom
A's 110 lowest springs are connected to ground.
The 6 014 springs in A can together absorb
part A weighs 349 200 tons and is 7.2 times
heavier than part C. If energy absorption is a
measure of 'strength', A is 77.1 times
'stronger' than C.
Spring details - same
Static force F in every spring is F = mg =
The spring may have cross area of about 0.5 m²,
if it is of steel. The static stress in the spring is
then about 70 MPa. Note that the static stress is same
everywhere in every spring - at top and bottom of the
Let's assume S = 3, i.e. the springs will
commence plastic deformation at 210 MPa stress. The
material of the spring is steel.
Note that the total cross area of removed springs
below top part C is 6.5 m² (13 springs each 0.5
m²) and that the total cross area of springs at
bottom of bottom part A is 55 m² (110 springs each
0.5 m²). The structure evidently gets
'stronger' with more springs added further
The critical force FC of a spring is
FC = 106.056 MN. After that it deforms
plastically absorbing more energy and soon has length
Let's assume that spring stiffness k = 3 GN/m
that is typical, when core and perimeter wall structure
of WTC is replaced by one spring (a bundle of steel
elements) with cross area 0.5 m² that can deform in
3-D. Actual k is easy to establish in a
xC = 0.037 m
ES = k(xC)²/2 =
x = 0.037/3 m
E = kx²/2 = 0.228 MJ
When the tower structure is in static equilibrium,
each spring is compressed 0.037/3 m and 0.228MJ energy
E is stored in it elastically. Our 110 floors
structure is therefore elastically compressed 1.357
meters. However a spring can be compressed elastically to
xC = 0.037 m (or 1% of L)
before it starts to deform plastically and the energy
EC required to compress it is then
From an energy absorption point of view factor of
safety is 9 (actually static S²). All 6 105
springs in our structure can elastically absorb 12.53
Top part C can absorb elastically totally 78 x
2.053 = 160.1 MJ. Bottom part A can absorb
elastically 6014 x 2.053 = 12.35 GJ.
By dropping top part C a certain distance, e.g.
L, a certain amount of potential energy
ED is released, where ED
= 13mLg = 1.7 GJ. It is 13.6% of
what the tower itself can absorb elastically.
By simple structural damage analysis you can establish
whether top part C can damage A, ground or
Structural damage analysis
experiment 1: Structure top part C
collapses from bottom up
In experiment 1 is described how something weak, top
part C, dropped on rigid ground will be affected
by the impact.
The top part C assembly of 13 m is
dropped on ground from distance
At the impact C/ground total 1.7 GJ is applied
to ground and C. The ground does not damp the
impact. It is rigid or solid and can absorb plenty
energy. Evidently top part C itself damps the
impact - it becomes compressed and damaged:
As top part C is 44.4 m tall and consists of 13
m separated by springs, it is the bottom floor
m of C that physically contacts ground and is
arrested by ground at the impact.
The remaining 12 m above continue to displace
down and compress the springs below. A certain damping
takes place, when the springs compress elastically and
As the dynamic forces acting on top part C and
ground at impact and later are equal and opposite (the
dynamic force is the energy applied
(0.5ED ) divided by the displacement of
the force - the structure is compressed and maybe
damaged), it follows that top part C will absorb
0.5ED and rigid ground will also absorb
0.5ED in the impact.
It would then appear that 0.85 GJ energy is applied on
top part C one way or another and as top part
C can only absorb 0.16 GJ elastically and 0.32 GJ
plastically (or 0.48 GJ totally), all springs in top part
C will fail.
Ground is rigid and undamaged and can easily absorb
Top part C is, as seen, not very strong, and it
is why its springs are 100% broken at impact with
In what order will the springs in top part C
It can be seen on videos of controlled demolitions of
buildings, where the bottom supports are destroyed first
and structure above drops and hits ground that
destruction is from bottom up, thus:
The bottom 12 springs fail first at impact with
ground, 12 floors m above then drop down
L, more potential energy is released, 11
springs fail at second impact, 11 m drop
down L, 10 springs fail, 10 m drop,
etc, etc, until the last one top spring fails and the
last roof m impacts ground from
Top part C, 12 L tall, is
destroyed from bottom up in 12 steps that take a certain
time. Total energy released due to failed springs after
initial impact C/ground, i.e. when C is
destroyed, is 72mLg or 8.82
GJ (it was stored in C prior impact) and all of it
is absorbed by the ground.
You can say that top part C is crushed-up by 13
impacts with solid ground. Bažant & Co suggest in
their ridiculous, peer reviewed papers published in
Journal of Engineering Mechanics that weak top part
C is rigid and remains intact, when impacting,
e.g. stronger bottom part A, and that is one false
idea of Bažant later adopted by NIST as true.
Structural damage analysis
experiment 2: Structure A loaded on
top deforms elastically
In experiment 2 is shown how bottom part A
behaves, when statically loaded from above.
Bottom part A is thus positioned on ground
A is, as already described, a tower structure
of 97 m and 97 L tall. It consists
of 6 014 springs that can absorb elastically totally
12.346 GJ. 1/9th of this energy is already stored
statically in the tower so another 8/9th or 10.974 GJ can
be applied and absorbed elastically.
Considering plastic deformations another 10.974 GJ can
The top floor m of A is supported by 14
springs located on the second floor m below that
is supported by 15 springs further below, etc.
The 14 top springs of bottom A can elastically
absorb totally 28.74 MJ but are statically only loaded by
one m. You can add another 13 m (i.e.
C) on the 14 springs and then they are under
original design load, like all the other springs in
Thus, just adding 13 m on top of A
nothing special will happen except that all springs in
A are again under original design, static
What happens if, in lieu of slowly putting 13 m
on A we drop 13 m on A from
L = 3.7 meter and we let 13 m impact
A! Will the dynamic force at impact crush A?
Structural damage analysis
experiment 3: Bottom part A,
impacted on top, damps impact due to elastic and plastic
In experiment 3 is shown what happens, when a rigid
mass of 13 m impacts A from above. 13
m are dropped on the top m of A from
L = 3.7 m.
The 13 m are connected together without any
interconnecting springs, and that assembly is here
called D. D is one rigid mass of 13
At impact D/A 1.7 GJ is applied to A and
D and, as in experiment 1 0.85 GJ is applied to
A can totally absorb 10.974 GJ so you would
expect rigid D to bounce on A. The dynamic
impact force if applied on A/D is, as
stated above, simply the energy applied divided by
displacement of force during impact/compression that
takes a certain time. The initial impact will be followed
by more impacts, if further m gets loose and
drops, but energy released in each impact will be
elastically and maybe plastically absorbed by intact
The beauty of a spring is that it can absorb energy
multiple times, when loaded in succession and that this
absorption takes time. The spring acts like a shock
absorber transmitting energy to other springs and
It is very strange that NIST suggests without
any evidence in its 10 000 pages 911-report that little,
weak top part C (or rigid D) can apply
energy on big, strong bottom part A that A
cannot absorb! The figures here indicate something
completely different! Why does NIST lie and spread
false information to the public. Is it in order to
It may be argued that the top 14 top springs and the
15 springs in the next layer of A below may be
destroyed locally in overload by the dynamic forces at
impact with D and that some extra energy released,
when D and one or more loose m displace
The 14 top springs of A can totally absorb
elastically 14 x 2.053 = 28.74 MJ and maybe plastically
totally say 57 MJ and the next 15 springs about the same.
The plastic destruction (failure) of springs takes time,
so in the mean time the dynamic impact force (i.e. energy
divided by displacement) can be absorbed elastically by
intermediate springs and transmitted to ground (as a
seismic wave). When one layer of springs is destroyed all
m above displace down L and more
energy is released - a second impact - and has to be
absorbed by intact springs like a shock
So D applies 850 MJ on A and about 114
MJ can be absorbed by destruction of the two top layers
of springs in A and the rest is absorbed
elastically by 95 other layers of springs in A and
transmitted to ground.
That D would destroy all 6 014 springs of
bottom part A is unlikely. The springs of A
will dampen the impact of D and further loose top
m of A dropping, while only some local
failures occur close to interface D/A.
It is quite easy to verify experiment 3 in a
laboratory. Just take the top C of any tower
structure, compress it to a rigid block D, and
drop D on the bottom part A and see what
Rigid D will always bounce and stop after
producing some local failures at top layers with few
springs of A, i.e. the weakest part of
Structural damage analysis
experiment 4: Small top C cannot
crush a bigger bottom
In experiment 4 is shown what happens, when top part
C impacts bottom part A from
Top part C is thus dropped on bottom part
A from L = 3.7 m. This is the
infamous WTC 1 event.
13 top floors m of WTC 1 drop on 97 intact
floors/columns m below and according videos of
suspect origin the 97 floors/columns below are destroyed
in a fountain of smoke, dust and debris - terrible - as
shown 'live on TV' September 11, 2001! In reality,
of course, it cannot happen.
At impact C/A 0.85GJ is applied to C
(with 12 springs at bottom) and 0.85GJ is applied to
A (with 14 springs at top) as explained
However, C does not impact rigid ground as in
experiment 1 and A is not impacted from above by
rigid D as in experiment 3.
In fact only the top m of A supported by
14 springs below and the bottom m of C
supported by 12 springs above contact each other in the
impact and the dynamic forces are then transmitted via
the springs to other m in A and C
via springs that behave elastically and dampen the
impact. The impact, like in experiment 3, will be split
in sub-impacts when/if further floors m gets loose
and drops but energy released in each sub-impact will be
elastically and plastically absorbed by the
So in experiment 4 the initial impact will really be
dampened, i.e. takes longer time, as both A and
will dampen (absorb the
energy of) the local impact C/A. It also means
that the dynamic forces are reduced. That small/weak
C will crush big/stronger A at increasing
speed and by gravity is impossible.
That C - that can absorb much less energy
elastically and plastically than A - can apply,
via dynamic forces at impacts, and release, via
structural/spring failures, more energy on A and
destroy A is impossible: C will destroy its
own springs first, before A is starting to get
destroyed and then C cannot apply or release more
energy to destroy A.
In 3D reality there will only be some local failures
at interface C/A at impact, elements of C
and A then get locally entangled, friction
develops and loose top part C will then just
bounce on top of A. A arrests C!
There is not enough energy for anything else. In 2D model
world the top line will evidently just pass by the bottom
As seen above 0.85 GJ energy is applied to 12 bottom
springs in C (and to remaining 66 springs above)
and 0.85 GJ energy to 14 top springs in A (and to
remaining 6 000 springs below) at impact
What happens if 0.065 GJ energy E is applied to one
spring with stiffness k = 3 GN/m and 3.7 m length
L? Answer: the spring will compress
x = 0.147 m (as x² = E/k) due
to the impact or 4% L. As one spring in our example can
only elastically compress 1% it means that the spring
plastically deforms and breaks at impact.
However, our spring is not alone but supported by
other springs above and below in the structure so you
have to consider that.
Evidently the 6 014 springs in A can easily
absorb totally 0.85 GJ energy elastically (as shown in
experiment 3). If the 78 springs in C can do it,
is another matter (as shown in experiment 1).
It is quite easy to verify experiment 4 in a
laboratory. Just take the top part C of any
structure, and drop it on the bottom part A and
register what happens. My experience is that C
always bounces on and is arrested by A, but I may
be wrong. I have only tested a limited amount of towers.
No smoke, dust, debris or ejections were produced though,
when dropping C on A.
Many people believe that scale or size of the
structure or the matters, e.g. that a small model of a
structure cannot crush itself but that a bigger structure
can or that material matters, e.g. that a structure of
brittle elements/connections will collapse but not a
structure of more ductile elements/connections. However,
to believe things like that is unscientific, terrorist
The experiment 4 impact, elastic compression of
springs and damping of parts can of course easily be
modeled mathematically using Finite Element Methods for
any size of tower springs
A linear spring-damper model of the form f(t) =
k*x(t) + c*v(t), where x = input displacement,
v = input velocity, and f(t) = output force
can be developed based on test data in the time domain of
The term k is the spring stiffness (N/m) and
c is the viscous damping coefficient
With k = 3 GN/m and c= 0.3 GNs/m the
tower parts A and C become flexible and
will visibly deform, compress, oscillate, be damped, for
several seconds after impact C/A.
Plastic deformation and it's time to develop failures
of a spring are more complex to model mathematically (but
it can be done).
That a 407 meter tower structure will explode in
smoke, dust and debris, rubble being formed and collapse
from top taking place in 20 seconds as shown 'live on
TV' Tuesday morning 11 September 2001 in the USA is
evidently not possible in reality.
What was shown 'live on TV' was just a stupid
movie made by disaster animators Hollywood style using
computer generated images, CGI! Imagine that!
The writer's attempts to crush a structure A by
dropping its top C on it have, naturally, always
ended up with no springs, elements or connections failing
in A and C and only bouncing/arrest of
C taking place.
The writer has never seen a top part C of a
tower structure impacting and destroying the bottom part
A due to gravity. Reason is that such destruction
is physically impossible! A always arrests
C. Anyone suggesting something else or shows it
'live on TV' or on below figure is a simple
M.Sc., Heiwa Co, European Agency
for Safety at Sea
November 12, 2012
Appendix - Comments by world famous civil engineers
not agreeing with above.
I have only a good university degree and 40+ years
experience of steel structural design, analysis,
construction, inspection, maintenance and repairs in the
marine and off shore fields, where structures are subject
to more severe conditions than the WTCs.
I have thus discussed my 2D findings above with many
world famous civil engineers that suggest that above is
not applicable to WTC 1 or 2 because they were particular
cases just bolted together in 3D and it wasn't the wall
supports - the springs - that failed but all the bolts
holding floors and walls together that sheared off floor
and the whole thing really collapsed as
seen live on TV. (http://www.cbsnews.com/2100-205_162-527157.html)
The WTCs were some sort of houses of cards structures
- with bolts though - just waiting for 30 years to be hit
by a hi-jacked plane up top, catch fire and suddenly
These world famous civil engineers better remain
I have therefore 2010 started a Challenge - the
Heiwa Challenge at http://heiwaco.tripod.com/chall.htm
. It is very simple.
Anybody that (1) can describe a structure A
with a top C and (2) show how top C crushes
bottom A by gravity receives € 1 000 000:-
I really looked forward to presenting above paper at
Evanston 5-7 August but the EMI invitation was withdrawn
July 28. If any engineer, though, finds that my conclusion
is wrong, she/he has the possibility to win my €1M
Challenge. Sorry I couldn't attend at Evanston!