Moderated Continuation - Why a one-way Crush down is not possible

you left an extra word in there that made your post nonsensical.

bill smith said:
So it is irrefutably established that the core columns will collapse solely under their own weight..
Click to expand...

fixed that for you. I realize what you were trying to say. Because you couldn't possibly be so stupid as to think the core columns would remain standing unbraced by the floors.
 
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bill smith said:
You may be cheap. I am not.
Click to expand...


Now, try something completely different. Take a deep breath and announce that you were wrong about the ability of unbraced columns to project 1,300 feet into the air without buckling.

C'mon, Bill, just do it! It will change your life.
 
Let's assume a massive 1000' tall steel core column cross-section: 60" x 36" x 1" thick.
b = 36"
h = 60"
t = 1"


It has a moment of inertia of:
I = (b * h^3 / 12) - [(b-2*t) * (h-2*t)^3 / 12]
I = (36 * 60^3 / 12) - [(36-2) * (60-2)^3 / 12]
I = 95182in^4

It has a radius of gyration of:
r = SQRT(I/A)
r = SQRT(95182in^4/188in^2)
r = 22.5in

It is fixed at the base and free at it's top
slenderness ratio = KL/r
K = 2.0 for this particular case.
slenderness ratio = 2.0 * 12,000in/22.5in
slenderness ratio = 1066

Let's also pretend that it is a compact section (which it's not) for ease of the critical buckling strength calc:
Pn = Fcr*A
Fcr = 0.877Fe
Fe = 3.14^2 * E / (slenderness^2)
E = modulus of elasticity = 29000ksi
Fe = 3.14^2 * 29000ksi / (1066^2)
Fe = 0.252ksi
Fcr = 0.877*0.252ksi
Fcr = 0.221ksi
Pn = 0.221ksi*188in^2
Pn = 41.5kip (or 41,500lb)

The selfweight of this particular column is 639,722lb.
Since the self-weight is more than the critical buckling strength (an order of magnitude I might add) the column falls over.
 
FYI, that's STRONG axis buckling, I would have used h = 36" and b = 60" if I was actually designing this in real life.

Edit: oh and all of the formula above can be found in the AISC Manual of Steel Construction (13th edition).
 
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Newtons Bit said:
Let's assume a massive 1000' tall steel core column cross-section: 60" x 36" x 1" thick.
b = 36"
h = 60"
t = 1"


It has a moment of inertia of:
I = (b * h^3 / 12) - [(b-2*t) * (h-2*t)^3 / 12]
I = (36 * 60^3 / 12) - [(36-2) * (60-2)^3 / 12]
I = 95182in^4

It has a radius of gyration of:
r = SQRT(I/A)
r = SQRT(95182in^4/188in^2)
r = 22.5in

It is fixed at the base and free at it's top
slenderness ratio = KL/r
K = 2.0 for this particular case.
slenderness ratio = 2.0 * 12,000in/22.5in
slenderness ratio = 1066

Let's also pretend that it is a compact section (which it's not) for ease of the critical buckling strength calc:
Pn = Fcr*A
Fcr = 0.877Fe
Fe = 3.14^2 * E / (slenderness^2)
E = modulus of elasticity = 29000ksi
Fe = 3.14^2 * 29000ksi / (1066^2)
Fe = 0.252ksi
Fcr = 0.877*0.252ksi
Fcr = 0.221ksi
Pn = 0.221ksi*188in^2
Pn = 41.5kip (or 41,500lb)

The selfweight of this particular column is 639,722lb.
Since the self-weight is more than the critical buckling strength (an order of magnitude I might add) the column falls over.
Click to expand...


There is an engineering lesson embedded in these arcane numbers. I can't follow this, and I'm probably not alone. Bill has no interest in learning anything, but some of us would like to understand what you've just written. Could you dumb it down a bit?
 
FineWine said:
There is an engineering lesson embedded in these arcane numbers. I can't follow this, and I'm probably not alone. Bill has no interest in learning anything, but some of us would like to understand what you've just written. Could you dumb it down a bit?
Click to expand...

I thought I was =[

I'll quote my own post later and walk you through it.
 
FineWine said:
There is an engineering lesson embedded in these arcane numbers. I can't follow this, and I'm probably not alone. Bill has no interest in learning anything, but some of us would like to understand what you've just written. Could you dumb it down a bit?
Click to expand...
Repost of post #1465. How does this slot in ?

' Look at it this way. Take just one outer core column and picture the floors runnng to at at about 12 foot intervals for 110 floors up to a height of 1300 feet.Do you think that puts a fair amount of vertical stress on the column ?

It still stands though. Now remove all the floors. Will the column now collapse under it's own weight ? (I'm not talking about stability here, only the fact that the column is strong enough to remain standing) '
 
bill smith said:
Repost of post #1465. How does this slot in ?

' Look at it this way. Take just one outer core column and picture the floors runnng to at at about 12 foot intervals for 110 floors up to a height of 1300 feet.Do you think that puts a fair amount of vertical stress on the column ?

It still stands though. Now remove all the floors. Will the column now collapse under it's own weight ? (I'm not talking about stability here, only the fact that the column is strong enough to remain standing) '
Click to expand...

Somehow, all of Bill's gedanken experiments lack something.
 
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Newtons Bit said:
Let's assume a massive 1000' tall steel core column cross-section: 60" x 36" x 1" thick.
b = 36"
h = 60"
t = 1"


It has a moment of inertia of:
I = (b * h^3 / 12) - [(b-2*t) * (h-2*t)^3 / 12]
I = (36 * 60^3 / 12) - [(36-2) * (60-2)^3 / 12]
I = 95182in^4

It has a radius of gyration of:
r = SQRT(I/A)
r = SQRT(95182in^4/188in^2)
r = 22.5in

It is fixed at the base and free at it's top
slenderness ratio = KL/r
K = 2.0 for this particular case.
slenderness ratio = 2.0 * 12,000in/22.5in
slenderness ratio = 1066

Let's also pretend that it is a compact section (which it's not) for ease of the critical buckling strength calc:
Pn = Fcr*A
Fcr = 0.877Fe
Fe = 3.14^2 * E / (slenderness^2)
E = modulus of elasticity = 29000ksi
Fe = 3.14^2 * 29000ksi / (1066^2)
Fe = 0.252ksi
Fcr = 0.877*0.252ksi
Fcr = 0.221ksi
Pn = 0.221ksi*188in^2
Pn = 41.5kip (or 41,500lb)

The selfweight of this particular column is 639,722lb.
Since the self-weight is more than the critical buckling strength (an order of magnitude I might add) the column falls over.
Click to expand...

Repost of post #1465. How does this slot in ?

' Look at it this way. Take just one outer core column and picture the floors runnng to at at about 12 foot intervals for 110 floors up to a height of 1300 feet.Do you think that puts a fair amount of vertical stress on the column ?

It still stands though. Now remove all the floors. Will the column now collapse under it's own weight ? (I'm not talking about stability here, only the fact that the column is strong enough to remain standing) '
 
bill smith said:
In my day to day life I trust experts while looking critically at whatever they are doing for me. I never ever trust blindly.

On the subject of 9/11 I trust absolutely nobody and so-called experts least of all. It's not complicated to work out why though I don't intend to have a long discussion about it.
Click to expand...

No you don't.

You can't look critically without some basic understanding and you lack the basic understanding.

You aren't looking critically at anything.
 
Newtons Bit said:
Let's assume a massive 1000' tall steel core column cross-section: 60" x 36" x 1" thick.
b = 36"
h = 60"
t = 1"
Click to expand...
This is the basic cross-section properties of the shape.


It has a moment of inertia of:
I = (b * h^3 / 12) - [(b-2*t) * (h-2*t)^3 / 12]
I = (36 * 60^3 / 12) - [(36-2) * (60-2)^3 / 12]
I = 95182in^4
Click to expand...

This is the moment of inertia of the shape, it's directly related to the members bending stiffness. See TFK's post above above this particular subject.

It has a radius of gyration of:
r = SQRT(I/A)
r = SQRT(95182in^4/188in^2)
r = 22.5in
Click to expand...

Radius of gyration is similar to the moment of inertia. It helps define an objects buckling stiffness.

It is fixed at the base and free at it's top
slenderness ratio = KL/r
K = 2.0 for this particular case.
slenderness ratio = 2.0 * 12,000in/22.5in
slenderness ratio = 1066
Click to expand...

The slenderness ratio defines a members length to stiffness for compression. The higher this number is the lower the critical buckling strength is. Most building codes limit this number to 200 for compression elements for various reasons I won't get into (they're good reasons).



This is a chart from AISC, it defines what K is for different compression cases. It's important to note that these fixed and pinned ends are actual reaction, not connections to other members (like in a moment frame or braced frame). There's a different set of charts for that kind of connection. The column that we are currently analyzing is case (e).



Let's also pretend that it is a compact section (which it's not) for ease of the critical buckling strength calc:
Pn = Fcr*A
Fcr = 0.877Fe​
Fe = 3.14^2 * E / (slenderness^2)​
E = modulus of elasticity = 29000ksi​
Fe = 3.14^2 * 29000ksi / (1066^2)​
Fe = 0.252ksi​
Fcr = 0.877*0.252ksi​
Fcr = 0.221ksi​
Pn = 0.221ksi*188in^2
Pn = 41.5kip (or 41,500lb)
Click to expand...

This was the meat of the calculation.

Fcr is the critical buckling stress (thousands of pounds per square inch) in which the column buckles and falls down.

Fe is the Euler Buckling Stress: the formula that our friend Leonhard Euler developed a long time ago for long slender columns. It's pretty darn accurate if the slenderness ratio is high (like it is in this case).

Pn is the nominal compression capacity again buckling. It's units are merely force (thousands of pounds).

Anything in particular you'd like me to go more into?
 
Newtons Bit said:
This is the basic cross-section properties of the shape.




This is the moment of inertia of the shape, it's directly related to the members bending stiffness. See TFK's post above above this particular subject.



Radius of gyration is similar to the moment of inertia. It helps define an objects buckling stiffness.



The slenderness ratio defines a members length to stiffness for compression. The higher this number is the lower the critical buckling strength is. Most building codes limit this number to 200 for compression elements for various reasons I won't get into (they're good reasons).

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_16329477c5ad0df89f.jpg[/qimg]

This is a chart from AISC, it defines what K is for different compression cases. It's important to note that these fixed and pinned ends are actual reaction, not connections to other members (like in a moment frame or braced frame). There's a different set of charts for that kind of connection. The column that we are currently analyzing is case (e).





This was the meat of the calculation.

Fcr is the critical buckling stress (thousands of pounds per square inch) in which the column buckles and falls down.

Fe is the Euler Buckling Stress: the formula that our friend Leonhard Euler developed a long time ago for long slender columns. It's pretty darn accurate if the slenderness ratio is high (like it is in this case).

Pn is the nominal compression capacity again buckling. It's units are merely force (thousands of pounds).

Anything in particular you'd like me to go more into?
Click to expand...


In the equation for Euler Buckling Stress, Fe = 3.14^2 * 29000ksi / (1066^2), the first number is pi, is it not? Could you explain how this equation was derived?
 
FineWine said:
In the equation for Euler Buckling Stress, Fe = 3.14^2 * 29000ksi / (1066^2), the first number is pi, is it not? Could you explain how this equation was derived?
Click to expand...

Yes, the first number is pi.

The derivation isn't common anymore, see: http://www.tech.plym.ac.uk/sme/desnotes/eulerderiv1.htm

The formula is not something that is merely theoretical. It is backed up by an obscene amount of lab testing.
 
Newtons Bit said:
This is the basic cross-section properties of the shape.




This is the moment of inertia of the shape, it's directly related to the members bending stiffness. See TFK's post above above this particular subject.



Radius of gyration is similar to the moment of inertia. It helps define an objects buckling stiffness.



The slenderness ratio defines a members length to stiffness for compression. The higher this number is the lower the critical buckling strength is. Most building codes limit this number to 200 for compression elements for various reasons I won't get into (they're good reasons).

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_16329477c5ad0df89f.jpg[/qimg]

This is a chart from AISC, it defines what K is for different compression cases. It's important to note that these fixed and pinned ends are actual reaction, not connections to other members (like in a moment frame or braced frame). There's a different set of charts for that kind of connection. The column that we are currently analyzing is case (e).





This was the meat of the calculation.

Fcr is the critical buckling stress (thousands of pounds per square inch) in which the column buckles and falls down.

Fe is the Euler Buckling Stress: the formula that our friend Leonhard Euler developed a long time ago for long slender columns. It's pretty darn accurate if the slenderness ratio is high (like it is in this case).

Pn is the nominal compression capacity again buckling. It's units are merely force (thousands of pounds).

Anything in particular you'd like me to go more into?
Click to expand...



Why would you compare the moment of inertia and buckling resistance of a single 36" x 60" x 1" wall x 1300 foot column to that of the entire central core?

The difference here is that the moment of inertia of the central core is tens of thousands of times greater than that of the column you calculated it for and the core itself did not weigh even one thousand times more than that column.

You need to do a calculation for the central core to try and prove it wasn't self-supporting.

It was self-supporting and the back of the envelope calculations I did (which you criticized as not being relevant) show that, with more than enough margin to make up for the lack of rigor. The geometry I used was much closer in it's moment of inertia to weight ratio to that of the core than what you have done here.

Out of curiosity last week I did an FEA model of a 10 story lattice structure, like the core, with columns and beams that were 10% the height of a story in the center and perimeter, and a hollow rectangular section of the same outside dimensions with solid walls as thick as the beams and columns of the lattice structure. The lattice structure with interior support from one side to the other had a moment of inertia 3 times greater than the hollow section and the simple hollow section geometry of the same outside plan dimensions as the core shows it to be self-supporting.
 
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