Saturday, February 23, 2008

Why the columns bowed in

Update 9/20/08: Rephrased a couple of sentences, referenced reduction in modulus, fixed some weird formatting.

Why did the columns bow in? There’s been quite a bit of speculation and misinformed opinion about the mechanics of the structure that caused that, so I hope to do a little bit of enlightenment. The first thing that needs to be looked at is the problem. Figure 1 shows a typical building section through a building such as the WTC towers.

Figure 1

Everything here is fairly straight forward. The gravity load path can be seen very easily. The floor trusses deliver the vertical floor loads to the columns which deliver them to the foundations. But what happens when a core column is severed as in Figure 2?

Figure 2

Things start to get a little bit more complicated. The first thing that should dawn on most people is that the floor is no longer being supported by the middle column which is going to cause some problems. The middle column will drop unless there is a force that can resist it, see Figure 3.

Figure 3

As the column drops, the top chord of the floor truss develops tensile forces (it is quite literally stretched). This tensile force has two parts, a vertical portion that pulls the column up and a horizontal force that pulls the rest of the structure in. This can be seen in Figure 4.

Figure 4

The other things to note here is that the left column is still under its full axial loading (P2, which will be important in the analysis). The left column is already shown pulled in to some degree, however it is not to scale. Further modifying the problem, we know that the fires caused the trusses to sag to some degree. If the fire is hot enough, the truss will become a tension only member. This means that the top chord of the truss will act something like a rope and pull inwards at its connections. This can be seen in Figure 5. There is another condition as well (which I have no illustrated) that will cause the heated floor to expand outwards without losing its bending capacity and thus not sagging. It is a condition that likely proceeded that of the sagging floor trusses.

Figure 5

NIST (1-6D) estimates that the total pull-in force at the exterior columns is roughly 6 kips (6,000lbs) at each column. The truss to column connection consisted of (2) 5/8” diameter bolts. Even non-structural grade bolts of this size will have a shear capacity of over 5kips each, so it is reasonable to assume that the top chord of the truss will not pull off of the columns at the connections due to a 6kip load.

Is this 6 kips enough to pull the column in several feet as seen in the photos of the tower?

Figure 6

Math is needed here. First some assumptions need to be made. For the purpose of this analysis, let the exterior column be HSS14x14x5/16 tubes at 25% of the maximum axial load prior to any damage. The column has the following properties: Similar to HSS 14x14x5/16
A = 15.7in^2
I = 739 in^4

S = 92.3 in ^3
Pn = 557k (from AISC LRFD 3rd, table 4-6 with an unbraced length, KL = 12’-4”)

Pu = ¼* 557k = 139k
Mn = 92.3in^3*46ksi = 4645 kip*in (Mn = maximum bending capacity)

The column itself will bend inwardly until it snaps if the column itself ever becomes inelastic. This can be defined by the ratio (I’ve simplified this a bit): Mu/Mn + Pu/Pn <>

Figure 7

Unbraced length = 37’-0”

Pu = 139k
Pn = 465k (from AISC LRFD 3rd, table 4-6 with an unbraced length, KL = 37’-0”)
Mn = 4645 kip*in
Mu = P*a

Mu = 6kip*1/3*37ft
Mu = 74kip*ft or 888 kip*in

Deflection = P*a *(3L^2 – 4*a^2)/(24*E*I) (Formula from AISC LRFD 3rd)
a = 1/3*L
= P * 1/3L *(3L^2 – 4/9*L^2)/(24*E*I)

= 23*P*L^3/(1296*E*I)

= 23*6k*(37ft*12in/ft)^3/(1296*29000ksi*739in) = 0.435in

Additional moment due to P-delta

Mu+ = 0.435in*139k = 61.02 kip*in

Additional Deflection
=Mu+*L^2 / (4*EI)

= 61.02kip*in*(37 * 12ft\in)^2 / (4*29000ksi*739in^3)
= 0.140in

Additional moment due to P-delta2

Mu++ = (0.435+0.140)in*139k = 79.93 kip*in

Additional Deflection
= Mu++*L^2 / (4*EI)
= 79.93kip*in*(37 * 12ft\in)^2 / (4*29000ksi*739in^3) = 0.184in

As seen, the first p-delta iteration results in an increased deflection of 0.140in. The second results in a deflection of only 0.184in. We can thus conclude that p-delta will eventually converge and that no further iterations are necessary. The 6kip pull-in force with no effect of fire will not result in the column becoming unstable. At 600C, the Modulus of Elasticity will have reduced to approximately 0.3 of its original value (see Figure 8), and the yield strength to 0.5 of its original value. The effect of the Modulus of Elasticity being so greatly lowered is of far greater important than the yield strength, however.

Figure 8
From AISC Facts for Steel Buildings


Unbraced length = 37’-0”
E = 0.3*29000ksi = 8700ksi
Pu = 139k
Pn = 465k*0.5 = 233k
Mn = 4645 kip*in *0.5 = 2323kip*in Mu = P*a
Mu = 6kip*1/3*37ft
Mu = 74kip*ft or 888 kip*in

Deflection = P*a *(3L^2 – 4*a^2)/(24*E*I) (Formula from AISC LRFD 3rd)
a = 1/3*L
= P * 1/3L *(3L^2 – 4/9*L^2)/(24*E*I)
= 23*P*L^3/(1296*E*I)
= 23*6k*(37ft*12in/ft)^3/(1296*8700ksi*739in)
= 1.45in

Additional moment due to P-delta

Mu+ = 1.45in*139k = 201.6 kip*in

Additional Deflection
= Mu+*L^2 / (4*EI)

= 201.6kip*in*(37 * 12ft\in)^2 / (4*8700ksi*739in^3) = 1.55in

Additional moment due to P-delta2

Mu++ = (1.45+1.55)in*139k = 417 kip*in

Additional Deflection
= Mu++*L^2 / (4*EI)

= 417kip*in*(37 * 12ft\in)^2 / (4*8700ksi*739in^3) = 3.20in

This results in the column becoming unstable due to p-delta. This can easily be seen in that the deflection due to P-delta2 is double that of P-delta1. It can therefore be concluded that it was necessary for both fire and damage to result in the collapse of the towers.

Saturday, January 5, 2008

The Peer Reviewers at J911

I've never really been generally impressed by the folks at the "Journal" of 911 Studies, but my latest encounter with one of their peer reviewers, Tony Szamboti, on the JREF forums leads me to believe that they're not only suffering from group think, but also egregiously incompetent on issues they proclaim to be experts on. Mr. Szamboti posited this question to a group of non-engineers:

What does the slenderness ratio of a structural steel column need to be to be in the inelastic buckling range?

What were the slenderness ratios of the central core columns at the collapse initiation sites of the 98th floor in the North Tower and 82nd floor in the South Tower?

I saw this question and jumped into the discussion, posting:
Inelastic buckling occurs for all slenderness ratios under the Euler limit. That's 4.71 * SQRT(E/Fy). Of course extremely stout members won't buckle inelastically, however none of the columns in the upper floors of the WTC were that stout.

Do you have any clue as to what you're talking about? I recommend picking up an AISC Manual of Steel Construction and see exactly how steel is designed these days. We're not in the 1940's, we know how steel fails now. Maybe you should update you knowledge to modern information.

Mr. Szamboti's replied:

How did I know you would come on.

You used an effective length factor of 1.0 in your letter to Gordon Ross, which is for a pinned connection, when you should have used .5 to .65 for fixed both ends connections for the tower columns. The 1.0 gave you larger slenderness ratios and they still weren't greater than 40. Now you are going to say the tower columns weren't in the short column category and would have been subject to inelastic buckling. The AISC equations you show here and which you used in your paper are conservative for design.

You want to say the tower columns would fail due to buckling. Well how about a test case were an I beam with a slenderness ratio of 20 or lower failed due to inelastic buckling. Do you have any test cases? I have an AISC manual right here. I am familiar with the equations and monograph. You want to go around asking others if they have a clue and you seem to be the one who should be asked that question Mr. Smarty pants.
This is where Mr. Szamboti shows his lack of knowledge as regards to structural engineering. The slenderness ratio that we are talking about, and what Mr. Szamboti struggles to understand, here is defined as K*L/r.

k = effective length factor
L = length of the column (in)
r = radius of gyratio of the column (in) - [This is a function of the geometric properties of a column]

For the columns that we are talking about, the variables L and r are very well defined and not argued. The effective length factor 'k' is where he slips up. In the commentary of the AISC Manual of Steel Construction (arguably the Bible of how to design steel in structures) this factor 'k' is defined. The first place is in table C2.2 (shown below).

Table C-C2.2 (click to enlarge)

Under column (a) it defines the theoretical k value of 0.5 and a recommended design value of 0.65. It's fairly easy to see that this is where Mr. Szamboti thinks the factor k is defined, as the tower exterior columns were moment frames, which means that the top and bottom portions of the columns were fixed. Column (a) shows a column element fixed at the top and bottom, so he used it. And he's very wrong. The commentary clearly explains how this table is to be used on the page before the table, "These range from simple idealizations of single columns such as shown in Table C-C2.2 to complex buckling solutions for specific frames and loading conditions". In his rush to prove me wrong, I can only surmise that he went through the commentary to find an answer to his question, and stumbling upon the first table that seemed to show an answer that confirmed his bias, he lept to a conclusion. An incorrect one not supported by the document he was referencing.

The correct way to calculate this effective length factor is with the
nomograph chart, shown below. The nomograph table is for frames which can translate horizontally, this contributes significantly to the stability of the frame.

Figure C-C2.4 (click to enlarge)

This table looks nonsensical, but it's fairly simple to use. First Ga and Gb need to be defined, which are simply a comparison of the stiffnesses of the columns to the girders. Ga is the comparative stiffness of the top point of the column and Gb is of the bottom. Then, to get the effective length factor, one merely needs to draw a straight line between these two points (see the figure below with Ga = 1.0 and Gb slightly stiffer).

Example nomograph

In this example, the k factor of a frame that has a column stiffness that is roughly equal to the girder stiffness is about 1.4.

It is very easy to see with this table that the lowest factor k that a column in a moment frame can have is 1.0, rendering Mr. Szamboti's statement that it should be 0.5 or 0.65 completely without merit. The purpose of this isn't to belittle or attack Mr. Szamboti as being an incompetent engineer. I'm sure he is an excellent mechanical engineer, however he is not an expert in structural engineering, far from it. He is most definitely unqualified to "peer-review" papers of a structural engineering focus for anyone.