Newton's Bit to Gordon Ross & The Journal of 9/11 Studies
Thanks to Newton's Bit. This article is originally produced here.
Gordon Ross Shows Collapse Progression
To Mr. Gordon Ross and the Journal of 9/11 Studies:
Regarding the reply to my criticism of the paper by Gordon Ross entitled, “Momentum Transfer Analysis of the Collapse of the Upper Storeys of WTC1”, I have demonstrated below the actual method in which to calculate the strain energy of the columns at the WTC.
For all compression members with eccentric loads or externally produced bending moments, the bending moment is magnified as the axial force approaches the ultimate compression capacity of the member. This is what is known as P-delta effects. As the column deflects due to the bending moment, the axial force becomes eccentric and creates another bending moment. This bending moment causes the column to deflect further which in turn means that the axial force creates yet another larger bending moment because the column has deflected outwards further. This continues until the column either reaches equilibrium or it fails. P-delta can be explained by the differential equation:
Mz = Magnified Moment (kip*in)
Mi = initial axial load (kip*in)
P = Axial force (kip)
y = initial deflection (in)
E = Modulus of Elasticity (29000ksi for steel)
I = Moment of Inertia (in4)
---Note, most of the calculations following are in the British Imperial Unit System. The numbers at the end will be converted to SI. For reference, a kip is 1000lb, and a ksi is 1000 psi.
A general solution of this equation will not be attempted here, as it is long and outside the scope of what this letter is intended to do. Following Timoshenko and Gere, the general solution is for a column with constant initial bending moments at the top at bottom is:
Pu = Axial Demand on the Column (kip)
Pn = Axial Capacity from Buckling (kip)
There is a limit to the interaction of bending and compression in a column. This is shown in the classic equation:
Mu = Bending Demand (kip*in)
Mn = Bending Capacity (kip*in)
e = eccentricity of vertical load (in)
Pu2 = Axial Load at an eccentricity e (kip).
In all real-world columns, there is always, at a minimum, a small but non-zero eccentricity between the axial force and the column. This is due to a variety of reasons, such as an unequal live load in two adjacent bays or a beam connection on the side of a column. For the purposes of this paper, let e equal a very small but non-zero number. As Pu/Pn approaches 1, the denominator of the term
approaches zero. The magnifier on the Mu/Mn term becomes infinity and the column stops being a compression member and becomes purely a bending member. In reality, the column would transition gradually at somewhere near 0.9 or 0.8 Pu/Pn depending on how eccentric the axial load was, however it is conservative in favor of collapse prevention, and easier math, to assume that this happens at Pu = Pn.
The strain energy of the column can then be calculated by separating the strain energy into two functions: the energy in compression and the energy in bending. These can be referenced from most engineering Strength of Materials books.
For the strain energy in bending, I am assuming 3 buckle points along the length of the column that rotate 30 degrees on each side. This is vastly conservative in favor of collapse prevention. In reality, columns will only go through an 8-12 degree rotation before failing. This of course varies depending on the column strength. The exterior columns were made of varying yield stress materials of up to 100 ksi. High strength steel such as this is more brittle than lower strength steel, such as the A36 wide-flanges, and will fail at a smaller rotation. Even then, a rotational fudge factor of 2.5 has been used to justify that the calculations are completely in favor of collapse prevention
To determine what the strain energy is, actual real column sizes are required. A guess on the cross-sectional area is not enough. To do this, columns based on the same data that Ross references are used to compute a cross-sectional area. The actual column size is then picked by using this cross-sectional area and the same nominal depth of the columns at WTC. The exterior columns were 14x14 box columns and the interior was made up of both box columns (of varying dimensions) and 14” nominal wide-flange shapes. Since the shapes of the interior box columns are not know, the interior columns are all assumed to be 14” nominal wide flanges.
Where (from Ross):
Weight Above = 59000tonnes
FudgeFactor = 4
There are 240 exterior columns. Assuming that the exterior is 50% of the total cross-section of steel, it follows that the cross-sectional area of steel of each is:
A similar shape is the HSS 14x14x1/2. The relevant properties to be used later in this paper are:
Ag = 24.6in2 r = 5.58in Z = 124in3 I = 743in3
There are 47 interior columns. The cross-sectional area of each is therefore:
A similar shape is a W14x370. The relevant properties to be used later in this paper are:
Ag = 109in2 Ix = 5440in3 Zx = 736in3 rx= 7.07in
Iy = 1990in3 Zy=370in3 ry=4.27in
The difference in strength between the two orthogonal axis of a wide-flange are shown with the x and y notation. In column failure, the column will always buckle in the weak-axis. Thus, the smaller numbers of the weak-axis are the relevant ones.
The calculations for the HSS14x14x1/2 follow, please note that the method for calculating the axial compression capacity is inelastic buckling, as seen in my previous letter:
The calculation for the W14x370 are performed in the same manner, the results are:
The total strain energy is:
Or a 269% increase in the absolute maximum. The real column sizes are required to determine what the actual strain energy in the columns are. This calculation is grossly in favor of higher strain energy and even then it is 2.69 times less than the calculation in the Ross paper.
Ross said in his response, “The conclusion of my article, that the energy would be dissipated in many more areas outwith the uppermost storey has been broadly accepted and no serious challenge to that conclusion has been forthcoming.” The summary of the energy losses from his paper:
Kinetic energy 2105MJ
Potential energy Additional downward movement 95MJ
Compression of impacting section 32MJ
Compression of impacted section 24MJ
Total Energy available 2256MJ
Momentum losses 1389MJ
Plastic strain energy in lower impacted storey 244MJ
Plastic strain energy in upper impacted storey 215MJ
Elastic strain energy in lower storeys 64MJ
Elastic strain energy in upper storeys 126MJ
Pulverisation of concrete on impacting floor 304MJ
Pulverisation of concrete on impacted floor 304MJ
Total Energy required 2646MJ
Minimum Energy Deficit -390MJ
Momentum loss and pulverization of concrete: This is another large mistake in Ross’ paper. When two objects collide, their energy is conserved. The typical method to easily to determine this is to assume a plastic collision, where M1V1 = M2V2. This is what Ross has done. He shows that in a perfectly plastic collision, there is an energy loss of 1389MJ. This is also true. However, there is a very important assumption in this that Ross has completely ignored. This loss in kinetic energy is transferred into the plastic deformation of the two bodies. In other words, the difference in energy from impact IS the “pulverization” of the concrete. Ross has calculated the damage of each concrete plate twice. In contrast, in an elastic impact, where the objects are not damaged by the impact, have zero loss in kinetic energy. Energy cannot be created or destroyed, however Ross has allowed 1389MJ (12*107 kip*in) to vanish. Where do you assume this energy goes Mr. Ross? From this alone, there is no longer an energy deficit in your paper, but rather an energy gain from floor to floor.
Mr. Ross, your conclusions and sums and methods have been proven wrong. In my previous letter I offered you the chance to fix and update your calculations out of professional courtesy. Out of respect for your abilities, I said it would be easy for you to do. I had hoped that you would take a harder look at that issue, and take another look at the rest of your paper, but you have chosen not to do so. Your response was nothing more than, “fake but accurate”. This is a disgusting manner for any engineer to respond. Sir, retract your paper or fix your calculations.