Already explained, by me and by Tony.

In B4, Nordenson computed the "Floor collapse potential energy". Here is the sentence in B4.1 that might contain the point that I am not getting: "

*The gravity load reaction at the former location of Column 79 under the 1.0DL + 1.0SDL + 0.25LL load combination was found to be 46 kips at Floors 13 through Floor 8*" - I am not sure what to make of the phrase "

*gravity load reaction*".

In B4.2 he constructs an idealized floor area that has "

*the equivalent 46 kip load as the corner reaction found in the SAP2000 model*". Looking at the geometry of this floor area, he determined that the center of gravity of this floor area droped a height of 83 inches.

In B4.3 he simply multiplies 46 kip * 83 inches = 3818 kip-inches initial

**P**E.

From this he subtracts several deformation energies (my "D1 to Dn"), listed in Table B6.1:

D1: South Bay Tensile Fracture Energy = 17 kip-in

D2: West Bay Tensile Fracture Energy = 14 kip-in

D3: Slab Boundary Rotations = 147 kip-in

D4: Slab Deformation = 70 kip-in

D5: Slab Shear = 14 kip-in

D6: Girder Impact Corner Yielding = 82 kip-in

That's Total Dissipated Energy Sum(D1..D6) = 345

And so

Remaining [kinetic] Energy at impact = 3818 kip-in - 345 kip-in = 3473 kip-in.

This is the NET energy in your post.

Please note that none of the terms D1..D6 takes care of how the falling girder bends!

Nordenson then assumes that all that NET Energy of 3473 kip-in must be dissipated by bending the girder below and ONLY the girder below! That's why he takes the stiffness of the girder below K = 7627 kips/in (derived in B5.2), and plugs into formulas in B5.3. Let me make those computations explicit. Nordenson writes:

"*Using the girder impact stiffness values, the impact energy was converted to a static force via deformation using the formula:*

PE = 1/2 K * D^2

where D is girder deflection and K is the girder spring stiffness

This equation can be rearranged and solved for deflection."

Nordenson skips this, so I write it out - and note that he should write KE instead of PE:

D = sqrt(2PE / K) = sqrt (2 * 3473 kip in / 7627 kip/in) = sqrt(0,9107 in^2) = 0,9543 in

Nordenson continues:

"*This deflection value can then be multiplied by the girder stiffness to find the equivalent static force using the equation:*

F = K * D"

Let me to this quickly:

F = 7627 kip/in * 0,9543 in = 7279 kip

(Nordenson writes in the text that this "

*yields a static force of 4133 kips*" - this is wrong, but Table B6.1 has the correct value)

So this proves that Nordenson wants to dissipate ALL of the KE established AFTER subtracting D1..D6 by bending the girder below ONLY.

Now look again at this drawing:

Your question: "

**What happens to the energy involved in falling beam flexure? Is it "lost"? If so - How?**"

My answer:

At the same time as the girder below is bent down by the impact force of the end of the falling girder, the inertia of the falling girder makes the falling girder bend, since it must react to the impact force of the girder below, which is equal in size and opposite in direction.

This bending builds up strain energy in the falling girder at the same time that straim energy build up in the impacted girder.

There is only one source of energy to bend both girders at the same time: The NET Kinetic energy of 3473 kip-in.

Therefore Nordenson is WRONG to imply that ALL of that net KE flows into the girder below - some flows into the falling girder at the same time.

Nordenon's formula

KE = 1/2 K(below) * D(below)^2

is wrong. It should be

KE = 1/2 K(below) * D(below)^2 + 1/2 K(falling) * D(falling)^2

The equivalent static force is still

F = K(below) * D(below)

but with a smaller D(below); and at the same time

F = K(falling) * D(falling)

I will have to think hard about how to re-arrange these formulas, but my math intuition tells me already that Tony's solution/formula for springs in series will solve the system.

In the meantime, you need to wrap your mind around the fact that the falling girder bends and thus uses some of the energy which is then no longer available to the girder below, resulting in less deformation below, meaning less equivalent static force.

Unless you can show me that the bending-down of the falling girder is already accounted for somewhere.

You can try to deny that the falling girder would bend and thus use up some of the available energy, but then you simply aren't getting it.