Engineering help requested: Calculating stiffness of a beam

[Oystein]

I am seeking help from someone with a good grasp on beam theory (Euler–Bernoulli, whatever), such as a structural engineer, to work out the stiffness of a beam in a particular situation:

Where I am so far:

A steel-frame building has long-span girders between two columns (c44, left and c79, right).
On one floor, the connection of the girder on c79 fails, and the girder (loaded with tributary beams and a floor slab) falls - or actually rotates around a hinge which is the seat at c44. The falling end then impacts the girder below, exerting a force on that girder:



(Nomenclatura: I will consistently speak of the "falling girder" and the "girder below", and will denote their properties with subscribt-_f) and subscript-_b in this thread)

This bending of the girder belowabsorbs some of the kinetic energy of the falling girder (and its load, which I'll not refer to explicitly in the remainder for simplicities sake).
At the same time, the falling girder, too, bends down as it experiences an equal upward force at the impact point, and that absorbs kinetic energy, too.

The following physical quantities are involved here:

• PE = Potential Energy differential of the falling girder between its original position and the height of its center of mass as it impacts the girder below. Unit: kip*in
• KE = Kinetic Energy of the falling girder at the moment of impact. Unit is kip*inches
• K = stiffness of the girders (K_b and K_f); a function of elastic modulus of the steel, it's size and shape, and it's boundary conditions unit is kip/inch
• F = the force of impact on the girder below; equal and opposite to the force of impact on the falling girder: F_b = -F_f. Unit is kip
• D = Deflection of a girder (D_b and D_f) due to force and elastic bending = F/K. Unit is inches.
• SE = Strain Energy of a girder (SE_b and SE_f) = 1/2 K D². Unit: kip*in

Now assume we have already worked out

• KE: is PE minus various losses due to inelastic deformation: Tearing and breaking of the floor slab, plastic deformation of the girder's end, etc = 3473 kip*in
• K_b: The stiffness of the girder below, given that it is a beam of length L supported on both ends with a point load only 10 inches from the support at c79 = 7627 kip/in.

This has been worked out in a paper that we are scrutinizing elsewhere. The author does some algebra to work out the equivalent static force on the girder below when the strain energy of that girder SE_b = KE.
This assumes that the girder below absorbs all the kinetic energy, but he forgets that the falling girder would absorb some energy, too:
KE = SE_b + SE_f

The two girders are springs in series, with an effective stiffness
K_eff = K_b*K_f/(K_b+K_f)
I have done all the algebra to work out the resulting force F at the moment that the two girders have absorbed all the KE:

F = SQRT(2*KE*K_eff)


What I want to learn:

How do I determine the stiffness of the falling girder?

It is attached at its left end to a hinge about which it rotates freely, and is loaded with F perpendicular to its long axis.

...Help...?

 

[rwguinn]

I always use the deflection under unit load equation for that. It's a bit tricky with a free-free beam, since applied force has to be done dynamically.
Come to think of it, w^2=K/m, so go from there. You can calculate the frequency in the direction you desire, you know the mass....

 

[Drewbot]

Why did the girder fall? Seems like a bad design.

 

[casebro]

Lots of variables in this thought process.

All of those physical qualities you listed could be changed by the deflection of the contributary beams, the stretching of the re-bar in the slabs, the air compression under the slab slowing fall speed, the actual center of momentum of the falling assembly, residual effects of whatever overloaded the beam to begin the cascade of failure. etc.

My answer would be "Enough so that the Towers came down and fell into their own basements."

 

[Oystein]

I always use the deflection under unit load equation for that. It's a bit tricky with a free-free beam, since applied force has to be done dynamically.
Come to think of it, w^2=K/m, so go from there. You can calculate the frequency in the direction you desire, you know the mass....

What is w?

 

[Oystein]

Why did the girder fall? Seems like a bad design.

It had beams framing into it from one side only. A fire on the floor below heated the beams, they expanded, pushed the girder laterally, and that failed the bolts of the seated connection. The girder then fell from the seat either when it was pushed off to the far side during heating cycle, or it was pulled off to the near side when the beams contracted on the cooling cycle.
(The details would affect how exactly, e.g. with how much excentricity, the falling girder would hit the girder below. This is not known and may have to be dealt with once we know conceptually how to analyse this)

 

[Oystein]

Lots of variables in this thought process.

All of those physical qualities you listed could be changed by the deflection of the contributary beams, the stretching of the re-bar in the slabs, the air compression under the slab slowing fall speed, the actual center of momentum of the falling assembly, residual effects of whatever overloaded the beam to begin the cascade of failure. etc.

My answer would be "Enough so that the Towers came down and fell into their own basements."

All understood.
It is assumed that the shear studs on the girder and beams that would provide for composite action had already failed. The girder would thus bend without the benefit of being stiffened by interaction with the slab.

The beams (four or five of them framing in at slighly <90° angle from one side, one framing in at very slender anggle (20° perhaps) from the other) are a problem as they are nearly point loads. I want to ignore that for the moment and treat the load as distributed evenly over the length of the girder.
The relevant stiffness, I believe, will be dominated by the girder alone, and I assume that it alone absorbs all the energy that the girder below does not.

As I said, plastic deformation of slab (including rebar) and beams up to the point where the girder impacts below are already accounted for (quantified as energy dissipated and deducted from PE).

The actual center of momentum of the falling assembly has been determined by a structural model in SAP2000.

I also keep in mind that, as the falling girder deflects downwards, this infused a bit more PE. I'll take care of that iteratively once I have an idea how much deflection to expect.

I think that the influence of air pressure and resistance is small compared to the margins of error we already have.



Essentially, I am looking at getting the resultant force on the girder below to within a factor of 2 or 3.

The paper this is from concludes that the force on the connection is about 10 or 11 times the vertical shear capacity -> connection fails (and that starts a cascade of floor failures). But again, the author didn't consider the deflection of the falling assembly (that's equivalent to infinite stiffness).
I have one debate opponent who has so far offered two ways to compute K_f, but I am rather certain his assumptions and thinking aren't right. I don't want to give you details as not to bias you. His values K_f are so low that the resulting force is lower than shear capacity by a factor of about 4-5 -> connection survives, collapse arrests there.
My hunch is that he's got the stiffness wrong by a factor of about 8 to 16, but struggle to make my case and be sure. This would put the resulting force so close to capacity that we'd have to conclude "maybe, or maybe not"

 

[casebro]

Steel from cold to molten only expands 3%. Would that be enough expansion to cause it to fall off it's seat?

And that expansion would be retained somewhat by the concrete slab, wouldn't it? I THINK concrete has a lower coefficient of expansion. But if higher, might be what pushed things around.

Though if it were heated to red hot, and that expansion resisted, it would have gotten thicker, then shrunk length wise. Still that 3%.

 

[casebro]

While heat rises by convection, would there be enough heat to also effect the bottom floor that same way? Maybe not as much, but the preload may have been so much that the margins of error push the conversation off it's seat? Distortion of the floor below might have a grater effect than upper beam deflection?

 

[Oystein]

Steel from cold to molten only expands 3%. Would that be enough expansion to cause it to fall off it's seat?

And that expansion would be retained somewhat by the concrete slab, wouldn't it? I THINK concrete has a lower coefficient of expansion. But if higher, might be what pushed things around.

Though if it were heated to red hot, and that expansion resisted, it would have gotten thicker, then shrunk length wise. Still that 3%.

Plausibly yes, these beams were more than 53 feet long, 3% of that is 1.6 ft. The seat of the girder was about 12 inches wide.
There is dispute over this - the max temperature of the beams is not know, and subject to several variables. Also, there is a dispute about whether or not the end of the girder would be pushed against the column flange, preventing a walk-off to the far side.
At any rate, the girder would be likely to be pulled off to the near side upon cooling if it didn't walk off first.

And anyway, we are assuming that the girder connection failed, either way.


But thanks for asking - I so far didn't think about the effect of heating on the elastic modulus. Would a hot girder be stiffer or less stiff than a cool one?

 

[Oystein]

While heat rises by convection, would there be enough heat to also effect the bottom floor that same way? Maybe not as much, but the preload may have been so much that the margins of error push the conversation off it's seat? Distortion of the floor below might have a grater effect than upper beam deflection?

All possible. We may have simulation data with average gas temperature low above the floor.
More important is that there was also fire on the floor below (and the floor below that), so it is quite plausible that the girder below wasn't at original, cold capacity.

Again, I am interested in order of magnitude.

 

[rwguinn]

Natural frequency, radians/second^2

 

[Oystein]

Natural frequency, radians/second^2

Ah.

Let's see. Just to get a first rough idea: It's a W33x130 girder (130 lbs/ft), 45.6 ft long
m_girder = 130 lbs/ft * 45.6 ft = 5926 lbs

Five W24x55 beams, each 53.66 ft long
m_beams = 5 * 55 lbs/ft * 53.66 = 14,758 lbs
Half of that loads the girder (the other half rests on the building's perimeter columns) -> 7.4 kips of effective mass

The paper determined that the mass of the falling floor assembly is effectively 46 kips; that would mean 46-7.4-5.9 = 32.7 kips of floor slab (reinforced concrete and metal decking).

I think I should work with the 13.3 kips steel, right?

As for stiffness ...
K_f = 3.7 kip/in is the lowest value claimed so far
K_f = 6.7 kip/in is the other low value claimed
K_f = 59 kip/in would result in F being about equal cold capacity
K_f = 7627 kip/in is the stiffness of the girder below (on account of it being loaded so close to its supported end)

Sooo with those values and

w = SQRT(K/m)

I get values of
w = 0.53 radians/s²w = 0,71 radians/s²w = 2.1 radians/s²w = 24 radians/s²
Now what?

 

[rwguinn]

I'm going to have to wait till I get to a real computer to see what the hills you really are going for, but I gave you a couple of ways to determine the stiffness of a beam. It all depends on the direction you are looking for, but if it's lateral stiffness, you can use the deflection equation Y=L^3/(12 EI) for a unit load, and then K=1/Y. That's if the 2 ends are in contact with something. I'll have to look up the equations for axial stiffness , as well as the exact values. ..
And yes, the beam only, unless the peripheral stuff adds stiffness. ..

 

[jaydeehess]

I'm going to have to wait till I get to a real computer to see what the hills you really are going for, but I gave you a couple of ways to determine the stiffness of a beam. It all depends on the direction you are looking for, but if it's lateral stiffness, you can use the deflection equation Y=L^3/(12 EI) for a unit load, and then K=1/Y. That's if the 2 ends are in contact with something. I'll have to look up the equations for axial stiffness , as well as the exact values. ..
And yes, the beam only, unless the peripheral stuff adds stiffness. ..

Axial stiffness is probably not required in this model. Its not at too great of an angle so its going to deform almost entirely laterally. The girder, as Oystein pointed out, is 50+ feet long and the end has dropped one storey, about 10 feet.
The left edge is sitting on a shelf, free to move horizontally or up, but not down. Think of it as hinged there. The other end has dropped and contacted the next lower floor pan.

 

[Oystein]

I'm going to have to wait till I get to a real computer to see what the hills you really are going for, but I gave you a couple of ways to determine the stiffness of a beam. It all depends on the direction you are looking for, but if it's lateral stiffness, you can use the deflection equation Y=L^3/(12 EI) for a unit load, and then K=1/Y. That's if the 2 ends are in contact with something. I'll have to look up the equations for axial stiffness , as well as the exact values. ..
And yes, the beam only, unless the peripheral stuff adds stiffness. ..

Why that equation?
Yes, the two ends are in contact with something:
- the left end is attached to a seat on a column, and I assume that it rotates around that point
- the right end touches the girder below that it just fell on

But that right end contact is the stress - I can't treat it as a support, or can I? Where is the deflection measured?

ETA: Yes, lateral stiffness.

ETA2: I think the fact that the girder has angular momentum - or that the lateral excentric force imparts torque on the beam - needs to be considered somehow. If the beam were totally free and hit by F laterally at the end, it would rotate around a pivot 2/3 L from the stressed end - right? There will be a force on the left end (the hinged one) different from F.


ETA3: In the meantime, some values to calculate with:
E = 29,000 kip/in^2
I = 6710 in^4
L = 547 in

So, your formula yields
K = 12 E*I / L^3 = 12 * 29,000 kip/in^2 * 6710 in^4 / 547^3 in^3 = 14.3 kip/in

 

[casebro]

Would a hot girder be stiffer or less stiff than a cool one?

I don't know about steel, but brass is strongest at 800, f? So steel strength is probably a curve too.

Thermal expansion is usually given as a rate, not a curve.

I'm inclined to think the first step is to figure how hot all this got before blaming expansion or contraction. And whether the contributary beams got so hot they buckled first, then their thermal changes become meaningless.

Dad, the carpenter, always said wood is stronger in a fire than steel.

I'm a machinist, welder, mechanic and wood carver. More logic than skill here.

 

[Oystein]

All good and right, casebro, but not what I am after.
I am taking the initial failure of the connection above as input, as premise.

Looking for the right way to calculate the stiffness of a beam supported on its far end around which it rotates, and experiencing a point load on the other, free end.

Still tying knots into my brain...

 

[rwguinn]

Why that equation?
Yes, the two ends are in contact with something:
- the left end is attached to a seat on a column, and I assume that it rotates around that point
- the right end touches the girder below that it just fell on

But that right end contact is the stress - I can't treat it as a support, or can I? Where is the deflection measured?

ETA: Yes, lateral stiffness.

ETA2: I think the fact that the girder has angular momentum - or that the lateral excentric force imparts torque on the beam - needs to be considered somehow. If the beam were totally free and hit by F laterally at the end, it would rotate around a pivot 2/3 L from the stressed end - right? There will be a force on the left end (the hinged one) different from F.


ETA3: In the meantime, some values to calculate with:
E = 29,000 kip/in^2
I = 6710 in^4
L = 547 in

So, your formula yields
K = 12 E*I / L^3 = 12 * 29,000 kip/in^2 * 6710 in^4 / 547^3 in^3 = 14.3 kip/in

Don't have time to mess with it tonight--session with the duty torturer at the Weatherford inquisition has me hurting.
I screwed up the equation--it's Y=L^3/48 E*I, not 12 EI, so you're actually at 57.2 kip/in.
I'm still unsure where you're going with this.
The deflection of the fallen girder generates a reaction force at each end of the girder, one of which is taken by the "Pinned" (Upper) end, the other taken by the girder on which it is resting.
since the angle of the fallen beam is not zero, the total reaction is indeed a function of the axial force on the fallen beam.
There are no simple solutions, and I'm not going to try to figure out the problem you're seeking an answer for.
You have the lateral portion of the fallen beam stiffness, which is what you asked for.

 

[Oystein]

Y=L^3/48 E*I is the formula for a beam simply supported on both ends for a point force in the middle. That is not the same force as the one acting on the free end, in fact it is half that force!

Now let's see if I can normalize that to the force on the free end...

To get F_end on the free end, I need to impose a lateral load in the middle of F_middle = 2F_end, and that gives me a deflection of

D = F_middle/K_middle = 2F_end/K_middle
<=> K_middle = 2F_end/D = 48EI/L^3
<=>F_end/D = 24EI/L^3 = K_end

Agreed? That would be 28.6 kip/in