Beam Above Another Beam - With Gap

[SlideRuleEra]

This is a situation related to thread507-110626 that I have had to deal with and have been puzzeled how to treat it.

1. Two equal length beams, one located directly above the other.
2. Assume each beam has it's own simple supports.
3. A small, uniform gap, say 1/4" between the two beams.
4. Apply a uniform distributed load to the top beam.

After the top beam deflects 1/4", it comes into contact with the lower beam (point of contact at the center of both beams).

As the beams start to share the load, it seems that things get more complicated.
Does the top beam suddenly become two-span (three supports) continuous?
Is the bottom beam treated as a simple beam with a point load at the center?

I will appreciate your input.

 

[SperlingPE]

If lower beam is stiff enough to have little or no deflection due to the upper beams's load, I believe the upper beam becomes a two span beam.
The lower beam is simple span with a point load.
Interesting.

 

[Lutfi]

This situation is not complicated at all. It is often presented in structural PE exams! I also recall solving it in my engineering mechanics class.

The top beam is simply supported till it deflects to the point where it will load the bottom beam. Then it becomes a two span beam. The reaction on the lower beam can be computed by equating the deflection to a point load, deflection= PL^3/48EI, (formula can be obtained from AISC manual or any static’s text book or even Roark’s). Solving for P, we get:

P = D48EI / L^3
Where D = Deflection

Watch the units and be consistent

That point load is the reaction on the bottom beam. (Keep in mind that in theory supports should have zero deflection).

The bottom beam can be analyzed as simple span with concentrated load at mid span, assuming that is where the intersection occurs, with M=PL/4.

I hope I recalled this correctly since I am at home and working from memory.

Regards,


Lutfi

http://www.cdeco.com

 

[JAE]

You all make sense here - but it does get more complicated in reality - as the upper beam deflects downward, at FIRST it comes into contact with the lower beam at a single point at midspan.

But the upper beam is in the shape of a parabolic arch and as the load gets distributed to the lower beam, the deflection of the lower beam takes up an arch shape as well.

Now I know that two parabolas will still only touch at a single point, but in reality, the top beam will most likely begin to lay down on the lower beam and you now have a dynamic situation in that the "point" load slowly becomes a short distributed load of progressively increasing length.

 

[Lutfi]

JAE,

For some reason I assumed the beams were orthogonal to each others. Re-reading Sliderule”s post the second time, I am not sure that is the case which renders my response to be invalid, in some way.

I need to scratch my head a little more on this one.

Regards,


Lutfi

http://www.cdeco.com

 

[SlideRuleEra]

Lutfi - Here is more detailed background:

This situation occured at an electric power plant flue gas desulphurization (FGD) scrubber. A W6, about 10' long supported ductwork that became coated with thick deposits during operations. When I was called in, the W6 had about 1" deflection (at the center). (The scrubber manufacturer had underestimated the weight & thickness of the deposits.) During an outage, the deposits were removed, the W6 then had about 1/4" deflection (at the center) from the weight of the "clean" ductwork.

We added another W6 under the original W6. (Because of other concerns, we could not replace the original beam or add one larger than a W6.) Of course the two beams now touched at the center but had a 1/4" gap at each end. I had plant operations jack up the the top beam and weld in a series of 1/4" shims (welds to the flanges of both beams) to create a kind of "funky" W12 "equivalent". (It has worked well as a composite member.)

However, I have always wondered how this system would have worked togther as two separate, unatttached W6 members when the deposits built up again.

 

[JAE]

You know ... I didn't even think of two beams orthogonal to each other but you're right...that would be a center point load then.

But SlideRuleEra's new description puts them parallel.

With the shims welding the two beams together, they would tend to act together, as a composite unit...assuming that the shim welds were enough to take the horizontal shear.

 

[jheidt2543]

Real problems are much more interesting than textbook problems, thanks for sharing SlideRuleEra!

 

[pba]

I think the responses both over simplify and over complicate the issue! As I see it, the top beam deflects so as to load the bottom beam and does at that point become a two span beam for any new loading. BUT the lower beam will also deflect as it is loaded and can therefore be considered to act as a spring of stiffness equal to the lower beam stiffness. Most people who can’t be bothered to work out the beams forces by hand can simply use basic software to get accurate answers. Now, given that the deflection equations are simplistic in that they do not accurately model differing material properties or partial end fixity, it would seem reasonable to seek a ‘simple’ engineering solution. Fitting shims between the beams gives the simplest solution and allows the combined stiffness (I1 + I2) of both beams to be used for any loading regime. Still open for debate is if the top beam needed to be jacked to an effectively straight form before placing the shims?