Beam on top of beam, not connected, and of different length. 7

[jgrady]

http://www.eng-tips.com/viewthread.cfm?qid=110626

Hi all,

I was reading this thread, and one of the key points for the load to be shared was that beams must be spanning the same length. My question is, what if the top beam is only over the first 1/2 of the entire span. Does the load sharing still apply for the loads placed over that portion of the span?

Regards

 

[TLHS]

If it's just sitting on the top and there's no real fixity at shared end, the top beam just rotates and doesn't do anything except add weight. It's not really clear what you're suggesting, though, so if you give a sketch it will likely help people answer your question.

 

[hokie66]

If the section on top does not extend for the full span, it just adds load to the bottom beam, and does not assist in carrying the load...if I understand your question.

 

[jgrady]

TLHS, I think you got what I was describing. Sorry for lack of picture, I figured it wasn't going to do anything but just needed to hear it from people wiser than myself.

Regards

 

[rb1957]

say the upper beam is only attached with a pin at each end, and the load is applied to the lower beam; so how will the upper beam deflect ?. i think it'll translate without particularly bending. it'll pick up some axial load, and so help carry some of the lower beam's moment that way.

say pinned upper beam and the load is applied to the upper beam, now the lower beam will deflect under pressure from the upper so both'll be carrying the moment (the lower providing an elastic foundation to the upper).

say fixed upper beam, and lower beam loaded. now the ends will drive some moment into the upper beam.

another day in paradise, or is paradise one day closer ?

 

[KootK]

The top beam carries load over its length. If it sees curvature, it sees strain. And if it sees strain, it embodies energy and therefore resists load. In this situation, it won't be a simple split based on relative moments of inertia however. If you make the top beam stiff enough, it will end up being simply supported by the lower beam.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[KootK]

I assumed that the load would be applied to the top of the two beam system. Hopefully that's the "given" that I thought it was.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[rapt]

Agreed with Kootk that the top beam will carry load based on its stiffness to its ends. It will then apply the load that it carries as point loads to the beam below, so the beam below will be designed tor the load that the beam above does not carry plus point loads from the beam above.

The result will be that the beam below will have lower mid span moments and higher moments at the point where the beam above deposits it loads, but its end shears will still be the same as if there was no top beam.

 

[SAIL3]

agree with KootK..but it is not a simple engineering problem , though...

 

[KootK]

If this is a real thing, I'd model it with a vertical gap and connect the beams with pin ended rigid links along the common interface.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[jfmann]

Some of the posts appear to be completely incorrect since they are (apparently) missing the key condition of the problem as stated, which is that the upper member is; (1) On top of and parallel to lower beam, and (2) Only one-half the length of lower beam (which spans between supports).

Unless the upper member is connected to lower beam......along entire length......with adequate connectors to make the two members act as one unified assembly......then the upper member is merely blocking (essentially just added weight) and does not reduce load that must be resisted by the lower (actual) beam.

If upper member was perpendicular (or skewed)........spanning between two completely different (additional) supports.....the situation then would be the classical shared-load condition, with load resisted in proportion to relative stiffness of the two members. However, that is not how I understand the problem as stated.

John F Mann, PE

http://www.structural101.com

 

[SlideRuleEra]

You fellows do come up with some interesting problems. Since I participated in the linked thread from years ago, was inspired to do a "science project" this morning. I modeled the current problem (not a software model in a computer... an analog model in my garage)

Below are photos of my 39" long, yellow pine beam.

The first is with no load: Deflection = 0", of course.

The second is a heavy load (can of nails) and on a stiff 19 1/2" beam: Midspan Deflection 1 19/32"

The third is the same load, plus the 19 1/2", beam at the midspan point: Midspan Deflection 2 1/4"

The fourth is a closeup of the two beams:

IMHO, rapt has put this into words better than I have seen.

Next step is to do the math and see if deflections are what they "should" be.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[Lion06]

I agree from a purely technical standpoint that the upper "beam" will reduce the midspan moment of the lower beam, but under most circumstances I don't see this providing enough benefit to justify the analysis.

Unless the lower beam is significantly more stiff than the upper beam, they will not share load over the overlapped length, based on stiffness. I see two different conditions that exist for this setup.

1. The upper "beam" is less stiff than the lower beam and the deflection and curvature of each is virtually identical. In this case you can say that the upper beam and lower beam "share" the load based on relative stiffness, but that doesn't mean the midspan moment is lower by that same %, since that load is still getting dumped into the lower beam at the ends of the upper "beam". It's essentially moving some uniform load over the midspan to a concentrated load closer to the supports.

I see a very modest benefit to the midspan bending moment in this case.


2. The upper "beam" is more stiff than the lower beam and the deflection and curvature of the two is no longer identical, so the beams won't "share" load based on stiffness any longer. So this condition allows the upper "beam" to essentially transfer all of the load at the overlap to the ends of the upper "beam".

I see this providing more of a benefit than condition 1.

Obviously, there are a lot of variables to this that affects how much help the upper "beam" actually provides. The length of the each beam, the loading, the relative stiffness of each. I, personally, would never take advantage of this and just know that qualitative benefit is in my back pocket as a little help.

 

[Lion06]

I have a project with very heavily loaded PT transfer girders that support walls above. They were aggressively designed for architectural reasons, so this discussion helps me sleep a little better now.

 

[KootK]

Well yeah, this is most definitely an academic musing, certainly for the responders if not for the OP. Two more important points:

1) The beams share load in absolutely all scenarios. See my energy argument above. It's just more complex than the ratio of Ix's as Sail suggested.

2) The beams will never share load simply based on Ix. One can see this by noting that the curvature at the end of the short beam is zero while it will always be a non-zero value at that same location in the full length beam.

@SR: admirable contribution!

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[KootK]

And low and behold... there's an excellent real world example. Two more now that I'm inspired:

1) I'm pretty sure that this is why brick angle lintels work as well as they seem to and;

2) In renovation, I've placed stiff steel beams over historic timber where there was a bending issue but sufficient shear capacity in the timber.

@Lion: in your example, I might actually be willing to take advantage of it if the load transfer out to the wall ends was designed by the book using strut and tie. Depending on the particulars, there could be some considerable efficiency gains.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[jfmann]

Very fine load test with great photos!.........however, this just demonstrates difference in deflection caused by different distribution of load........which you can see by referring to basic beam formulas (se Table 3-23 in ASIC Manual).

White board is not "sharing" load and does not reduce load that is being resisted by the main beam. Visually......picture supporting each end of the white board on blocking (which could be however high you want; and zero-weight relative to beam) and you should see this essential point.

Also.......picture the white board as being almost as long as the beam.........and you will then see deflection reduced to almost zero again.

John F Mann, PE

http://www.structural101.com

 

[KootK]

I'd model it with a vertical gap and connect the beams with pin ended rigid links along the common interface.

Correction: compression only pin ended rigid links.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[jfmann]

Concept of "load sharing" is subject to shades of meaning and interpretation. However, in general, I submit that it means that load resisted by a member is somehow reduced.

However, for practical purposes......and as demonstrated (though with caveats noted below) by the simple though fine load-test illustrated above....the "board on top of beam" can be used to reduce deflection of, and bending moment in, the beam......... even though total load resisted by the beam is not reduced at all......and deformation of the two independent members is significantly different.

Based on standard calculations.........point load (P) at midspan of simple span causes midspan deflection that is 16-percent greater than midspan deflection caused by two equal loads (each P/2) placed at third-points of the beam.

As for the greater increase of deflection (compared to theory) that appears to have occurred in the above load test..........several factors are likely relevant ......including additional weight of white board (though that may be minor) and behavior of this specific wood beam. If same test is performed using several, more substantial (deeper) beams, deflections should average out to values resulting from standard calculations.

When length of "white board" member is equal to span of the lower beam........the 2 beams will share load similar to the condition for an upper beam oriented perpendicular to, and placed on top of, a lower beam. However, in such case, ends of the upper beam are supported by entirely different supports.

John F Mann, PE

http://www.structural101.com

 

[hokie66]

While SRE's demonstration is interesting, it does not depict the problem posed by the OP. The OP clearly stated that the upper 'beam' goes from the support to midspan. In that case, I still believe the upper beam does not assist in sharing the load.