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

 

[Brad805]

I know hokie. I was just joking around while waiting for an analysis to run. It is always interesting to see how we engineers delve into subjects.

 

[jfmann]

hokie......belated response to this;

jfmann,
Do you disagree with my statement of 1 Feb, 01:51?

"Interesting again, but the loadings are not the same. For a simple spanning lower beam and also a simple spanning upper beam over the half span, both with the same I, and loaded uniformly but ignoring self weight, the bending moments are the same, wl^2/8".

Do not understand what you were getting at .......however, for stated case (or for any position of upper beam (length L/2) along lower beam, of length L) bending moments are not the same. Span is of course different for each beam......and, for stated problem, there is no load on lower beam other than load distributed to it by upper beam.

For upper beam centered within span of lower beam, uniform load on upper beam only and stiff upper beam (contact only at ends of upper);
M-upper (max) = wL^2/32 M-lower (max) = wL^2/16
For completely flexible upper beam (with full contact).....which is same as having no upper beam......M-lower (max) = 3wL^2/32.......which equals sum of the 2 maximum moments for separate beams with contact only at 2 points.
For stiff upper beam........moment in lower beam is 67-percent of moment without any upper beam.

For configuration with one end of upper beam at midspan of lower beam;
For stiff upper beam (contact at ends only)........moment in each beam remains the same: M-upper (max) = wL^2/32 M-lower (max) = wL^2/16
For completely flexible (or no) upper beam;
M-lower (max) = 9wL^2/128
For stiff upper beam.........moment in lower beam is 89-percent of moment without any upper beam.

For any stiffness of upper beam (located entirely within span of lower beam), lower beam must still support entire load......such that maximum shear at support is not reduced, which is not entirely consistent with notion of load sharing (if defined as "taking load away" from some other element).

Consider upper beam as header beam several feet above lower beam, supported on 2 posts, with some load on this header beam only. Not including weight of posts, this setup results in the same loading condition as for upper beam on lower beam without posts. Yet it is somewhat of a stretch to consider configuration of header beam to be "load sharing" condition, even though, compared to same load supported directly on lower beam, bending moment and deflection of lower beam is reduced...........as thoroughly documented by discussion.....and SRE model (with fine documentary photos) above.

John F Mann, PE

http://www.structural101.com

 

[SlideRuleEra]

SRE, Don't knock yourself out, but sounds like you are having fun.

You are right. I have a lot of hobbies, one is woodworking another is "building" (i.e. assembling parts) for desktop computers for family members. Had never thought of actually making a working (analog) wooden computer - it fits right in to my other interests. I don't expect it solve the problem at hand, but working with it and showing the photos may help both me, and some ET members get a better feel for the problem. Besides, as you know, an analog computer's output is only as good as the operator's, and in this case the builder's, skill. I enjoy the challenge of seeing how well I can construct and operate this one.
P.S. I did get some material for better beams at the computer store... uh, that is the lumber yard. When the time comes, hopefully JFM can help me calibrate them.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[hokie66]

John,

I was really only interested in the stiff upper beam over first half of span case, and for uniform loading. I think you will agree that that results in a point load on the lower beam of wL/4, and the uniform load w extends over the rest of the lower beam. Therefore, the moment on the lower beam is wL^2/8, just as it would be if the upper beam didn't exist.

 

[jfmann]

My analysis is for load on upper beam only. However, if same uniform load is extended along uncovered half of lower beam, there remains a difference (compared to case without upper beam) since total load on lower beam is only 3wL/4 (not wL), which reduces shear.........and changes moment distribution.

This type of condition (though more complex) sometimes occurs when designing multiple floor systems (one above other).......or even for long-span roof trusses with interior bearing walls that are neglected by everyone involved..........but all that is for another story.

John F Mann, PE

http://www.structural101.com

 

[hokie66]

It does reduce the shear on half the beam because half the upper beam load goes directly into the support. Thus the shape of the moment diagram changes, but the maximum moment is at centre span, and is still wL^2/8.

 

[SlideRuleEra]

Hokie - I'm pretty sure maximum moment is not as straight forward as wL^2 / 8. Based on my efforts on this thread, consider the sketch shown below:

Beam "L" deflects from two fixed end points, "A" and "C"
Deflection of Beam "T" exactly follows the deflection curve of Beam "L". Then things get more complicated.
One end of Beam "T" is fixed at point "A". However, the remainder of Beam "T" is not fixed. It's location dependent of how much Beam "L" deflects.
How much Beam "L" deflects depends on the what Beam "T" does.
The deflection of Beam "T" has to be measured from Straight Line "A - B". The shape of Beam "T" deflection curve is complex.
Since the Beam "T" deflection curve is complex, then both the forces and the moment in Beam "T" are complex, too.
What happens in Beam "T" directly effects to forces, moment and deflection of Beam "L"......... and round and round we go.

IMHO, the sketch is simple, accurately shows what is really happening to the two beams, yet is devilishly difficult tie down mathematically.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[SlideRuleEra]

I need to reword the line that reads: Beam "L" deflects from two fixed end points, "A" and "C"

It should say: Beam "L" deflects from two simple supports, "A" and "C". These simple supports are at fixed locations.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[hokie66]

My answer is about the situation where the T beam has the same I as the L beam, so the deflection of T does not follow L. A point load at centre span from the top span reaction gives the same moment in the bottom beam as does a uniform load over the half span. The deflections would be different, but not the moments.

 

[jfmann]

For upper beam with one end at midspan of lower beam.....and with uniform load on both upper beam and lower beam (which I was not considering in my analysis).....hokie is correct that maximum moment in lower beam is wL^2/8......just basic statics.

Loading shown in SRE diagram is what I have been considering (though with only 2 points of contact for upper beam)..........hokie is adding uniform load to remainder of the lower beam.

Although ambiguous.....key issue / question raised by thread referenced at very beginning of this thread....was whether two beams (even though not connected) could be considered to act together to resist load. The original question implied (but was not clear enough) that perhaps the two beams could be considered to form a unified beam, with increased section properties compared to just the two separate beams. Of course, the answer to any such implication is no.

However.....for the cases that I was considering, with load only on upper beam ........that upper beam does redistribute load......which, depending on relative position, can result in lower stresses for lower beam compared to the same load applied (in same position) to lower beam only. This is not all that much of a surprise......although, it is of some interest to see differences, especially for deflection.

John F Mann, PE

http://www.structural101.com

 

[Lion06]

I'm coming back into this late, but did I read that the upper beam helps to resist load that is applied to the lower beam? That can't be correct if they're not connected, can it? At the most basic level, if you hang a load from the lower beam and that's the only applied load, how will the upper beam help resist that load (regardless of where the load is placed on the lower beam)?

 

[KootK]

That started with me Lion06. See rb1957's response as he explained it for me perfectly. It only applies if you've also got a load on the upper beam as well.

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]

Lion (geez, sensing "Oz" strain with that intro).........for load on upper beam.......the upper beam merely redistributes load (compare to case without any upper beam) as long as upper beam has adequate stiffness such that there will be a gap between two beams except at ends of upper beam.

Of course for load applied only to lower beam.........unconnected upper beam just follows along for the ride and does not alter behavior of lower beam in any way.

John F Mann, PE

http://www.structural101.com

 

[Lion06]

I took a look at rb's post. I see the merit. I have to think about it more, but here are my initial thoughts. The strain energy increase in the upper beam only occurs because the upper beam has load on it and it also requires that the load on the upper beam is placed first. If you start with the load on the upper beam then add the load to the lower beam I recognize that the new load on the lower beam is not the sum of what it was seeing from the upper beam plus the new load on the lower beam. This "feels" more like a redistribution of load than the upper beam helping. When you apply the hanging load on the lower beam, the upper beam picks up more of the load from the load on it and redistributes it to the support and a different location on the lower beam. That, for some reason, feels fundamentally different than helping to resist the actual load in question. I'm struggling to come up with a good analogy.

 

[rb1957]

"A point load at centre span from the top span reaction gives the same moment in the bottom beam as does a uniform load over the half span. The deflections would be different, but not the moments."

sorry, but ??

for a single beam with a UDL over partial span (lets say semi-span to keep the numbers simple), Rb = w*L/2*(L/4/L) = wL/8 and Ra = 3wL/8, and M(x) = Rax-wx^2/2, Mmax, at x = Ra/w, = Ra^2/2w = 9wL^2/64

for a beam with a point load, wL/4 (lets neglect the wL/4 at a, the load from the semi-span beam balances the reaction at a), Mmax = wL/8*L/2 = wL^2/16 (at L/2)
for the beam, length L/2, with UDL, w ... Mmax = wL/4*L/4-wL/4*L/8 = 3wL^2/32 (at L/4).



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

 

[Archie264]

rb1957,

Unless we're discussing something different I get that Ra = wL/4 and Rb = wL/2. Accordingly, the moment at midspan is WL^2/8, as noted. Shear and deflections are different, of course, but that was also noted.

 

[jfmann]

Lion...considering this- "When you apply the hanging load on the lower beam, the upper beam picks up more of the load from the load on it and redistributes it to the support and a different location on the lower beam. That, for some reason, feels fundamentally different than helping to resist the actual load in question. I'm struggling to come up with a good analogy."
Making it much too complicated.......this is basically a simple determinate statics condition. The only issue at all occurs when upper beam has enough stiffness (which I have calculated and reported in several posts) that it contacts the lower beam only at ends (of upper beam). Even then, all that happens is distribution of load (on upper beam) to lower beam at those ends. There is no indeterminacy; "strain energy" is going off on unnecessary tangent........this is just Statics 101.
Your statement above is apparently based on an assumption that the upper beam is in continuous contact with lower beam along entire length of upper beam......which is not the case for load applied to the upper beam, unless upper beam is essentially nothing more than a membrane (paper) that conforms to shape of lower beam.....in which case there is really no issue to be discussed.

rb157.....moment values are for cases with load only on upper beam.......and with upper beam having adequate stiffness so that upper beam is in contact with lower beam only at ends of upper beam (since lower beam deflects and takes curved shape due to loads).

John F Mann, PE

http://www.structural101.com

 

[rb1957]

archie ? w applied to the semi-span beam only; reactions sum to wL/2. with the semi-span beam attached at end a, reactions are the same as for a beam with a semi-span UDL, 3/8wL and wL/8. no?

@jf ... your posts confuse me, i appreciated the two point contact simplification. you said "A point load at centre span from the top span reaction gives the same moment in the bottom beam as does a uniform load over the half span."
"A point load at centre span from the top span reaction" ok, wL/4 ...
"gives the same moment in the bottom beam" ... Mmax = wL/8*L/2 = wL^2/16 ... no?
"as does a uniform load over the half span" ... I took to mean a semi-span UDL applied directly to the lower beam; I calc'd Mmax = 9wL^2/64.

of course, for reactions, you can replace a UDL with a point load at the CG.

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

 

[Lion06]

jfmann - I think we agree. I was responding to a much earlier part of the discussion that said that the upper beam will resist load that is place on the lower beam only, with certain conditions being met (one of them was that the upper beam was initially in full contact with the lower beam under load). I don't see that. My comment about strain energy was in response to rb's post on that part of the discussion.

 

[jfmann]

rb157.....you were referring to statement by hokie66.........which is correct (moment in lower beam being wL^2/8)

For uniform load on half span of lower beam.....reactions are not equal (as you assumed).......they are wL/8 (far end) and 3wL/8 (near end)

Moment at midspan of lower beam;
Due to point load of wL/4 at midspan (from inside end of upper beam) = wL/8 x L/2 = wL^2/16

Due to uniform load on (other) half of lower beam; = wL/8 x L/2 = wL^2/16

Total moment at midspan of lower beam = wL^2/16 + wL^2/16 = wL^2/8


John F Mann, PE

http://www.structural101.com