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

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

(OP)

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

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

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

Regards

## RE: Beam on top of beam, not connected, and of different length.

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 ?

## RE: Beam on top of beam, not connected, and of different length.

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.

## RE: Beam on top of beam, not connected, and of different length.

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.

## RE: Beam on top of beam, not connected, and of different length.

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.

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

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.

## RE: Beam on top of beam, not connected, and of different length.

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

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

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.

www.SlideRuleEra.net

www.VacuumTubeEra.net

## RE: Beam on top of beam, not connected, and of different length.

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.

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

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!

## RE: Beam on top of beam, not connected, and of different length.

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.

## RE: Beam on top of beam, not connected, and of different length.

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

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Correction:

compression onlypin ended rigid links.## RE: Beam on top of beam, not connected, and of different length.

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

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

Since my analog model is based on the OP's stated conditions (short beam length = 50% long beam length) the equal point loads are at the

quarter points, not thethird-pointsper your assumption. I have run the math (attached). Theoretical midspan deflection of the quarter point loaded beam should be 1.46 x midspan loaded beam. My empirical results give a factor of 1.42. Well within 3% of the correct (1.46) value... not bad for a garage experiment with material on hand.The values of theoretical deflections are somewhat different than my empirical. This is most like caused by deviation in the thickness of the 39" beam - I had ripped this board, with a hand circular saw, some months ago. A small variation in the thickness (measured at 0.28" average, will have a huge effect on the value of the Moment of Inertia). The ratio given above is what's important - not the values themselves.

Also, I believe the analog model is a good bit more sophisticated than just a "load test". I'll explain:

Within the limits of an analog simulation, i.e. for precisely one set of conditions, the model demonstrates how to mathematically determine the absolute bounds of deflection (or by extension, shear/moment, beam bending stress, or similar properties).

The "Best" case (lower bound for deflection in my model) is where the "short" beam is stiff, say that its Moment of Inertia (I) approaches infinity. Then lower bound deflection is calculated by putting the two point loads on the "long" beam.

The "Worst" case (upper bound for deflection) is where the short beam is "flexible", "I" approaches zero. In my field test assume that a 19 1/2" strip of thin paper was the "short" beam, with its "I" really equal to almost zero. Then the upper bound deflection is calculated by putting the single point load on the "long" beam.

I submit that for ANY value of "I" for the "short" beam the deflection will be between these to bounds. Of course, all of the above applies to my 39" long beam. BUT, the results can be extrapolated to any two "real" beams, of any length, properties, or loading patterns (not just point loads).

Bounding a problems comes from an older time before today's sophisticated software and methods made it archaic, but in many real problems the lower bound and the upper bound are so close together that finding the "exact" answer is not necessary.

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## RE: Beam on top of beam, not connected, and of different length.

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## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

The OP asks the question about a full length simple span beam with another beam on top extending only half-way across.

Assume the upper beam has some stiffness, and assuming that the load is applied to the top of the half-beam along its length and then applied to the top of the full beam along the remaining half-length of span, then there certainly would be some load sharing between the two beams.

But the actual behavior depends on the relative stiffnesses of the two beams.

Upper beam very stiff (i.e. rigid):If the upper beam's stiffness is high enough, then the upper beam will be essentially spanning its length with one half of the load going into the end support and the other hslf load becoming a point load on the lower beam at its midspan point.

Upper beam very flexible (i.e. wet noodle):If the upper beam is much more flexible, then it will tend to simply distribute the applied load down to the lower beam.

Upper beam somewhere in between:The upper beam will take some of the load and deposit it at its ends like end reactions of a simple span beam.

But as the upper beam deflects onto the lower beam (i.e. the upper beam span-stiffness condition is lower than the lower beam) it will deposit some of the load through direct contact along some length at each end with varying degrees of pressure. There would be a gap - similar to SlideRuleEra's photos above - but there would also be a length of contact at the ends. A very complex problem.

I think it would be easily modeled in an analysis program with two beams and multiple joints along their lengths with rigid compresion-only links between them.

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## RE: Beam on top of beam, not connected, and of different length.

Question posed by original post (from 2006) referenced at top of this thread was;

"How do you calculate Section and Moment of Inertia of 2 beams, one on top of the other beam, without being connected

to each other?"

Basic answer is......unconnected beams remain independent such that section properties are not modified for either beam. However, the question was so general as to leave open all sorts of options.

New question posed at start of this thread actually does not state how the "half-length beam" is positioned relative to underlying (main) beam.

Key issue though is "load sharing". After further reflection, definition I offered in previous post is not entirely satisfactory since a fully-connected, partial-length coverplate is reasonably be considered to share load with the main beam, reducing stress and deflection of the main beam (without coverplate).

Key difference however.........which goes back to original (2006) question....is that the fully-connected coverplate (and similar assemblies) increase section properties of the main beam...........whereas the "upper beam" condition discussed in this thread remains an independent element and does not increase section properties of the main beam.

Evenso.......considering that "upper beam" in this case (with adequate stiffness) reduces bending stress (but not shear stress at supports) and deflection of the main beam.....it may be fair to conclude that a form of "load sharing" has occurred.

Of course the more important issue is how effective such assembly is at reducing stress and deflection in the main beam, considering cost of the upper "beam".......compared to other methods. For practical installation, it is unlikely that this type of installation would be more beneficial than simply installing another beam alongside, though this discussion is more about analysis than practicality.

As for orienting "half-length beam" so that one end is at midspan of main beam.............reduction of stress and deflection in main beam still occurs (compared to point load at midspan of main beam alone) as long as point load is applied to upper beam away from inside end of upper beam.......since, in that case, some load is distributed (by upper beam) over to support without loading the main beam.

For an independent upper beam to be effectively useless.......it must simply contact the main beam at midpsan........for then, entire point load is applied to main beam, with upper beam than merely acting as blocking. This occurs when deflection of upper beam (taken as spanning between ends) is greater than deflection of main beam (with two point loads at end of upper beam). For case with ends of upper beam at quarter points of main beam......such condition is the case when EI of upper beam is 9-percent (1/11) or less of the EI of lower (main) beam.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

"first 1/2 of the entire span" is a good enough description for me.

## RE: Beam on top of beam, not connected, and of different length.

First photo shows the stiff "short" beam putting two equal point loads on the "long" beam, one at x = 0, the other at x = midspan. Measured midspan deflection = 1.13" (Lower Bound of Deflection)

Second photo assumes a very flexible "short" beam (explained in my previous post) putting single point load at the "long" beam at the quarter point. Measured midspan (yes, midspan to be consistent) deflection = 1.59" (Upper Bound of Deflection)

Again, as above, to verify the analog model checked ratio of theoretical deflections (1.38) to measured deflections (1.38). Calculations attached.

I agree that this problem is surprisingly complex, but working with my analog has really helped to get a grasp on what is happening. Been a very fascinating day!

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## RE: Beam on top of beam, not connected, and of different length.

across its end for a short distance. Then it lifts off the lower beam as its stiffness results in a lower deflection than the lower beam.

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## RE: Beam on top of beam, not connected, and of different length.

As for when upper beam becomes totally "useless"........actually it is looking like stiffness of upper beam would have be even lower than the 9-percent that occurs for "first contact" (for conditions noted), since, after contact at midspan, the main beam would now deflect more......so that upper beam would deflect more, such that the midspan support for upper beam would be a spring-type support. Equilibrium position is very likely reached sometime before zero stiffness, but looks like upper beam would continue to distribute some load out to ends (of itself) even with stiffness less than 9-percent of main beam.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

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.

## RE: Beam on top of beam, not connected, and of different length.

Exactly. When the stiffness reaches 0%, that is what I am calling the "Upper Bound of Deflection".

John, I think that you, JAE, and I are saying the very same thing. Just that each of us is looking at it differently. The way I present it is certainly the least sophisticated, but it does give a numerical "feel" to the problem. I hope that someone will set up and run the software to verify if we are correct. Another way would be to use a spreadsheet program to do a Monte Carlo simulation of a large number of upper beams with various "stiffness".

www.SlideRuleEra.net

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## RE: Beam on top of beam, not connected, and of different length.

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## RE: Beam on top of beam, not connected, and of different length.

There are actually loading scenarios where loads

notplaced over the upper beam would still be partially resisted by the upper beam.## RE: Beam on top of beam, not connected, and of different length.

Huh?

## RE: Beam on top of beam, not connected, and of different length.

as shown above it depends on many factors ...

is load applied to the upper beam ? (for clarity I'd restate your questions as ... how much load does the upper beam react load?)

if no load (it is all applied to the lower beam), then the upper beam will pretty much go along for the ride, if it is not continuously attached (which seems implied, 'cause if continuously attached the beams will work together).

if all the load is applied to the upper beam, then it'll share with the lower if it's deflection causes it to bear up against it; the lower beam is deflecting due to load applied to it beyond the span of the upper beam.

if load is applied to both, then it'll depend on how this distribution is happening ... fixed proportion or stiffness related.

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

## RE: Beam on top of beam, not connected, and of different length.

Case as stated in the initial question........(1) Lower beam with length L, (2) Upper beam of length L/2 positioned over "first" half of span of lower beam, (3) Loads placed anywhere on upper beam.......is just special case of the more general case of upper beam placed anywhere within length of lower beam.

For the general case........results can be classified into two subsets; (1) When upper beam is in contact with lower beam at ends only, and (2) When upper beam is in contact with lower beam at each end (of upper beam) and at some other point or points between ends (of upper beam).

For earlier (also special) case discussed above...with upper beam centered within span of lower beam...my initial calculation was off by factor of 2.....correct result is that upper beam will remain in contact at ends only when upper beam has stiffness (EI) that is 18-percent (2/11) of stiffness of lower beam. For greater stiffness (of upper beam), upper beam is in contact with lower beam only at ends of upper beam.

For upper beam with stiffness (EI) less than 18-percent of lower beam.......some redistribution of load (applied by upper beam to lower beam) will occur since upper beam will now be in contact with lower beam at midspan (of both beams)......resulting now in 3 locations (instead of just 2) where load is applied by upper beam to lower beam....whew!

Key to solution is "simply" that there are 2 unknowns and 2 equations. Unknowns are conveniently defined as; (1) Deflection of lower beam at midspan (Dm)....and (2) Reaction force applied to lower beam by upper beam at each end of lower beam; Pend. Other "unknowns" are easily calculated.......such as , force applied to lower beam at midspan (Pm)is equal to (P - 2Pend). All other deflections of lower beam may be calculated once the two forces (Pm, Pend) are known. Deflections of upper beam may be calculated once Pm is known (which acts upward against upper beam).

Net result (for equal modulus, E)...........Pend = 8P I-upper / (2 I-lower + 5 I-upper}

When I-upper = (2/11) I-lower...........Pend = P/2........such that Pm = 0....(no force applied by upper beam to lower beam at midspan)

For greater I-upper......there is no contact between beams except at each end of upper beam.......and Pend = P/2

For lesser I-upper.......Pm > 0.......Pend < P/2.......upper beam is in contact with lower beam at 3 points

When I-upper =0.......Pend =0.......Pm = P.......and it is as if there is no upper beam (which must be case if I-upper =0)

For general case (with upper beam located anywhere along lower beam).........calculations just get more complex..........such that numerical solution using spreadsheet is the way to go!

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

(2) Reaction force applied to lower beam by upper beam at each end of UPPER beam; Pend.Sketch would help I know

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

@Hokie: is that "huh", KootK's statement is unintelligible and requires clarification? Or "huh", KootK's statement is unbelievable and requires proof? I'll tailor my response accordingly.

## RE: Beam on top of beam, not connected, and of different length.

"Limiting" stiffness of upper beam can be calculated, similar to (but more complex) than above.

When stiffness of upper beam is greater than such limit......bending stress in upper beam and deflection of upper beam (between ends of upper beam) is not reduced by any property of lower beam..........although downward movement at end of upper beam that is on lower beam is of course a function of the stiffness of lower beam.

When stiffness of upper beam is less than limiting value.........upper beam will contact lower beam at some point between ends of upper beam.........and then the lower beam will have some effect that reduces stress and deflection of upper beam (compared to case with no interior contact).

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

When lower beam is entirely within span of upper beam (inside of supports)......load is not necessarily shared in the sense of load being apportioned to each beam in accordance with relative stiffness or some other criteria. As shown above.......when stiffness of upper beam is greater than some limiting value, the upper beam must resist all the load by itself, spanning between ends of upper beam. For such condition, the lower beam also must resist all the load, with some part of total load applied to lower beam at each end of upper beam. However, some effects (bending moment, deflection) of those 2 loads on lower beam are reduced compared to the case of load being applied to lower beam (at same location or locations along lower beam) without any upper beam.

To summarize.......key to the question is whether or not upper beam contacts lower beam between ends of upper beam.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

I just couldn't figure what load a beam not loaded would need to resist. So I suppose 'unintelligible' is the way I would describe your statement, but maybe I'm just thick.

jfmann,

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

## RE: Beam on top of beam, not connected, and of different length.

1) imagine a 10' beam.

2) throw a 5' beam over the left half of the 10' beam.

3) place point load A at 2.5', on the upper beam.

4) imagine that stiffness are such that the beams remain in continuous contact.

5) add point load B at 6', on the lower beam.

6) Imagine that point load B causes the beams to separate vertically.

In this case, I propose that the upper beam helps to resist point load B even though B is not actually applied to the shorter beam.

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

now along comes load B, on the lower beam. this'll affect (reduce) the contact between the beams, causing the upper beam to deflect more, and react more load at it's ends. So as a consequence of load B, I see an increase in strain energy in the upper beam (due to load A).

does any of this help the OP ?

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

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

(1) Same modulus for both beams......and uniform section for each beam

(2) For any point on upper beam.....downward movement due to deflection of lower beam ...is taken as straight-line proportion (between ends of sloped upper beam)

(3) Between ends, deflection of upper beam calculated using standard deflection formulas

(4) Single point load (P) applied at midspan of upper beam.......this results in load of P/2 applied at midspan of lower beam (and at rigid support for both beams)

For any point on upper beam........total downward movement is the sum of (2) and (3).

At rigid support, deflection of upper beam due to (2) is of course zero

At inside end (which is at midspan of lower beam), deflection of upper beam due to (2) is equal to deflection of lower beam (since upper beam is in contact), while deflection of upper beam due to (3) is of course zero.

To ensure that there is no contact between beams other than at ends of upper beam.........moment of inertia of upper beam must be at least 75.22 percent of the moment of inertia of lower beam.

For lesser moment of inertia (of upper beam)......there will be contact between ends of upper beam.........the location of which is a function of the moment of inertia of upper beam. Such contact then causes some redistribution of load onto the lower beam, although the effect is relatively minor unless upper beam is much more flexible.

For point load applied to upper beam at different location..........results are expected to change somewhat (would have to extend routine for that)

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

First photo: A single beam with UDL over half its length. Analog mid span deflection = 0.81"

^{+}0.03" Theoretical mid span deflection = 0.80"Note that I do not use a true UDL, but 9 equal & equally spaced point loads. I did the math, 9 point loads is better than 99% the same results as a true UDL.

Second Photo: Same long beam, with similar closely matched short beam spanning half of the long beam. Same UDL. Analog mid span deflection = 0.72"

^{+}0.03"Here are the detailed inputs:

Long Beam: 39" clear span, simple supports, I = 3.4 x 10

^{-3}In^{4}Short Beam: 19.5" total length, I = 2.9 x 10

^{-3}In^{4}Uniform Distributed Load = 0.317 Lb / In, It covers the top of the 19.5" beam.

Jfmann - Please see if you can confirm, or disprove these results. Then give us the numbers that your calculations reveal for the above inputs.

The more I work with the analog model, the more intricate and complex I find a "digital" solution will be. To be sure, I don't know exactly what is going on with these two beams, but they are definitely sharing load. I have come across a clue for the reason... but we'll get back to that later.

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## RE: Beam on top of beam, not connected, and of different length.

^{6}psiwww.SlideRuleEra.net

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## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

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

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

As stiffness of upper beam becomes less....contact occurs near midspan of lower beam, with initial point of contact moving closer to support as stiffness is reduced.

This behavior is somewhat different with point load at midspan of upper beam (discussed previously)......for which case initial point of contact occurs about 3/4-span of upper beam.

SRE.......for specific conditions listed...calculated deflection is 0.64 inches (PL^3/48EI-lower.....with P = wS/2 (S=span of upper beam)

Although, as noted in one prior post, this beam-arrangement may be considered "load sharing", it is significantly different than the condition when two elements are connected to form a unified assembly with greater section properties.

Upper beam is simply redistributing load on lower beam (compared to case without any upper beam).......resulting in reduced deflection and bending stress in lower beam........of course at the cost of using upper beam. For the current case, half the total load is effectively removed from lower beam for deflection and bending stress (but not for bearing at support). Of course deflection is not reduced by half since remaining load is shifted out to midspan of lower beam, increasing deflection for that half of the load compared to condition without upper beam.

For current case (using listed data per SRE).......without any upper beam.........midspan deflection of lower beam should be 0.80 inches (just as measured which I find quite amazing given many potential sources of variation with inputs).

If the two beams were connected together to form a composite section, effective stiffness would be greatly increased and deflection would be much less (to be continued).

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

I did an error analysis on the model and was able to greatly improve accuracy, specifically by testing to get a better estimate for the two beams' moment of inertia. Also, helps that as time goes by I've had a number of occasions to solve problems using analog methods. Long ago, even had the opportunity to get a brief introduction to a "real" electronic analog computer - just before it was retired. They were very impressive machines and amazing to watch the results being drawn as a graph on a large X-Y plotter. Of course, they were no match for a digital computer, even then... except for rare, unusual problems. I'm beginning to believe that may be the case here.

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## RE: Beam on top of beam, not connected, and of different length.

I think you may be getting some composite action with your short beam. Would be interesting to see what happens if you grease it...may have to pin one end to prevent it sliding downhill.

## RE: Beam on top of beam, not connected, and of different length.

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## RE: Beam on top of beam, not connected, and of different length.

This demonstrates how connecting beams together (for true load sharing) is much more efficient than just using 2 independent "stacked" beams.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Of course, we know that, but that is not the subject here. We are arguing about beams that are not composite in any way.

SRE,

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

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

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

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

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.

www.SlideRuleEra.net

www.VacuumTubeEra.net

## RE: Beam on top of beam, not connected, and of different length.

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.

## RE: Beam on top of beam, not connected, and of different length.

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

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

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.

www.SlideRuleEra.net

www.VacuumTubeEra.net

## RE: Beam on top of beam, not connected, and of different length.

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.www.SlideRuleEra.net

www.VacuumTubeEra.net

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

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

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

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

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

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 ?

## RE: Beam on top of beam, not connected, and of different length.

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.

## RE: Beam on top of beam, not connected, and of different length.

...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

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

@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 ?

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

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

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

The scenario I considered had w on the upper beam and w on the portion of the lower beam without a beam above it. Upper beam extends to midspan. Therefore the loading on the lower beam is P = wL/4 at midspan and a uniform load of w on half the span. And "L" is the length of the lower beam.

## RE: Beam on top of beam, not connected, and of different length.

I hope I'm not too late to throw another spanner in the works.

With a stiff upper beam the load is transferred to the ends of the beam, and calculation of shears, moments and deflections is quite straightforward (making all the usual simplifying assumptions), but what happens as the stiffness of the upper beam is reduced, so that there is contact between the two beams along their length?

I have come to the conclusion that for a point load the upper beam will contact either at the two ends, or in the middle, but there will never be continuous contact. For a uniform distributed load there may be contact at the ends and in the middle, but there will still be no continuous contact.

My argument is as follows:

Consider two thin weightless beams, no shear deflections, plane sections remain plane, zero friction between the two beams (and anything else I may have forgotten).

For simplicity consider the short beam to be half the length of the long beam, placed symmetrically with a point load at mid span.

The upper beam is sufficiently flexible to contact the lower beam.

From symmetry both beams have zero slope at mid-span.

To remain in contact both beams must have the same curvature, so the bending moment at mid-span must be in proportion to their stiffness.

To remain in contact the slope and curvature must change at the same rate, so the bending moments must reduce in the same proportion, but the shear force is decreasing in the top beam and increasing in the bottom beam, so their bending moments cannot remain in the same proportion as we move along the beam.

It follows that the two beams can only be in contact at mid-span.

With a UDL the top beam may deflect to re-contact at its ends, but at this point it will have a different slope to the lower beam, and zero curvature, so this can only be a point contact.

Hence the upper beam may contact the lower at both ends, in the middle, or at all three points, but there can never be a continuous contact.

Of course in a real beam there is no such thing as a point load, the contact "point" will be distributed by non bending behaviour, and shear deflections will probably be significant, but even so I would expect the load transfer to be much more localised than a simple analysis might suggest.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

I suspect that you're right that, strictly speaking, there is no continuous contact. However, I think it likely that there are more than three discrete points of contact. Depending on relative, stiffness, perhaps enough to approximate continuous contact.

## RE: Beam on top of beam, not connected, and of different length.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

"To remain in contact both beams must have the same curvature, so the bending moment at mid-span must be in proportion to their stiffness.

To remain in contact the slope and curvature must change at the same rate, so the bending moments must reduce in the same proportion, but the shear force is decreasing in the top beam and increasing in the bottom beam, so their bending moments cannot remain in the same proportion as we move along the beam.

It follows that the two beams can only be in contact at mid-span."

Discussion to this point has varied between 2 different configuration of upper beam.........for now, per your stated conditions, we go back to configuration with upper beam centered within span of lower beam .....and with point load at midspan of upper beam.

First, there is no valid reason to limit conditions to "flexible" lower beam........one end of the spectrum is infinitely stiff lower beam and infinitely flexible (zero stiffness) upper "beam" (actually membrane)........for which case certainly upper beam simply takes shape of lower beam and there is complete continuous contact.

For conditions somewhere in middle of spectrum .....with both beams having stiffness between zero and infinite, but not necessarily equal .....and for condition when midspan of upper beam (supported at each end on lower beam) has deflection just enough to contact lower beam .........curvature of each beam is function of stiffness as well as loading..........both of which are different for each beam.

Note that initial instance of this condition occurs without any reaction force occurring at point of interior "contact"; picture upper beam just a micron above lower beam.

For upper beam to be in contact with lower beam only at midspan (of itself) ......but not at each end....is not possible since the following inconsistent conditions would have to exist at the same time;

1) Point load transferred directly through upper beam into lower beam.......such that there are no upward loads anywhere else against upper beam .....so that upper beam could only be straight, without any curvature.

(2) Lower beam would deflect due to point load and would therefore have curvature.........so that ends of straight upper beam would have to be in contact with lower beam.

More importantly, ....since slope of any beam is function of moment of inertia (I-value), length, load and position along beam.......it is clear that there are many combinations for two beams to have different slopes.

For example.......for configuration considered, slopes at x=L/4 (of main beam) are calculated using any method (conjugate beam, virtual work);

Upper beam, slope = PL^2/64EI-upper............which is standard PL^2/16 (for generic "L") for slope at end of beam, with L/2 input for actual length of upper beam

Lower beam, slope = PL^2/32EI-lower

So that, clearly.......slope of each beam is different, even for same I-value, which of course can vary from zero to infinite for each beam.

Bottom line.......in general, beams do not have same curvature

Indeterminate Behavior

To clarify previous post.......configuration is statically determinate when upper beam contacts lower beam at ends only.

When upper beam contacts lower beam at some interior point (between ends up upper beam), behavior changes to indeterminate.........meaning that equations of geometric compatibility must then be used to obtain solution for reaction forces at each point of contact. Interior support (provided by lower beam) for upper beam, between ends of upper beam, is then a spring-type support.........not a rigid (non-yielding) support.......since lower beam continues to deflect at that location.

As discussed in previous posts...........spreadsheet routine is very useful for such indeterminate conditions to illustrate behavior for range of beam stiffness and support configurations.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

I agree with this point. To have any curvature there must be some shear transfer at points other than mid-span, so my suggestion of the point loaded beam contacting only at mid-span clearly does not work. There must be sufficient shear transfer at the ends of the short beam so that its curvature at mid-span is greater than that of the long beam. It will then have three points of contact, two ends and mid-span.

I will respond to the other points later; I have to go now.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

Infinitely stiff lower beam and/or zero stiffness upper beam: Yes, in this case the two beams will remain in contact for any loading, and the upper beam will have no effect on the load distribution in the lower beam (assuming it is weightless). So my comments only apply to all real beams.

Curvature depends on stiffness as well as loading: Yes, but it remains true that for the two beams to be in contact over any finite length they must have the same slope and curvature over that length, so the ratio of bending moment to flexural stiffness must remain constant over that length.

..it is clear that there are many combinations for two beams to have different slopes: yes, in general the slope of the end of the upper beam will be different to the slope of the lower beam at that point. That is why they can't remain in contact. In fact they must have different slopes, because the curvature of the upper beam is zero, but the lower beam has a positive curvature, so if they had the same slope at the end point the upper beam would have to pass through the lower beam because it's downward slope was constant, whereas the lower beam was curving upwards.

Bottom line.......in general, beams do not have same curvature: That is the point. To remain in contact they must have the same curvature over the contact length, but due to the nature of the load transfer it is not possible for the curvatures to remain the same.

Regarding the behaviour of a beam with uniform load, I am now confident that this is the same as for a point loaded top beam; indeed the lower beam has no information about what the loading on the top beam is.

Starting with a comparatively rigid top beam, all the load will be applied as point loads at the top beam ends. If the stiffness of the top beam is now progressively reduced until the beams contact at mid-span, there will now be a load transfer a mid-span, as a point-load, and this behaviour will be the same no matter whether the load applied to the top beam is a uniform or a point load. The load at which contact occurs will be different, but the behaviour after contact will be similar, i.e. the contact point will remain a point.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

For a point load at mid-span there are always 3 point contacts, no matter how flexible the upper beam is (greater than zero), but for a UDL as the upper beam stiffness is reduced the contact will spread out from the centre, and the contact pressure will adjust so that the curvature of the two beams is equal over the contact length. At some point the contact pressure will reduce to zero, and the beams will separate until recontacting at the end of the shorter beam.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

Addressing following statement by Doug;

"In fact they must have different slopes, because the curvature of the upper beam is zero, but the lower beam has a positive curvature, so if they had the same slope at the end point the upper beam would have to pass through the lower beam because it's downward slope was constant, whereas the lower beam was curving upwards."

With load on upper beam, slope (curvature) at end of upper beam is not zero. Also, per my previous post, at end of upper beam, slope of each beam is different........for equal I-values, slope of upper beam is half the slope of lower beam for that specific configuration (upper beam centered on lower beam).

For future reference, I designate this configuration (upper beam centered on lower beam) as Case 2.

Per previous post....for Case 2....and based on assumption that contact at each end of upper beam is a point-contact (which is not correct for relatively low values of I-upper as discussed below).......contact first occurs when I-value of upper beam is (2/11) of the I-value of lower beam. Of course we are assuming constant I-value for each beam (and same modulus). We now calculate slope for each beam, at end of upper beam (x-lower = L/4). Actually, those are the same values as listed in most recent previous post since....for I-upper just less than (2/11) I-lower.....no force transfer occurs at this "contact" point. So now we just have to input the new I-upper;

Upper beam, slope = PL^2/64EI-upper = PL^2/[64E(2/11)I-lower] = 11PL^2/128EI-lower = 0.086 PL^2/EI-lower

Lower beam, slope = PL^2/32EI-lower = 0.031 PL^2/EI-lower

If correct, slope of upper beam would then be greater than slope of lower beam.....at end of upper beam. This can not occur.......such that there must then be some length of contact (instead of only point of contact) at end of upper beam. Such length of contact will change behavior (for both beams)......since assumed shape of upper beam will be different compared to only point-contact at end.

So now this becomes even more interesting;........Is end of upper beam "forced" to take on same curvature as lower beam for some length?.......If so, how does that alter behavior of each beam compared to assumption of only point of contact?

First thought is to assume, that yes, upper beam will then be forced to take on same curvature........and that this is "accomplished" by force transfer taking form of distributed load (whether constant-uniform or some other distribution) instead of point load at end of upper beam.

Of interest now is.......for what I-value does slope at end of upper beam first equal slope of lower beam?

This is easily answered by equating the two expressions above, so that (for Case 2).........I-upper = (1/2) I-lower

For lower value of I-upper, end of upper beam must then take on same curvature as lower beam. We must then calculate; (1) Location where slope of the 2 beams is no longer equal, and (2) Shape of distributed load at end of upper beam.........which I suspect (for now anyway) will be getting tad beyond "scope" of basic analysis since distributed load (at end of upper beam) must very likely be varying (non-constant)........and may even be non-linear. It may even be that gap occurs under end of upper beam, with contact occurring inboard of end of beam.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Yes, they can be in point contact, but they cannot be in contact over any finite length unless both beams have the same slope at any section through the contact length. Therefore the curvature (rate of change of slope) must also be equal over the contact length, so the bending moments must be in proportion to the beam flexural stiffness values, as must the shear forces.

I have now satisfied myself that the way this works for an upper beam with a UDL is:

- There is a point contact at the ends of the beam.

- At mid-span they are in contact with zero shear and equal curvature.

- Moving away from mid-span they remain in contact with the UDL distributed such that the shear force and bending moments remain in the ratio of their flexural stiffness, hence the curvature of the two beams remains equal at any cross section.

- At some point the beams separate, such that the upper beam can have zero bending moment at the end, and the lower beam has a moment equal to the end reaction x the distance from the end of the upper beam to the support.

- At this point there is another point load transfer to maintain equilibrium.

I have modelled this behaviour in a spreadsheet, producing the graphs shown below. In the deflection graph (left) the red and blue lines show the beam deflections measured from the end supports (the blue line is hidden because the separation is very small). The green line shows the separation of the two beams (right hand scale), showing they are in contact for about 1 metre, then separate and re-join at the upper beam end. The bending moment diagram shows that the bending moments factored by the stiffness ratio (10) are exactly equal over the contact length, then diverge.

I have also uploaded the spreadsheet (see SSSpanU Example Tab), which uses the Excel Solver to find the end reaction, and uses a quadratic formula to find the contact length. But beware, it is very sensitive to changes in these two values, and needs a good starting guess, or it won't find a solution.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

First.......lets clarify that configuration being analyzed is for upper beam (length L/2) centered within span of lower beam (length L).....and for uniform load on upper beam only.

For more general condition length of upper beam could be different (within limits so as not to result in drastically different configuration that would not be applicable) and loading on upper beam could be whatever you want.

Taking your conclusions in order;

(1) At end of upper beam......clearly, at lowest theoretical limit of stiffness for upper beam (I-upper = 0)......and with I-lower > 0.......upper beam is in full contact along entire length of upper beam. For positive, but very low value for I-upper, some gap eventually develops. However, without at least some mathematical basis, there is no basis for claim that contact at end of I-upper is limited to point contact. In fact, for uniform load on upper beam ...and for relative stiffness used for your analysis (I-upper = (1/10) I-lower) there must be some length of contact at each end of upper beam, as discussed below.

(2) For specific case you are considering, there is contact at midspan.........but this is not always the case, as discussed below, and as discussed at length in previous posts.

(3) This appears to be key flaw in your analysis. Moment (and shear) in each beam is not proportional to relative I-values (stiffness is not correct term since stiffness is function of E, I and L) wherever curvature is the same; see below for explanation. However, your moment graph already proves that this is not the case...........since, where red line crosses blue line, graph shows equal moments at point where the beams are not in contact and therefore do not have same curvature.

(4) Zero moment at end of upper beam is not caused (as implied by your statement) by separation of two beams somewhere between end and midspan of upper beam. Moment at end of upper beam is zero since that is always the case at end of beam that is free to rotate. Your description of moment in lower beam is correct only for x = L/4 (at end of upper beam). However, there is no obvious meaning to such statement.

(5) Meaning of "At this point" is not clear. More important is lack of free-body diagram to demonstrate "equilibrium" of upper beam.........which is most important flaw in your analysis.

Since your deflection graph lists absolute values.........you must be using specific moment of inertia values for each beam (which is fine for spreadsheet routine)........however such limited analysis is of course not useful for understanding general behavior for this setup.

Sounds like your analysis is for I-lower = 10 I-upper.........which is same as I-upper = (1/10) I-lower.

Red line in moment graphs must be for upper beam (with M = 0 at left end).

Explanation that solution is "very sensitive to starting value" is strong indicator that your analysis is not quite correct.

Based on your "timeline" of events, it sounds like you have started from an incorrect assumption that both beams start off with contact at midspan (only?)........and that the ends of upper beam mysteriously rise up above lower beam and then "reconnect" just as mysteriously at some future (unspecified) time.

If, as should be the case, analysis is for any moment of inertia values for upper beam (I-upper) and lower beam (I-lower)........then your claim that there must be contact at midspan of lower beam is not correct unless I-lower is infinite (in which case it is bedrock, not a beam!) or, more usefully, large enough.......relative to upper beam........such that deflection of upper beam results in such contact. Obviously, for a wide range of I-upper values (starting of course with theoretical "infinite"), deflection of upper beam (acting as simply-supported beam spanning between ends which are in contact with lower beam) is not enough to result in contact at midspan of each beam (for stated configuration).

For real-life demonstration of this behavior.......just review photos provided by SRE (above) which clearly show ......for point load and uniform load on upper beam.......continuous gap between both beams except at ends of upper beam.

As reported in previous posts........it is relatively straightforward to calculate value of I-upper that results in contact at midspan.........which is I-upper = (2/11) I-lower.

However, with I-upper = (2/11) I-lower, there is not yet any redistribution of load (compared to contact only at ends of upper beam) since upper beam has barely made contact (picture upper beam one micron above lower beam for I-upper infinitesimally less than (2/11) I-upper).

As value of I-upper is reduced below "first contact" limit (FCL; (2/11)I-lower ].......redistribution of load onto lower beam occurs, with lower beam providing an effective spring-type support for upper beam at midspan (of both beams). However, as discussed in my previous post, this is when behavior gets complex since contact at ends and midspan of upper beam is now over some length (instead of just point)......although, for at least some I-upper values near FCL, an assumption of point contact remains accurate enough for engineering purposes certainly and very likely for pure-analysis purposes as well.

As for relative slopes

at end of upper beam(again, for stated case);Slope of I-upper = wL^3/[192 E I-upper].........with "L" of course defined as L-lower.........and L-upper = L-lower /2 or L/2

Slope of I-lower = wL^3/[64 E I-lower].........obtained most conveniently via conjugate beam

Equating these two expressions results in.........I-upper = I-lower / 3

So that...........when I-upper is 33.3% of I-lower........slope (curvature) of each beam is equal at end of upper beam

Since this value of I-upper is greater than (as previously reported) value of I-upper that would result in first contact at midspan [ I-upper = (2/11) I-lower] if there were only point contact at each end......it is reasonable to conclude that slope of each beam is then the same at end of upper beam, before there is any contact at midspan.

As I-upper is reduced below (I-lower / 3)....but is greater than (2/11) I-lower (meaning no contact at midspan).......one of the following must be the condition at end of upper beam;

(1) Length of contact at each end must then be greater than zero

or

(2) Point of contact occurs inboard of end of upper beam........and there is no contact (gap) outboard of this new point of contact

For I-upper just less than I-lower / 3......slope of upper beam (at end) would have to be greater than slope of lower beam, except for the "obstruction" caused by contact with lower beam. The lower beam effectively causes an opposite rotation for end of upper beam............which is effectively the same as would occur if a moment is applied at end of upper beam causing opposite rotation. By calculating the difference in rotation for upper beam (that is, the "lost" rotation compared to condition without such obstruction), we can calculate amount of effective moment that has been applied by the upward force from lower beam.

Having now a modified expression for rotation of upper beam. If we know where to apply this moment (in opposite sense) to lower beam, we can then also determine modified expression for rotation of lower beam........which is necessary since force applied by upper beam is now distributed instead of point. We may then calculate modified slope of each beam along length of lower beam (which is of course the baseline).

Problem of course is determining where along lower beam this effective moment occurs (for lower beam)........or, stating another way, shape of distributed force along length of contact ...constant uniform, or uniformly varying or non-linear?

For relatively small length of contact......we might "reasonably" assume constant uniform distribution. However, as length of contact increases, shape of this distribution becomes critical for accurate analysis.

More development of necessary expressions is required for complete analytical solution........and makes me wonder whether anyone has bothered to do this before! (very likely yes, somewhere, sometime)

However, basic behavior is not as proposed by Doug above.

Finally.....claim that moment in each beam is proportional to relative I-values ......when beams have same slope (curvature)......is not correct. Moment graph by Doug shows that is not correct as previously noted. However, for more analytical demonstration........we can start by calculating moment in each beam at midspan, since (for this case) slope of each beam is of course the same value; zero.

M-upper = w(L/2)^2/8 = wL^2/32

M-lower = (wL/4)(L/2) - (wL/4)(L/4) = wL^2/8 - wL^2/16 = wL^2/16

So M-lower = 2 M-upper..........for all I-values for each beam...........that is, moment in each beam is not dependent on relative I-values

Also.......when slope at end of upper beam first equals slope of lower beam........moments are clearly not equal since M-upper = 0 and M-lower = (wL/4) (L/4) = wL^2/16

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

With distributed contact at each end of upper beam........what would otherwise be full rotation at end of upper beam is not allowed to occur (due to obstruction of lower beam being in the way)........resulting in some "effective moment" at end of upper beam that causes upward deflection of upper beam, relative to condition with only point contact at ends.

Therefore, since such distributed contact occurs for I-upper = (1/3) I-lower.....that is, for I-value greater than (2/11) I-lower....contact at midspan will not occur for I-upper = (2/11) I-lower.........there will still be some gap between beams at midspan.

At what value of I-upper this gap at midspan is finally closed is yet to be determined analytically (by myself anyway)........though, for now, I suspect there is some such value (which would of course have to be relatively low).

Also.......my conclusion (stated in previous post)........that contact at midspan occurs when I-upper = (1/10) I-lower, which is condition that Doug analyzed........may not in fact be correct since resistance to end-rotation of upper beam might even then be great enough (due to distributed contact at end of upper beam) to prevent such contact at midspan.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

1) the left end condition and the relative stiffnesses of the two beams are critical to understanding the solution.

2) if the left end of the top beam is on the support and the top beam is stiffer than the bottom beam such that there is essentially two-point contact between the two beams, fully half of the load on the top beam simply passes into the support and other than additional shear never makes it to the bottom beam. In this case, it does help the bottom beam if the bottom beam is not end-shear limited. Essentially, this can be generalized to say that the upper beam transfers some of the load towards the end of the lower beam, which is good for the lower beam until it becomes shear limited. So, in all cases as long as the upper beam carries some load to the end reaction it has to benefit the lower beam.

3) to the degree that either or both beams extend past or are short of the left support, the moment distribution in the bottom beam changes and so would it's deflection, stress, etc. Again, if any of the load is moved towards the end of the lower beam, it will help it.

4) if the upper beam is of great stiffness then it is a simple situation with two points of contact, with half the load now on the lower beam at the free end of the lower bean and no load sharing along the gap between the two beams.

5) if the upper beam is of minimal stiffness then it is again a simple situation this time with matching deflections along the beam interface.

6) when relative stiffness between the upper and lower beams is in somewhat comparable, the contact patches at the ends of the lower beam grow from zero (high upper beam stiffness) to some intermediate length of contact patch to perhaps fully in contact. This does not mean that the contact forces are equal, in fact I would expect them to vary. In these situations, the lower beam would feel a variable load distribution along some length at both ends of the lower beam. Again, if any of the load is moved towards the end of the lower beam, it will help it.

7) where there is load sharing along the entire length of the beam interface (non-uniform for any realistic stiffness of the upper beam), any additional load on only the lower beam causes additional lower beam deflection, a P-delta analysis will reveal where the next geometrical equilibrium exists and what the new extent of the contact areas are between the beams and how the contact forces vary within the contact patches. This new status changes the interaction between the two beams and must move the interaction along the continuum from uniform to two-point loads which in every case has to be worse for the lower beam.

8) a FEA analysis would clarify this easily

I'm probably full of sh.t but it was fun!

## RE: Beam on top of beam, not connected, and of different length.

Where they are not in contact the slopes and curvatures can be anything, but if the beams are in contact with different slopes and/or different curvatures then it can only be a point contact.

It follows that two beams of any finite stiffness can only be in contact over the full length of the shorter beam if the longer beam has zero moment at both ends of the upper beam (assuming transverse loading, and no friction between the beams), which is not the case for any of the examples discussed here.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

In fact if we are talking about steel beams of normal dimensions I think the variations from simple beam theory would be fairly small. The point contact at the ends would extend to over a small length, and there will be some shear deflection effects, but that's about it.

By the way, is anyone else getting the display area here extended beyond screen width, or is it just me? Makes reading very difficult.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

(6) For all practical purposes (and even for analytical purposes) point-contact occurs at each end of upper beam unless I-value of upper beam is low enough such that slope at end of upper beam (calculated for condition of point contacts) is equal to slope of lower beam.........or would be greater than slope of lower beam if free-rotation of upper beam were not prevented from occurring by lower beam. For what I have designated "Case 2", which is upper beam centered on lower beam, "low enough" means that I-upper <= I-lower/3. See below for discussion of this issue for Case 1.

(7) For Case 1......point-contact at each end may not be worst case (especially for shear).........since, for distributed contact at outer end of upper beam, some load that would (for point contact) be transferred (by upper beam) directly to support is now applied to lower beam, even though this load is very near end of lower beam.

(8) Though Finite Element Analysis might be of some use at micro-level near supports.......especially if bearing stress were critical.......I doubt such method is going to be of any benefit for understanding of the problem or the solution!

For Case 1........comparing slope of each beam is complicated by difference in elevation between ends of upper beam........due to deflection of lower beam at midspan.......which causes undeflected line of upper beam to rotate slightly (about end at support) with respect to initial level position.

For uniform load on upper beam........before considering macro-rotation of upper beam, we have the same result as for Case 2; I-upper = (1/3) I-lower

Slope of upper beam = w(L/2)^3/(24E I-upper) = wL^3 / 192E I-upper

Slope of lower beam = (wL/4)(L^2)/16E I-lower = wL^3 / 64E I-lower........with Point Load, P(at midspan)= wL/4

Initial (undeflected) line of upper beam rotates through an angle, the tangent of which is equal to midspan deflection of lower beam divided by length of upper beam (L/2). At support, this angle must be added to slope of upper beam (calculated for initial level position) to obtain actual slope of upper beam.

Midspan deflection of lower beam = (wL/4)L^3/48E I-lower = wL^4 / 192E I-lower

Tangent of angle = [wL^4/192E I-lower] / (L/2) = wL^3/96E I-lower

In usual parlance of analysis.......we take tangent of small angles to be equal to the angle (which of course is in radians already), so that;

Slope of upper beam = wL^3 / 193E I-upper + wL^3 / 96E I-lower

Now setting expressions for slope of each beam equal, we obtain the following (perhaps surprising) result.......I-upper = I-lower

This means that distributed contact occurs at outer end of both beams when I-upper is equal to or less than I-lower.

This can be understood by considering that half the total load has been transferred off of lower beam by upper beam........but even moreso due to deflection of lower beam, which adds to outer-end rotation of upper beam........such that upper beam can have relatively large I-value (compared to Case 2).

However, at inner end of upper beam.........we must subtract macro-rotation of upper beam, and set expression for slope of upper beam equal to zero-slope of lower beam at midspan, resulting in the following; ......I-upper = I-lower/2

This means that distributed contact occurs at inner end of upper beam when I-upper is equal to or less than 50-percent of I-lower.

For Case 1........point-contact will occur at each end of upper beam as long as I-upper is at least half of I-lower.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Anyway.....believe my analysis addresses all combinations of I-values for each beam

Essentially, for Case 2.....as I-upper is decreased below limiting value (I-lower / 3)........contact "starts" at each end ......and increases toward midspan as I-upper continues to decrease.

Analytically at least......contact along full length of I-upper only occurs for I-upper = 0.

Up next (if feasible without resorting to high-powered analysis program, which, again I do not believe will be necessary) is to work out shape of distributed pressure when finite-length contact occurs.

Expanding on that issue;

When upper beam is flexible enough such that lower beam prevents full rotation of upper beam at ends of upper beam......lower beam does this by exerting (redistributed) upward force on upper beam that effectively can be considered to have the same effect (on upper beam) as a concentrated moment applied to end of "free" upper beam. That is because what would otherwise have been rotation of upper beam at end........and deflection of upper beam at midspan.......are now reduced. Another (and perhaps more insightful) way to consider this effect is to see that effective "span" of upper beam has been reduced.

At same time, center-of-gravity of force applied to lower beam (at each end of upper beam) is now moved closer to midspan of upper beam. Total force at (near) each end remains the same.

For Case 2.......this condition then results in greater (even though relatively small) deflection and rotation of lower beam......so that we get cycle of movements that converge to solution.

For Case 1........effect is more complex since effect (on lower beam) at one end of upper beam is opposing effect at other end.

Analytically.......this might devolve into some horrendous higher-order equation.........well, hopefully not!

Numerical solution may then be order of the day.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

We then find that effective span length is reduced quickly;

For example.........if we consider I-upper = I-lower / 4..........which is less than limiting value of I-upper / 3

Effective span for I upper becomes 0.908 of L/2.....0.908 is cube root of (3/32)........meaning that, at each end, center-of-gravity of "reaction" force is moved inward about 5-percent of original span for upper beam (which is L/2).

Of course, now we have to "recalibrate" by calculating "refined" slope of lower beam at this new location........and the beat goes on!

Clearly there is higher-order equation to be developed.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

However, as I have been describing (at length!) in most recent posts........it is this very interaction which must be addressed (and which I have been addressing with equations) to obtain complete solution.

At each end of upper beam.....and of course this is only when I-upper is sufficiently less than I-lower for distributed contact to occur.....lower beam will "force" upper beam to maintain same slope as lower beam (although secondary effect of interaction is that there will then be some additional effect on lower beam itself).

Due to this "forcing".......effective length of upper beam is reduced as length of contact increases towards midspan of upper beam.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

No, I mean under any circumstances whatsoever.

For two surfaces to remain in contact over any finite distance they must have the same curvature, by definition.

It follows that for the two beams to have any more than point contact at the end of the short beam they must have the same curvature, so either the top beam must have a non-zero moment, or the bottom beam must have zero moment, but both of these are ruled out by statics, so it must be a point contact.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

When I-upper is greater than limiting value (which depends on configuration).......there is only point contact at each end. Each beam has different curvature with slope of upper beam necessarily less than slope of lower beam.

When I-upper is less than limiting value.......there is some length of contact, resulting in distributed pressure, since lower beam prevents what otherwise would be greater end-rotation of upper beam. For such condition, both beams must have the same curvature along that length of contact, however short it may be. I have stated exactly that in several posts now. Solution to determine exact length of contact and shape of pressure-distribution along length of contact must be based on a geometric-compatibility condition of equal-curvature along length of contact.

When I-upper is greater than or equal to limiting value......any specific configuration of this general configuration (short beam on top of long beam) is a relatively simple determinate problem.........which may be solved using statics alone.

When I-upper is less than limiting value........any configuration of this general configuration becomes an indeterminate problem........requiring use of additional calculations to ensure geometric compatibility. Of course this means that statics alone does not provide a solution.

Your conclusion about resulting moments is not correct. Moment at end of upper beam (and lower beam) is always zero and then increases along length of beam. Although your statement that "top beam must have a non-zero moment" does not specify a location, it appears that you are trying to claim that.......if beams have same curvature at end of upper beam........then moment at end of upper beam must be greater than zero (based on your incorrect assumption that moment in both beams must be equal where they have same curvature). However, such conclusion is not correct.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

As an example........for simply-supported beam, with uniform load on entire length..........slope at end of beam is the classic ....wL^3/24EI

It should be clear then that the same curvature value may occur for different combinations of w, E, I and L.

Note also that the expression for slope (curvature) is not a function of bending moment.........which is not a function of E or I.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

First....in mechanics......the so-called "rigid" joint is more accurately termed a continuous joint (similar to behavior of continuous beam at interior support)........allowing rotation (for all members at joint) greater than "fixed" joint (zero rotation) and less than "hinge" joint (no-resistance to full rotation as simply-supported element). Another expression for a continuous joint is a spring-type joint, relative to rotation.

.

The analogy then is that.......finite-length contact at end of upper beam (with lower beam) results in an effective "continuous joint", similar to (but not exactly like) a continuous ("rigid") joint of a rigid frame (or continuous beam).......for which the column (similar to lower beam in this case) prevents what would otherwise be full hinge-rotation of the girder (upper beam)......but allows some rotation.....which of course is the same as rotation at top of column (lower beam).

Primary difference (between this analogy and actual condition for 2 stacked beams) is that moment at very end of upper beam is of course always zero. Also, it remains to be determined whether or not some discrete analogous "end-moment" could be calculated to result in the same actual reduction of end-rotation and deflection of upper beam (due to prevention of upper-beam end-rotation by lower beam).

As first thought, it may be that such effective (though fictitious) analogous end-moment is equal to end reaction R (wL/4 in this case) times distance between center-of-gravity of R and the end of upper beam.......where location of center-of-gravity for R depends on shape of pressure-distribution along length of contact.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

1) There are three sets of plots below:

1a) Shear, which I believe to the the best indicator of contact zones.

1b) Moment

1c) The axial force in the rigid links.

2) Within each plot set, there are six cases:

2a) Uniform 30 kN/m load applied to entire length of upper beam. Ix_upper = 10 x Ix_lower.

2b) Uniform 30 kN/m load applied to entire length of upper beam. Ix_upper = 1.0 x Ix_lower.

2c) Uniform 30 kN/m load applied to entire length of upper beam. Ix_upper = 0.1 x Ix_lower.

2d) Concentrated 150 kN load applied to center of upper beam. Ix_upper = 10 x Ix_lower.

2e) Concentrated 150 kN load applied to center of upper beam. Ix_upper = 1.0 x Ix_lower.

2f) Concentrated 150 kN load applied to center of upper beam. Ix_upper = 0.1 x Ix_lower.

3) Lower beam span = 10m; Upper beam span = 5m; 1.0 x Ix_lower = W18x35, 50 ksi steel.

My conclusion: there only seems to be a narrow band of relative stiffnesses where the interaction is very complex. If the relative beam stiffness ratio exceeds that range, the upper beam simply ends up simply supported. If the relative beam stiffness ratio is less than that range, the upper beam simply serves as "blocking". All this would seem to be pretty consistent with a lot of people's intuition here.

## RE: Beam on top of beam, not connected, and of different length.

In geometry a curved line segment can be defined by a starting point and an equation for the slope, which will also define the curvature. It follows that any two coincident line segments of finite length must have the same curvature and slope at any point, and hence two beams following that line will have bending moments in proportion to their EI values.

Any analysis that ignores this indisputable fact will be wrong.

If you can find some fault with that statement please let me know. Otherwise any analysis must take it into account, which mine does, also satisfying equilibrium and all other compatibility requirements.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

"Finally.....claim that moment in each beam is proportional to relative I-values ......when beams have same slope (curvature)......is not correct. Moment graph by Doug shows that is not correct as previously noted. However, for more analytical demonstration........we can start by calculating moment in each beam at midspan, since (for this case) slope of each beam is of course the same value; zero.

M-upper = w(L/2)^2/8 = wL^2/32

M-lower = (wL/4)(L/2) - (wL/4)(L/4) = wL^2/8 - wL^2/16 = wL^2/16

So M-lower = 2 M-upper..........for all I-values for each beam...........that is, moment in each beam is not dependent on relative I-values

Also.......when slope at end of upper beam first equals slope of lower beam........moments are clearly not equal since M-upper = 0 and M-lower = (wL/4) (L/4) = wL^2/16"

As noted in more recent posts....."stiffness" is a function of E, I and L (not just E & I)......so that, when moment is proportional to stiffness, the L value must also be considered for each member.

However......more importantly....moment is only proportional to stiffness when all members involved participate in resisting moment, acting together (as opposed to acting independently). The total moment remains unchanged. Such participation occurs for "rigid" joints.......and for crossed beams that must deflect together.

For this general configuration.......there is no participation at all (in the sense of moment-sharing) when I-upper is large enough such that it deflects independently of lower beam (other than at ends of upper beam). That behavior occurs of course because there are no moment-resisting joints connecting the two beams.

The interesting part is when I-upper is low enough such that some indeterminate behavior occurs........due to lower beam preventing full end-rotation of upper beam (which does occur when I-upper is large enough). However, even then, there is no real moment joint connecting these two beams.

Furthermore, relative to your analysis....for Case 2, there is no way that upper beam contacts lower beam at midspan unless I-upper is zero. Otherwise, for uniform load on upper beam.......which would then have to be pushing upward as well (as uniform load)......moment in upper beam at every point along beam would be zero........which is completely inconsistent with having curvature of a moment-resisting beam element. The only way for an element to have zero moment along entire length is for the element to be a cable-element that does not resist moment at any point.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Assuming two beams, the top one shorter than the bottom, arranged symmetrically about mid-span, with the top beam subject to a UDL of its full length.

- Check if the beams contact at mid-span. If they do not, or if it is a point contact, find the moments in both beams from statics. If they do contact over a finite length:

- Find the total moment at mid-span from statics

- Distribute this moment in proportion to the beams EI values.

- Assume some contact length, symmetrical about mid-span.

= Starting from the mid-span moments, find the moments in both beams at the start of the contact length

- Calculate the point force transfer at the start of the upper beam required for static equilibrium with the calculated moments.

- Find the point force transfer at the start of the contact length required for overall static equilibrium.

- Apply these loads to both beams assuming point contact of the ends of the upper beam on the lower beam. Calculate the difference in deflection at mid span, relative to the ends of the upper beam.

- Adjust the contact length until the difference in deflection is zero.

I have set up the spreadsheet to use the solver, but if you don't have that activated you can also use the goal-seek function. You need to make cell L13 equal to zero by adjusting cell L2.

The spreadsheet works on the assumptions listed above, but it will work with any UDL, any ratio of beam lengths up to 1:1, and any finite ratio of EI values.

Using this spreadsheet with a range of values supports what I have said previously; assuming both beams have finite EI, and the top beam is shorter than the bottom:

- If the top beam has an EI greater than some limiting value then it will span between contact points at its ends, with no other contact.

- For any non-zero EI lower than this value there will also be contact over some length, symmetrical about the mid-point of the beams.

- The contact zone will increase in length as the upper beam EI reduces, but it will never extend over the full length.

- The pressure distribution over the contact length will be a fixed proportion of the applied UDL, such that the shear force in the beams is in the same ratio as their EI values.

- In addition there will be a point force transfer at the ends of the contact zone to maintain static equilibrium.

The graphs below show deflections and factored bending moments for a 10 m lower beam, 5 m upper beam, with a range of relative EI values (lower/upper = 5, 50, and 500).

http://interactiveds.com.au/images/Stacked%20Beams...

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

For the stiff upper beam I agree, but for the lower end of the range, as shown above, the contact zone never extends over the full length, there will always be a finite length of separation over which the beams can have a different curvature. In your case with the top beam resting on the left half of the lower beam the contact zone will spread out from the left hand end (where they both have zero moment), and the gap will be at the right hand end only. Other than that, I think your results are in line with mine.

John - I'm not sure where to start. Are we in agreement the M/EI = 1/R and that 1/R is curvature?

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

Frankly, I'm amazed the agreement is so good.

All Strand7 results are directly from the output, other than the upper beam moments, which are multiplied by the EI ratio (10).

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

Understanding curvature compatability here really does seem to be the key to understanding the behaviour. While I was setting up my model, there was a frustrating span of time where I couldn't get the rigid links to stop absorbing tension. Those results were interesting too for the case of a flexible upper beam.

A hand full of links at the right end of my model would pick up a ton of alternating force quickly which dissipated fast. A moment couple. In essence, the system was attempting to enforce curvature compatability at the right end as quickly as possible. Of course, in hindsight, it is obvious that should be the case.

## RE: Beam on top of beam, not connected, and of different length.

Upon more careful checking (at least for Case 2)........turns out midspan contact occurs before end-rotation of upper beam can be obstructed (resisted) by lower beam.

Summary of revised analysis (Case 2....upper beam centered on lower beam.....uniform load on upper beam only);

First "contact" (at midspan) occurs for I-upper = 41.67% of I-lower......however, no midspan force is developed since equal deflection of each beam occurs without any force applied at midspan (by upper beam to lower beam and vice-versa).

Lower beam provides spring-type support for upper beam. As I-upper is reduced as percentage of I-lower......greater resistance is provided by this spring-support action at midspan of lower beam.

For values of I-upper down to about 30-percent of I-lower.....length of contact remains relatively short, around midspan.

For I-upper = 33.3% of I-lower (when slope at end of I-upper would equal slope of I-lower, if contact at midspan did not occur......however, contact does occur at midspan, such that slope of I-upper remains less than slope of I-lower).

Reaction force at midspan is about 9-percent of total load on upper beam (wL/2.

In area of contact.......moment in upper beam (M-upper) is about 21 percent of moment in lower beam (M-lower).

For I-upper = 25% of I-lower....reaction force at midspan is about 20-percent of total load (wL/2)....in area of contact, M-upper is 15 to 16 percent of M-lower.

For I-upper = 20% of I-lower.....length of contact increases......total reaction force at and near midspan is about 27-percent of total load.... ..in area of contact, M-upper is about 12 to 13 percent of M-lower

For I-upper = 10% of I-lower.......length of contact extends well towards end of upper beam (essentially as shown by Doug in graph; though do not see graph posted now)......M-upper around midspan is about 5-percent of M-lower, while maximum M-upper is about 11-percent of I-lower.

In general .....within areas where beams are in contact......moment in upper beam is less than ratio of EI values and there is no fixed relation.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

I-upper = 41-2/3 percent of I-lower

M-upper = 50-percent of M-lower (in area of contact, at and near midspan of both beams)

I-upper = 40-percent of I-lower

M-upper = 49-percent of M-lower

I-upper = 33-1/3 percent of I-lower

M-upper = 45-percent of M-lower

I-upper = 30-percent of I-lower

M-upper = 42-percent of M-lower

I-upper = 25-percent of I-lower

M-upper = 38-percent of M-lower

I-upper = 20-percent of I-lower

M-upper = 33-percent of I-lower

I-upper = 10-percent of I-lower

M-upper = 18 to 23 percent of I-lower

Moments in upper beam are greater than reported in previous post. However, basic result remains the same ........which is that ratio of moments in area of contact (generally at or near midspan except for values of I-upper less than about 15-percent of I-lower) is not the same as ratio of EI values for two beams.

Note that;

(1) Maximum moment in lower beam remains the same (wL^2/16).......occurring at end of upper beam.

(2) Between ends of upper beam.......maximum moment in lower beam is lower for lower values of I-upper

For I-upper = 33-1/3 percent of I-lower.....M-lower (midspan) = 91-percent of M-lower (max)

For I-upper = 20-percent of I-lower.........M-lower (midspan) = 73-percent of M-lower (max)

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

I-upper = 41-2/3 percent of I-lower

M-upper = 50-percent of M-lower (in area of contact, at and near midspan of both beams)

I-upper = 40-percent of I-lower

M-upper = 47-percent of M-lower

I-upper = 33-1/3 percent of I-lower

M-upper = 37-percent of M-lower

I-upper = 30-percent of I-lower

M-upper = 32-percent of M-lower

I-upper = 25-percent of I-lower

M-upper = 25-percent of M-lower

I-upper = 20-percent of I-lower

M-upper = 18-percent of I-lower

I-upper = 10-percent of I-lower

M-upper = 6 to 8 percent of I-lower

Moments in upper beam are now less than reported in previous post, though similar for higher values of I-upper. Basic result remains the same ........which is that ratio of moments in area of contact (generally at or near midspan except for values of I-upper less than about 15-percent of I-lower) is not necessarily the same as ratio of EI values for two beams. For one condition (I-upper = 25-percent I-lower, the ratios do end up the same. And adjacent to that value, you see I-upper diverging, in different directions heading away from the I-upper,25 value.

Note that;

(1) Maximum moment in lower beam remains the same (wL^2/16).......occurring at end of upper beam.

(2) Between ends of upper beam.......maximum moment in lower beam is GREATER (not lower as incorrectly reported in previous) for lower values of I-upper

For I-upper = 33-1/3 percent of I-lower.....M-lower (midspan) = 73-percent of M-lower (max; without upper beam)

For I-upper = 20-percent of I-lower.........M-lower (midspan) = 85-percent of M-lower (max; without upper beam)

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Upon comparing my "20-percent" solution (I-upper = 20-percent I-lower) to what is your graph of same configuration (you have ratio just expressed other way, as multiple of 5).......I get similar results, though one qualitative difference at midspan.

First....in your graph, it appears that heading for red line (representing normalized moment in upper beam) should have the ratio reversed........should be "Top x EI-top / EI-bottom"

Actual red-line moment values are then 20-percent (1/5) of values shown in graph.

At midspan (and within about 2-percent of span either side for length of contact), I derive total reaction force that is about 27-percent of total load (that is 0.27 of wL/2). This redistribution of load at midspan of course reduces each end reaction (by half that amount).

From your graph......for upper beam, maximum moment (at peak of red-line; about 42-percent of span) ........when divided by 5........is about 22-percent (.20 x 164/150) of moment in lower beam at same location. For my analysis, I get about 22-percent of lower beam moment, at similar location. As noted above however, at and near midspan, I get upper-beam moment that is somewhat less......about 18.2 percent of moment in lower beam......which is itself about 85-percent of what would be midspan moment of lower beam without any upper beam (for same loading).

So now......to address fundamentals.......that is.......Curvature = M/EI..........which of course is a fundamental "building block" of structural analysis, no doubt.

Doug.....as2 for the "1/R" expression you are using......I am not sure if you are meaning to refer to a radius (producing constant curvature), which I do not believe should be the case since shape of curvature is in general not constant.

There is difference between curvature and slope. Looking at this case for example.........at midspan of both beams, slope is the same (zero) yet curvature is not the same. We see this by looking at the total shape of each beam. This is one reason I remain skeptical about "forcing" each beam to have the same moment where they are in contact.

The other reason is results of my calculations........even though it has taken an inordinate amount of time to get to fairly reasonable results (mostly due to me going off on "tangent", though it was somewhat of a refresher). I would be much more persuaded to agree to position advocated by Doug if, for example, ........for the case for I-upper = 40 percent of I-lower .......if M-upper would be equal to 40-percent of M-lower at midspan, since the two beams have some length of contact (even though short) at and adjacent to midspan. My results, which I believe are "good" now (and not all that complex anyway)........show M-upper being 47-percent of M-lower at and adjacent to midspan, which is not particularly close to 40-percent (ratio of I-upper to I-lower).

For the 40-percent condition (I-upper = 40-percent of I-lower)........I calculate small reaction force at midspan equal to 1.786-percent of total load (wL/2). Though short, length of contact is approximately 2-percent of L-upper (1-percent on each side of midspan).

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

There is difference between curvature and slope. Looking at this case for example.........at midspan of both beams, slope is the same (zero) yet curvature is not the same. We see this by looking at the total shape of each beam. This is one reason I remain skeptical about "forcing" each beam to have moment in proportion to EI-values where they are in contact.

Also.......if curvature were equal to slope........then, for Curvature = M/EI........we would have to say; Slope = M/EI........yet this is clearly not true since when slope is zero (such as at midspan of simply-supported beam), moment would have to be zero........which of course is not the case.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

## RE: Beam on top of beam, not connected, and of different length.

I think that's a good summary of what happens.

It's also instructive to look at what happens from first contact, as the upper beam EI is reduced. Using the example of the upper beam half the length of the lower beam, symmetrically placed about the centre, with a full length UDL.

The upper beam first touches the lower when its EI reduces to 1/2.4 of the lower beam. At this stage there is no load transfer at midspan, and the upper beam moment is half that of the lower beam. Its EI is less than half, so its curvature is greater, and there is only a point contact.

The point contact continues until the EI factor reduces to 1/4. At this stage 20% of the UDL is transferred to the lower beam, the lower beam moment has increased, and the upper beam moment reduced such that the ratio is 4:1, so the curvatures are equal.

Any further reduction of the upper beam EI would further increase the curvature of the upper beam, but it is constrained by the lower beam, so the point contact spreads to a contact over some finite length.

From this point the contact length increases as the relative EI of the upper beam is reduced, as has been described before, but so long as the EI is greater than zero there must always be some non-contact zone at the ends of the beam.

Finally it should be said that we are talking ideal beams here, and the gap size as the contact zone extends is tiny. In any real beams the deviation of the beam surface from a smooth curve would result in multiple point contacts, rather than a single contact zone with two well defined gap lengths at the end. Also the "point contacts" would also be "small area contacts", because a point contact would result in infinite stress.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

However there is no obvious reason to then claim that point contact (only) continues for all I-upper values down to 25-percent of I-lower........or that distributed contact then suddenly occurs

when I-upper = 25-percent of I-lower.

For I-upper values less than 41-2/3 percent of I-lower, it is much more logical to conclude that distributed contact occurs and becomes longer with lower values of I-upper.

Which then leads back to my example of I-upper = 40-percent of I-lower........for which moment in upper beam, M-upper = about 47-percent of M-lower (per my calculations)

Key to choosing the 40-percent value is that length of contact is short so that there is not much (none, I think) risk of shape of pressure distribution drastically changing results. However, if the claim of beams sharing moment in proportion to M/EI values when they have contact over some (any) finite length is valid in this case, then it should be valid for the 40-percent configuration.

If the 47-percent moment I calculate is essentially correct, then either the M/EI relation is not valid for this case (and I agree that would take some explaining!) or the curvatures are not the same (also problematic, but perhaps more likely)......or there is some other explanation that is not obvious now (most likely calculation error).

Considering that I calculate the exact relative I-upper value (41-2/3 percent) as Doug is reporting for initial contact provides some confidence level that my routine is on track......however, I acknowledge the possibility of calculation error.

My routine does not calculate length of contact exactly......however, for the I-upper-40 configuration, only short (or very short) length of contact is expected. For relatively "wide" range of contact lengths about midspan (up to 4-percent of span length for I-upper-40), changes in moment for each beam are very small. Most important is reaction force at midspan, which I calculate as 1.78-percent of total load (wL/2)........and then of course reduce the end reaction by half that amount.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

For continuous beam.......moment for each beam segment is of course the same at interior support. Slope is also the same at end of each beam segment at this support (that is how segments are "continuous"). Yet curvature for each segment (at support) is function of stiffness which is function of E, I and L for each segment.

Curvature is defined as change in slope over distance.

In 'my' book (vintage, though I suspect fundamentals remain yet relevant!).......change in slope over distance is equal to area under the M/EI diagram over that same distance.

"Beam on elastic foundation" requires differential equation to solve.

If my calculations are correct (and that is yet to be proven), my guess at this point is that the lack of physical connection between beams may be involved.

However........of interest (and perhaps confusion) is the classic example for "shear flow".......where one upper "plank" slides relative to lower "plank". Only problem there is that these planks are always shown to be in full contact along entire length.........and shown to be same length (at least before deflection of the assembly).

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Using only point load reaction at midspan.......I calculate that "perfect" contact occurs only at midspan, down to I-upper-25 (that is I-upper = 25-percent of I-lower).......for which configuration the midspan reaction is (somewhat amazingly) exactly 20-percent of total load on upper beam (wL/2). Also, at midspan, M-lower (moment is lower beam) is very near to being exactly 25-percent of I-lower........which, interestingly, is not the case for greater values of I-upper.

At I-upper-24........small amount of load distribution (along lower beam, near midspan) occurs........and, yes indeed, M-upper = 24-percent M-lower

So there it is (well not entirely, but mostly!).......and I have learned something along the way.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Reaction at midspan = 21.01 percent of total load

Reaction at 0.49 span (that is, one-percent of upper-beam-span to left of midspan)......and at 0.51 span (mirror image) = 0.36 percent of total load (very small)

Reason that some load distribution must be included is that, when total reaction is attempted only at midspan, with otherwise "perfect" contact (with lower beam) at midspan......some "over-deflection" occurs adjacent to midspan, meaning that some contact (and load distribution) must occur with (to) lower beam adjacent to midspan.

However.........although M-upper is 24.01 percent of M-lower at 0.49 span..........M-upper is only 23.59 percent of M-lower at midspan.......and, this 23.59 percent value is very insensitive to changes in "trial" load distribution pattern around midspan (which occurs during "convergence" process of calculation).

Of course (and as indicated by my evolving calculation results throughout) this may be calculation "tolerance". However, it is consistent with previous results showing that, as I-upper is reduced further, M-upper as percentage of M-lower drifts further away from the I-upper to I-lower ratio........such that I am going to reserve complete judgment on the overall solution until I can check out more refined results for lower I-upper values more thoroughly.........AND double-check that end-rotation of upper beam is still "unobstructed" at ends.

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

I think something you may be missing is that when the EI of the top beam is low enough for a finite contact length at mid-span there is still a point load transfer at the ends of the contact zone.

In the case of the top beam EI being 24% of the lower the contact length is 0.0505103 m either side of mid-span. For an applied UDL of 20 kPa the contact pressure is 16.129032 kPa (20 x 1/1.24). This gives a contact force of 1.63 kN from the UDL over the full contact zone. The required force for equilibrium is 21.38 kN, so the missing force is supplied by a point force of 9.88 kN at each end of the contact zone. If the top beam EI was increased towards 25% of the lower beam the contact length reduces towards zero, and the point force at the ends increases towards 10.

The point forces at the ends of the contact zone allow all the requirements of equilibrium and compatibility to be met precisely.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

Although I have been "tripped up" on first-pass at logic with this analysis........it appears that distributed load should not be uniform, but should taper off away from peak pressure at midspan.

I submit that supposed "point loads" for I-upper between 24 or 25ish and 41-2/3 percent of I-lower are in fact very short distributed loads. Length of distribution appears to be increasing exponentially.

The only reason that "point load" appears to provide solution, is that length of distribution is so small (for higher values of I-upper in this range) that we do not "see" distribution with precision of accounting we use for values.

Any greatly different distribution does not appear consistent with both beams being in continuous contact, which is apparently driving proportionate load sharing in contact region.

Have to look back to find total length of beam you are using........for I-upper-24, what is contact length you calculate in terms of percentage of total span?

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Over the length of the contact zone compatibility requires that the applied load over any length is distributed to the two beams in the ratio of their EI values. If the applied load is uniform, the load transfer between the two beams must also be uniform. This leaves a force to be transferred to maintain equilibrium. It can only be transferred where the beams are in contact, but it cannot be transferred anywhere within the contact zone, so it must be applied at the ends of the contact zone, where the curvatures of the two beams, moving away from the CL, start to differ.

This is only exactly true for our ideal beams. In real beams there will be some local deformation so that "point" loads are distributed to bring the contact stress down below the yield stress. There will also be some transverse strain which this analysis ignores. Curvature of the two beams will be exactly the same at their contact surface, but the curvature of their neutral axes will be slightly different. But this non-linear behaviour is equally true of a point load at mid-span with no contact length.

For my last quoted example I used a 10 m lower beam and 5 m upper, so the total contact length was 2.0204% of the upper beam length. I used a lower beam EI of 10,000 kN.m2, but this does not affect the contact length.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

I have checked reaction force for I-upper values between upper limit of I-upper = 41-2/3 percent of I-lower (first contact)........and I-upper of 25-percent of I-upper, which for some inexplicable reason (for now) is a clear break-point between; (1) Point-only contact (and reaction force) and, (2) Distributed-pressure contact for lower values of I-upper. I have verified this by moving edge of potential distributed contact zone closer and closer and closer to midspan to try and find another point of contact (adjacent to midspan) .....and, amazingly, it does not happen for any values as close to midspan as you care to get. Also, the midspan reaction (applied by lower beam) for I-upper-25 is an exact 20-percent of total load, which also is surprising.......though I suspect that for fixed dimensions of this configuration the "stars" just align perfectly for that relative value of I-upper.

For I-upper less than 25-percent of I-lower.......some length of contact and distributed contact-pressure occurs. It is reasonable that character of this contact pressure occurs in a way that demonstrates a transition consistent with behavior for point-load-only contact ......that is, contact pressure should have large peak pressure at midspan that reduces down to zero at edge of contact area. In fact, that is what I find for I-upper-24.5 and I-upper-23, with results listed below. Of interest is that rate of pressure-reduction (from peak to zero) is not straight-line.......in fact reduction is very rapid, such that pressure most likely conforms to an exponential function (which is not surprising at all).

For I-upper = 24.5 percent of I-lower

Length of contact (centered at midspan) = 2.08 percent of L-upper (which is L-lower /2)

Total reaction force = 20.683 percent of total load (wL/2)

Peak node force at midspan = 99.40 percent of total reaction force

M-upper at edge of contact = 24.72 percent of M-lower

M-upper at midspan = 24.29 percent of M-lower

For I-upper = 23 percent of I-lower

Length of contact (centered at midspan) = 5.50 percent of L-upper (which is L-lower /2)

Total reaction force = 22.77 percent of total load (wL/2)

Peak node force at midspan = 94.30 percent of total reaction force

M-upper at edge of contact = 23.26 percent of M-lower

M-upper at midspan = 22.21 percent of M-lower

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Why of course not? For an ideal beam (plane sections remain plane, etc) that is exactly what I would expect. For a real beam that is a close approximation to what I would expect.

At the contact points both beams have the same slope and curvature, but a different rate of change of curvature. There must be a point load at this location so that the rate of change of curvature after contact is equal for the two beams. Working the other way, the beams while they are in contact must have equal slope, curvature, and rate of change of curvature. For them to separate their rates of change of curvature must change at a point, so there must be a point load transfer.

For your second example my spreadsheet gives:

Top moment/Bottom moment at both locations: Exactly 0.23 (to machine precision)

Gap at both locations: 0.000% of total deflection (1.09E-14 m at both locations to be exact).

Contact length = 4.083% of upper beam length

Contact pressure = (1/0.23)/(1/0.23 + 1)

Point loads at end of contact zone = WL/2 x 0.19495

Entering your loads into my spreadsheet I get almost the same moments as you (differences in the last S.F.) and a gap of -0.28% and-0.27% at the ends of the contact zone and mid-span (i.e. the top beam is calculated to be below the lower beam).

Similar findings for the first example, but the difference is less because the contact length is less.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Beam on top of beam, not connected, and of different length.

You seemed to be claiming that shape of load on upper beam should define shape of pressure within contact zone........and I was suggesting that does not make sense if you consider distributed load with highly irregular shape (like up-down roller coaster shape) or even just series of highly unequal point loads. It seems most logical for pressure to develop smoothly as for any spring-type support.

If your model (like mine) requires that assumed shape of pressure distribution be input, then solution may not be completely accurate (of course we are talking minute details here, though the principles at work are of interest).

Do not see how rate of change of curvature for each beam would be equal at edge of contact zone since such rate must be different outside edge of contact zone for beams with different curvatures to merge with the same curvatures. Only way they could have same rate of change for curvature......at edge of contact zone...... would be if they already had same rate of change outside of contact zone, and such rate remained the same within contact zone.

After further refinement of my model.......for I-upper (23-percent)......I get contact length of 5.2-percent of upper beam length (only slightly reduced from previous results). Total reaction force remains the same at 22.77-percent of total load on upper beam. I am using discrete points along beam (close spacing near midspan) with peak reaction force at midspan, tapering away. Solution requires highly concentrated reaction force at midspan to ensure smooth contact along contact zone. There are no gaps. However, total reaction force is remarkably constant within variations of node point locations.

May take this up again during some blizzard, but have to take a break for now........has been fun!.......oh and if you are near Belmar NJ sometime, will be happy to buy you that brew!

John F Mann, PE

www.structural101.com

## RE: Beam on top of beam, not connected, and of different length.

Likewise if you are ever passing through Sydney.

Not a bad discussion for such a "simple" question!

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/