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

[jgrady]

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

Hi all,

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

Regards

 

[hokie66]

KootK,
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?

 

[KootK]

@Hokie:

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.

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.

 

[hokie66]

I don't see how that can be. The upper beam is supported on the lower beam, and point load B is also supported on the lower beam. The upper beam can't assist in carrying a load which is not applied to it.

 

[rb1957]

for KootK's scenario, load A is being carried by both beams, due to the continuous contact.

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 ?

 

[KootK]

@rb1957: that's exactly what I was thinking regarding the example. And of course it helps OP. We're discussing his issue, in great detail, and developing a deeper understanding. Besides, at some point, a good thread becomes less about the OP's specific question, and more about the fun issues that the responders tease out of it for debate. I wouldn't even bother if my responses hat to be limited to "yes, there will be load sharing".

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

 

[jfmann]

For stated question..........with upper beam (having length of L/2) positioned so that one end is at midspan of lower beam, we obtain the following results, after crunching numbers, and based on the following conditions;

(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

http://www.structural101.com

 

[jfmann]

Also........the algebra for generalized solution to this problem is not all that bad, though tedious as you might imagine........spreadsheet routine illustrates behavior of the 2-beam assembly much better

John F Mann, PE

http://www.structural101.com

 

[SlideRuleEra]

Hokie - I have carefully gone over your posts, including all other postings through today. Also have worked with my analog model to refine the input values. Bottom line... looks like KootK is on the right track, a similar sized short upper beam (no connections just sitting on top of the lower beam) does contribute to the load carrying, even with a uniform distributed load applied to it. Here are my analog results:

First photo: A single beam with UDL over half its length. Analog mid span deflection = 0.81" [sup]+[/sup] 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" [sup]+[/sup] 0.03"

Here are the detailed inputs:

Long Beam: 39" clear span, simple supports, I = 3.4 x 10[sup]-3[/sup] In[sup]4[/sup]

Short Beam: 19.5" total length, I = 2.9 x 10[sup]-3[/sup] In[sup]4[/sup]

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.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[SlideRuleEra]

One additional input for the analog model: Modulus of Elasticity for both beams = 1.75 x 10[sup]6[/sup] psi

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[Archie264]

What is under discussion is simply how a load is applied to a beam. The bottom beam is loaded by the reaction(s) from the top beam unless the top beam is so flexible that there is contact between the two beams, in which case it becomes a uniform loading, or a proportion thereof.

 

[rb1957]

sorry Archie, not the way i see it. i thought the question was how would two beams work together to carry a load ? the original question was very open and left room for much interpretation.

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

 

[Archie264]

Draw a free-body diagram of just the lower beam and the whole thing becomes a lot simpler.

 

[jfmann]

For uniform load condition.......moment of inertia for upper beam must be at least 50-percent of lower beam to ensure no contact between ends of upper beam.
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

http://www.structural101.com

 

[SlideRuleEra]

JFM, thanks for taking a look. I'll start with your 0.64" deflection and see if I can back figure what is going on with the analog system.

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.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[hokie66]

SRE,
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.

 

[SlideRuleEra]

Hokie - Your are probably right, since they were hand ripped they have rough surfaces. I had though about putting a sheet of poly between them. I'll be going by a lumber yard tomorrow, may stop in to see if they have something better to use as beams. What a place to go to get parts for your DIY computer.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[jfmann]

For stated configuration as before (half-span)........if upper beam is connected to lower beam so that they act as unified assembly (several wood screws would likely work for example above)........moment of inertia is increased by a very large multiple of 72 (assuming depth of 3/8-inch for each beam and backfiguring width to match I-values listed by SRE).......and deflection at midspan is reduced to 0.328 inches........or about 50-percent of the deflection using independent (not connected) beams.

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

http://www.structural101.com

 

[hokie66]

John,
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.

 

[Brad805]

Is it not about time to bring up the subject of friction? I think the horse is dead and has started to decay.

 

[hokie66]

We already did, Brad. But in the great scheme of things, this is not a particularly long thread, and nobody is required to participate.