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

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jgrady

Mechanical
Jan 30, 2015
34

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
 
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jfmann said:
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.

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 the third-points per 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.

[idea]
[r2d2]
 
We look forward to your further demonstration, but I think the load application would have to be a bit more sophisticated...
 
I see I was involved in the referenced thread as well.

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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Interesting discussion......and for such a "simple" issue.

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
 
John,
"first 1/2 of the entire span" is a good enough description for me.
 
Ok, have finished the next analog computer run. Results indicate the "short" beam, covering the first half of the "long" beam does help:
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)
Two_Beam_01.jpg


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)
Two_Beam_02.jpg


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!

[idea]
[r2d2]
 
See the attached - close up of SlideRuleEra's first photo - note the flexure in the upper beam has created a measure of contact
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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 http://files.engineering.com/getfile.aspx?folder=a7d2129a-dfcd-4086-b89e-c342d6f4f046&file=Upper_beam_end.jpg
hokie, yes I did miss that "first", ha

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
 
SRE,

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.
 
jfmann said:
...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.

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".

[idea]
[r2d2]
 
OP said:
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?

There are actually loading scenarios where loads not placed over the upper beam would still be partially resisted by the upper 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.
 
"Does the load sharing still apply for the loads placed over that portion of the span?"

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 ?
 
Any load applied to lower beam (outside, or even "under" lower beam) has no effect on upper beam (since it simply moves down with lower beam as a block of wood, or a bird!) and there is no load sharing. This is true no matter where upper beam is positioned along lower beam.

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
 
Oh heck.......too many "upper beam", "lower beam".......one correction to text (that I see now anyway)

(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
 
If any load at all is applied to the upper beam, it will share in providing resistance to that load. Here, strain = participation in load resistance. As I mentioned above, this is direct consequence of energy conservation. Specifically, it's the balance between strain energy and potential energy. SRE's experiment has validated this nicely.

@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.

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.
 
Just to be clear for the stated question.......effect of load on upper beam is not reduced by lower beam unless upper beam is so flexible that it contacts lower beam at some point between ends of upper beam (similar to discussion above for the special case of upper beam centered with length of lower beam).

"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
 
As noted in one previous post........the meaning of "share" in this context requires clear definition since without such definition it can be misleading.

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
 
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