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

[mecheng24]

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

 

[jfmann]

Doug.......when you say "have" different curvatures........you mean in the absence of interaction.

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

http://www.structural101.com

 

[IDS]

Doug.......when you say "have" different curvatures........you mean in the absence of interaction.

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/

 

[jfmann]

Doug.......he basis of your claim appears to be that, if beams have same curvature at any location, then they must have the same bending moment........which is simply not correct .....as I have already demonstrated in recent post.

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

http://www.structural101.com

 

[jfmann]

To elaborate and clarify...........curvature of a flexural element (beam) is a function of E, I and L, as well as loading.......as is seen in all expressions of curvature (slope).

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

http://www.structural101.com

 

[jfmann]

Realizing there is another "vantage point" (analogy) from which to visualize behavior once the indeterminate "phase" of this general configuration is reached. Of course, as with just about any analogy, the analogy is not a perfect resemblance of the actual condition........just another viewport shall we say to understand behavior.

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

http://www.structural101.com

 

[KootK]

I ran a SAP model as JAE and I recommended above: using rigid, compression only links. I've attached the model in case anyone would like to tinker with it. Here are the details:

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.

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.

 

[IDS]

John - you do not seem to have read what I have written. I have never suggested that beams that have point contact or no contact must have the same slope and curvature, in fact I have explicitly said the opposite. Also I have never said that beams with the same curvature will have the same moment. The moment will be in proportion to their EI values.

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/

 

[jfmann]

Doug......the following is from end of my long post of yesterday (19:47)......this is for Case 2 (upper beam, with Length of L/2, centered within lower beam, with Length of L)

"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

http://www.structural101.com

 

[IDS]

I have modified my spreadsheet (copy attached), and it is now much more stable, working with a wide range of relative beam lengths and beam EI values. The procedure used is now:

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 Beams2-3.PNG

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

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.

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/

 

[IDS]

Spreadsheet results compared with a Strand7 analysis using contact elements with friction set to zero:

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/

 

[KootK]

I think that our results are in full agreement Doug. When I said that at some ratio of relative stiffness the results became less complex, that was a poor choice of phrasing. It would be better to say that at some point, the macroscopic behaviour is no longer significantly affected by localized complexities.

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.

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]

Must admit my analysis went off-track ......forgetting to verify key assumption, which was (incorrectly) that end-rotation of upper beam would be obstructed before contact at midspan occurred. Initially (as seen in early posts) I was focusing on deflections (and should have stayed with that line).....but was tempted by curves, ha!

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

http://www.structural101.com

 

[jfmann]

Ah, clicking tad too quickly through formulas..........revised moments for Case 2;

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

http://www.structural101.com

 

[jfmann]

Alrighty then.....my bad......updating yet again......going to get this!

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

http://www.structural101.com

 

[jfmann]

Doug......see your graphs again........quite informative

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

http://www.structural101.com

 

[jfmann]

Hazards of long posts.......one important correction;

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

http://www.structural101.com

 

[IDS]

John - I will have a look at your numbers later (doing some real work now ), but for now, I haven't suggested that slope and curvature are the same. The point is that for two lines to be in contact over any finite length their slope must be equal at every point along the contact. It follows that their curvature must be equal, so M/EI must be equal. If it isn't the beams will deform until the moments are equal, or they separate, or the contact length reduces to a point.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

"moments are equal" should read "M/EIs are equal" of course.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IFRs]

Can it be that if the natural curvature of the upper beam is at a greater radius (flatter shape) than the lower beam that the forces on the upper beam cause it to conform to the lower beam's shape. The forces in the upper beam are modified by the lower beam, the upper beam no longer feels just forces from above. ???