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Shear Flow with a Uniform Moment
2

Shear Flow with a Uniform Moment

Shear Flow with a Uniform Moment

(OP)
If a composite member has a uniform moment along it's length (dM=0), how do I calculate the required shear flow capacity between it's bonded components for composite behavior?

I understand this is a little academic for these boards, but separating the theory conversation from the application should keep the discussion on topic.

Structural, Alberta

RE: Shear Flow with a Uniform Moment

Quote (wadavis)

I understand this is a little academic for these boards
I don't think that's a thing... pipe

Interesting question... Although it feels intuitive to think there should be some longitudinal stress to resolve, I think the answer is as the formula would suggest - zero.

RE: Shear Flow with a Uniform Moment

Doesn't it all depend on the shear across that section of member where the constant moment occurs?

If you have a simple span member with two concentrated loads at the 1/3 points, the moment is constant between the two loads but the shear varies from load to load in a "bowtie" pattern.

q = VQ/I right? So per the V there would be some type of shear that would have to be some bonding between the components.

If you have a member with two concentrated end moments - perhaps no shear along the member and no bonding required?

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RE: Shear Flow with a Uniform Moment

I don't think the shear will look like a bowtie without a uniform force, if the two loads are equal it's actually zero for the portion of the beam between them. From a purely academic point of view, if you have way of applying moment with no shear, all you are doing is applying axial forces to the two halves of the member and the connection between the two is irrelevant. However, in a real world case, I don't know how you can apply only moment, and our beam buckling formulas all rely on the tension and compression regions being tied together such that buckling is restricted, and needs to occur laterally and torsionally.

RE: Shear Flow with a Uniform Moment

I'm confused, wouldn't shear flow be VQ/I? Wouldn't the support reactions introduce shear due to the applied moment. Seems like you would use this shear to design for composite connection.

RE: Shear Flow with a Uniform Moment

Quote (BowlingDanish)

I think the answer is as the formula would suggest - zero.

I think this too.

I believe that the trick to this is recognizing that, at the ends of the member, the moment has to be applied in such a way that it instantaneously creates a strain profile common to both materials. That's a pretty tall order for any real world loading situation other than where the "member" is really just a segment of consideration within a larger member. Especially so if one or more of the materials is a thing that cracks and creeps and generally behaves in a complex, non-linear manner.

Where the instantaneous development of a common strain profile is not realistic, shear flow capacity would need to be provided needed near the ends of the member to bring things in line, so to speak. In a lot of practical situations, I think that it would be prudent and conservative to assume that the moment originates in one of the materials and has to migrate to the other via localized shear flow.

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: Shear Flow with a Uniform Moment

Quote (OP)

I understand this is a little academic for these boards

Pfft. Get the attention of the right cadre here and we'll bury you alive in esoteric voodoo.

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: Shear Flow with a Uniform Moment

Maybe I am misunderstanding but how do you have uniform moment without shear introduced by your reactions? If there are no reactions doesn't the object just spin? In this case I don't think you would need a tie between the elements but it also wouldn't be statically stable.

RE: Shear Flow with a Uniform Moment

Quote (jd)

Maybe I am misunderstanding but how do you have uniform moment without shear introduced by your reactions?

Opposing end moments. It's a rare thing but not unheard of. The central segment of JAE's 1/3 point loaded beam is a good example. A column assumed to be loaded, and supported, eccentrically by the same amount is another.

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: Shear Flow with a Uniform Moment

Nevermind. I realize I'm wrong. Sounds like application would be providing a moment at the two ends of the member (similar to grabbing a board by the edges and bending it). Interesting question.

RE: Shear Flow with a Uniform Moment

So say you stacked 10 1" plates together and bent them vs 1 10" tall plate. Would they behavior fundamentally different? Typical logic would be yes the 10" plate would have much larger moment of inertia than the individual plates. But maybe in this instance it doesn't matter. Not really sure. Maybe curvatures are restrained and locked to be the same so it doesn't matter? Interesting question.

RE: Shear Flow with a Uniform Moment

I think I didn't stop to think - with two point loads the shear is zero between the loads.

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RE: Shear Flow with a Uniform Moment

Quote (jd)

So say you stacked 10 1" plates together and bent them vs 1 10" tall plate. Would they behavior fundamentally different? Typical logic would be yes the 10" plate would have much larger moment of inertia than the individual plates. But maybe in this instance it doesn't matter. Not really sure. Maybe curvatures are restrained and locked to be the same so it doesn't matter?

The moments would have to be applied as a bunch of independent, linearly varying axial stresses that, when summed, would simulate the stress and strain profile of a composite member. Like I said, that's a tall order but, if it can be accomplished, there is no VQ/I demand between plies. Really, the end result would be a bunch of disconnected layers with linearly varying, mostly axial loads applied to them such that the aggregate condition would simulate the strain profile of a composite member.


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: Shear Flow with a Uniform Moment

(OP)
In my case I have an eccentric axial load applied to a member, that's where the uniform load without shear is coming from. Although VQ/I = 0, the deflection of that member depends on I, the I is different for a composite member than for a bundle of fibers with no shear flow. So there must be a way to calculate the required cohesion for composite behavior from the moment.

Quote (KootK)

The moments would have to be applied as a bunch of independent, linearly varying axial stresses that, when summed, would simulate the stress and strain profile of a composite member. Like I said, that's a tall order but, if it can be accomplished, there is no VQ/I demand between plies. Really, the end result would be a bunch of disconnected layers with linearly varying, mostly axial loads applied to them such that the aggregate condition would simulate the strain profile of a composite member.



So with a symmetrical profile or a profile proportioned to the load shape the local dP/dA = Axial Stress. But with a non-symmetrical member that is not the same shape as the load profile, dP/dA =/= Axial Stress. So some sort of composite behavior is required.

Structural, Alberta

RE: Shear Flow with a Uniform Moment

Quote (wadavis)

So some sort of composite behavior is required.

I don't believe it is.

Consider this;

- A rigid clamp that matches the member cross-section (any shape) is installed at each end of the beam
- It rotates about a perfect pin at the neutral axis of the member
- By applying a uniform moment at each end without shear, you are effectively dialing in a certain final rotation of the clamps. That rotation is governed by the stiffness of the axial compression and tension of each longitudinal fibre - which doesn't depend on I.

I can't say I understand your sketch though... The 'load' you are applying is not that of a uniform moment?

RE: Shear Flow with a Uniform Moment

Quote (wadavis)

the I is different for a composite member than for a bundle of fibers with no shear flow.

This statement is only true when shear flow is required. That, of course, is the overwhelming majority of the time. When shear flow is not required, Ix is identical for members with and without laminar shear connection. This recent thread is highly related to your question and may well resolve this for you: Concentric Tube Bending



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: Shear Flow with a Uniform Moment

(OP)
That composite behavior vs shear load is what I'm struggling with.

When I apply my eccentric tension to the member (triangle load distribution in the above sketch), convert the eccentric load to a Force and Moment at the centroid, and calculate the stress in the extreme fibers (P/A + My/I); I get a different stress result at any given location than I get when I slice the member and load into small slices and distribute a fraction of the load over the corresponding fraction of the member area.

In a member without laminar connections the method of slicing (dP/dA) should get the same result as any other method but if you calculate the bending stresses as a composite member (P/A + My/I) you get slightly different stresses. See the sketch, I'm afraid my notation is very rust, sorry.


Sorry about the unclear sketch earlier. The right most diagram is a simplification of my problem, the left and center sketches were my brainstorms on the rigid clamping approach, circumstances where I get the same stress result for shapes with or without laminar connection.

I'm not trying to be stubborn here, just this problem has me doubting.

Structural, Alberta

RE: Shear Flow with a Uniform Moment

OK... So it's not just a uniform moment at each end - you're also introducing additional axial forces.

My view would be this - based on nothing but intuition;

- Take the same member supported on perfect rotational pins, and supported vertically.
- Take the same rigid clamp and apply your moment, giving you a certain rotation and curvature of member
- Now apply your axial load (its concurrent but meh)
- I would guess that due to the curvature of the member, the axial load you are applying would either try 'straighten' the member (tension) or curve it further due to eccentric load (compression).
- This 'straightening' or 'additional curving' would result in some tangible vertical shear force at the support.
- This shear force is what you would use to calculate the required shear flow

Pure speculation pipe

PS - I still dun get your sketch. calculus is beyond me now.

RE: Shear Flow with a Uniform Moment

I'm still struggling a bit with the thought that the members don't need any connection between layers in this case. I understand the theory and agree without V the equation says it's not necessary, but take the example of a phone book. If you grab it by the ends and twist your wrists it's fairly flexible. Don't you think it would be more rigid if it was a solid piece?

RE: Shear Flow with a Uniform Moment

I don't believe that the presence of the axial stress changes anything. By negative superposition, you can strip it out and again deal with the uniform moment as before.

Quote (jd)

I'm still struggling a bit with the thought that the members don't need any connection between layers in this case.

Consider the absurdly trivial yet altogether salient analog of an axially loaded member. Say, 2-2x6 that you'd like to resist an axial load of 1000 lb compositely. Don't sweat buckling or the moments due to eccentricity. Neither is germane to the example. You've got three possibilities:

1) Put all 1000 lb on one of the 2x and do not nail the plies. One ply resists, the other does not, the axial strain profiles of the two sticks do not match, and you have the analog of the non-composite case.

2) Put all 1000 lb on one of the 2x and do nail the plies. Load transfers between plies, each ply resists 500 lb, the axial strain profiles of the two stick match, and you have the analog of the composite case where shear transfer between plies is required.

3) Put 500 lb on each of the 2x and do not nail the plies. No load needs to be transferred between plies, each ply resists 500 lb, the axial strain profiles of the two stick match without any interconnection, and you have the analog of the composite case where shear transfer between plies is not required.

The uniform moment example here is very much like #3.



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: Shear Flow with a Uniform Moment

In the case of the latest sketch by wadavis, in order to act compositely, every horizontal section through the profile has a shear flow wherever the area under the load diagram above or below the section differs from the area under the stress diagram above or below the section. The shear flow is equal to that difference.

BA

RE: Shear Flow with a Uniform Moment

But KootK what about the phone book analogy or do you not agree with that analogy?

RE: Shear Flow with a Uniform Moment

Quote (KootK)

Where the instantaneous development of a common strain profile is not realistic, shear flow capacity would need to be provided needed near the ends of the member to bring things in line, so to speak.

Quote (BAretired)

every horizontal section through the profile has a shear flow wherever the area under the load diagram above or below the section differs from the area under the stress diagram above or below the section. The shear flow is equal to that difference.

I think that we're probably speaking to the same phenomenon here. I've explored it a bit below for the case of a rectangular section with a load eccentricity of B/6. I think that it's worth noting that this shear would be of a different, and lesser, order of magnitude that one would expect from a true flexural application.


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: Shear Flow with a Uniform Moment

Quote (jd)

But KootK what about the phone book analogy or do you not agree with that analogy?

I agree with the phone book analogy but feel that it is missing an important feature in this instance. Namely, it's missing little axial loads at the end of each page that increase linearly towards the outside edges of the phone book and simulate the linearly varying stress that would normally produce linearly varying strain.

At the end of the day, all you need to make use of composite section properties is for all of the cross section to flex according to a common strain diagram. One way to achieve that is to force them to strain together via shear flow. Another is to load them in such a way that their strains match at the boundaries without the need for shear flow capacity.

I made the sketch below to study your phone book analogy as I see it. The pages are separated by by rolling pins and each page is loaded independently. Not sure if this will clarify things or further confuse them. Let me know if I've made things worse.

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: Shear Flow with a Uniform Moment

Quote (jd)

but take the example of a phone book. If you grab it by the ends and twist your wrists it's fairly flexible. Don't you think it would be more rigid if it was a solid piece?

I also believe that three additional things are producing that flexibility:

1) the pages in compression are buckling.

2) the pages are not held together vertically.

3) your hands are not providing perfect shear slip restraint at the ends to apply the moment as I've described.

Obviously, I've never seen you bend a telephone book. You might be the hulk and doing an incredible job of it.



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: Shear Flow with a Uniform Moment

The main problem with the phone book analogy is that the pages on the compression side have virtually no resistance to buckling.

BA

RE: Shear Flow with a Uniform Moment

A laminated beam with two concentrated loads placed symmetrically on the span has zero shear between loads and requires no shear resistance between plies. The plies must be adequately fastened to each other to prevent individual plies from buckling under load.

BA

RE: Shear Flow with a Uniform Moment

The only shear flow required between components under pure bending (ie imaginary) would be what is required to prevent one component from buckling differently than adjacent components. This would be similar to the required longitudinal connections between elements in a built up steel member in AISC.

Keeping our phone book analogy - we'd need to transfer the resultant axial loads due to the end moments for each page - I figure dipping the edges into 1/2" of glue. I can't think of away to keep the compression pages from buckling though.

RE: Shear Flow with a Uniform Moment

A little late to the party, but back to the case the OP is describing - as long as there is no change in moment/axial force there is no shear flow in the situation you have described. However, the two members need to be connected at the end receiving the eccentric load such that the triangular stress profile develops. What does the connection look like at this end?

RE: Shear Flow with a Uniform Moment

Wadavis, I think you may be confusing classical shear flow with shear lag. When you load a column eccentrically, in order for the load to "spread out" there needs to be shear between the fibers. This is not your classic VQ/I. To be honest, I'm not sure how you would compute this so for grins I made an FEA model of an eccentrically loaded column. Once the load spread out there was hardly any shear as would be expected when the moment is not changing. Here's a screen shot of the shear distribution. This is a 10" wide column with just the right 1" loaded. It's 100" long. It only took about 11" before the shears went practically speaking to zero.

RE: Shear Flow with a Uniform Moment

(OP)
Thanks everyone for the great input. I'm with the phonebook example, as long as there is some way to create even strain where the load is applied (shear flow resistance or rigid clamping) the member wont have a shear flow demand and therefor Ix wont be dependent on it. Bend your phonebook with the spine providing shear flow, then cut a 2" notch out of the spine and try again, should have similar behavior.

Quote (Kootk)

2) Put all 1000 lb on one of the 2x and do nail the plies. Load transfers between plies, each ply resists 500 lb, the axial strain profiles of the two stick match, and you have the analog of the composite case where shear transfer between plies is required.
This member should behave the same if you glue the plies together, or if you just nail the the last 1' of each end together with no shear connections along the center span.

I was barking up the wrong tree with my problem. Looks like my problem is a local shear flow issue as the loading distributes across the member profile, well shown in KootK's vertical load sketch (edit: and in Dozer's FEA). I'm going back to the drawing board: seeing if I can find any literature on the shear load on the connections between atmospheric tank shells of different thicknesses under hydro-static stress.

Structural, Alberta

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