## Shear Flow

## Shear Flow

(OP)

I am sure most here are familiar with the concept of shear flow as it related to horizontal shear stresses in beams.

When designing a I-Shaped plate girder, most references, if not all, will design the weld between the flange and web using the shear formula VQ/I to determine the force on the welds.

My question is, isn't there bending stress on the weld as well in the form of MC/I?

If one had a simply support girder with a uniformly distributed load, the shear at the center of the beam would be = zero and the moment at a maximum. This would imply that no weld would be needed at the center, yet this is the section where having the entire cross-section engaged in bending is most critical.

What am I missing here?

When designing a I-Shaped plate girder, most references, if not all, will design the weld between the flange and web using the shear formula VQ/I to determine the force on the welds.

My question is, isn't there bending stress on the weld as well in the form of MC/I?

If one had a simply support girder with a uniformly distributed load, the shear at the center of the beam would be = zero and the moment at a maximum. This would imply that no weld would be needed at the center, yet this is the section where having the entire cross-section engaged in bending is most critical.

What am I missing here?

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Maybe I am misunderstanding my texts.

How is the horizontal shear calculated?

VQ/I is the horizontal shear due to vertical shear forces on the beam. internal equilibrium states that this is true. Horizontal shear forces due result from vertical.

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What is the horizontal shear force in the center of the beam I described in the original post...i.e., where the moment is max?

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Your explanation seems perfectly logical and agrees with what the texts say.

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The horizontal shear stresses are a result of the differential moment. If you think about a simply supported beam with a uniform load, the moment diagram takes on a parabolic shape. It increases rapidly from 0 at the ends, and levels off to a maximum at the center. As you get closer to the center, the differential moment decreases which, in turn, causes a decrease in the horizontal shear flow. This continues to the point where the moment is at a maximum (the slope is 0) in which case the differential moment is 0, hence your horizontal shear flow is 0.

This matches what you will get if you plot the horizontal shear flow of a uniformly loaded, simply supported beam. The horizontal shear flow will be a maximum at the ends (where the slope of the moment diagram, or the differential moments, are highest) and taper down to 0 at the center (where the slope of the moment diagram, or the differential moments, are zero).

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Internal equilibrium in the beam would dictate that.

For the case of the simply supported beam with uniform loading, the shear at the support is max and moment is ZERO, and, horizontal shear is VQ/I.

So how can the horizontal shear exist where no moment does if what you guys are saying is correct?

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Mike McCann

MMC Engineering

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Taking your simply supported beam idea - at midspan, the moment is maximum and vertical shear is zero.

At this point, the horizontal shear between the flanges and webs is zero (i.e. VQ/I = 0 because V=0). As you move outward towards the supports, the vertical shear V increases and so does the horizontal shear.

What is basically happening is that as you move from the support towards midspan, the shearing stresses are gradually transfered into longitudinal stresses in the flanges through the horizontal shear at the flange/web interface.

At midspan, there is Mc/I stress in the weld, but it is NOT a shear through the weld but rather a axial force in the cross section of the total weld itself.

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I'll use a simple working example to make it easier to talk about. Let's talk about a simply supported, 20' long beam with a uniform load of 1klf. Let's look at the "differential element" from 0' - 1'. The moment at 0' = 0, and the moment at 1' = 9.5k-ft. This would be the differential moment for that element (9.5-0 = 9.5)

Now let's look at the element from 9' - 10'. The moment at 9' = 49.5k-ft, and the moment at 10' = 50k=ft. The differential moment for that element is only 0.5k-ft.

Now, if you take infinitesimally small elements (instead of the 1' elements discussed above) what you will find is that the differential moment is equal to the slope of the moment diagram. This makes sense if you think about it from a mathematical point of view - dM=slope of the moment function.

Now, with the above talked through, you should be able to see that the differential moment at the midspan (point of maximum moment) is zero, therefore the horizontal shear flow is zero.

Now just for a little more practical look at the situation. If you imagine the traditional picture that is used to illustrate shear flow (It was mentioned above), of several elements lying on top of each other with no connection between them. You load the system and they deflect together, but they also slide with respect to each other. That sliding is a maximum at the ends, and is 0 in the middle. This also supports the fact that horizontal shear flow is 0 where the slope of the moment diagram is 0, or decreases as the moment decreases.

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This is precisely what I was picturing.

As you say, at midspan, there is no vertical shear force on the weld, however, if were to calculate a stress on the weld cross-section by MC/I (c being = to the distance to the centroid of the weld cross section from the n.a.), I could then determine the longitudinal force on that weld. Correct?

This would be a longitudinal force that results from bending, no? But since the force in the surrounding material (flange, web) would have only infintesimally higher and lower (varying by distance from the n.a. only ever so slightly), the weld really sees no stress in terms of being a connection between web & flange.

agreed?

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in my opinion JAE is wrong.

The welds which stitch together a flange and a web transfer longitudinal force from web to flange. This is what we call shear flow. The shear flow is maximum at the ends, and zero at midspan. By the time you reach midspan, all of the flange force has developed in the flange, because of all the shear flow which has been transferring to the flange. So Mc/I defines the stress in the flange at midspan, but not the force in the weld--because that force is already in the flange.

I am really just repeating what graybeach said.

DaveAtkins

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I dont think you are disagreeing with JAE too much here.

There has to be be some bending stress in the weld definded by MC/I at any point along the beam.

If you were to calculate MC/I at midspan and a obtain a stress at a point C above the n.a. that conincides with the centroid of the weld and then sum that stress about the area of the weld, the result would be a longitudinal FORCE on the weld, No?

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DaveAtkins

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Stillirz-

I agree with you that the weld sees longitudinal stress more or less equal to the longitudinal stress in the adjacent flange. It is ignored in the weld analysis because it doesn't contribute to the failure of the weld. The weld will fail in shear, due to the shear flow, so that is what is checked.

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I'm trying to get a full understanding.

So i guess it would follow then that the weld can be designed for VQ/I.

A good example might be a simply supported beam with constant moment and no applied shear forces.

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Thanks all

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It appears that the conclusion is that the axial stress in the weld due to bending is not significant, but there is no explanation why it's not.

Vertical shear can be significant if it coincides with a high bending moment.

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But at the midspan, there IS axial force in the weld itself. The weld is a part of the cross section just like the flanges and web. If the beam bends, then there will be axial stress in the weld that will be trying to shorten it (on the compression side) or lengthen it (on the tension side). This is axial stress and in no way conflicts or counters SHEAR stress that comes through the weld (zero at midspan and max. at ends).

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Why is it that we never check horizontal beam shear when designing regular hot rolled beams? Mechanics of materials tells us that the horizontal and vertical shear stresses will be the same on any differential element.

We don't check horizontal shear because we can count on that shear to be resisted by the web along the entire shear span of the member (L/2 for simple support and uniform load). Essentially, shear stress is redistributed along the length of the member such that it is shared by more lightly loaded segments of beam web.

Since this resisting length in horizontal shear is something to the tune of 10 X the resisting length of the web in vertical shear, we acknowledge that horizontal shear will not govern and therefore do not bother to check it. Similar assumptions are made when we design studs for composite beams and chord welds for joist reinforcement.

In ductile systems, horizontal shear can be redistributed or "smeared" to the locations of horizontal shear resistance. This has a small impact on our "plane sections remain plane" assumption but that is generally ignored in practice.

Returning to the welded plate girder, if a fabricator told you that they welded the web and flanges together such that the welds could develop the full yield strength of the web plate in shear (full pen for example), would you bother to compare the weld shear capacity to your VQ/IT demand? Probably not because you'd know by inspection that, if your web works in vertical shear, then you've got oodles of capacity in horizontal shear.

In this case, your Mc/I stresses would likely be the dominant stresses affecting the weld. Of course, those stresses would be less than the stresses at the extreme fiber of your beam flange so you probably still wouldn't bother to check them.

In a more typical plate girder scenario, the welds are not designed to develop the full shear capacity of the attached web plate. Not even close. The connection will be made with relatively small welds that may be continuous or intermittently spaced.

As designers, we specify the welds such that our horizontal shear capacity is relatively close to our horizontal shear demand. Essentially, the web/flange welds are shear critical because we deliberately design them to be that way.

Another reason why axial weld stress is typically ignored has to do with the origins of the stresses involved. In general, a web/flange weld will experience both VQ/It shear AND Mc/I axial stress simultaneously. To be highly rigorous, one would need to check the interaction of these stresses using Von Mises criteria or something along those lines.

However, VQ/It stresses are large when vertical shear is large and Mc/I stresses are large when moment is large. And, as we all know, moments are generally maximum when shears are at a minimum and vice versa. Therefore it is rarely necessary to check the interaction.

In reality, we make this same assumption whenever we check flexural stress and vertical shear stress independently for a beam design. There IS an interaction between flexural stress and vertical shear stress that we ignore when we do this.

As others have suggested above, there are situations where it is not okay to ignore the interaction. A short, heavily loaded cantilever comes to mind...

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You get units of ksi for VQ/(It), the classic shear stress formula.

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(1) The answer is yes, there would be bending (axial) stress on the weld if the weld is continuous. If the weld is not continuous, it attracts no axial stress.

(2) It is true that no weld is needed at midspan to carry shear stress. However, intermittent welding is needed to brace the compression flange to prevent it from buckling about its minor axis. Having the entire cross section engaged in bending does not include weld metal, particularly if the weld is not continuous. Even when continuous, it is a very small part of the area of the section and is normally ignored when calculating section properties.

The development of the horizontal shear formula VQ/Ib may be found in "Elements of Strength of Materials" by Timoshenko and MacCullough and many other sources.

BA

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Neither is horizontal shear not considered because of plastical overall distribution of the length, most will remember the distribution of shear stresses on a section of a double tee beam when trying to justify the uniform shear stress on the web approach; this is still an elastic VQ/Ib figure and an elastical one, to be checked against respective shear yield limit.

The VonMises part is actually true. Our rules are in more than able to use some devised calculations to design something, also empirical in if that if doing so, problems do not develop. Even if two more factors may contribute in the compressive zone for the weld stand the shear force stresses without consideration of the axial forces, namely the bigger strength of the weld material itself and the fact of that if not heat-treated significant tensile stresses may develop and get locked along the welds, there is the less known realm of the Heat Affected Zone anywhere around welds, and the locked stresses turn against strength in the tensile flange. But, as indicated, other controls are applied when designing these welds, namely, a minimum size for the welds, depending upon the flange and web thickness, and the resulting minimum weld size may easily be 3 or 4 times what required just to pass at typical safety the horizontal shear; and this additional safety factor seems to be more than enough to deal with any present interaction effect, for these failures that I know are not popular; and if not, anyway, the interaction equation is at hand always ready to be used.

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Think about a composite beam. If one uses elastic design, the force in each shear stud is not Mc/I, but rather (VQ/I) times the stud spacing. Similarly, the longitudinal force in a weld which fastens a web to a flange is VQ/I per inch.

Or calculate Mc/I at midspan for a plate girder constructed of 50 ksi steel. Assuming this is 30 ksi, the weld is no good--its capacity is 0.3*70 = 21 ksi.

DaveAtkins

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BARetired-

I have to disagree with you there. A beam with a bending moment will have a flange in compression. If it has a compressive stress in the elastic range, Hooke's law says the flange must have a corresponding strain, ε=σ/E. The weld is fused to that flange, and so will have the same strain as the flange. The weld then, must have a longitudinal stress corresponding to that strain, and it's elastic modulus. If it has a cross sectional area, which it must, then it must carry a longitudinal force.

What happens to that force when the weld terminates? It gets transfered back to the flange and web through shear in the weld/base metal interface.

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What would one do in this theoretical situation (it would most certainly never happen)?

How would one design the weld?

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I agree with you. Each small length of a stitch weld will be axially strained as it tries to match the strain of the parent metal. However, the axial capacity of the weld is not a useful strength as it does not exist in the gaps.

I should have stated "If the weld is not continuous, it attracts no useful axial stress".

BA

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I'm trying to wrap my mind around what miecz is saying. It kind of makes sense, but I guess we just assume that the weld has zero cross sectional area. I have designed many such welds and have never included axial force. But then we don't really combine the bending stress and horizontal shear for the web and flanges either.

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I agree the longitudinal force in the weld is not at all useful. But some are saying it's not there. When I read your post, I thought you were in that camp.

ishvaaag-

I also don't entirely agree with assuming a completely distributed action (plastically) of horizontal shear. Seems to me that some of those welds are working a lot harder than others. I guess it's all based on laboratory testing, but my brain short circuits on the idea.

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I guess, since there is no differential strain or relative movement between the weld and the web/flange there is no force imparted by one to the other.

It would be similar to cover plates on a column; surely there is axial load in the cover plate, but that axial load is not exerting a shear force at the interface b/t column and plate. I think the same can be said here.

This is one of those things that is hard to explain, yet seems to be intuitively understood.

...Like why the Steelers beat the Browns everytime they play.

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OK. I guess I didn't write exactly what I meant either. When I wrote that the weld will have the same strain as the flange, I should have wrote that the weld will have

nearlythe same strain as the flange.Except for the special case of a beam with moment and no shear, I believe there will be force transfer from the web to the weld, and from the weld to the flange. It will transfer in shear through the weld.

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The VQ/I is then the change that happens in that axial stress along the length of the cover plate.

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Isnt MQ/I really just MC/I x Area?

I just found an old 1988 Engineering Journal Article that i think lays this out nicely.

Fourth Qtr/1988.

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MQ/I which in essence is the average stress (axial stress) in the cover plate times its area (Mc/I x area).

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I think you are confusing two different things.

The Mc/I stress we are all discussing is an axial stress on the weld cross section. So if you had a 10 ft. long simple span beam, the weld between the flange and web at the top would be compressed axially...like a long slender column.

This is no different than the small fillets on rolled wide flanges where the flange and web meet. These definitely have axial stress just like the flange and weld adjacent to them.

This axial stress is totally different than shear stress though the weld due to the VQ/I.

The axial stress is simply shortening the weld length due to the beam bending. The shear stress is trying to rip the weld into two long half-welds by failure through the throat thickness.

Your analogy using 0.3 x 70 is an AISC maxium allowable SHEAR stress. It is not the maximum allowable axial stress on the weld.

Check out AISC table J2.5 - For fillet welds the

shear stressis listed as you indicate....but right after that is the line for "Tension or Compression parallel to weld axis". Here it states "Tension or compression in parts parallel to a weld need not be considered in design of welds joining the parts". So the 0.3 x 70 doesn't apply to the Mc/I stresses.What others have said above is accurate - we normally don't check the axial forces on a weld such as this, only check the shear through the weld which gets larger towards the ends.

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DaveAtkins

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I would agree with you that it is hard to visualize perpendicular stresses on the weld. The bottom flange is going to try to force itself upward - toward the web - and you'd have compression there. The top flange is the loaded surface usually and would also force itself down on the web.

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DaveAtkins

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When the beam bends, the bending stress in the fillet weld varies slightly over its height because the term 'c' varies. The welds carry a shear force of VQ/I but that is not the principal shear. The maximum shear can be found by using the Mohr's circle.

At the midspan of a beam uniformly loaded by gravity loads, the upper welds are in pure axial compression and the lower welds are in pure axial tension. They have zero shear stress at that point.

BA

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Graybeach had a good model, Suppose the flange and web are not welded but put under beam loading and deflected. The middle of the flange will be in contact with middle of the web, but the ends of the flange will have moved over the ends of the web.

Think of it like putting a cover plate on a beam. AISC makes you develop the force in the cover, close to the ends of the plate.

Michael.

Timing has a lot to do with the outcome of a rain dance.

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http://

Michael.

Timing has a lot to do with the outcome of a rain dance.

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I bet he could explained this in less than 30 words.

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Nice try, but that was more than 30 words.

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Classic Miecz!!

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That website I posted tells about the same story as Timoshenko, but in simpler terms.

In working stress, I'm sure you know the normal shear distribution in beams, zero at the extreme fibres and maximum in the middle. The complimentary shear stress is longitudinal and in proportion to the normal shear stress. That is what the welds have to resist. At the ends, there is maximum shear, normal and complementary, and minimum compression or tension due to bending. At the point of zero shear, there is no shear but maximum compression/tension due to bending. Besides the two are not additive.

Sorry for the pedantry.

Michael.

Timing has a lot to do with the outcome of a rain dance.

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In the above analysis for calculating the horizontal shear, in the equation MC/I, the parameter C approaches zero towards the N.A. and maximum at the flanges. This assumes the beam is subjected to pure flexure. Timoshenko titles this expression as a the "flexure formula" and gives good examples of the theory (Gere and Timoshenko: Mechanics of Materials: Third ed SI: pp260). In pure bending, shear stresses on the N.A. are zero so a weld not required but try to make the beam stay together. Other factors arise if the beam is subject to non-uniform bending.

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Horizontal shear is at a maximum at the NA, not zero.

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_{b}/2. Draw the Mohr's circle to confirm.BA

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Is it possible, and how?

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BA

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I can't yet visualise how a beam can be bent, but have zero shear flow.

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BA

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That was a pretty dumb sounding answer I gave for shear flow when I said a weld would not be needed. I saw the problem clearly when I wrote it.

I assumed that if the beam were a series of plates, like a cart-spring, pinch the middle tight beneath the load (axle) and the plates slide away from the pinch - zero stain beneath the axle.

If the plates were pinched at each end and torqued equally with the centre of the moment on the line of the N.A., then the plates would slide over each other but remain static at the N.A.

Maybe

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My tuppence worth...I got involved in this when trying to understand Staad country code calculation to the French Code. I was finding beams that passed every other country code checks without issue, fail under the french code. It came down to Von Mises and the programming assumption of a single shear centre, that was producing misleading results and I see that in this thread regarding the classical view. If the check was done by hand, it worked fine.

Basically, an open thin-walled section like a deep I-beam, involving Von Mises,should be modelled with two shear centres. Imagine a thread throughout the length of the beam, at the top and bottom of the web/flange intersections. Without stiffeners and loading our uniformly distributed beam: at midspan it is possible that the shear centres which must be complementary can behave very differently and lead to flange or web buckling as further loading is applied. The top thread is in compression axially and may experience different shear flow behaviour as it becomes unstable. At the centreline of the web there is no shear flow. I think of castellated beams. At the top and bottom there is potential for separate regions of horizontal shear flow as the subsections are mobilised around their respective shear centre, with its own polar moment of inertia value on the compression shear centre. The top flange/shear centre can become unstable and we cannot use the I-value for the full section.

So what is the horizontal or vertical shear flow in this situation? The weld is part of a surface area under torsional stress.

The classic formulae work for the most part but do not always apply to all details. Was that gobbleydook?

Robert Mote

www.motagg.com

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Have to disagree with you there. As apsix recently pointed out, for a normal beam (no special cases), shear (horizontal and vertical shear stress for any element are equal) is highest at the neutral axis, or the centerline, of the web.

A castellated beam does not cut out the entire web, and is generally used for beams with high bending moment and low shear. Those remaining sections of web are working hard.

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If a beam was under pure flexure (ignoring self-weight and other gravity effects), is there any horizontal shear stress along the N.A.?

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The answers are:

(1) The horizontal shear is VQ/I per unit length which must be transferred by the weld from web to flange.

(2) There is bending stress in the weld, Fb = Mc/I which acts over the cross sectional area of the weld.

(3) No weld is needed at midspan for the purpose of transferring horizontal shear, but weld may be needed to prevent the web from buckling sideways.

(4) I'm not sure that you are missing anything. If the entire cross section is engaged an inch or two each side of midspan, you do not need any weld at midspan to engage it for that short length. The only purpose of the weld at the point of zero shear would be to prevent the web from moving away from its position at the center of the flange.

BA

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Your statement of total horizontal shear between a support and midspan is true if and only if the moment at the support is zero. If the support moment is not zero, it is not true. Moreover, if the beam is subjected to constant moment throughout its length, the horizontal shear at every section from end to end of the beam is zero.

That is easily verified, because at every point along the span, V = 0. Therefore, VQ/I = 0.

BA

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What is the point? The only point, so far as I can see is to determine how much weld is required. If there is horizontal shear to resist, the weld must be capable of resisting it. If the weld is only for the purpose of holding the web and flange together, then a stitch weld may be sufficient.

In the case of a simple span with symmetrical cantilever each end, when the cantilevers are identically loaded and the span is unloaded, the span has a constant moment and zero shear. The welds between web and flange in the span carry no horizontal shear but they are required to hold the pieces together...a nominal weld should be adequate.

BA

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- Members in pure bending will have shortening of the compression face and lengthening of the tension face.

- This will occur in your example above, at the flange/web interface, unless there is a (shear) connector forcing the elements to act compositely, ie. both elements will have the same tension or compression at the interface.

- The force this connection is resisting could be called horizontal shear and will occur wherever there is curvature, as I see it.

- I understand that my argument isn't compatible with the horizontal shear = VQ/I version.

Feel free to point out my mistake.

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However the tensions perpendicular to the axis of the beam at the compression side are tensile, and if you do not show physical opposition to the tensile force there provided by a weld, the flange may buckle away from the interface web-flange.

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DaveAtkins

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Shear flow is consistent with "flexural displacements between segments within a section". If two or more segments stacked together without bond, under load, which produces vertical shear, the segments deflect and the relative displacement is greatest at the end, zero in the middle. THe horizontal shear is therefore required to prevent the relative displacements with magnitude varying from maximum at the ends, zero in the middle.

Make sense?

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Michael.

Timing has a lot to do with the outcome of a rain dance.

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It is true that there are principal shear stresses even at locations where there is only bending stress, but those stresses are on a plane that is 45 degrees from the horizontal, right? These wouldn't be considered in a weld design.

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Not true! In the case of a simple span of length L with cantilever of length 'a' each end, if load P is placed at the end of each cantilever, M = P*a at each support. The same moment occurs throughout the span L. Each support carries a vertical reaction of P.

Weld is required between flange and web throughout the length of each cantilever to resist horizontal shear and develop the combined section. Weld is not required in the span because every fiber of the beam has been strained to P*a*c/EI immediately outside each support, where 'c' is the distance from the neutral axis to the fiber.

If the beam was actually built without weld in the span, it would fail, but not because of horizontal shear. Compression elements would buckle because they lacked lateral support, so a nominal weld is required across the span just to hold everything together.

Curvature in the cantilevers varies from 0 at the ends to P*a/EI at the supports. Curvature throughout the span is P*a/EI with no help from the welds.

BA

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Yes, I agree.

BA

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I am not sure that horizontal shear should not be considered in addition to vertical shear at the flange weld. I design ply web-beams and I would certainly check the glue line for a combination of both. I think in the Eurocodes, a test for a combination of both stresses would have to tried.

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The horizontal shear is a result of the vertical shear? That's like saying you should check for the moment and the normal stresses associated with that same moment. Am I missing something?

I've never seen any standard spec, code provision, or textbook advocate doing that.

Can you describe the beams you're referencing and exactly what you're checking for?

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Horizontal shear stress are the direct result of applied vertical shear forces, no?

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In timber, bending and shear stresses are considered together - its the nature of the materials. Maybe not necessary in steel unless you castellate the beam - maybe. Check out shear deflection.

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I had the same problem when I started learning. Until I saw a heavy riveted steel girder, the kind you find in old bridges.

Have a look at that, some time.

By the way, JAE,StructuralEIT,BAretired, ishvaag and all, thanks for the great posts

respects

IJR

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After this exhuastive chat, it's pretty clear to me now.

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Sorry, I had an experience to share but apparently failed to express properly(real sorry mentioning learning):

You visit an old factory or a recent great structure and see details and configurations and your analytical mind jumps in and you get inspired. Right when you think welds you see rivets and the contrast works your mind. You think draining systems and you visit a football pitch, and the draining system there makes you stop and think. Your interpretation of theories and practices gets reinforced.

Yes, the thread is long, yet great. I quit now.

respects

ijr

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Michael.

Timing has a lot to do with the outcome of a rain dance.

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Do you calculate the shear flow at the support and distribute those welds evenly accross the length of the beam? or would you put all the weld length required to develop the flanges closest to the support without any weld in the center.

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BA

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a) you calculate the weld you need for the max shear at the ends. And then apply that weld on a repeated basis along the length of the member? (i.e. if you need 20" of weld, you apply 20" of weld at the end, then a space (not sure space size?), then 20" more weld, repeated accross the beam)

or

b) If you need 20" of weld you apply 10 welds that are 2" long equally distributed along the length of the beam.

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I also meant to write that I typically provide enough weld at the ends to develop the reinforcement (are at a minimumn like 12" of continuous weld).

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Try a weightless warren truss with the end bays cantilevered and only the cantilevers loaded.

Michael.

Timing has a lot to do with the outcome of a rain dance.

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This Paragraph reads:

"High strength bolts or welds connecting flange to web, or cover plate to flange, shall be proportioned to resist horizontal shear resulting from bending forces on the girder...."

Should this not read:

"...horizontal shear resulting from DIFFERENTIAL bending forces on the girder..."

As stated before, a beam theoretically in pure bending will have no horizontal shear forces "resulting from bending".

Please comment.

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However, that doesn't make the terminology correct.

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I, to this point, have used an engineering journal article from 1988.

In order to come up with a "V" to use in VQ/I, the paper uses "V=equivalent shear" based on the sections maximum allowble moment using a simply supported beam with a point load at the center for the analogy.

Or,

M=Pl/4

therefore

V= P/2 at each end-->

P=2V

sub back into M eqn.

M=2VL/4

M=VL/2

V=2M/L

M= allowable moment of column section.

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I am not sure the statement is needed at all as it seems to be stating the obvious, but if included, I would prefer "...horizontal shear resulting from loading of the girder..."

The only forces acting on the girder are applied loads, including self weight. There is no such thing as a "bending force".

BA

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BA

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We've established that the welds in question experience both shear and flexural stress as a result of differential bending stresses along the length of the member. Essentially, the felxural stress manifests itself as axial stress along the longitudinal axis of the welds.

The flexural / axial stress is a no brainer for a chunk of steel (the weld) that is braced every which way. However, it can be argued that the horizontal shear in the weld produced by differential flexural / axial stress along it's length should add to the VQ/IT shear burden of the welds.

I propose that a simple way to account for this extra shear burden would be to include the cross sectional area of the welds in the Q term of the VQ/IT expression. This would increase your shear demand by a tiny -- probably negigable -- amount. At the same time, it would be a conservative way to account for any flexural / axial stress imposed on the welds themselves.

How about it?

I can think of no practical situation in which a responsible engineer would conclude that he or she could reliably count on a zero moment gradient in a member under consideration. Additionally, the welds serve a number of other important purposes as pointed out above. While the theory behind and constant moment scenario makes for interesting discussion, I don't think that it warrants any serious attention in practical design.

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Are you actually retired?

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Engineering Journal, Fourth Quarter 1988.

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Affirmative. I retired on June 1, 2008.

BA

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Michael.

Timing has a lot to do with the outcome of a rain dance.

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If I understand correctly, I can calculate the axial force in the plate. I can then design the welds to have a shear capacity that exceeds the axial force in the plate (between the support and point of zero shear). In a simply supported beam, with a point load in the center, I would need to develop the axial force in the beam on each side of the point load.

To further clarify, the amount of welds you need along the plate should be equal on each side of zero shear (max moment). And the welds on each side of the zero shear should have a shear capacity greater than the axial force in the plate established from your max moment calc.

Am I correct in this thinking? If so, what would be the benefit of establishing the instantaneous shear flow. Why not skip shear flow and base your weld design on the tension or compression in the plate.

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If you want to place the weld where it is needed you need to know the magnitude of shear between flange and web all along the beam. Your example of a load at midspan happens to have constant shear at all points.

Consider instead a beam with loads at the third points. The shear in the middle third is zero, so you don't need much weld in that location. If the section is to be developed, the weld must be situated where it is needed, in this case, in the outer thirds.

In the case of uniform load, the shear varies from a maximum at the support to zero at midspan. If you want to place the weld where it is needed, you would have more weld near the ends than in the middle.

BA

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But...do we need to calculate shear flow then? or can we just find the tension/compression in the plate? In your third point loading example above, am I able to find the tension/compression in my plate and design my welds in the outer thirds to have a shear capacity greater than the tension?

...i guess i'm not seeing the value of the actual shear flow calc unless im missing something

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The horizontal shear per unit length between web and flange in the outer thirds is the maximum flange force divided by a, or P*y*Af/I which is the same as the shear flow, VQ/I since Q at the weld is defined as Af*y.

For any load case, including a uniform load, knowing the flange force at a point does not give you the magnitude of horizontal shear and hence amount of weld per unit length at any given point. You could calculate the flange force at two points along the beam separated by distance d. The difference would give you the total force going into the weld within the distance d.

Although you could do the above, it seems more direct to draw the shear diagram, then multiply the shear at any point by Q/I, which is normally constant for the entire length of beam.

BA

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Think about concentrically loaded column that consisted of stocky web and flange plates (none slender), the column can support significant load before seperation of the plates(due to buckling). However, the flanges and web will be dislocated once an horizontal force is introduced, now you need to weld/glue the pieces according to VQ/I, which obviously does not equal to P/Af, or P/Aw. Make sense?

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I learn't that shear flow was the way perpendicular flows around a section in unsymmetric bending and that the thing that you were referring to is called shear stress.

It appears that some people would use them interchangeably.

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Please see the attached PDF for terms I meant.

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I think I got that, shear flow stress was drawn with length varies to indicate the change in intensity along the beam (from center towards the ends). Note the beam was a cut out from W-shape to domonstrate location of stresses and terminology of my understanding.

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Don't you mean "a few kips per lineal foot" of shear flow... :)

This thread is sheer madness... :)

tg

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I know you are THE trainguy, but are you a real train guy? Somehow connected with railroads and that industry. I have spent a lot of years designing and building railcars, mostly special, heavy duty railcars, but other types too; and doing work on various railcar problems, derailments, running gear and HWAH load handling.

No, I meant what I said. I got an oversupply of kips, some left over from the last job. But, am not sure yet exactly how to allocate them kips to the various welded joints, am working on that right now. But without the shear I don't know how to get them kips to flow into the joints. Thus, my only need is for lineal feet or inches of shear flow. Will buy them in units of 1000 lbs. too, if the price is right.

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I would not have named myself so if it weren't true. I am a structural engineer who first worked 8 yrs designing buildings and then 12 yrs designing rail vehicle structures (passenger coaches/ locomotives & some freight cars).

For the sake of humanity (or at least the portion tuned in here), I have no further comments on shear flow.

tg

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I was the Chief Engineer at The Maxson Corp. in St. Paul, MN in the 70's; we designed and build special and heavy duty railcars and transport and lifting equip for many of the major railroads and other industrial customers. I started my own consulting business in 1977 and did much of the same design work, and then helped find builders for the equip. We also did a fair amount of work for various transit authorities, mostly work cars and equip. Without knowing it, I suspect you have seen some of our equip. I have worked for or with many of the large railcar builders and been in their shops over the years. Either helping them with a problem on one of their car designs or helping my clients buy lots of cars from them. That business has really changed over the years and seems to a pretty well dried up.

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Would like to correspond, but I'm not sure how while maintaining some degree of anonimity(sp?) in this very public space. FYI, I work in engineering consulting, clients being primarily Eastern Canadian rail vehicle re-work shops, the passenger rail authority, and some vehicle OEM's on occasion.

tg

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Final effort to answer your OP question.

wrt Plane Stresses from Strength of Material - there are two types of stresses acting on a plane element taken out from a beam, namely "Normal Stress & Shear Stress". The former came from Mc/I, that result in compressive or tensile stresses acting normal to the sides of the plane element; the latter came from VQ/I, the result is shear stresses acting on the sides of the plane element.

I think it is simpler to treat the shear stress as glue, or bond, necessitated for things stay together as a whole, otherwise the adjacant pieces could slide freely if no friction in between. On the other hand, the normal stress pushes things together, or pulls things apart, quite different phenomenon, isn't it?

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I'd probably have to push Timoshenko out of the way.

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You beat me to it! I was hoping someone would look in a strength of materials text and scan that page or two. Surprise, surprise, my texts say exactly the same thing, although the ones by Timoshenko are called theory of elasticity texts, talk about stress at a point, and have a lot more ∫∫'s and dx/dy's in them than I like to deal with in my old age. Seems we agree again, I certainly couldn't have said it any better, although I am sure it would have taken me far more verbiage, my hat's off to you. Except, your 14:27 post, I might have worded it: yes they do coexist.... now mind your own business and get the hell back to work, or I'm gona tell your boss your playing with his expensive computer. I think we are all getting pretty stressed out, beam web or not and who ever's in there, on this one. Let's give Stillerz the award for longest thread, would you please second that, so we can all get back to work. And, so the world wide supply of shear flow will come back to normal, because this thread and that supply can not coexist in a state of "statical equilibrium."

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I concur. I am running short on supply as well. Wish Santa bring some with him this time around. Cheers! :)

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In this example, the shear flows horizontally along the top flange, then down the web and finally horizontally to the tip of the bottom flange.

The value of shear flow varies from one point in the cross section to another. Shear flow in the web carries the shear V. Shear flow in the flanges creates a pair of equal and opposite horizontal forces which create a moment. The shear center of the channel may be determined by finding the point where the moment is zero.

Except for csd72, this point appears to have been missed by contributors to this thread.

BA

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If you are illustrating flexural and shearing stress, f = Mc/I is a stress but VQ/I is not. The shear stress in your sketch is VQ/Ib where b is the width of the rectangular beam.

BA

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Isn't this shar stress a part of the shear flow? I think Timoshenko's book has similar illustration shown in the chapter dealing with shear.

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Make a book on this thread, it could be the best seller of the year, and you can sell to whoever has shortage of it, such as dhengr :) Cheers.

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For the 2nd question, try to bend two chopsticks with space in between, then you will see what is missing.

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Cntw has this covered, but if you want to see what is the least amount of web you can get away with I recommend you look at castellated beam design.

Arguing with an engineer is like wrestling with a pig in mud. After a while you realize that they like it

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For what it's worth, I think that your assessment is bang on. For the highly idealized thought experiment that you proposed, there would be no need for web/flange welds nor for the web plate itself.

In residential wood floor trusses, often the diagonal web of a panel near the middle of the truss will be omitted to create a mechanical chase. This is justified in a similar fashion. I'm not sure that I agree with it (assumed loading vs. real loading) but it is done, without incident at far as I know. I imagine that the trusses are probably saved by some minimal vierendeel truss action across the

affected panel.

KK

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The tension flange could be replaced by a cable in the middle half. The compression flange could be replaced by a compression strut. If the connections are hinged at the quarter points, the structure has four hinges and is unstable, so a diagonal brace is needed for stability.

BA

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A assembly of two same size rectangular beams subjected to pure bending, one simply stacks on top the other, zig-zag would result at the beams' ends. However, if we glue/weld/nail these two beams together, the zig-zags at the ends would disappear, now the ends would be two slanted straight lines.

My question is: what is the stress in the bonding/binding agent (glue/weld/nails), if the joint is right on the N.A.?

(Ignore friction between the beams)

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cntw1953: that should just be straight up VQ/It

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The application of the end moments in manner consitent with these assumptions is a big part of what makes that scenario so hypothetical.

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Now, still under pure bending, we all know the stress distribution differs for the 2 beams assembly and the single (smae size) beam, because "I" differs for the two system.

My question now is what makes the latter stronger than the former if there is no stress at the N.A. ?

Am I missing something again?

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To get the most out of your combined beam cross section with respect to flexural capacity, you want a stress distribution where the fibers furthest from the neutral axis are the most heavily stressed. Of course, strain compatibility must still be maintained.

In the case of general loading, this can only be accomplished by providing a horizontal shear connection between the constituent parts of the cross section. It is precisely that shear connection that allows the axial stresses and strains to be transferred to the outskirts of the cross section where they will be most effective.

In your end moment scenario, there is only ONE way to apply those moments without inducing localized horizontal shears at the ends and thus violating the assumptions inherent in the problem. The end moments have to be applied as axial stresses distributions consistent with Mc/I of the composite section. This will yield linearly varying stresses and strains across the depth of the section at the ends.

With the end moments applied in this way, the stresses don't need to be transferred to the outskirts of the cross section because that's where they have originated from in the first place. Consequently, no horizontal shear stresses are required to facilitate the transfer and it does not matter if the constituent parts of cross section are connected.

In summary, under the specific case of equal end moment loading, you get the composite stress distribution for "free". You get it without having to do do the work of providing a mechanism (horizontal shear) for transferring axial stresses to the outskirts of the cross section.

Having the stress distribution of a composite section also gives you the bending strength of a composite section.

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Makes sense?

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I'm afraid that I don't follow with the clamping force. Can you elaborate?

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I was trying to depict the action of "hold the beam ends and bend", obviously "clamping force" was a false expression for that. Sorry.

Thank you very much for walk with me on this seemly non-sense excercise. Your comments/insights on this is sincerely appreciated.

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The beams that are welded at the neutral axis will have 4x the moment of intertia of the stacked beams, so for the same end moment the welded system will see only 1/4 of the deflection of the stacked system.

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There is a shear across the "weld" at the neutral axis of the composite beam. But the shear is not a result of VQ/It. The shear is a result of the applied moment, which must be applied as a couple.

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I agree it is the case for general loading cases. But for the rare event of pure bending, there maybe more things need to be sorted out before we can confidently claim it follows the pattern of the general case.

Please review the 9 posts above yours, pay close attention to KootenayKids' responses on the question I posted on 22 DEC 09, 12:08.

Have a joyful holiday season.

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Sorry to have your post hijacked, I think the phenonmenon of the current discussion and the understanding on your case are somehow interrelated. Hope you don't mind. Thanks.

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I think the post is great.

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I have read the previous posts. My posts apply to the special case. That's why I mentioned an "end moment." I was so intrigued by your question that I went ahead and built a finite element model of stacked beams, with and without a weld at the interface, with end moments. The shear pattern in the composite beam indicated a shear across the weld, and the non-composite section deflected 4x that of the composite section.

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Have fun in the holidays.

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This has been a long drawn out discussion and maybe I'm missing your intent. Suppose we have a beam of span L with a cantilever of length A at each end. The beams are stacked but the cantilever sections are welded together so they act as a composite unit for a distance A at each end. The load consists of a point load, P at the tip of each cantilever.

Are you saying this beam would behave differently than the same stacked beam in which welding is continued across the entire length, L + 2A? If you are, I disagree.

BA

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Isn't that (shear) intriguing? Pure bendind - shear at N.A. ?

Thanks for the update. I would need time to get my head kick-start :)

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That's not what I was saying. The model I had in mind came from cntw's 22 Dec post that had the zigzag pattern at the beam end, and that's not the case for a cantilevered beam. I don't honestly know how a cantilevered beam would behave. I'll see if I can model it up.

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You're right, for the case that you describe, the beams would behave the same.

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The cantilevered, partially welded system is the closest thing to a practical situation in which the end moments would be applied as I have suggested above (applied linear varying stress instead of a force couple). This is why, in this scenario only, the non-composite beam segment is as effective (I, Mr) as a composite beam.

Miecz: if you modeled the beam using force couples to simulate the moments at the beam ends, your FEM model is likely giving you answers that are not consistent with the assumptions of our discussion. The only way to model the situation purely is to apply the end moments as linearly varying stresses.

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