Horizontal shear in a beam or plate girder is mobilized by bending, so horizontal shear is not 0 at the center as would be for vertical shear.  It is actually at a maximum with the maximum moment. The shear is created by the tension/compression differential in bending.

You should not design this weld for MC/I.  The entire section resists the moment if the welds are adequate to transfer the shear.  I remember a demonstration in statics class where a load is applied to a stack of unattached plates and to a stack of plates attached together.  The unattached plates slip relative to each other resulting in a large deflection.  The attached plates are much stiffer.  The attachment is meant to prevent the slip and allows I to be calculated based on the whole section, not just the sum of the I's of the individual plates.  At the middle of the beam you describe, the shear has already flowed into the flanges.  Theoretically, no weld is needed at that infinitesimal point.

Ron- This was my thought as well.
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.
 

...."do result"

 

Ron-
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?  

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

Here are my thoughts.

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).

Can you explain what you are calling "differential moments"?
 

Am I wrong in saying that horizontal shear forces in a beam are the direct result of vertical shear forces being applied to the beam?
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?  

As you increnentally walk along any beam from point A to point B, it is the difference in the moment at A and the moment at B due to the applied vertical load.

Mike McCann
MMC Engineering

Stillerz.

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.

Yes.

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.

JAE-
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?  

Good work EIT, JAE.

Shear is the derivative of moment along length and is at maximum for ordinary beams near supports; the rate of variation of moments is then also at maximum there, at ends, and so is horizontal shear in web-flange welds or vertical sections in flanges. Remember the old practice of setting rigid welded stubs at ends of top flanges of composite beams, that proceeds just of these considerations.

I should make clear that my point here is trying to justify to myself why the longitudal force in the weld due to bending stresses can be ignored.  

....at the center.

The reason that the flange-web weld is critical at the mid point of the beam is that the compressive load on the flange is maximum at that point. The purpose of the web is to support the flange to keep it from buckling out-of plane. Although the theoretical loads are zero the effect on "I" is what is important to prevent column buckling.

I think this is the first time I have ever said--

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

Dave-
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?
 

No.  The force in the weld is the SHEAR FLOW, not the longitudinal bending stress.  Mc/I defines the stress in the flange, which has been building up from the end of the span to midspan, not the longitudinal force in the weld connecting the web to the flange.

 

DaveAtkins

I think part of the disagreement might be the assumption of the shape of the weld.  If you assume the weld has a triangular shape in cross section, then I see how one could say there is some axial component to it (I disagree with that).  If you assume the weld to be a plane (at the juncture of the flange and web), then I can see how one would say there is no axial component (I agree with this).

Quote:

I should make clear that my point here is trying to justify to myself why the longitudal force in the weld due to bending stresses can be ignored.  

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.

I guess I am not really disagreeing with anyone.
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.  

I was almost afraid to post this question because I thought I should have a full understanding of this subject given my experience. But, apparently it wasn't a "dumb question". I just hate doing stuff in a cook-book manner. I like to know WHAT I am designing not just the procedures.

Thanks all

It's a valid question.
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.

DaveAtkins - you don't disagree with me at all. What you said was simply a clearer way of saying what I said rather clumsily.

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).   

I believe that there is something important missing from the discussion so far.  Namely, the distributed nature of horizontal shear resistance in typical flexural members.

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...

Nice discussion. Just want to discuss the units being used in the equation VQ/I. V is shear in pounds, Q is the first moment about the neutral axis in inches^3, and I is the moment of inertia in inches^4. It looks like there is an extra inch in the denominator - unless V is actually the vertical shear x 1", which would make it a moment and get rid of that extra inch, leaving only pounds which is the shear we are looking for. Right? The "I" term is a dead give away that this is a flexural equation.

 

VQ/I should give units of k/in of shear flow.

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

Stillerz said:

Quote:

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

(2)  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?

(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

I don't entirely agree with assuming a completely distributed action (plastically) of horizontal shear. In my view this approach should include some limit as grip of bolts on plates, or number of rows of bolts in shear. Simply, not all fasteners are going to work the same even at extreme deformation excursions, and to neglect this can result in some "zip" failure that may progress uncontrolled.  The more common case of studs on flanges in composite beams where plastical distribution is assumed is one where various controls and allowances are implicitly introduced, the main one the ability to pass the brute yield (and even overstrength) capacity of the steel member.
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.

Yes, an interesting discussion.  But I still think I am right

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

Back to what Stillerz observation about a plate girder carrying only moment and no shear, i.e. pure bending.  This case proves that the web to flange welds carry horizontal shear only (theoretically).  With pure bending, there are no forces in this weld.

Quote:

If the weld is not continuous, it attracts no axial stress.

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.

graybeach-
What would one do in this theoretical situation (it would most certainly never happen)?
How would one design the weld?  

miecz,

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

In the theoretical situation of pure bending, I would just put the minimum size weld.  There is no force you could calculate for it.

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.  

 

BAretired-

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.
 

I think what Miecz is saying is logical....it was the same thing i was thinking when i posed this question in the post.
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.  

I believe that a cover plated column does use shear at the interface to get axial load into the cover plates.  How else is the load getting there?

yes....shear only
 

Quote:

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.

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 nearly the 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.

disregard my last couple posts....

The formula that we use for the end development force that needs to be transferred into the cover plate (at its cutoff point)is MQ/I which in essence is the average stress (axial stress) in the cover plate times its area (Mc/I x area). Some engineers will conservatively use 0.6 Fy x area of cover plate.

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

Jike-
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.

 

Yes. I believe that is what I said...

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

I guess you'd want to use the yeild moment in he MQ/I equation (something like 0.66FY for the old ASD).  

...in order to closely approximate and average stress.  

DaveAtkins


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 stress is 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.



 

I still have a hard time believing there is a tensile stress trying to pull the weld apart (perpendicular to its length) at midspan, where the web is connected to the bottom flange.  I think there is tension in the flange and in the lower half of the web, but not in the weld.  I think the weld is just transferring the tension from the web to the flange via shear along its length.

DaveAtkins

Dave - not perpendicular to its length...parallel to its length.

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.

 

What I meant is that an axial force in the weld would create a failure plane perpendicular to the longitudinal axis of the weld, and I just don't believe that kind of force exists in the weld.  I think there is just a shearing force along the length of the weld.

DaveAtkins

Well, if the beam around it is compressing in strain, the weld must go with it and compress as well.  Hooke's Law means that where you have strain, you must have stress.

 

If top and bottom flanges are welded to the web with a fillet weld each side of the web, the fillet becomes part of the beam.  Its area contributes to the beam area and the moment of inertia.

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

I think we all agree that "an axial force in the weld (that)would create a failure plane perpendicular to the longitudinal axis of the weld" doesn't exist.

To understand, read up on complementary shear stresses.

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.

A not bad piece on complementary shear stress

http://www.codecogs.com/reference/engineering/materials/beams/shear_stress_in_beams.php

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

Where's Timoshenko when you need him?
I bet he could explained this in less than 30 words.  

The thing with many of the good researchers of the XIX and XX century is that they have high respect of the mathematical basis of mechanics, and not as many professors to get the bunch of them not to be brilliant. I lacking much the mathematics I would love to know, appreciate very much the overspread reach of those people able to adjudicate symbolical models for everything subject to their consideration at ease. It should also be matter of reflection to see how much the mathematical advances themselves were tightly tied to the advances in mechanics always targeted to science of course but then immediately to engineering. Said another way, at those societies these gents got intelligent numbers of them to be fully armed to mathematically describe phenomenons. I don't think now that is true to the same extent, if only by overtextension of the related "trades".

ishvaaag-

Nice try, but that was more than 30 words.

HAHAHAHA-
Classic Miecz!!

You know, I can't count, today was giving 70 euros to pay 21.
 

I always found Timoshenko complicated.
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.

Dear All

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.

The equation MC/I calculates flexural stress, not horizontal shear.
Horizontal shear is at a maximum at the NA, not zero.

A beam in pure bending has no shear.  So the shear stress, VQ/Ib is zero at all points along the span.  The maximum principal shear at any section, however is f_b/2.  Draw the Mohr's circle to confirm.

 

BA

I did think about that (a little bit) but couldn't think how you could get bending into a beam without shear.

Is it possible, and how?

Yes, it is possible.  Place an equal and opposite bending moment at each end of the beam.  You have a case of pure bending throughout the span with no end reactions and a deflected shape in the form of a circular arc with radius of EI/M.

BA

You're right! Thanks.

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

Sleep on it.  It will come to you.

BA

Dear apsix

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

Fascinating discussion.

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

rtmote-

Quote:

At the centreline of the web there is no shear flow.

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.

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

No. Moment=constant, Shear=0. Each fiber is compressed or elongated as to satisfy the plane-remains-plane in the same radians per unit length and no shear develops between fibers, along any height of the constant section.

Simply talking, moment is the sum of area of the shear diagram, therefore, using the max moment for a simply supported beam at the mid span to calculate the horizontal shear flow at a given plane should give the total horizontal shear in that plane up to the section considered. This methodology is simple to use, however, since it's the total shear does not given you the higher shear flow that should occur near the support.

This is an interesting discussion but I'm starting to wonder what is the point that is being made. An example that I have often found useful is comparing a 2x4 to a phone book. A phone book cannot carry shear loads between the pages. If you bend a phone book the pages slide but stay the same length. The example presented above of a beam in pure bending where there is no shear is interesting but what is the point? The welds are still absolutely critical to prevent buckling of the flanges and web. Try bending a phone book in pure bending. The pages in compression will buckle.

Let's get back to the original questions:

Quote:

(1)  When designing an 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.
(2)  My question is, isn't there bending stress on the weld as well in the form of MC/I?
(3)  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.
4  What am I missing here?

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

shin25,

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

Compositepro,

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

Which leads us back nicely to my problem of last week.

- 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.
 

No. For a segment with the shape of a part of a ring, the longitudinal forces along the tangent are compression (or tension for the tensile face), where the circumferential faces sustain in any case normal forces to them. For these, at the limit, faces, you have a triaxial state of tensions where the stresses are principal. The need of the welds doesn't come from any variation along the fiber of tangential (along the shape of the curved beam) longitudinal stresses, where in the varying moment, yes, because the difference in resultant  for a cut rectangular segment in that case exacts that equilibrium be restored by a shear force in the cut along the paralell to the beam's axis.

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.

 

Can we all agree that the welds which connect a web to a flange (in a bending member with shear) are designed for VQ/I, and nothing else?  The weld does experience some longitudinal stress due to bending, and does prevent buckling of parts which are in compression, but is only designed for VQ/I.

DaveAtkins

I didn't read through the entire discussion, my thoughts below might cross someone's before.

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?

DaveAtkins, I  agree, except sometimes it is governed by the vertical force transfer from flange to web, typically from a point load on the flange.

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

BA-
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.


 

apsix said:

Quote:

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

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

StructuralEIT,

Yes, I agree.

BA

Dave Atkins

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.

hemis-

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?

E.I.T.-

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

Yes, I don't know why I have a question mark.  That should be a period.  I was making a statement and following it up with an example.

i see...makes more sense....i started this post and even i am sick of it!!!!!

I looked but couldn't find the "beating a dead horse" gif icon.

 

But there were a lot of good posts here...don't get me wrong.

 

StructuralEIT

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.

stillerz

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

I guess I never really had a problem with the concept of horizontal shear...that was basic stuff from mechanics of materials. I just had a hard time justifying ignoring the bending stresses in the weld design.
After this exhuastive chat, it's pretty clear to me now.  

Stillerz

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  
 

AISC has, or used to have, a statement that the horizontal shear stress in the weld was not additive to flexural stress.

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

if your shear flow is maximum at the supports, and zero at midspan (which I agree with), how do you design  your welds at the web/flange interface.

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.

I determine the max shear flow at the ends, then design a skip weld based on that load and apply it to the entire length.  It's a little on the conservative side, but there other things to consider like making sure there is no local buckling for lack of welding to tie it all together, and I think that is how most people do it.

I don't think I have used a welded plate girder over the last fifty years, but if I did, I would do it the way StructuralEIT stated.

BA

StructEIT, to clarify, did you mean:
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.
 

I vote for option (b).  Usually you calculate it on a mm/m basis or something akin.  Depending on the code in effect and the designer's preference, often there's some extra weld at the ends to be extra certain that slip is prevented where the shear is high.

No, it's determine by k/in comparisons.  You have a k/in capacity of weld, and a k/in demand of shear flow.  Iwhatever the max shear flow is at the ends is what the skip weld gets designed for.  If it's 0.8k/in=9.6k/ft, then I size a skip weld to provide 0.8k/in.  A 1/4" fillet 3" @ 12" is good for 0.928k/in or 11.14k/ft

k/in of weld is fundamentally the same as mm weld / m isn't it?  It's just a matter of whether or not you've already got your weld size picked out.  What junior suggests above is essentially stitch welding.

Sorry, the "no" was referenceing the post before yours, kootenaykid.  I agree with you.  I started typing and got distracted by work, then finished.

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).

Ahh.. darn real work.

StructEIT i belive you are describing option b?

 

Yes, option b, but I always provide a continuous weld at the ends.  Sometimes enough to develop the plate (or whatever we're talking about), sometimes 12".  It really depends on how critical it is and how much load it's really seeing.

Apsix. In the beam subject to perfect bending, there is no change of curvature over the span. No change in curvature means no change in the force in the flanges, and, consequently, no shear to pass from through the web from flange to flange.

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.

Can I please refer all to the very last paragraph on page 16.1-62 of the 13th AISC Code.

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.

 

Stillerz, I wonder if the typical nature of beams in construction USUALLY have differential bending and the theoretical bending just doesn't occur all that often?

I agree...you can't have a beam with no self-weight...i.e., zero shear forces.
However, that doesn't make the terminology correct.  

This makes me think of a whole other question...How to properly size welds for column cover plates? Or, in other words, compression members with primarily axial load.  
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.  

Quote:

"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".

Stillerz,

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

well stated BA

In the case of column cover plates, the force to be carried by welding is the entire axial force in the cover plate. If the force is A*Fs where A is the area of cover plate and Fs is the maximum stress therein, you need enough weld to carry that force at each end of the cover plate.  Between the end welds, you need enough weld to prevent the plate from buckling as an individual member.

BA

Stillerz: could you provide the author and title of that paper that you mentioned above?  I might like to get my hands on a copy.

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.

BA-
Are you actually retired?
 

Reinforcing Loaded Steel Compression Members by Jack H. Brown.
Engineering Journal, Fourth Quarter 1988.  

Stillerz,

Affirmative.  I retired on June 1, 2008.  

BA

Stillerz, VQ/I gives you the instantaneous rate at which load is being transferred to the flange. The sum of VQ/I from beam end to the point of zero shear, is the force in the flange. It should agree with the force in the plate calculated from the bending moment.

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

paddingtongreen, this concept is bothering me...I understand your post, however, based on that logic, there would be little reason to calculate the instantaneous shear flow.

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.

JrSEng,

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

Valuable post BA. thanks.

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

In the case of third point loading, you could do as you suggest.  The moment is constant in the middle third and equal to P*a where a is the distance from load to support at each end.  The average stress in the top flange is P*a*y/I, where y is the distance to the middle of the top flange.  The total force in the top flange is P*a*y*Af/I where Af is the area of the flange.

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

Shear flow is the minimum stress required for weld/glue to hold discrete elements/parts together under forces, such allows the combined elements/parts to act as a single composite structure, or built-up shape.

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?

It seems that it is a difference in terminology:

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.

csd72:

Please see the attached PDF for terms I meant.

• http://files.engineering.com/getfile.aspx?folder=e06be012-743b-4635-a7e3-cb

cntw, your diagram is what BA and I are referring to as well...however, our discussion had to do with conditions where instantaneous shear flow is different at different locations along the length of the member, and thus, the proper way of welding such a section.  
 

Jr:

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.

I've watched this thread with some considerable interest, and will have to read the middle 50% of the posts to get back in the shEEr flow of it.  That is, to see if I think I still understand it, or maybe not after the additional reading.  Seems to be a mighty interesting topic, with several different variations in the ways of explaining it, and some considerable misunderstanding by some of the populous of its meaning and effect.  In any case,  I am working on a fairly large and complicated steel fabrication, and was finally getting around to starting to design a bunch of the welds, and low and behold I couldn't find any shear flow.  I riffled through my calc. pages and couldn't find any, I went out to the shop and they didn't have any in the welding supplies room, checked with our welding suppliers, they don't have any.  Seems there is a world shortage of shear flow, I think it's all been used up by this thread.  Could you guys/gals spare me a few thousand lineal feet of shear flow, so I can get this design done?    :- ), I think I know what that symbol means.

Come on dhengr,

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

This thread is sheer madness...  :)

tg

trainguy:

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.

dhengr,

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

Trainguy:

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.

Does this post get the blue ribbon for Eng-Tips' all time longest forum discussion?

  

dhengr,

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

Stillerz:

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?

Indeed they are quite different, however, one cannot exist without the other in statical equilibrium.  

Yes, they do coexist while minding their's own business.

I'd like to crawl inside a stressed out beam web and see what the hell is really going on in there.
I'd probably have to push Timoshenko out of the way.  

cntw1953:

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."  
 

dhengr:

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

Shear flow is illustrated in this page from "Elements of Strength of Materials" by Timoshenko and McCullough.  

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

• http://files.engineering.com/getfile.aspx?folder=5b9299ac-0e2f-49ca-9135-97

Hope the attached sketch makes the cut.

• http://files.engineering.com/getfile.aspx?folder=d73978fa-3041-4c1c-8279-d7

cntw1953,

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

Yes, I should have b under the denominator. Thanks.

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.

I think maybe there should be a new forum for "Shear Stress/Flow Engineering" and universities should work on new ABET accredited curriculums for Shear Engineers.  

Stillerz:

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.   

In the interests of making this thread even longer: If a simply supported beam experiences zero shear load (ie constant moment) over say the central 50% of its span, is there any need for connection between the web and flanges in this region (apart from the need to avoid buckling of the compression flange)? Is there any need for a web at all in this part of the span if the compression flange is stable?

Then you have seperated parts rather than an unit beam, that affects moment of inertia.

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

tigermoth,
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

Thanks for your replies. I wouldn't seriously consider removing the mid-span web in this way; it was more of a thought experiment. If we assume a beam or length L simply supported at each end, with point loads at say .25L in from each end, what function is the web actually performing doing in the central 50% of the span? It can't be resisting shear load because there is none. It will be resisting a small bending moment (but not much compared with the flanges). If there were air gaps between web and flanges over the central 0.5L only, would those gaps tend to close as load is applied? I would imagine they would, perhaps due to the web being stiffer in bending than the flanges (like a ruler on edge vs chopsticks). If so, there must in a 'normal' beam (with no gap between flanges and web) be a vertical load transfer between flanges and web: hence a need for welds. Not sure how I would quantify this by manual calculation.

Tigermoth,

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

In the tigermoth example, the web carries no shear in the middle half of the span, but it adds stability to the structure.  

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

I think many have getting tired on this, but I couldn't hold on to my DUMB question:

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)

Yes, the idealized model would have to preclude buckling failures as well.

cntw1953: that should just be straight up VQ/It

Under pure bending (end moments), ignore selfweight, does VQ/It still hold?  

Sure does.  It'll just yield a value of zero for your case is all.  You have to remember: if your member is loaded the way that you say it is, you shouldn't have ANY "zig-zag" at the ends.  Every differential element of every cross section will wind up exactly where it ought to be via axial strain alone.  There will be no need for shearing stresses to keep plane sections plane.

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

The explanation on beam ends makes sense. Thanks.

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?

Now THAT is an interesting question.  Again, I think that the crux of the matter is in how those end moments are applied.  Try this:

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.

Would there be a possibility that the two systems have identical "I" when subjected to pure bending? The reason is when bending is introduced at the ends, there must be a clamping force couple that introduce strains on the extreme fibers. Due to linear elasic behavior of the beams, that would resulting in identical linear strain and stress distribution for both systems.

Makes sense?
 

Yes, given the moment application stipulations described above, the composite and non-composite systems would have identical "I" values under pure bending. For a system to  be composite with respect to moment of inertia, all that is required is that plane sections remain plain across the combined section.  The system doesn't care whether this is enforced through horizontal shear or serendipitous loading.

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

In my mind, the bending can be introduced by several different devices, all will result in linear strain and stress in the systems. One is the simplest - hold the beam ends by two fingers and apply bending force, now the top most fiber would be in axial compression, and the lower extreme fiber would be in axial tension. If we can control the forces precisely, through inear elastic behavior, both beam systems would deform identically under the same amount of force (bending moment) applied.

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.

cntw1953-

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.

cntw1953.-

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.

Miecz:

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.

This post has taken over the entire forum!!!!!

Stillerz:

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.

No need to apologize as i am only kidding.
I think the post is great.  

cntw1953-

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.

As I have suggested before, you may want to compile it into a book dedicated to dhengr, so he would never run out supply again :)

Have fun in the holidays.

miecz,

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

miecz:

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 :)

BA-

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.

BA-

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

BA:  I agree with your comparison using the cantilevered beam example.  It is a perfect example for this particular discussion.  

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.

 

Merry Christmas and Happy Shear Flow!

The beam assembly with double cantilever ends analogy concludes this discussion perfectly. Thanks to all involved. Merry X'mas & Happy New Year.