Shear Flow 15

[Stillerz]

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?

 

[Stillerz]

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?

 

[DaveAtkins]

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

 

[Lion06]

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).

 

[miecz]

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.

 

[Stillerz]

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.

 

[Stillerz]

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

 

[apsix]

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.

 

[JAE]

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).

 

[KootK]

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...

 

[mudflaps]

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.

 

[Lion06]

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

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

 

[BAretired]

Stillerz said:

(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

 

[ishvaaag]

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.

 

[DaveAtkins]

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

 

[graybeach]

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.

 

[miecz]

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.

 

[Stillerz]

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

 

[BAretired]

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

 

[graybeach]

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.

 

[miecz]

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.