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?

 

[IJR]

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

 

[paddingtongreen]

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.

 

[JrStructuralEng]

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.

 

[Lion06]

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.

 

[BAretired]

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

 

[JrStructuralEng]

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.

 

[KootK]

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.

 

[Lion06]

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

 

[KootK]

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.

 

[Lion06]

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).

 

[KootK]

Ahh.. darn real work.

 

[JrStructuralEng]

StructEIT i belive you are describing option b?

 

[Lion06]

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.

 

[paddingtongreen]

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.

 

[Stillerz]

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.

 

[JAE]

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?

 

[Stillerz]

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.

 

[Stillerz]

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.

 

[BAretired]

"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

 

[Stillerz]

well stated BA