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?

 

[BAretired]

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

 

[KootK]

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.

 

[Stillerz]

BA-
Are you actually retired?

 

[Stillerz]

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

 

[BAretired]

Stillerz,

Affirmative. I retired on June 1, 2008.

BA

 

[paddingtongreen]

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.

 

[JrStructuralEng]

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.

 

[BAretired]

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

 

[JrStructuralEng]

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

 

[BAretired]

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

 

[cntw1953]

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?

 

[csd72]

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.

 

[cntw1953]

csd72:

Please see the attached PDF for terms I meant.

 

[JrStructuralEng]

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.

 

[cntw1953]

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.

 

[dhengr]

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.

 

[trainguy]

Come on dhengr,

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

This thread is sheer madness...

tg

 

[dhengr]

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.

 

[trainguy]

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

 

[dhengr]

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.