Shear Flow with y' = 0?

[Lion06]

I am having a bit of a problem with a shear flow problem. I have a WF with strong axis bending (and some torsion) that is getting a vertical plate welded to (1) side of it.
This plate is to take out the torsion. When I am calculating the shear flow, the A' is the Area of the plate, but y' is the distance from the centroid of the composite section to the centroid of the connected piece (plate). In this case the centroidal axes are coincident, and y' = 0. This is obviously problematic because I can not calc shear flow with this. I have considered calcing the shear flow at the juncture of the web/flange and distributing it based on relative thickness between the flange and plate; this is not at the EXACT juncture that I am seeking, but I feel it is close enough since the weld will most likely be overdesigned by a significant factor.
Does anyone have any suggestions.

 

[xxpegasus11]

not sure I understood yr problem correctly - are you welding a plate to side of web, thickening the web?

 

[Lion06]

No, I am welding the plate to the edges of the flanges in an effort to simulate an HSS section.

 

[WillisV]

See this thread thread507-168270 which had a similiar situation. See my response mid-way down - I believe this same type thought process would work for this problem.

 

[Lion06]

Willis- Will that work? It appears to me that the w you are calculating is a vertical line load that will produce the required deflection. I agree with that. How does that vertical line load of w get transferred into a horizontal shear flow resisted by the bolts (in the case of thread you referenced)? If, for example, the w turned out to be 9.3 kips/ft - would you say (1) - 3/4" A325N bolt/12" is ok?

 

[WillisV]

Yes.

 

[miecz]

MYerges

I think your approach is correct.

If you think of this as welding the flange to the plate and the web, instead of the plate to the flange, then your weld will take a portion of the total shear flow. The question is, what portion? Distributing it based on the relative thickness of the plate and the flange makes sense to me. After all, if the web and plate deflect the same, then the shear in each will be proportional to the relative thickness, as opposed to the relative size of the welds.

 

[WillisV]

jmiec..I'm not so sure on that - if you are looking at deformation compatibility the load in each will be proportional to the moment of inertia of the plate vs. that beam - not the areas of plate to flange.

 

[miecz]

WillisV

Oops, I meant to say the distribution should be based on the relative thickness of the web and the additional plate. That would make the shear flow proportional to the thickness of the plate vs. that of the web of the beam.

Here's how I got to that conclusion. Imagine that the section is made up of four plates, two flanges and two webs. Now weld the flanges to the web plate, and to the other plate. All the welds will have shear flow, proportional to the thickess,(and moment of inertia) of the vertical plates.

 

[miecz]

WillisV

I don't believe the thinking that went into your response to thread507-168270 applies to this case. The compatible deformation reasoning only applies to the case when the connectors are located at the neutral axis.

 

[vonlueke]

MYerges: I assume your I-beam y axis is vertical, with bending about the x axis, so your statement y' = 0 doesn't seem to make sense. The section cut for shear flow is generally on (or close to) only one plane, not two different planes. So there's no way to completely section off a free body of your attached plate and, at the same time, have a meaningful y' value.

Instead, your section cut would be, e.g., near the inner face of your I-beam flange, cutting the web and your attached plate (or weld) simultaneously. So A' would basically be the cross-sectional area of one flange, not of the attached plate. Then, the shear stress can be roughly assumed to be evenly distributed across this section cut, meaning it is distributed approximately in proportion to the web and attached plate thicknesses.

 

[Lion06]

Vonlueke -
I said in my original post that my idea was to do basically what you suggested. I am not sure you are understanding the problem correctly if you think that y'=0 doesn't make sense. I have a WF oriented vertically (like a normal beam) with an eccentric pt. load causing strong axis bending and torsion. The plate is being welded to one side only at the top and bottom flanges. The plate is sized 1/2" larger than the depth of the beam to allow for the weld at the top of the top flange and bottom of the bottom flange to the plate.
This being the case, when you determine the plane on which you want to calc the shear flow, the orientation of that plane to the strong or weak axis is irrelevant. I initially tried to make the plane a vertical plane cutting both of the weld locations to get an accurate shear flow. When you do that, however, the centroidal axis of the connected part (plate) and the centroidal axis of the composite section (WF w/plate) have a separation of zero (0). I then decided to do the approximation, which pretty much amounts to what you described. I was just trying to find out if anyone else had a different suggestion that might yield more accurate results.

 

[miecz]

MYerges

I believe vonlueke and I are thinking alike. Let me try to explain this again, without the typos. You need to think of the flange as the "connected part", not the vertical plate. Then, in the shear flow equation, A' becomes the area of the flange, and y' becomes the vertical distance from the horizontal neutral axis to the weld.

The calculated shear flow is distrubuted to the two vertical elements, the web of the beam and the vertical plate, in proportion to their thickness.

That the weld is not in the same plane as the web fillet muddies up the solution some. Could you fit the plate between the flanges?

I'm not sure how the antisymmetry of the composite member effects the distribution, but I believe it would be conservative to ignore it in the design of the weld.

 

[COEngineeer]

MYerges, your problem is pretty complicated. Go open your mechanics of material text book and look for "shear center". Your calc will be 10X easier if you put the plates on both sides so you can assume the welds on both sides will take the same amount of shear and the shear center will be the center of gravitiy.

 

[vonlueke]

MYerges: Thanks for the clarification. I agree with you and jmiec that your second method, jmiec's posts, and my post all describe the same approach.