BTH-1-2014 Shear ? 2

[ColinPearson]

Hello folks!
It's been a long day in an already very long week and it's only Monday so maybe some sleep would help, but I have a question about ASME BTH-1-2014 Eq 3-37. It's turned up before on the forum but not really in a way that answers my questions. Really, I think I jsut need a little hand-holding here...

fcr = SQRT( fx^2 - fxfy + fy^2 + 3fv^2) <= Fy / Nd EDIT: original post said "+ fxfy", fixed here

where:
fx = computed normal stress in X direction
fy = computed normal stress in Y direction
fv = computed shear stress

Now... say my XY plane is "flat", and I've got forces Fx, Fy and Fz. Let's say further that all forces act at some moment arm such that there are bending stresses induced by all three forces. Now, I would consider the compression and tension resulting from those bending stresses to be acting normal to the XY plane, but wouldn't you say they acted in the Z direction? I'm confused when the explanation of the variables offered by ASME talks about the normal stresses in both the X and Y directions; do they mean the normal stresses caused by the forces acting along X and Y? I'm looking at BTH-1-2014 Chapter 3-2.5; same chapter for the 2011 version.

Thanks in advance, cheers! (sketch attached)

 

[ColinPearson]

Thanks for the visual! Appreciate all of y'all's help today.

 

[ColinPearson]

I've actually submitted two separate inquiries to ASME regarding BTH-1; I'll post back and let you folks know what they say.

 

[WinelandV]

Colin,

What exactly are you trying to design? The lug, or the thing that the lug attaches to? When I designed padeyes in the past, our company decided that the item being lifted wasn't covered under BTH-1. The item being lifted would still be checked, but only to whatever code to which it was designed - for me, this meant checking a lot of J4 checks in AISC 360.

Now then, for the padeye itself, it was basically major axis bending, minor axis bending, and tension in combination (eqn 3-35). Weld at the base gets sized for the bending(s), tension, and the shear(s). I'd only drop in to eqn 3-37 if I did a FEA of the lug as plate elements, and I can't imagine analyzing a lug with FEA.

Just my .02

 

[ColinPearson]

winelandv - Just looking at the lug/padeye itself. I agree with you that the thing it attaches to is not covered by BTH-1 other than IMO you could make a case that the section 3-3.4 on Welded Connections would require a little checking of the thing that gets the lug. J4 is good stuff for the thing that gets the lug as you say.

Also agree that 3-35 must be checked, but we have seen many lugs that get loaded at quite shallow angles with respect to their base. That is, if the lug is made out of b" (long) x t" (thick) plate that is h" (tall), then the angle from the "bt" plane is often fairly low; say less than 45 degrees is not uncommon. Therefore, there is a significant amount of shear, but at the loading is arranged, there is also bending about both the major axis (about the axis through the "t" dimension in my example), bending about the minor axis(here, the axis through the "b" dimension), and some axial force (along the axis parallel with the "h" dimensions). To cover all bases, there are also times when there is torsion about the "h" axis as well. I know that torsion is never good, and maybe those are some shallow angles to be loading something, but that is how I see them on numerous jobs so I'd like to be able to correctly address the calcs. Anyway, the presence of that worth-investigating amount of shear along with at least bending and axial (though not always torsion) got me looking at Eq.(3-37) for the lug plate itself; do you think I'm off base here? Thanks for your input, have a good eve!

 

[WinelandV]

Less than 45 degrees? Your lifters are really putting the shackles through their paces. Also torsion... yikes. I see why you're struggling for a BTH-1 solution here - it was written for the majority of cases for lifters and lifter attachments. Usually, loads stay in one plane and rotate around the pin. Sometimes, you get the out of plane loading during uprighting. Torsion, though, I've never seen it addressed, and I'm struggling to come up with a scenario for it. If you need to get all these forces through your lug, though, may I suggest lengthening the lug, and adding four triangular stiffeners (both sides of the hole, both sides of the lug). Larger I for the out of plane bending, and can resolve the torsion into a force couple into those stiffeners

|​

----|----

|<------hole assumed in center of lug​

----|----

|​

(lug plan view)

On the other hand, while possible to try and design for the way things are being done, maybe some push back the other way is needed, saying, "we can go to these limits, but not beyond."

 

[ColinPearson]

An example where you'd get some torsion and have such shallow angles would be lugs located radially on a hemispherical head being used to lift vertical pressure vessel; no uprighting necessary as the vessel would remain vertical. In these cases, let's say 6-8 lugs spaced equal (angular) distances around the head, the angle of slings w.r.t. horizontal may be about 60 degrees or better, but due to the curvature of the head where they are welded, the angle w.r.t. the base of the lug is considerably less. So, the rigging is not necessarily being aggressively loaded, but the lug itself sees a pretty flat angle. More on the torsion part in a bit.

Maybe I can ignore torsion, but until I had my calcs running smoothly and had an opportunity to see the actual numbers, I was hesitant to do so. I've got two situations that are in mind:
1) a lug welded to some piece of structure used to pick up "stuff" - generally much lighter duty. These are the ones where significant weak axis bending is to be expected b/c boilermakers are boilermakers and they will make it happen, engineer's nerves be damned. In all of these cases, I would put the lug hole located axially, so there is biaxial bending and tension, but never the torsion. Simpler case although I did have the question about normal+shear and Eq.(3-37)that I'm working through.

2) Lugs welded to the head of a vessel as described above. If the length of lug req'd (due to length of weld req'd) is fairly long, we would slide the hole to one end to allow clearance for the shackle/turnbuckle without having to make the lug excessively tall. Be amaze by my keyboard artwork!!! The single quote marks and periods are just for spacing, I couldn't get it quite right but I think they convey the message.
' __________
' |..O...........| <--------- lug with hole to one end
' |...............|
~~~~~~~~~~~ <--- weld to vessel

...instead of...

''''''''''' ____
'''''''''''/..O..'''''''''/.........\ <--------- lug with hole at lug CL but tall enough to allow enough taper to provide adequate clearance for shackle
'''''''/............'''''/...............~~~~~~~~~~ <--- weld to vessel


If all the rigging came to a single point, then even with this longitudinal offset of the hole w.r.t. the "LUG_WIDTH x LUG_LENGTH" axis, then there would be no torsion. However, since the rigging does not come to one exact single point, then some over the lugs are loaded across the minor axis to a relatively small degree; this is the torsion I was trying to capture. winelandv, it sounds as if you are familiar with the subject but if anyone is not, one reason the rigging would not come to a single point is that the crane hook would often be a "sister" or "duplex" or "ram's horn" - a pair of hooks with the distance between CLs of the hooks being something like 18"-30", more or less depending on size. Heck, even with a single hook, I've seen situations where some of the lugs could be loaded out of plane by 10 degrees or more. To illustrate a bit further, say we had 6 equally-spaced lugs arranged concentrically and aligned radially; the lugs at 0* and 180* may be loaded in major axis only, but the lugs at 60*, 120*, 240* and 300* would all have some minor axis loading as well, due either to the rigging collecting at one or the other of two hooks, or even just due to the multiple slings having to lay next to each other and not being able to a=occupy the same physical space.

I like the idea of the stiffeners, and definitely will look at putting that into my spreadsheet. Thanks again for your time and thoughts.

 

[WinelandV]

Ok, I'm keeping up with you now.

Sounds like for case 1 you're good to go.

Case 2 - I say that there is no torsion at each lug. Thinking through the FBD for each lug individually, there's still just one resultant force that's going through (ideally) the center of the shackle pin - thus no twist. Now, if there's too much slop in the shackle in the hole (pin length > lug thickness), the shackle could shift to one side moving the resultant off of the "vertical" centerline. We prevented this by welding cheek plates to the lug so that width of the lug + cheek plate was at least 90% of the pin length. You probably know this, though. But that's my reasoning behind no torsion, or at least, small enough to be ignored compared to the bi-axial moments and tension.

 

[ColinPearson]

Maybe I'm getting the old panty-bunch over nothing. I should have just provided a better picture to begin with , see the attachment and you can why I'm thinking about torsion. My life would definitely get one step simpler if I decide it's OK to ignore!

You can see the lugs at 120* and 300* (* = degrees) are loaded straight on, but 0*, 60*, 180* and 240* are all loaded at about 11* w.r.t. their longitudinal axis. The hole is well offset like my upper "sketch" above. Maybe I'm worrying about nothing? I'm just worried that 11* or so created a moment about the axis that would be basically normal to the vessel surface; would you think the effect is small enough to discount?

Thanks!

 

[WinelandV]

Well...

And I'm sure the answer is "always been done this way", but I don't like the turnbuckle being used like that. I haven't opened my Crosby in a few years, but I have no memory of an allowable out of plane loading on a turnbuckle. Any chance of slapping a shackle on the lug, and using a super short, high capacity nylon to connect the shackle to the turnbuckle?

Another thought, any chance of getting the lug turned so as to line up with the rigging?

 

[ColinPearson]

Yep, thought about both of those. I ended up using a shackle in between the lug and the turnbuckle b/c I didn't like it either. We also used cheek plates such that the shackle was bearing on about 85% of its pin width. That eliminated the lousy loading no the TB but left the loading on the lug still a question in my mind. I had thought about cutting the lugs differently so that they all aim directly where they 'should', but decided it wasn't worth the increased fabrication cost, layout/fit-up time, or the possibility of using the wrong lug in the wrong place. Based on previous experience and the homework that had been done (once we fixed the TB loading with extra shackles), I had no concern about the lug, rigging or vessel itself; I just want to improve my calcs. All in all, it flew well, but I made a mental note to revisit the issue when between projects and I had time to tune up my calcs. It's kind of a pet project that I've been working on in my spare time but I'm enjoying the exercise.

For the record, Crosby states (in caps) that, "TURNBUCKLES RECOMMENDED FOR STRAIGHT OR IN-LINE PULL ONLY".

And, again, thanks for the input.

 

[ColinPearson]

If anyone is interested...

QUESTION TO ASME (WITH RESPONSES):
For members that are subject to combined normal and shear stresses, Equation (3-37) limits the critical stress to fcr <= Fy/Nd. If the member is loaded in the X direction (in such a way as to induce bending, average shear and torsion), Y direction (in such a way as to induce bending and average shear) and Z direction (in such a way to induce axial stress and bending), could the Committee please clarify the following:
1) Is "fv" a combination of all shear stress due to Fx, Fy and Fz, whether an average shear stress or shear stress due to torsion?
[highlight #FCE94F]YES[/highlight]
2) Is "fx" a stress normal to the XY plane, induced by the bending moment of Fx about axis YY?
[highlight #FCE94F]YES[/highlight]
3) Is "fy" a stress normal to the XY plane, induced by the bending moment of Fy about axis XX?
[highlight #FCE94F]YES[/highlight]
4) Does Equation 3-37 correctly apply to a situation where there are loads in three directions?
[highlight #FCE94F]YES[/highlight]
5) If the answer to 4) is "no", then could the Committee please provide a recommended form of an appropriate equation (such as a full 3D version of a Von Mises yield criteria) that is consistent with the nomenclature used in BTH-1?
[highlight #FCE94F]NOT ANSWERED SINCE 4) WAS "YES"[/highlight]

More to follow when I get their response about some question w.r.t. the weld design in these cases.

 

[WinelandV]

Hmmm...

A few thoughts:

1) Try it using average shear stress with the max bending stresses and tension stress. If it checks out, no worries.

2) If it doesn't check out, then I would refine the analysis to use the actual shear stress VQ/It. This would probably entail checking the critical stress at tenth (or hundredth) points in both axis (x and y) which just jumped the complexity of the spreadsheet.

3) Or just hand wave it away by noting that your max normal and shear stresses don't occur at the same location - which is why we don't do a combined axial/moment/shear check in beams (at least beams without torsion).

Hopefully this post is helpful.

 

[ColinPearson]

The answers from ASME led me to believe that the fx and fy were stresses in "native" or "obvious" Height direction ("Z" axis), and that they were due to about the "native" or "obvious" Length x Width directions ("X" and "Y" axes). I hadn't checked the Commentary in BTH because in the middle of writing this sheet, I switched from 2011 to 2014 which moved the Commentary to the back, and, I suppose therefore out of mind. Despite what my informal answers from ASME appeared to say, I found:

"Equation (3-37) is the Energy of Distortion Theory relationship between normal and shear stresses (Shigley and Mischke, 2001). The allowable critical stress is the material yield stress divided by the applicable design factor, Nd. For the purpose of this requirement, the directions x andy are mutually perpendicular orientations of normal stresses, notx-axis and y-axis bending stresses."

It does appear that KootK and TehMighty were correct (again!). Had they used some other subscript, I probably wouldn't have stumbled so badly on this one, but less than 3" away on the same page, there are multiple other topics of discussion where "x" and "y" are talking about the "native" axes of the thing you're designing. Perhaps a poor choice of subscripts, but I still wish I had snapped to a little (lot) quicker. Thanks KootK, TehMighty and winelandv for your time and valuable input!!!